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User arupinski - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T17:40:36Z http://mathoverflow.net/feeds/user/12301 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/133449/signless-stirling-numbers-of-1st-kind-and-probabilistic-descent/133585#133585 Answer by ARupinski for Signless Stirling Numbers of 1st Kind and Probabilistic Descent ARupinski 2013-06-13T01:55:26Z 2013-06-13T02:14:10Z
Start with an arbitrary permutation in $S_n$ written in cycle form:

\begin{align*} \psi := (\pi(n-1),\ldots,\pi(n_1))(\pi(n_1-1),\ldots,\pi(n_2))&\ldots(\pi(n_{k-1}-1),\ldots,\pi(n_k)) \end{align*}

Next, for $\ell = 1\ldots k$, define the sets $N_\ell$ as the collections of elements in each cycle of this permutation: $N_\ell := \{\pi(n_{\ell-1}-1),\ldots,\pi(n_\ell)\}$

Now impose the following condition on the $N_\ell$:

$\max\left(\nu|\nu\notin\bigcup_{i<\ell}N_i\right)\in N_\ell$

Clearly for a given permutation in cycle notation, this inclusion condition defines a unique order on the cycles (the leftmost cycle is the cycle containing the maximal element and from there one chooses the cycle with the largest remaining element to be the next cycle, and so forth). Then, given the ordered cycles one obtains a unique descent associated to $\psi$ by taking the lengths of each cycle to be the step sizes starting at $n$, so if \psi satisfies the max element condition on its ordered cycles, we associate it to the descent $[n=n_0,n_1,\ldots,n_k=0]$.

Finally, one notes that the count of permutations whose cycle structure leads to a given descent is exactly the numerator of the probability for the original descent; since every permutation leads to some descent by this process, one gets the bijection you sought.

Another way to view the above bijection is that when you choose your first number, make a cycle of length represented by the difference of the number you chose from the start number, place the number $n$ at the start of your cycle and fill the remaining slots of the cycle however you like. Then choose your next descent step at random, form another cycle whose length is the difference between this step and the previous one, place the largest number not yet used at the start of the cycle, and fill the remainder any way you like with the unused numbers. Repeat until you exhaust your descent; at this point every number will have been put into some cycle and you will have a permutation corresponding to your particular descent.
 http://mathoverflow.net/questions/130660/picturing-a-certain-torus-and-klein-bottle Picturing a Certain Torus and Klein Bottle ARupinski 2013-05-15T02:01:59Z 2013-05-15T11:56:28Z
The other day I was explaining orientability to someone and we were walking through some of the statements about orientability on the  Wikipedia page on the topic. While I was able to satisfy his curiosity, one statement on that page (which I did not even attempt to delve into with him) has been nagging me since then:

"For example, a torus in $K^2\times S^1$ can be one-sided and a Klein bottle in the same space can be two-sided."

Because this statement bothered me (since it runs counter to normal intuition about orientable surfaces in Euclidean spaces), I have been thinking about it more over the last few days. I have been able to determine which copies of these submanifolds should have the stated properties and convince myself how the non-orientability of the ambient space $K^2\times S^1$ allows for the submanifolds in question to twist back on themselves in unusual ways, but nevertheless I still cannot form a decent picture of what this really means.

The real issue with my understanding what is going on with these submanifolds seems to be that although these phenomenon occur in a non-orientable space, this space can itself be embedded in an orientable space and so it seems that these odd tori and Klein bottles should therefore embed in an orientable space as well and so I should have some chance of visualizing these phenomena when I project down to $\mathbb{R}^2$ or $\mathbb{R}^3$

Question: Does anyone have a good picture or other approach to help visualize what a one-sided torus or two-sided Klein bottle looks like?

So while it may be too much to hope for a projection that accurately reflects the sidedness of these creatures, I am hoping someone may have a decent projection of either of these creatures to the plane or 3-space that shows some manifestations of their odd behaviour in their ambient space. Or, barring an actual picture, perhaps someone who has thought about this more has some other way of thinking about them which at least gives a better intuitive sense of how to look at them in their ambient space and 'see' (whatever that may mean when you think about them) these counterintuitive features.
 http://mathoverflow.net/questions/129909/does-this-linear-algebra-construction-have-a-name Does this Linear Algebra Construction have a Name? ARupinski 2013-05-06T23:15:17Z 2013-05-11T16:33:15Z
Let $\mathcal{R}$ be a ring and let $v^0,\ldots,v^{k-1}\in\mathcal{R}^m$ with $m \geq k$. Suppose we wish to find $w\in Span(v^0,\ldots,v^{k-1})$ such that $k-1$ specified coordinates of $w$ vanish (say $w[j_0] = \ldots = w_[j_{k-2}] = 0$). Then the following determinantal construction for $w$ does the trick:

$w = \sum_{i=1}^k (-1)^k\cdot\det\begin{pmatrix}{v^0}[j_0]&{v^0}[j_1]&\ldots&{v^0}[j_{k-2}]\\{v^1}[j_0]&{v^1}[j_1]&\ldots&{v^1}[j_{k-2}]\\\vdots&\vdots&\ddots&\vdots\\\widehat{v^i[j_0]}&\widehat{v^i[j_1]}&\ldots&\widehat{v^i[j_{k-2}]}\\\vdots&\vdots&\ddots&\vdots\\{v^k}[j_0]&{v^k}[j_1]&\ldots&{v^k}[j_{k-2}]\end{pmatrix}\cdot v^i$

The individual coordinates of $w$ also have a nice determinantal form (which follows from the above expression):

$w[i] = \det\begin{pmatrix}v^0[i]&v^0[j_0]&v^0[j_1]&\ldots&v^0[j_{k-2}]\\v^1[i]&v^1[j_0]&v^1[j_1]&\ldots&v^1[j_{k-2}]\\\vdots&\vdots&\vdots&\ddots&\vdots\\v^{k-1}[i]&v^{k-1}[j_0]&v^{k-1}[j_1]&\ldots&v^{k-1}[j_{k-2}]\end{pmatrix}$

For example, suppose we start with the 3 vectors $v^0 = \langle 2,3,5,7,11,13\rangle$, $v^1 = \langle17,19,23,29,31,37\rangle$ and $v^2 = \langle 41,43,47,53,59,61\rangle$ and wish to find a vector $w$ in their span whose final 2 coordinates are 0. Then the above formulas give us the following vector:

\begin{align*} w &= \det\begin{pmatrix}31&37\\59&61\end{pmatrix}\cdot v^0 - \det\begin{pmatrix}11&13\\59&61\end{pmatrix}\cdot v^1 + \det\begin{pmatrix}11&13\\31&37\end{pmatrix}\cdot v^2\\ &= -292\cdot v^0 + 96\cdot v^1 + 4\cdot v^2\\ &= \left\langle\det\begin{pmatrix}2&11&13\\17&31&37\\41&59&61\end{pmatrix},\det\begin{pmatrix}3&11&13\\19&31&37\\43&59&61\end{pmatrix},\det\begin{pmatrix}5&11&13\\23&31&37\\47&59&61\end{pmatrix},\right.\\ &\;\;\;\;\;\;\;\;\;\;\left.\det\begin{pmatrix}7&11&13\\29&31&37\\53&59&61\end{pmatrix},\det\begin{pmatrix}11&11&13\\31&31&37\\59&59&61\end{pmatrix},\det\begin{pmatrix}13&11&13\\37&31&37\\61&59&61\end{pmatrix}\right\rangle\\ &= \langle 1212, 1120, 936, 952, 0, 0\rangle \end{align*}

These formulas are easily derived/proved using Cramer's rule and/or other methods involving exterior products (which is how I came up with them when I was trying to construct such a vector), and like Cramer's Rule are rather beautiful, so I would be surprised if they are not already written down/used somewhere. Nevertheless, I don't recall having ever seen such constructions in any Linear Algebra books.

 Main Question: Has anyone seen either of the above equivalent formulas before? If so, do they have a name?

 Secondary Question: Whether or not they have a name, has anyone seen these formulas used as part of any other proofs?

 Update: After thinking more about Question 2 over the last few days I realized this sort of construction $could$ arise in some group representation problems where one wants to explicitly find a character which is zero on certain conjugacy classes, such as when one is calculating induced representations. But I have definitely never seen it in any such character-theoretic arguments, and it is not even immediately clear to me whether it would necessarily give non-virtual representations in general. On the other hand, studying for a given group $G$ which sets of characters paired with which sets of conjugacy classes give rise to pure representations via this construction (aside from the trivial cases of pairs obtained from induction of representations) sounds intriguing enough that I will probably take a look at it. Still not clear if the construction has actually been used in any such proofs elsewhere though...
 http://mathoverflow.net/questions/129345/n-in-a-row-game-on-mathbbr2/129371#129371 Answer by ARupinski for $n$-in-a-row game on $\mathbb{R}^2$ ARupinski 2013-05-02T04:12:34Z 2013-05-02T04:12:34Z
So now Im thinking about the following approach. Let $H$ be the discrete hypercube $[1,...,kn]^d$ with $k,d>> n$ (im not sure how big I need them yet, it just seems that I need them to be sufficiently large compared to $n$ for this to work). I can project $H$ onto $\mathbb{R}^2$ via a linear transformation which sends the unit basis vectors of $\mathbb{R}^d$ to $\mathbb{Q}$-linearly independent vectors in $\mathbb{R}^2$. The point of this is that the only straight lines of points in the projection come from straight lines in $H$. Now P1 plays all his moves on these projected points and I want to argue that the total number of lines of $n$ points in a row contained in $H$ is orders of magnitude greater than the number of lines P2 could have blocked (even accounting for the fact that if P2 played on grid points it would block several potential $n$-in-a-rows simultaneously). The approach then boils down to a slightly stronger version of Van der Waerden's theorem for a bichromatic coloring of $H$. I just cannot figure out the estimates necessary to show for sure that once all of $H$ has been colored (with only the assumption that P1 only colors points of $H$ on each of his plays) there must be enough monochromatic $n$-in-a-rows for sufficiently large $d,k$ that P2 could not possibly have blocked them all. (I assume $k>>n$ to give the possibility of $n$-in-a-rows running at slanted angles through $H$ so that P2 cannot simply block by playing on lines parallel to the coordinate axes). Anyone have ideas on how to rigorously complete this argument or see any fatal flaws I overlooked?
 http://mathoverflow.net/questions/120148/edge-coloring-of-the-complete-graph-without-any-rainbow-paths/120174#120174 Answer by ARupinski for Edge-coloring of the complete graph without any rainbow paths ARupinski 2013-01-29T04:07:59Z 2013-01-30T00:04:41Z
It seems that one can use a similar trick to that used in your construction on $K_{2^k}$ for any $K_{4n}$: label the vertices by elements of $G = (\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2n\mathbb{Z})$ and assign edge colors by the difference of the two endpoints (EDIT: as noted in Ilhee's comment, this won't be well-defined; am thinking about possible modifications). Since the sum of all the elements of $G$ is the identity, the same argument which shows your coloring of $K_{2^k}$ is rainbow-free applies to this coloring of $K_{4n}$.

Unfortunately this argument fails when we try to use it to label $K_{4n+2}$ by elements of $(\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/(2n+1)\mathbb{Z})$ since the sum of the elements of the group is not the identity in this case.

My gut tells me there is some clever way to label the vertices in the $K_{4n+2}$ so as to define colors on the edges which force your condition, but I don't see it; I'll have to think about it some more.
 http://mathoverflow.net/questions/119377/automorphism-of-finite-groups-and-hurwitz-spaces/119379#119379 Answer by ARupinski for Automorphism of finite groups and Hurwitz spaces ARupinski 2013-01-20T00:25:12Z 2013-01-20T00:25:12Z
Such automorphisms need not extend to inner automorphisms of $S_n$. Take the dihedral group $D_8$ sitting transitively inside $S_4$, say as the following set of permutations: $\{(1)(2)(3)(4), (13)(2)(4), (1)(3)(24), (12)(34), (14)(23), (1234), (13)(24), (1432)\}$. Now the sets of reflections $\{(13)(2)(4), (1)(3)(24)\}$ and $\{(12)(34),(14)(23)\}$ are equivalent to one another via an outer automorphism of $D_8$, but clearly this automorphism is not inner in $S_4$.
 http://mathoverflow.net/questions/117151/does-this-poset-have-a-unique-minimal-element Does this poset have a unique minimal element? ARupinski 2012-12-24T16:35:44Z 2013-01-17T22:47:52Z
Recently I have been thinking about the following poset: the underlying set is $\mathcal{AFT}$ consisting of all (finite) automorphism-free undirected trees (with at least one edge to exclude the trivial cases pointed out by Joel) and the poset relation $\leq$ is defined by $T \leq U$ if $T$ can be obtained from $U$ by successively deleting one leaf node at a time in such a way that each intermediate tree is also an element of $\mathcal{AFT}$.

The smallest element of $\mathcal{AFT}$ is the seven node tree which is the Dynkin diagram of $E_7$ (and which I will therefore refer to simply as $E_7$ from herein):

So $E_7$ is certainly a minimal element in the above partial order.

Question: Does $(\mathcal{AFT},\leq)$ have a unique minimal element, namely $E_7$?

There are several equivalent formulations of this question which I have considered, hoping one of them might lead somewhere useful:

Question 2: Can every element of $\mathcal{AFT}$ be obtained by starting at $E_7$ and successively adjoining leaf nodes so that we remain in $\mathcal{AFT}$ at every stage?

>

Question 3: Is there an element of $\mathcal{AFT}$ besides $E_7$ such that deleting $any$ single leaf node results in a tree not in $\mathcal{AFT}$.

Question 3 in particular seems simple enough that it must have been answered somewhere before, but alas almost any search for 'automorphism-free' and 'trees' results in papers about the result that almost every tree has an automorphism or results on the fixed vertices/edges of trees under automorphisms.

Trying to construct a minimal example larger than $E_7$ for Question 3 keeps leading to near-misses where all but one leaf nodes' removal takes us outside $\mathcal{AFT}$, but after removing this one leaf node, there is still a sequence of removals remaining in $\mathcal{AFT}$ that leads back to $E_7$ which is the best evidence I have so far that Question 1 is true.

So, does anyone know where this might have been considered already, and if so, if the answer to Question 1 is affirmative?
 http://mathoverflow.net/questions/111684/does-every-bipartite-graph-with-512-edges-have-an-induced-subgraph-with-256-edges/111701#111701 Answer by ARupinski for Does every bipartite graph with 512 edges have an induced subgraph with 256 edges? ARupinski 2012-11-07T04:06:04Z 2012-11-07T04:06:04Z
It's possible I messed up in my calculations, so by all means check it to be sure, but I think the following is an example of what you seek:

Start with a copy of $K_{5,103}$ with vertices $v_1\ldots v_5,w_1\ldots w_{103}$ and remove the edges $v_iw_i$ for $1\leq i\leq 5$. Add two copies of $K_2$ to make a total of 512 edges. Then unless I messed up somewhere in my calculation, no induced subgraph can have exactly 256 edges in it.

There were another of other near misses I found (similar initial setups which had exactly 1 way (up to relabelling of vertices) to remove vertices leaving exactly 256 edges), so it shouldn't be hard to modify this construction to find an example if it turns out the above example fails after all.
 http://mathoverflow.net/questions/110609/a-simple-question-about-the-center-of-a-finite-group/110610#110610 Answer by ARupinski for A simple question about the center of a finite group ARupinski 2012-10-25T02:57:35Z 2012-10-25T02:57:35Z
As it is not entirely clear where the containments are meant to be, I will assume you are asking about the case $N\triangleleft K\triangleleft G$; in this case $Z(G) = Z(G/N) = 1$ is not sufficient in general to guarantee $Z(G/K)$ is also trivial. As an example, take $N$ to be the Klein-4 subgroup of $S_4$ and $K = A_4$. Then $G/N\cong S_3$ so $Z(G) = Z(G/N)$ is trivial, but $G/K$ is abelian (isomorphic to $\mathbb{Z}/2\mathbb{Z}$) and so $Z(G/K)$ is not trivial.
 http://mathoverflow.net/questions/108726/getting-a-bound-on-the-coefficients-of-the-factor-polynomial/108753#108753 Answer by ARupinski for Getting a bound on the coefficients of the factor polynomial ARupinski 2012-10-03T21:50:54Z 2012-10-03T21:50:54Z
Any such bound will depend only on $n$ and $m$ since once can take for instance $f(x) = x^n-1$ (so $M = 1$). Then depending on $n$, the cyclotomic factor $\Phi_n(x)$ has coefficients as large as you like: for example $\Phi_{105}(x)$ has coefficients of modulus $2$, $\Phi_{385}(x)$ has coefficients of modulus $3$, for further references see OEIS sequence A013594 (http://oeis.org/A013594).
 http://mathoverflow.net/questions/107567/subwords-of-cube-free-binary-words/107646#107646 Answer by ARupinski for Subwords of cube-free binary words ARupinski 2012-09-20T03:17:44Z 2012-09-20T03:17:44Z
Try this: start with the infinite Thue-Morse word $TM = 0110100110010110...$ which is known to be cubefree and make the following transformations:

$0 \Rightarrow x0$

$1 \Rightarrow x1$

This gives the the infinite word:

$TM^\prime = x0x1x1x0x1x0x0x1x1x0x0x1x0x1x1x0x...$

Next make the following transformation on $TM^\prime$:

$x\Rightarrow 11$

$0\Rightarrow 0$

$1\Rightarrow 00$

This gives the following infinite word:

$C = 110110011001101100110110110011...$

Clearly $C$ does not contain the subword $010$, and unless I overlooked something in my analysis, the cube-freeness of $TM$ implies $C$ is also cube-free, which would mean there is no upper bound to the size of cube-free binary strings not containing all six of the given strings.

(Incidentally, the composition of these two transformations is equivalent to performing the transformation $0\Rightarrow 110$, $1\Rightarrow 1100$ on $TM$ to get $C$, but I figured the intermediate step makes it a bit clearer, to me at least, why $C$ should be cube-free).

If anyone sees any flaw in the reasoning that $C$ is cube-free, please do point it out.
 http://mathoverflow.net/questions/104939/induction-from-cyclic-sylow-subgroup-are-there-any-nice-properties/104963#104963 Answer by ARupinski for Induction from cyclic / Sylow subgroup are there any nice properties ? ARupinski 2012-08-18T02:02:13Z 2012-08-18T02:09:51Z
This is a set of observations that should help you understand your inductions of non-trivial characters. If I read Marty's answer correctly, this should be a concrete way of understanding what he is saying using your particular example.

First note that each non-trivial character $\chi$ has a conjugate character $\overline{\chi}$ and that the inductions of $\chi$ and $\overline{\chi}$ must also be conjugate. Next, irreducible characters occur in sets which are related to one another by outer automorphisms of the dual group, these sets are defined by those irreducible characters with common kernel. Finally, note that the induction of the regular representation of the subgroup is the regular representation of the big group.

How does this help? Well in your example, lets look at what remains from the regular representation of $GL_3(\mathbb{F}_2)$ after we take away the induction of the trivial character of the Sylow 7-group. The remaining pieces are:

$3V_{3a} \oplus 3V_{3b} \oplus 6V_6 \oplus 6V_7 \oplus 6V_8$

Each induction of the 6 nontrivial characters must have dimension 24, and since the nontrivial characters are all more or less equivalent (up to outer automorphisms of the dual group to be technical), all their inductions should look more or less the same as well. This means that each induces to either $W = V_{3a}\oplus V_6 \oplus V_7\oplus V_8$ or its conjugate $\overline{W} = V_{3b}\oplus V_6 \oplus V_7\oplus V_8$.

With a bit of extra work, you would then find that if a character $\chi$ of the Sylow 7-subgroup induces to $W$ then so do the characters of $\chi^{\otimes 2}$ and $\chi^{\otimes 4}$ while the characters of $\chi^{\otimes 3}$, $\chi^{\otimes 5}$, and $\chi^{\otimes 6}$ will induce to $\overline{W}$, although the particular split on the exponents don't matter to your particular question (as currently stated).
 http://mathoverflow.net/questions/104162/failure-of-a-basic-fact-from-representation-theory Failure of a basic fact from Representation Theory ARupinski 2012-08-06T23:19:43Z 2012-08-07T17:01:26Z
Recently I have been working with a certain subgroup of $GL_{10}(\mathbb{F}_2)$ and for various reasons was fairly sure it contained a normal subgroup isomorphic to $A_5$. Today I was able to affirmatively show that this presumed copy of $A_5$ does indeed exist in the bigger group. Now the matrices composing this copy of $A_5$ are themselves $2\times 2$ matrices over a certain commutative 5-dimensional algebra $\mathcal{S}$ (necessarily with zero divisors) over $\mathbb{F}_2$; this gives rise to a 2-dimensional representation of $A_5$ over $\mathcal{S}$ which I am fairly sure is irreducible (as opposed to just indecomposeable).

Now it is a fairly standard exercise early in learning representation theory to show that a simple group cannot have an irreducible representation of dimension 2. As the proof of this simple fact relies only on considerations involving the characters of elements of order 2, does the existence of the above representation rely only on the fact that I am working in characteristic 2, or is it related to the fact that I am working with rings with zero divisors as opposed to an algebraically closed field, or something else more subtle? I am primarily curious as to whether it is solely related to the characteristic, as I am also working with some related groups defined over commutative rings in other characteristics which I suspect also contain subgroups isomorphic to $A_5$. Hopefully someone with more background in modular representation theory than I can shed some helpful light on this situation.
 http://mathoverflow.net/questions/103882/minimal-blocking-objects-with-shadows-like-a-cube/103907#103907 Answer by ARupinski for Minimal blocking objects with shadows like a cube ARupinski 2012-08-04T02:14:26Z 2012-08-04T02:20:12Z
For $C_3(n)$ an absolute upper bound is $3(n-1)^2+3$ which can always be attained by taking the cubes on three faces adjacent to a given corner, removing the corner itself, and removing all but one cube on each edge incident to the corner. Unfortunately, this gives a bound of 15 in the $n=3$ case., so does not improve your particular case. Somewhat generalizing this construction to higher dimensions, for $d\geq 4$ one gets an absolute upper bound for $C_d(n)$ of

$[n^d-(n-1)^d]-d(n-1)+1$

To prove this always works, consider $C_d(n)$. We will replace each cube by its center point and consider the lattice as in Joseph's previous question which he references.

WLOG, let one corner of $C_d(n)$ be centered at the origin. The lines we want to block have $(d-1)$ coordinates fixed and one coordinate which varies. Now consider the set of all cubes whose center has at least one coordinate equal to 0; this is exactly the set of all cubes lying on one (or more) $(d-1)$-dimensional faces adjacent to our origin cube. There are $[n^d - (n-1)^d]$ such cubes. Now remove all cubes along the edges incident to our cube centered at the origin, this removes $d(n-1)+1$ cubes leaving the set $S_d(n)$ which consists of all cubes with at least one and at most $(d-2)$ coordinates equal to 0. The only lines which do not intersect this set are those which have $(d-1)$ fixed coordinates equal to 0, i.e. the lines incident to our cube centered at the origin. Since cubes are adjacent iff they differ in exactly one coordinate, it is easy to check that the $S_d(n)$ is connected for $d\geq 4$ but not connected for $d=3$. So we need only add our cube centered at the origin, and one cube adjacent to it to block the remaining lines while ensuring the entire collection is connected.

Note that if the above held for $d=3$, one could use it to push our bound down from Joseph's 15 to 14; however the set $S_3(n)$ consists of 3 disconnected pieces and so the reasoning leading to this formula fails. However this formula does imply that as $d$ or $n$ gets large, one can block all lines with an arbitrarily small fraction of all available cubes.

There is probably a way to do some more inclusion-exclusion to further eliminate some particular pieces of $S_d(n)$ as unnecessary for blocking purposes and thereby further reduce our bound, but offhand I don't see it.
 http://mathoverflow.net/questions/103719/lattice-cube-minimal-blocking-sets/103741#103741 Answer by ARupinski for Lattice-cube minimal blocking sets ARupinski 2012-08-01T23:26:32Z 2012-08-01T23:36:09Z
This is somewhat of an expanded comment on Eoin's answer and his follow-up question of whether such sets correspond to hyperplanes in the discrete torus.

In the $C_2(n)$ case minimal blocking sets correspond to permutation of ${1...n}$ (consider a permutation matrix; then a minimal blocking set can be formed by choosing the points corresponding to the positions of the 1's in this matrix). Conversely, every minimal blocking set in the $d=2$ case arises this way, so for $n\geq 4$ not every blocking set corresponds to a hyperplane on a torus.

In the $d=3$ case this can be extended to see that every minimal blocking set corresponds to a set of $n$ permutations whose permutation matrix sums to the all 1's matrix (think of stacking the permutation matrices up to form our cube); again it is clear that not all such sets can correspond to hyperplanes if $n \geq 4$.

For $d \geq 4$, one can apply an analogous stacking approach using sets in $C_{d-1}$ to generate all minimal blocking sets although it is not immediately obvious to me how to relate these sets to permutations easily.

For $n = 2$, it is easy to see that there are exactly two minimal blocking sets in any dimension and both always correspond to toric hyperplanes, and it seems that all minimal blocking sets are also toric hyperplanes for $n=3$ in any dimension due to how little room there is in $S_3$ to form such sets (this should follow from induction on the dimension; once one lays down the first $d-1$-dimensional layer and one point in the next layer, this should determine where all the remaining points must go in the last two layers). If anyone can easily formalize this idea for the $n=3$ case (or provide a counterexample) I would love to see it.
 http://mathoverflow.net/questions/95572/does-this-knot-invariant-distinguish-trefoil-chiralities Does this knot invariant distinguish trefoil chiralities? ARupinski 2012-04-30T12:59:34Z 2012-05-04T10:47:35Z
Let $C_N$ denote the labelled configuration of $N^{th}$ roots of unity with $p_J = e^{\frac{2\pi iJ}{N}}$ for $J = 1\ldots N$.

As a corollary of something else I was playing around with, I recently proved the following:

Theorem: Every tame knot (or link) $K$ has a (not necessarily minimal) stick presentation which such that the sticks can be projected onto the set of chords of some $C_N$ with the following crossing condition: Whenever the projection has chords $p_{J_1}p_{J_3}$ and $p_{J_2}p_{J_4}$ such that $1\leq J_1 < J_2 < J_3 < J_4\leq N$ then $p_{J_1}p_{J_3}$ crosses in front of $p_{J_2}p_{J_4}$. (In other words for any two intersecting chords in the projection, the chord which has the lowest numbered endpoint passes in front of the other chord).

Germane to this question is the fact that the theorem leads to a knot invariant which, for lack of a better name, I will call the circular stick number of $K$ and which is defined to be the minimum $N$ needed to obtain a projection of $K$ with the above properties.

My proof of the above theorem was very non-constructive, so I wanted to see some concrete realizations of such projections. And after some work, I was able to find that for the left trefoil knot, the circular stick number is 7. After much more playing around on $C_7$ and not finding a projection of the right trefoil, I moved on to $C_8$ where I was able to find a projection of the right trefoil.

Main Question: Does this invariant actually distinguish chiralities of the trefoil, or can someone find a projection of the right trefoil with the above properties on $C_7$? If it doesn't distinguish chiralities in this case, is it possible it distinguishes chiralities of some other pair of knots, or does anyone see a slick way to see that it cannot distinguish chiralities?

Secondary Question: (Mainly for knot theorists, or anyone with a deeper knowledge of knots than I) Has this invariant been studied before, and if so, what is the terminology used for it?

I am fairly sure that the general answer to the main question is strongly related to the characterization (and in particular the disjointness) of the sets of forbidden minors of the following two sets (which are almost certainly not known in general as the Robertson-Seymour theorem is non-constructive):

$\bullet$ Graphs which are $L$-lessly embeddable for a given chiral knot $L$

$\bullet$ Graphs which are $L^+$-lessly and $L^-$-lessly embeddable for a given chiral knot $L$ with chiralities $L^+$ and $L^-$

Note that graphs in the second set may admit embeddings in $\mathbb{R}^3$ containing either $L^+$ or $L^-$, but have at least one embedding which does not contain both $L^+$ and $L^-$. As of yet, I have not been able to flesh out a proof using this approach however.

For those who don't want to search for the solutions I found, for the left trefoil the points of $C_7$ are connected in the following order:

$p_1 \rightarrow p_3 \rightarrow p_5 \rightarrow p_7 \rightarrow p_2 \rightarrow p_4 \rightarrow p_6 \rightarrow p_1$

For the right trefoil the points of $C_8$ are connected in the following order:

$p_1 \rightarrow p_3 \rightarrow p_7 \rightarrow p_5 \rightarrow p_2 \rightarrow p_8 \rightarrow p_4 \rightarrow p_6 \rightarrow p_1$

 EDIT: Here are pictures of these two projections to help clarify the situation:

https://docs.google.com/open?id=0B5BVGcL23IkoSTlNNGZKZnZtT1E

With regards to Dylan's question about how this projection arose, I was considering the parametric curve $S(t) = (t,t^2,t^3) \subset \mathbb{R}^3$. In a comment or answer on a past MO post which for the love of me I cannot find now, someone had shown in a very simple way that no two chords of $S$ intersect one another anywhere in $\mathbb{R}^3$ (unless they share an endpoint). Thus, for each $n$, $K_n$ can be embedded in $\mathbb{R}^3$ as the set of chords connecting the points $(1...n)$. Now for a fixed tame knot or link $K$, the Robertson-Seymour theorem implies there is a finite set of forbidden minors for graphs which are $K$-lessly embeddable. Hence, every embedding of $K_n$ for $n$ sufficiently large contains $K$, so one may ask (like I did) "what $n$ is sufficient for a given $K$, and what order must I connect the points $(1...n)$ in order to realize $K$?" The crossing condition comes from looking at the chords projected onto the $yz$-plane as viewed from the $-x$ direction (which I think is the direction I did my crossing calculations from). Finally, I moved the $yz$-projections of the integral $t$-valued points around so that they lay on the unit circle to make the pictures easier/clearer.

 UPDATE: I was just skimming through a paper on some results on knots which were unrelated to this one. That paper referenced the Ramsey number $r(L)$ of a link $L$, and upon following its references I found this paper which proves the existence of $r(L)$ using essentially the representation I layed out above:

http://www.ams.org/journals/tran/1991-324-02/S0002-9947-1991-1069741-9/S0002-9947-1991-1069741-9.pdf

So it would appear the answer to Question 2 is this is known (in at least one paper) as a plat representation (much to my surprise, Negami derives the existence of such representations using essentially the argument I gave above). So that would seem to settle this thread entirely...
 http://mathoverflow.net/questions/94086/partitions-of-an-interval/94088#94088 Answer by ARupinski for Partitions of an interval ARupinski 2012-04-15T03:14:41Z 2012-04-15T03:14:41Z
Unless I'm misunderstanding your defintion of having an 'next' interval to the left or right, what about a Cantor set like construction: divide (0,1) into three equal intervals. Let $(\frac{1}{3},\frac{2}{3})$ be included in your decomposition, then divide $(0,\frac{1}{3})$ and $(\frac{2}{3},1)$ into three equal intervals, keeping the middle one of each and further subdividing the left and right pieces. Repeating this process ad nauseum gives a decomposition of (0,1) such that no interval has a nearest neighbor on either side but whose closure is all of (0,1).
 http://mathoverflow.net/questions/86477/simple-question-in-the-representation-of-sl2-c/86529#86529 Answer by ARupinski for Simple question in the representation of SL(2,C) ARupinski 2012-01-24T12:44:52Z 2012-01-26T04:00:18Z
This is a bit of expansion on the answer of Mike Skirvin; in particular it gives one way of explicitly calculating the combinatorics involved. My previous answer, although correct in its result, is horribly roundabout and overly computational, a mathematical Rube Goldberg machine if you will; so after waking up this morning I realized there is a much easier approach using a recursion on Symmetric powers.

Define:

$L = U^3 + U + U^{-1} + U^{-3}$

$M = U^4 + U^2 + 2 + U^{-2} + U^{-4}$

Now define a sequence $S_i$ by:

$S_{-2} = 0$

$S_{-1} = 0$

$S_0 = 1$

$S_1 = L$

$S_k = L\cdot S_{k-1} - M\cdot S_{k-2} + L\cdot S_{k-3} - S_{k-4}$ for $k\geq 2$.

Note that the exponents of $U$ in $S_1$ are exactly the weights mentioned by Mike in his comment (1). It turns out the same is true for all the $S_k$: the exponents of $U$ in $S_k$ are exactly the set of weights of $Sym^k(Sym^3(V))$ and the coefficient of $U^\ell$ is exactly the multiplicity of the weight $\ell$ in $Sym^k(Sym^3(V))$.

From this, you can pick out the subrepresentations by looking at where coefficients change; since the weights of any $Sym^\ell(V)$ occur with multiplicity 1, the only time the coefficients change is when a new summand occurs.

For example, working out $Sym^3(Sym^3(V))$ one gets the following expression:

$U^9 + U^7 + 2U^5 + 3U^3 + 3U + 3U^{-1} + 3U^{-3} + 2U^{-5} + U^{-7} + U^{-9}$

For the module corresponding to the leading coefficient, subtract 1 from each exponent giving a copy of $Sym^9(V)$ and leaving:

$U^5 + 2U^3 + 2U + 2U^{-1} + 2U^{-3} + U^{-5}$

Repeat this process to pull out a copy of $Sym^5(V)$ and finally a copy of $Sym^3(V)$; there are no more terms left, so this is the complete decomposition of $Sym^3(Sym^3(V))$. In general, the expression for $Sym^k(Sym^3(V))$ in $U$ so obtained is of the form:

$a_0U^{3k} + a_2U^{3k-2} + a_4U^{3k-4} + ... + a_4U^{-3k+4} + a_2U^{-3k+2} + a_0U^{-3k}$

Then $a_0 = 1$ by Mike's comment (2) and the multiplicity of $Sym^\ell(V)$ for $\ell\geq 0$ in the decomposition is just $(a_\ell - a_{\ell+2})$ and the multiplicity of $Sym^{3k}(V)$ is 1 since $a_{-2} = 0$.

As for the recursion, it ultimately expresses symmetric powers in terms of lower symmetric powers and exterior powers; this can be proven using multiplication of Young diagrams and inclusion-exclusion although I don't have a good reference at hand.
 http://mathoverflow.net/questions/85593/decompose-tensor-product-of-type-g-2-lie-algebras/85625#85625 Answer by ARupinski for Decompose tensor product of type $G_2$ Lie algebras. ARupinski 2012-01-14T00:20:38Z 2012-01-14T05:02:41Z
The search term you want to look for is "Klimyk's Formula." This formula boils down to the following:

Fix $G$ a compact complex semisimple Lie group. Suppose $V(\lambda)$ and $V(\mu)$ are irreducible representations with highest weights $\lambda$ and $\mu$ respectively. Let $W_\lambda = {\lambda_1,\lambda_2,\ldots \lambda_d}$ be the multiset of weights of $V(\lambda)$ with $d = dim(V(\lambda))$. Then the irreducible components of $V(\lambda)\otimes V(\mu)$ are given by ${V(\mu+\lambda_i)}_{i=1}^d$.

To apply this in practice, you need to be comfortable with the concept of defining $V(\lambda)$ when $\lambda$ is not a dominant weight (which sometimes causes modules to cancel when they appear with both positive and negative signs in the sum), but it applies to lots of groups (even beyond the scope of compact complex semisimple in some cases if im not mistaken), and Littlewood-Richardson is just the special case of this formula in type $A$.

An example for $G_2$ (since that is also my favorite compact semisimple Lie group) is to let $\lambda = [1,0]$ be the highest weight of the 7-dimensional representation and $\mu = [0,1]$ the highest weight of the 14-dimensional adjoint representation.

The seven weights of $V(\lambda)$ are $[1,0]$, $[-1,1]$, $[2,-1]$, $[0,0]$, $[-2,1]$, $[1,-1]$, and $[-1,0]$ so Klimyk tells us the 98-dimensional tensor product decomposes as:

$V([1,1]) \oplus V([-1,2]) \oplus V([2,0]) \oplus V([0,1]) \oplus V([-2,2]) \oplus V([1,0]) \oplus V([-1,1])$

This is where familiarity with interpreting modules with non-dominant highest weights comes in; $V[-1,2]$ and $V[-1,1]$ turn out to be 0-dimensional modules, while $V([-2,2]) \cong -V([0,1])$*. Thus the terms which do not disappear are $V([1,1])$ which is a 64-dimensional module, $V([2,0])$ which is a 27-dimensional module, and $V([1,0])$ which is the 7-dimensional defining representation, a total of 98 dimensions.

If you had instead chosen to switch $\lambda$ and $\mu$ and add the 14 weights of $V([0,1])$ to [1,0], you would have obtained 14 modules, but as before, some would have been zero and others would have cancelled in pairs ultimately leading to the same three modules as above being the only things left over. In my opinion, this reflexivity always holding is the coolest thing about Klimyk's formula.

One neat corollary to Klimyk's formula is that a tensor product of two irreducible modules cannot decompose into a sum of more than $d$ irreducibles where $d$ is the minimum of the dimensions of the two modules.

*EDIT: After posting, I decided to add a bit more about modules with non-dominant highest weight. Basically, the weights of a group $G$ are permuted via the Weyl group action on the weights. Weights are determined by integer $r$-tuples where $r$ is the rank of $G$; tuples containing a -1 lie in the walls of the Weyl chambers and so the modules with these highest weights end up being 0. There are a few other subspaces which also correspond to walls; weights $\mu$ not lying in the walls satisfy $w(\mu) = \lambda$ for some dominant weight $\lambda$ (all coordinates nonnegative) and a unique $w$ in the Weyl group (i.e. only one $w$ in the Weyl group will take $\mu$ to a dominant weight, so $\lambda$ is also uniquely determined). Then $V(\mu)$ is defined by the following relationship:

$V(\mu) = (-1)^w\cdot V(\lambda)$

Here $(-1)^w$ is the sign representation which appears with all Weyl groups; in the $A$-series whose Weyl groups are the $S_n$'s this is the ordinary sign representation.
 http://mathoverflow.net/questions/83572/generalization-of-a-result-on-solvable-groups Generalization of a Result on Solvable Groups ARupinski 2011-12-15T22:35:37Z 2011-12-16T00:26:15Z
This question concerns finite groups.

It is a well-known fact that every subgroup of a solvable group must again be solvable; this is easily proven by looking at the derived series of a given subgroup.

What I have been thinking about for awhile now is if/how this generalizes to arbitrary finite groups? Specifically, given some set $S$ of finite simple groups, consider the set $\Gamma_S$ of all finite groups whose composition series only includes groups in $S$. Then the above statement on solvable groups can be rephrased as:

THEOREM: Let $S = \{ \mathbb{Z}/p\mathbb{Z}\}_{p\in P}$. If $G\in \Gamma_S$ then $\forall H \leq G$ one has $H\in\Gamma_S$.

Trying to generalize to arbitrary $S$, the obvious generalization to try is:

CONJECTURE (1): For $S$ an arbitrary set of finite simple groups, if $G\in \Gamma_S$ then $\forall H \leq G$ one has $H\in\Gamma_S$.

Unfortunately, this is easily seen to be false; let $S = \{A_6\}$. Then $A_6\in \Gamma_S$ but $A_5\leq A_6$ and $A_5\notin \Gamma_S$. The failure in this example then leads me to the following second attempt at a generalization:

CONJECTURE (2): Let $S$ be an arbitrary set of finite simple groups, and $c(S)$ denote the set of all finite simple groups which appear as composition factors of some subgroup of some element of $S$. If $G\in \Gamma_S$ then $\forall H \leq G$ one has $H\in \Gamma_{c(S)}$.

I have been trying to figure out how to prove Conjecture (2). One thought is to use some appropriate analogue of the derived series for the general case, although coming up with the right analogue seems elusive. I have also thought about using the characters of elements of $S$, but this too does not lead to any immediate insights.

So, does anyone know whether Conjecture (2) is indeed true, and if so have either a (short enough for a post) proof or reference to where this question or similar ones might have been considered before?
 http://mathoverflow.net/questions/79373/some-questions-concerning-a-random-number-process Some questions concerning a random number process ARupinski 2011-10-28T11:36:06Z 2011-10-31T02:36:24Z
Consider the following Markov process: Start with an integer $N = N_0$. Now repeatedly choose an $N_i$ uniformly at random in the range $[1...N_{i-1}]$ until $N_i = 1$ at which point one terminates the process. This produces a nonincreasing integer sequence ${N_0,N_1,\ldots,N_{k-1},N_k = 1}$.

Experimental evidence shows that as $N_0$ grows large, the expected length $E(N_0)$ of such a sequence seems to approach $ln(N_0)$. Equivalently, one expects that the average over many steps of $N_i/N_{i+1}$ is approximately $e$. Convergence to this expectation is slow however; for example if $N_0$ is a 1000-bit integer one finds that $E(N_0)$ satisfies roughly $2.71^{E(N_0)} = N_0$ and in particular the base agrees with $e$ to only around 2 decimal places.

Because the $N_i$ were chosen uniformly at random, for any given $i$ the expectation of $N_i/N_{i+1}$ is 2, so this seems to contradict the above observation that the average of $N_i/N_{i+1}$ is approximately $e$. To understand this discrepancy, consider a toy example where $N_1 = qN_0$ and $N_2 = (1-q)N_1$ for some $0\leq q\leq 1$. One sees that $N_2 \leq \frac{1}{4}N_0$ with equality iff $q=\frac{1}{2}$; clearly the average of the step ratios $q$ and $1-q$ is equal to the expected single step ratio $\frac{1}{2}$, but the composition of the steps has led to an overall decrease in the sequence at a rate faster than division by 2. Hence the above observation that the overall step decrease rate is approximately division by $e$ is plausible.

Main Question: How does one understand the appearance of $e$ in the expected step down rate (as opposed to some other constant)? Presumably it should appear as a result of some averaging of all possible step ratios, but I can't seem to see what the correct average to be considering is.

Secondary question: At the risk of being vague, does anyone know what an inverse to this process looks like? That is, a process where $M_0 = 1$ and at each step one chooses an $M_i\geq M_{i-1}$ at random such that the expected growth rate is roughly multiplication by $e$ but such that the expected growth at any step should be multiplication by 2. Clearly one cannot choose the next number uniformly at random since this would lead to infinite step and expected growth, so what probability distribution (if any) can be put on the integers greater than $N_i$ so that choosing a next element will lead to a process which looks roughly like the original process in reverse?

 http://mathoverflow.net/questions/75859/is-there-an-easy-description-of-the-structure-of-this-infinite-group Is there an easy description of the structure of this infinite group? ARupinski 2011-09-19T15:35:39Z 2011-09-20T07:47:17Z
Let $S_\infty$ denote the full symmetric group on countably many points and index the points by $\mathbb{N}$. For any weight (for lack of a better name) function $w:\mathbb{N}\rightarrow\mathbb R^+$, define a subgroup $S_\infty(w)\subset S_\infty$ as follows:

$$S_\infty(w) := \{\pi\in S_\infty|\sum_{\pi(k)\neq k} w(k)<\infty \}$$

The intuitive idea is that given a weight function $w$ a permutation $\pi\in S_\infty$ is in $S_\infty(w)$ if the sum of the weights of all points moved by $\pi$ is finite.

There are two well-known examples of such subgroups:

$S_\infty(1)$ is the group of finitely supported permutations of $\mathbb{N}$ or equivalently the direct limit of the finite symmetric groups.

$S_\infty(\frac{1}{n^2})$ is simply $S_\infty$ since the series $\{\frac{1}{n^2}\}$ is convergent.

So in this context, I have been thinking about the group $G := S_\infty(\frac{1}{n})$. It is clear $S_\infty(1)\not\cong G$ since $G$ contains many subgroups isomorphic to $S_\infty$ (for example the set of all permutations of the points indexed by $b^i$ for $b$ fixed and $i\in\mathbb{N}$).

On the other hand,  $G$ contains a proper subgroup isomorphic to the direct sum of countably many copies of $S_\infty$, so in particular it is easy to see $S_\infty\not\cong G$ since in $G$ elements of infinite order always have nontrivial centralizers while in $S_\infty$ this is not always the case.

Question: Aside from the description of $G$ as $S_\infty(\frac{1}{n})$, is there any easy to describe/understand presentation of this group?

Primarily I am torn on whether $G$ is itself isomorphic to a direct sum of countably many copies of $S_\infty$; every time I attempt to describe such an isomorphism, there are elements of $G$ not accounted for.
 http://mathoverflow.net/questions/75698/examples-of-seemingly-elementary-problems-that-are-hard-to-solve/75717#75717 Answer by ARupinski for Examples of seemingly elementary problems that are hard to solve? ARupinski 2011-09-18T02:06:43Z 2011-09-18T02:27:09Z
One simple to state problem in number theory is the Collatz Conjecture, and I'm kind of surprised no one has mentioned this one yet (considering the examples you cite, I suspect it is because this one does not require anything beyond grade-school math to understand and so may seem below the level of this question). Nevertheless I add it to the list because it is amazingly addictive to think about.
 http://mathoverflow.net/questions/75232/to-what-extent-can-one-prescribe-degrees-of-irreducible-representations-of-a-grou To what extent can one prescribe degrees of irreducible representations of a group? ARupinski 2011-09-12T15:58:37Z 2011-09-12T19:57:34Z
Suppose one starts with an (infinite) multiset of positive integers $\mathcal{A} = \{a_i\}_{i\geq 0}$ such that:

$1=a_0\leq a_1\leq a_2\leq\ldots$

Can one always find a (necessarily infinite) group $G$ such that the set of degrees of its finite dimensional irreducible complex representations is exactly $\mathcal{A}$?

Clearly the answer to the above question is no for arbitrary $\mathcal{A}$; for example if $a_1 = N>1$ (i.e. the trivial representation is the only 1-dimensional representation of the potential group), then a simple argument involving the decomposition of the tensor square of this $N$-dimensional representation shows that $N\leq a_2\leq \frac{N^2+N}{2}$. Using more general forms of such arguments involving tensor powers of decompositions of $GL_n(\mathbb{C})$-representations, one can find other restrictions ruling out certain potential degree sequences.

$\bf{Question:}$ Given $\mathcal{A}$ such that there are no obstructions arising from tensor power considerations, is it always possible to find/construct a group $G$ whose irreducible degree sequence is exactly $\mathcal{A}$?

I assume that in general this is still a hard question, so I am also interested in partial results. I would also be interested in negative results along the lines "Even though this sequence $\mathcal{A}$ has no obstructions, it still cannot be the degree sequence of any group for some deeper reason."
 http://mathoverflow.net/questions/74902/free-group-of-finite-rank-can-contain-free-groups-of-infinite-rank-as-a-subgroup/74917#74917 Answer by ARupinski for free group of finite rank can contain free groups of infinite rank as a subgroup ARupinski 2011-09-08T18:34:33Z 2011-09-08T18:34:33Z
In case an explicit description of such a subgroup would help, let $F_2 = \langle a,b|\rangle$ be the free group on 2 generators. Then the subgroup generated by the elements $\{a^nb^n\}_{n\in\mathbb{N}}$ is free since it is a subgroup of a free group. Next, it is easy to check that the generator $a^kb^k$ is not an element of the free group generated by the $a^\ell b^\ell$ for $\ell\neq k$ so one concludes that this subgroup has infinite rank.
 http://mathoverflow.net/questions/74471/multiplying-functions-on-the-unit-square-as-generalized-matrices/74493#74493 Answer by ARupinski for Multiplying functions on the unit square as generalized matrices ARupinski 2011-09-04T04:28:01Z 2011-09-04T04:28:01Z
This is a partial answer concerning the determinants of such objects. While Will Jagy is right that such objects will have an uncountable spectrum, there still may be some hope for assigning a determinant in some cases as follows. First there is one obvious candidate of a trace for such objects: $Tr(F) = \int_0^1 F(t,t)dt$ (Note that the Identity does not have trace equal the dimension of the matrix under this definition, but this definition parallels your approach of replacing all finite summations with integrals). Then in analogy with finite-dimensional matrix theory, the $k^{th}$ powersum of the spectrum of $F$ should be $Tr(F^k)$. Now using the Newton identities, one can recursively construct the sequence of elementary symmetric functions of the spectrum of $F$.

On the one hand, for ordinary matrices $M$ acting on a vector space $V$ of dimension $d$, the elementary symmetric function $e_i$ evaluated on the eigenvalues of $M$ is equal to the trace of the operator defined by the action of $M$ on $\Lambda^i V$. In particular, $\Lambda^d V$ is one-dimensional and therefore the action of $M$ on this space is simply a scalar, this scalar is exactly $e_d = det(M)$.

In your case, your generalized matrices act on an infinite-dimensional function space $V$ and $\Lambda^\infty V$ does not make sense. Nevertheless, combining the earlier definition of the trace of $F$ with the interpretation of the elementary symmetric functions above, we therefore have the following candidate for a determinant of such a generalized matrix $F$ (provided the limit exists):

$Det(F) = \lim_{i\rightarrow\infty}e_i(F)$

Where:

$e_1(F) = Tr(F)$

$e_2(F) = \frac{1}{2}\left(e_1(F)\cdot Tr(F)-Tr(F^2)\right)$

$e_3(F) = \frac{1}{3}\left(e_2(F)\cdot Tr(F)-e_1(F)\cdot Tr(F^2)+Tr(F^3)\right)$

$\ldots$

Of course the difficulty in this approach lies in efficiently calculating the $e_i(F)$ and showing the above limit exists. Thus you might need to assume some additional constraints to make this approach work.
 http://mathoverflow.net/questions/67199/subdivision-of-triangles-into-congruent-triangles Subdivision of triangles into congruent triangles ARupinski 2011-06-08T02:02:03Z 2011-06-08T19:29:00Z
Recently some old notes of mine have gotten me to thinking about the problem of subdividing a triangle into $N$ smaller triangles, all congruent to one another. A little thought shows the following are possible values of $N$:

$\bullet$ If $N$ is a perfect square then any triangle can be subdivided into $N$ such smaller triangles

$\bullet$ If $N$ is a sum of two squares, say $N = a^2+b^2$ then a right triangle with side lengths in the ratio $a:b:\sqrt{a^2+b^2}$ can be subdivided into $N$ smaller triangles

$\bullet$ If $N=3$ then a $30-60-90$ triangle admits a decomposition into 3 congruent triangles

$\bullet$ Furthermore, by iterating the subdivisions indicated above, one can also obtain such subdivisions for any $N$ of the form $3^k\cdot m^2$

Question 1: Are the values of $N$ listed above the only ones possible if one requires the subtriangles be similar to the original triangle?

Question 2: In each of the examples above, the subtriangles formed are always similar to the original triangle. Does the answer to Question 1 change if one no longer requires that subtriangles be similar to the original triangle. (For example, if one does not require that the subtriangles are congruent to the original triangle, then an equilateral triangle may be subdivided into 3 congruent copies of a 30-30-120 triangle)

 http://mathoverflow.net/questions/65459/appearances-of-exotic-compact-lie-groups Appearances of 'exotic' compact Lie Groups ARupinski 2011-05-19T19:06:40Z 2011-05-21T15:25:16Z
The structure theorem for compact Lie Groups states that all compact Lie groups are finite central quotients of a product of copies of $U(1)$ and simple compact Lie groups. And yet, as easy as arbitrary compact Lie groups are to describe, most Lie Groups one encounters are the various quotients of simple compact Lie groups and maybe some products of these groups (I will herein refer to such groups as standard Lie groups). The only compact examples which one encounters regularly that are not standard Lie groups are the unitary groups $U(n)$ (which are quotients of $U(1)\times SU(n)$) and $SO(4)$ (which is the diagonal $\mathbb{Z}/2\mathbb{Z}$ quotient of $Spin(4) \cong Spin(3)\times Spin(3)$).

I am currently trying to further expand my knowledge and understanding of compact Lie groups, so I am wondering:

Question:  Has anyone encountered examples of non-standard Lie groups (other than the $U(n)$'s and $SO(4)$) in their research, as the autormorphism group of some object they were studying, or in some other way? If so, would you give a bit of description of the setting you were working in as well as a description of the non-standard group which appeared?

Although given a non-standard group, one can easily construct algebraic objects for which it is the automorphism group, I am more interested in instances of the reverse of this process wherein a non-standard group appears in the course of thinking about some other problem.

Edit: Since there still seems to be some misunderstanding of the intent of the question, to clarify the situation I am interested in, I am looking for groups of the form $G_1\times\ldots\times G_k/H$ where each $G_i$ is a compact simple Lie group, $k\geq 2$ and $H\subsetneq Z(G_1\times\ldots\times G_k)$ is not of the form $h_1\times \ldots \times h_k$ with $h_i\subseteq Z(G_i)$. So examples with multiple factors such as the the Structure Group of the Standard Model described by Theo are the sort of thing I'm looking for.
 http://mathoverflow.net/questions/64925/occurence-of-trivial-representation-in-a-tensor-square Occurence of trivial representation in a tensor square. ARupinski 2011-05-13T19:21:53Z 2011-05-13T20:23:55Z
Suppose $G$ is a group and $V$ an irreducible representation of $G$. One has that $V\otimes V\cong \Lambda^2(V)\oplus Sym^2(V)$. It is well-known that if the trivial representation appears as a subrepresentation of $\Lambda^2(V)$ then $V$ is of quaternionic type; while if the trivial representation appears as a subrepresentation of $Sym^2(V)$ then $V$ is a of real type. From this approach, it is clear that the trivial representation cannot appear in both $\Lambda^2(V)$ and $Sym^2(V)$.

What I am curious about is as follows:

Question: Is there is some (relatively easy) way to see why the trivial representation cannot appear in both $\Lambda^2(V)$ and $Sym^2(V)$ without introducing the machinery of real/quaternionic types?

As a bit of motivation, if one looks at other subrepresentations, then for example if $G = G_2$ and $V_n$ is an $n$-dimensional irreducible representation of $G_2$, then $V_{64}$ appears as a subrepresentation of both $\Lambda^2(V_{27})$ and $Sym^2(V_{27})$. In particular it is possible for the intertwining number of $\Lambda^2(V)$ and $Sym^2(V)$ to be nonzero, but by the real vs. quaternionic characterization, the trivial representation is somehow special in that it cannot contribute to the intertwining number.
 http://mathoverflow.net/questions/63589/the-first-eigenvalue-of-a-graph-what-does-it-reflect/63702#63702 Answer by ARupinski for The first eigenvalue of a graph - what does it reflect? ARupinski 2011-05-02T14:20:37Z 2011-05-02T14:20:37Z
The few times I have ever worked with eigenvalues of graphs, it has been in relation to the path algebra of the graph; each path in the graph is an element of this algebra. At any rate, the characteristic polynomial of the graph gives exact recurrences for calculating the number of paths of a given length and in particular the largest eigenvalue will give the asymptotic growth rate of the number of paths of different lengths. This asymptotic growth is true if the largest eigenvalue has modulus larger than all other eigenvalues of the graph. If there are multiple eigenvalues of maximum modulus then the asymptotics of the path-growth function will change although offhand I don't know if having multiple eigenvalues of the same maximal modulus is actually possible (perhaps someone who does more graph theory than I would care to comment on this).
 http://mathoverflow.net/questions/133793/pell-numbers-and-the-primes Comment by ARupinski ARupinski 2013-06-15T01:27:45Z 2013-06-15T01:27:45Z Note that your second sequence $\\{2, 0, 2, 4, 10, 24,\ldots\\}$ is just the doubles of your first sequence, shifted right by one step. So it probably would not appear in the OEIS. http://mathoverflow.net/questions/133449/signless-stirling-numbers-of-1st-kind-and-probabilistic-descent/133585#133585 Comment by ARupinski ARupinski 2013-06-13T01:56:53Z 2013-06-13T01:56:53Z Note that this is a rewrite of my original answer to correct the cycle condition which I had misstated as pointed out by the OP. http://mathoverflow.net/questions/133449/signless-stirling-numbers-of-1st-kind-and-probabilistic-descent/133452#133452 Comment by ARupinski ARupinski 2013-06-13T01:08:38Z 2013-06-13T01:08:38Z Ahh, I see the issue. When I originally was constructing the bijection, I thought it was necessary to have the fixed points. But what it looks like I really want for the bijection is just that the steps of the descent mark the right endpoints of the cycles of $\psi$. Will edit the answer to remove this issue. http://mathoverflow.net/questions/129873/square-and-reversed-integer Comment by ARupinski ARupinski 2013-05-08T00:04:12Z 2013-05-08T00:04:12Z @Gerry Myerson: thanks for clearing that up. Obviously I read the formulation through too quickly without thinking about what it was asserting. http://mathoverflow.net/questions/129873/square-and-reversed-integer Comment by ARupinski ARupinski 2013-05-06T23:28:31Z 2013-05-06T23:28:31Z You claim that this is true for $m < 10^8$, but why does for example $m = 32$ work? I have that $f(32\times 32) = 4201 \neq f(32)\times f(32) = 529$ even though for 32 one has $a_0,a_1\in\{0,1,2,3\}$. Is there an extra assumption that is missing here? http://mathoverflow.net/questions/129345/n-in-a-row-game-on-mathbbr2 Comment by ARupinski ARupinski 2013-05-04T00:37:30Z 2013-05-04T00:37:30Z @Benjamin: I had thought about that sort of approach too, but how do you ensure that your 4-in-a-rows so that they align with each other (i.e. why can P2 not see this strategy and play to block the 4-in-a-rows that you need, forcing you to form one not aligned with your previous 4-in-a-rows)? Is there a good way to force P2 to only block certain potential 4-in-a-rows so that P1 eventually gets a 5-in-a-row? http://mathoverflow.net/questions/129345/n-in-a-row-game-on-mathbbr2/129347#129347 Comment by ARupinski ARupinski 2013-05-02T22:06:17Z 2013-05-02T22:06:17Z Also by playing his points so that intersections of any of the lines he forms do not coincide with any of P2s already played points, it is obvious that by move 5 P1 has too many double 3-in-a-rows lined up for P2 to stop them all with just one point. http://mathoverflow.net/questions/129345/n-in-a-row-game-on-mathbbr2/129347#129347 Comment by ARupinski ARupinski 2013-05-02T22:04:42Z 2013-05-02T22:04:42Z @Ricky: So maybe he needs one more move then... if P1 always plays his points such that they are not collinear with any pair of points already on the plane, then even if P2 has blocked 2 of the potential 3-in-a-rows on the Fano configuration, P1s fifth point will form several new partial Fano configurations; by this point there are 10 different lines formed by the pairs of P1s points; each of these lines intersects one another somewhere in the plane and at most 3 of P2s points are on any of these lines (P2s first point does not help P2 in any way because of P1s non-collinearity strategy). http://mathoverflow.net/questions/129345/n-in-a-row-game-on-mathbbr2/129347#129347 Comment by ARupinski ARupinski 2013-05-02T02:03:42Z 2013-05-02T02:03:42Z So no matter how P2 responds now, at least one of the edge points of this Failed Fano configuration is open, so P1 forms two 3-in-a-rows simultaneously, thereby foiling any further attempt by P2 to win. I don't see any good way to extend this approach to $n>4$ (on account of the rather special nature of the Failed Fano configuration), but maybe this sparks an insight by someone else for the general case... http://mathoverflow.net/questions/129345/n-in-a-row-game-on-mathbbr2/129347#129347 Comment by ARupinski ARupinski 2013-05-02T01:57:27Z 2013-05-02T01:57:27Z For $n=4$ a similar approach should work. P1 can always play his 2nd and 3rd points into a triangle with at most one of its edges containing one of P2s first two points. Now no matter how P2 chooses to block one of these edges, P1 can find a point such that none of the lines connecting that point to his previous 3 points contain any of P2s points thusfar and which is not collinear with any pair of his previously placed points. Note that P1s points now form the 3 vertices and center point of the Failed Fano configuration. http://mathoverflow.net/questions/120277/factorization-of-permutations Comment by ARupinski ARupinski 2013-01-30T04:38:01Z 2013-01-30T04:38:01Z Your question seems to be covered by the comments and answers to [mathoverflow.net/questions/62088/…. And unless I misunderstand what you are asking in (2), there is no way $\Phi$ can be surjective for an entire range of $\alpha$'s if it fails to be surjective for some $\alpha$ in the given range. http://mathoverflow.net/questions/119679/confirm-refute-fx-where-fx-x-th-mersenne-prime-m-p-where-x-is-1 Comment by ARupinski ARupinski 2013-01-23T22:51:21Z 2013-01-23T22:51:21Z See primes.utm.edu/notes/faq/NextMersenne.html and some of the linked pages; in particular there are conjectures about the log distributions of Mersennes and some supporting evidence. http://mathoverflow.net/questions/119474/does-this-system-of-equations-admits-a-solution/119476#119476 Comment by ARupinski ARupinski 2013-01-21T16:35:05Z 2013-01-21T16:35:05Z The covering sequence can then be taken as the $a_i$ and the sequence of cofactors can be taken as the $b_i$ which then gives a solution to your particular problem. http://mathoverflow.net/questions/119474/does-this-system-of-equations-admits-a-solution/119476#119476 Comment by ARupinski ARupinski 2013-01-21T16:34:15Z 2013-01-21T16:34:15Z @Antisha: Ronald Graham's 1964 paper "A Fibonacci-like sequence of composite numbers" shows how to construct a linearly recurrent sequence all of whose entries are composite numbers using the Fibonacci recurrence. By modifying his approach it should be possible to find a covering sequence for this recurrence, (although there may be some complications I don't immediately see due to the fact that here the associated polynomial $x^2-3x+2$ factors while for the Fibonacci sequence the associated polynomial $x^2-x-1$ does not factor). http://mathoverflow.net/questions/117151/does-this-poset-have-a-unique-minimal-element Comment by ARupinski ARupinski 2013-01-20T22:16:14Z 2013-01-20T22:16:14Z There also seem to be examples with infinite degree vertices, so perhaps total classification might be out of reach, but classification of those with only finite degree vertices might be doable. I am definitely going to think about this some more.


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# The Knight's Sword

3,276pages on
this wiki
 Release date: 6 April 2001 (Update) Start: Talk to the Squire in the White Knights Castle in Falador Difficulty: ***** Length: Short/Medium Requirements: 10 Mining, 10 Cooking (optional), 15 Smithing (optional), another player to distract Sir Vyvin, food and armour (recommended) Items needed: Redberry pie, pickaxe, 2 Iron Bars Monsters to kill: None, but you must be able to evade aggressive level 32 Hobgoblins, level 68 Ice Giants, and level 57 Ice Warriors (easily done at Combat level 30)

## WalkthroughEdit

To start the quest, talk to the Squire in the southern room on the ground floor in the White Knights Castle. He explains he lost Sir Vyvin's sword, a sword that was passed down for five generations. Offer to help, and he'll explain Imcando Dwarves made the sword, a race he believes no longer exists. However, he thinks Reldo the librarian at Varrock Castle might know more about the Imcando Dwarves and tells you to talk to him.

Head to the Varrock Library, the north-most room in the castle, and talk to Reldo. He'll tell you that he believes some Imcando Dwarves may be living in Asgarnia on the southern plateau, and that their favourite food is redberry pie. Before you leave, grab the Pie Dish in the room on the eastern side of the castle.

### Making a redberry pieEdit

Unless you have one already, you will now need to bake a redberry pie. Gather one bundle of redberries from south-east of Varrock, and then head to Lumbridge to create flour. Use it on a bucket of water and create pastry dough. Use the pastry dough on the Pie Dish, and then use the redberries on the pie dish. Now simply cook the pie, and you have created a redberry pie.

### Thurgo and the portraitEdit

Now, head to Thurgo's Peninsula, south of Falador, and give him the redberry pie. He'll eat it and be much more responsive. He offers to make the sword for you, but he needs a picture of the sword to make an exact replica. Go back to the White Knights Castle in Falador and speak to the Squire again. He will tell you that the portrait of the sword is in his cupboard upstairs. Go up to the second floor, then go up the ladder in the south-western corner to go up to Sir Vyvin's quarters.

This is where you'll need somebody else to help you out. You'll have to search the cupboard nearby to get a portrait of his sword, but you can't do this unless you have another player talking to Sir Vyvin at the same time. Once you have the portrait, grab a pickaxe, some food, and some armour from your bank. You should also create or buy two Iron Bars now.

### Blurite oreEdit

Head back to Thurgo and he will take the portrait from you. Now, in order to create the sword, he needs two Iron Bars and one Blurite ore. Blurite ore can only be obtained from the dungeon beneath the cliff beside his house. Enter the dungeon and go past the Muggers, Pirates, and Hobgoblins to where the Ice Giants and Ice Warriors are.

Near the Ice Giants, you'll some some blue coloured rocks. These are blurite rocks containing blurite ore. Mine one ore and then leave the dungeon. Mine a second blurite ore if you want a sword for yourself. Go talk to Thurgo and he can now make the sword. He'll hammer a sword and give you the Faladian Knight's Sword. Once you have it in your inventory, return to the Squire in Falador who will take it from you to complete the quest.

### Creating a sword for yourselfEdit

It is possible to have your own Faladian Knight's Sword, though you can only get it before the quest is completed. Bring another two Iron Bars and one blurite ore to Thurgo. Drop the one sword he made, and then ask him to make you another sword. When you have it, pick up the one you dropped and you'll now have two swords. The Squire will only take one, leaving the other in your hands.

## RewardsEdit

• 1 Quest Point
• Faladian Sword, ONLY if an extra one is made during the quest
• Smithing experience:
• ${Level \times 375 + 350 XP}$
• 725 xp at level 1 (instantly reach level 7)
• 37,475 xp at level 99

## TriviaEdit

• When this quest was first released, it was simply titled "Sword Quest."

format: plain


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## 4. Quantization of Symplectic Quasi-States

Cite this as: AimPL: Non-Hermitian quantum mechanics and symplectic geometry, available at http://aimpl.org/nhquantumsymp.


text:
# Create iid normal random vectors

Le us create N independent and identitcally distributed (iid) normal random vectors with zero mean and specific covariance matrix. Each vector is of size p. Let $\Sigma$ be the specific covariance matrix.

    p = 20;
N = 50;
M = chol($\Sigma$)
Matr =  M' * randn(p, N)

In fact I want to know why in all references, they use chol?. Why they don't simply write Matr = randn(N, p) * ($\Sigma$) ? So what is the advantage of chol (or cholcov) in this case?

I made some results, and I noticed that they significantly differ when I remove the chol !! Even it takes more time for the computation..why??

Any help will be very appreciated!


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IMR Press / JOMH / Volume 17 / Issue 4 / DOI: 10.31083/jomh.2021.025
Open Access Original Research
Effect of a combination of aerobic exercise and dietary modification on liver function in overweight and obese men
Show Less
1 Research Academy of Grand Health, Ningbo University, Ningbo City, Zhejiang, China
2 Faculty of Sport Science, Ningbo University, Ningbo City, Zhejiang, China
J. Mens. Health 2021 , 17(4), 176–182; https://doi.org/10.31083/jomh.2021.025
Submitted: 11 January 2021 | Accepted: 8 February 2021 | Published: 30 September 2021
Abstract

Background: Obesity is not only associated with cardiovascular diseases but also a primary cause of liver dysfunction and other related diseases. This study’s aim was to determine the impact of a combination of dietary modification and aerobic exercise on liver function in overweight and obese adult males.

Methods: 45 overweight or obese men were randomly divided between the control group (n = 22) and intervention group (n = 23). Subjects in the intervention group were provided with dietary modification and aerobic exercise programmes. Dietary modification is a diet which restricts calorie intake and balances nutrients. Before and after 12 weeks of intervention, participants’ anthropometric characteristics and biochemical parameters relating to liver function including aspartate aminotransferase (AST), alanine aminotransferase (ALT), alkaline phosphatase (ALP) and gamma-glutamyl transferase (GGT) were measured.

Results: 12 weeks of aerobic exercise and dietary modification resulted in average weight loss of 10.6%, and body mass index, waist circumference and fat percentage decreased by 10.2%, 9.4% and 14.5% (p $<$ 0.05). AST, ALT, GGT and ALP in the intervention group reduced by 20.6%, 18.1%, 37.7% and 6.1% (p $<$ 0.05). Compared to the control group, AST, ALT, GGT and ALP in the intervention group were markedly lower (p $<$ 0.05). Furthermore, there was a markedly positive relationship between the reduction rates of body weight and GGT (p $<$ 0.05).

Conclusion: 12 weeks of aerobic exercise and dietary modification caused significant weight, waist circumference and body fat percentage reduction in overweight and obese men and their liver function was improved. The findings can provide a scientific reference for the improvement of liver function and prevention of liver diseases among overweight and obese people.

Keywords
Aerobic exercise
Dietary modification
Liver function
Overweight and obesity
Weight reduction
1. Introduction

Due to high energy intake and low levels of physical activity, the prevalence of obesity has increased rapidly [1]. Obesity, as a chronic disease, has become one of the most serious public health threats of the 21st century [2]. Many studies have noted that obesity and being overweight are generally accompanied by one or more risk factors of cardiovascular disease [3-5]. Obesity not only has an association with cardiovascular diseases but also a major cause of liver dysfunction and other related diseases. Recent studies have proven that being overweight or obese can result in liver dysfunction including impaired hepatic mitochondrial function, liver cirrhosis and fibrosis, and can even lead to the occurrence of non-alcoholic fatty liver disease [6-8]. Therefore, for overweight and obese people, improving their liver function and preventing fatty liver diseases is of great significance.

The liver plays a crucial role in lipid metabolism. Lipids can accumulate in the liver due to an imbalance that exists between the delivery of fat derived from adipose tissue stores or food intake to the liver and the consumption of fat as an ingredient of lipoproteins [9]. This partly explains why such a close relationship exists between liver diseases and obesity. At the same time, several epidemiological research projects have demonstrated that almost all patients who suffer from non-alcoholic steatohepatitis are between 10% and 40% over their ideal weight. Although adults with normal body weight can suffer from non-alcoholic steatohepatitis, it is more frequently detected in those who are overweight or obese [10].

It is essential for overweight and obese individuals to choose an appropriate method for reducing their body weight. Generally, aerobic exercise and dietary modification are considered to be effective and non-medical treatments for the monitoring and management of body weight. Previous studies have proven that aerobic exercise intervention is better for improving cardiopulmonary function and decreasing the risk of cardiovascular disease in comparison to dietary modification intervention [11,12]. At the same time, dietary modification intervention may be more effective than aerobic exercise intervention for facilitating a reduction in body weight and body fat [13,14]. Therefore, a combined aerobic exercise and dietary modification intervention is more frequently used to facilitate weight reduction among overweight and obese people. However, there is still debate surrounding whether this combined intervention can improve liver function in different individuals [15-17] and the impact the combined intervention has on the liver function of overweight and obese adults remains unclear.

Liver function is measured by several biochemical parameters including aspartate aminotransferase (AST), alanine aminotransferase (ALT), alkaline phosphatase (ALP) and gamma-glutamyl transferase (GGT) [18,19]. This study’s aim was to examine the effect of a combined aerobic exercise and dietary modification intervention on weight reduction and liver function in adult overweight and obese males. The study’s hypothesis was that 12 weeks of dietary modification and aerobic exercise intervention could reduce body weight and improve the liver function of overweight and obese adult males.

2. Methods
2.1 Study design

The study was a randomised controlled trial which used a control group. By utilising a random number generator, participants were randomly assigned to either the control group (n = 22) or the intervention group (n = 23) and participants were provided with aerobic exercise and dietary modification for 12 weeks. Participants’ anthropometric characteristics and blood biochemical indicators relating to liver function were measured both before and after the 12-week intervention period. The purposes and procedures of the research were all explained to participants and it was requested that they read and sign an informed consent form before participation. The research protocol was in full compliance with the latest modification of the Ethics Guidelines of the Declaration of Helsinki which was reviewed and approved by the Human Ethics Board of Ningbo University.

2.2 Participants

In this study, overweight was defined as a body mass index (BMI) of 24.0 to 27.9 kg/m${}^{2}$ and obesity was regarded as a BMI $\geq$ 28.0 kg/m${}^{2}$, based on the guidelines for preventing and controlling obesity in Chinese adults [20]. Subjects were recruited using local newspaper advertisements and posters for this study. Subjects who fulfilled the following inclusion criteria were included in our study: 1) adult men aged 20 years or older; 2) a BMI now more than 24 kg/m${}^{2}$; 3) body weight change of no more than 5 kg in the previous 6 months; 4) no exercise habits, meaning a total exercise time of less than 150 minutes per week; 5) no current or past disorders relating to the respiratory or cardiovascular systems; 6) no injuries or musculoskeletal disorders which affect participation in physical activity. A total of 45 men fulfilled these criteria. The subjects’ baseline characteristics are shown in Table 1.

Table 1.The baseline characteristics of the subjects
 Control group (n = 22) Intervention group (n = 23) P-value Age (yrs) 49.5 ± 11.4 50.8 ± 10.9 0.92 Height (cm) 167.2 ± 5.6 167.9 ± 6.8 0.36 Weight (kg) 80.0 ± 9.8 81.1 ± 10.6 0.52 BMI (kg/m${}^{2}$) 28.3 ± 2.1 28.7 ± 2.6 0.14
2.3 Measurements
2.3.1 Anthropometric characteristics

Subjects were required to wear light clothing with no shoes when their body height and weight were measured. BMI was then calculated as body weight in kilograms divided by the square of body height in metres (kg/m${}^{2}$). Each subject’s waist circumference was measured with a measuring tape. Bioelectrical impedance analysis was used for assessing body fat percentage using the bipolar foot-to-foot technique (BF-689; Tanita). Before measurements were taken, participants were required to fast for 10 to 12 hours and they had to have an empty bladder. Accuracy was 0.1 kilogram (kg) for the measuring of fat mass and 0.1% for the assessing of body fat percentage with a 50 kHz intensity of frequency of induction. It was reported that the repeatability coefficient of the measurements was 0.985 and the technical error was 0.639 [21].

2.3.2 Liver function

Following 10 to 12 hours of fasting, each participant’s venous blood sample was obtained from the antecubital vein and delivered directly to the central laboratory for the analysis of liver function parameters, including AST, ALT, GGT and ALP. Liver enzymes were tested enzymatically using an automatic analyser (Ci8200; Abbott Architect) which used standard protocols based on the manufacturer’s instructions. All laboratory assays were tested without knowing any of the subjects’ information.

2.4 Interventions
2.4.1 Aerobic exercise

The intervention was a moderate-intensity aerobic exercise programme consisting of a 90-minute session for 12 weeks (three times per week). Each 90-minute session included 15 minutes of warm-up and stretching exercises, 60 minutes of jogging or brisk walking and 15 minutes of cool-down and stretching exercises. Exercise intensity was expressed by the percentage value of maximal oxygen consumption (VO${}_{2}$max) and each subject’s VO${}_{2}$max and corresponding heart rate were obtained before the intervention. The linear relationship between the VO${}_{2}$, ratings of perceived exertion and heart rate has been proven [22]. Subjects exercised at 50% to 60% of their VO${}_{2}$max for the first four weeks of the intervention and this was gradually increased. For the final four weeks, the exercise intensity was set 60% to 70% of the subject’s VO${}_{2}$max. Indoor exercises were conducted on rainy days (seven times) using stair-stepping or ergometric cycling and the intensity of the indoor exercise matched that of the outdoors exercise as closely as possible. Several principal researchers and experienced fitness trainers were involved in the aerobic exercise intervention.

2.4.2 Dietary modification

The dietary modification programme consisted of one 90-minute session each week for 12 weeks and included theoretical and practical knowledge of diet management. During the session, participants were taught how to restrict their calorie intake to 1,680 kcal per day (a mean of 840 kcal of carbohydrates, 420 kcal of fat and 420 kcal of protein). The dietary modification programme’s main objective was to help subjects restrict their calorie intake and obtain a well-balanced intake of carbohydrates, protein, fat, vitamins, minerals and amino acids. Our research team had more than 10 years of experience in instructing subjects with this dietary modification. Their experience has proven that the dietary modification intervention is incredibly effective and safe for decreasing body weight and helping subjects form healthy eating habits. Additional detailed information relating to the dietary modification programme has been previously published [23].

2.5 Statistical analyses

In this study, the test data were presented as mean $\pm$ SD. Within-group differences between baseline and follow-up changes were tested using a paired t-test. Univariate analyses of variance (ANOVA) were conducted to assess between-group statistical differences. Pearson’s correlations were used for determining the relationship of reduction rate between anthropometric characteristics and liver function. Data analysis was conducted using the IBM Statistical Package for Social Sciences (SPSS, version 22.0). For statistical analysis, p $<$ 0.05 was considered to be statistically significant.

3. Results
3.1 Changes in anthropometric characteristics following the intervention programme

The results of differences in anthropometric characteristics following 12 weeks of the intervention programme are shown in Table 2. No significant changes in body weight, waist circumference, BMI and body fat percentage were observed after 12 weeks in the control group (p $>$ 0.05). However, in the intervention group, body weight decreased from 81.1 $\pm$ 10.6 kg to 72.4 $\pm$ 9.2 kg (a decrease of 10.6%, p $<$ 0.05), BMI decreased from 28.7 $\pm$ 2.6 kg/m${}^{2}$ to 25.7 $\pm$ 2.3 kg/m${}^{2}$ (a reduction of 10.2%, p $<$ 0.05), waist circumference decreased from 99.0 $\pm$ 8.4 cm to 89.6 $\pm$ 7.8 cm (a decrease of 9.4%, p $<$ 0.05) and body fat percentage decreased from 25.0 $\pm$ 4.3% to 21.5 $\pm$ 5.0% (a reduction of 14.5%, p $<$ 0.05) after 12 weeks of intervention. The intervention group’s body weight, waist circumference, BMI and body fat percentage showed significant decreases in comparison to the control group (p $<$ 0.05).

Table 2.Changes in anthropometric characteristics following 12 weeks of the intervention programme
 Control group Intervention group Pre-intervention Post-intervention Pre-intervention Post-intervention Weight (kg) 80.0 ± 9.8 80.2 ± 9.5 81.1 ± 10.6 72.4 ± 9.2${}^{*,\}$ BMI (kg/m${}^{2}$) 28.3 ± 2.1 28.3 ± 2.4 28.7 ± 2.6 25.7 ± 2.3${}^{*,\}$ Waist circumference (cm) 98.5 ± 9.0 98.8 ± 9.7 99.0 ± 8.4 89.6 ± 7.8${}^{*,\}$ Body fat percentage (%) 25.4 ± 4.6 25.6 ± 5.1 25.0 ± 4.3 21.5 ± 5.0${}^{*,\}$ Note: ${}^{*}$ means p $<$ 0.05 vs. pre-intervention; ${}^{\}$ means p $<$ 0.05 vs. control group.
3.2 Changes in liver function parameters after the intervention programme

The results of changes in liver function parameters following 12 weeks of the intervention programme are shown in Table 3. In the control group, no significant changes were observed in AST, ALT, GGT and ALP after 12 weeks (p $>$ 0.05). However, in the intervention group, AST decreased from 26.4 $\pm$ 9.7 U/L to 19.7 $\pm$ 5.6 U/L (a decrease of 20.6%, p $<$ 0.05), ALT decreased from 32.4 $\pm$ 22.7 U/L to 23.0 $\pm$ 10.4 U/L (a reduction of 18.1%, p $<$ 0.05), GGT decreased from 40.5 $\pm$ 19.0 U/L to 23.2 $\pm$ 9.9 U/L (a decrease of 37.7%, p $<$ 0.05) and ALP decreased from 203.0 $\pm$ 46.0 U/L to 188.7 $\pm$ 42.4 U/L (a reduction of 6.1%, p $<$ 0.05) following the intervention programme. The intervention group’s AST, ALT, GGT and ALP showed significant decreases in comparison to the control group (p $<$ 0.05).

Table 3.Changes in liver function parameters following 12 weeks of the intervention programme
 Control group Intervention group Pre-intervention Post-intervention Pre-intervention Post-intervention AST (U/L) 27.1 ± 10.4 27.3 ± 9.8 26.4 ± 9.7 19.7 ± 5.6${}^{*,\}$ ALT (U/L) 32.8 ± 20.7 34.4 ± 19.3 32.4 ± 22.7 23.0 ± 10.4${}^{*,\}$ GGT (U/L) 39.5 ± 18.4 41.4 ± 19.2 40.5 ± 19.0 23.2 ± 9.9${}^{*,\}$ ALP (U/L) 200.7 ± 50.6 207.2 ± 45.7 203.0 ± 46.0 188.7 ± 42.4${}^{*,\}$ Note: AST, aspartate aminotransferase; ALT, alanine aminotransferase; GGT, gamma-glutamyl transferase; ALP, alkaline phosphatase. ${}^{*}$ means p $<$ 0.05 vs. pre-intervention; ${}^{\}$ means p $<$ 0.05 vs. control group.
3.3 Correlations of reduction rate between anthropometric characteristics and liver function parameters

The results of correlations of reduction rate between anthropometric characteristics and liver function parameters are shown in Table 4. It can be seen from the table that only the GGT reduction rate displays a marked correlation with the decline rate of body weight, waist circumference, BMI and body fat percentage (r = 0.56 to 0.79, p $<$ 0.05). No significant correlation was found between the decline rate of obesity and the reduction rates of AST, ALT and ALP (p $>$ 0.05).

Table 4.Correlations of reduction rate between anthropometric characteristics and liver function parameters
 AST Δ ALT Δ GGT Δ ALP Δ Weight reduction -0.12 -0.09 0.76* -0.10 BMI reduction -0.11 -0.06 0.72* -0.16 Waist circumference reduction 0.03 0.04 0.79* 0.04 Body fat percentage reduction -0.26 -0.28 0.56* 0.28 Note: AST, aspartate aminotransferase; ALT, alanine aminotransferase; GGT, gamma-glutamyl transferase; ALP, alkaline phosphatase. $\Delta$ means reduction rate; ${}^{*}$ means the significance level p $<$ 0.05.
4. Discussion

Being overweight or obese makes individuals more vulnerable to liver dysfunction. Previous studies have shown that liver cirrhosis and fibrosis, impaired hepatic mitochondrial function and non-alcoholic fatty liver disease are all related to obesity [6-8]. Weight reduction can be helpful in facilitating the improvement of liver function. A combined aerobic exercise and dietary modification intervention is frequently used to aid weight reduction in overweight and obese individuals, but its effect on liver function remained unclear. Therefore, the aim of this study was to conduct an exploration of the impact of a combination of aerobic exercise and dietary modification on liver function in overweight and obese adult males. The results showed that 12 weeks of aerobic exercise and dietary modification led to significant weight, waist circumference and body fat percentage reduction in overweight and obese men and improved their liver function.

Obesity has become an increasing global public health problem in recent years. According to World Health Organization statistics [24], the prevalence of being overweight and obese among individuals aged 18 years and older is 39%, among which the prevalence of obesity is 13%. The fundamental cause of being overweight and obese is believed to be an increased intake of energy-dense foods and low levels of physical activity. Being overweight or obese can lead to a variety of metabolic disorders, including hypertension, hyperlipidaemia, type 2 diabetes and non-alcoholic fatty liver disease [25-27]. Many studies have shown that weight reduction significantly prevents the occurrence and development of the aforementioned metabolic disorders [28,29]. Therefore, it is both beneficial and healthy for overweight and obese adults to control their body weight or lower their BMI.

The liver is an important organ which is involved in maintaining the balance of lipid metabolism. When the body is in a state of illness, there are abnormalities in the lipid metabolism and a large number of lipid components enter the liver cells in order to make the liver synthesise the fat which has increased and accumulated, causing swelling, degeneration and even apoptosis in the liver cells and thereby resulting in impaired liver function [30,31]. Being overweight or obese can cause abnormal lipid metabolism, which can result in liver dysfunction. Verrijken et al. [32] reported that obesity is positively and significantly related to liver function parameters, including AST, ALT, GGT and ALP, and participants with high levels of visceral adipose tissue ($\geq$ 113 cm${}^{2}$) have a poorer liver function than those with low levels of visceral adipose tissue ($<$ 113 cm${}^{2}$). This study further demonstrated that weight reduction among overweight and obese men can dramatically enhance the values of AST, ALT, GGT and ALP. The study’s findings were consistent with those of Skrypnik et al. [15], who reported that exercise intervention causes a significant improvement to liver function among individuals with abdominal obesity. Based on the above results, it can be concluded that 12 weeks of a combination of aerobic exercise and dietary modification significantly improves the liver function of overweight and obese adult males.

Obesity and fatty liver have a close relationship. Previous research has highlighted that being overweight or obese is an independent factor that affects liver fibrosis in those with non-alcoholic fatty liver disease [33,34]. There is a greater likelihood that obesity will cause non-alcoholic steatohepatitis by disturbing Kupffer cell function and sensitising oxidant stress and hepatocytes to endotoxin [35]. Hannah and Harrison [36] demonstrated that a 3% to 5% weight reduction is related to decreased steatosis, while weight reduction of 7% to 10% is associated with fibrosis regression and non-alcoholic steatohepatitis remission. In this study, it was discovered that body weight decreased by 10.6% and the AST, ALT, GGT and ALP decreased by 20.6%, 18.1%, 37.7% and 6.1% following the combination of aerobic exercise and dietary modification intervention. The results show that liver function improvement may be concerned with weight reduction in overweight and obese adult males.

Aerobic exercise has been reported to decrease liver fat and insulin among obese individuals [37,38]. Keating et al. [39] documented that 45 to 60 minutes of aerobic exercise with 50% VO${}_{2}$max (four days per week) can significantly decrease 28% of intrahepatic lipid and that 30 to 45 minutes of aerobic exercise with 70% VO${}_{2}$max (three days per week) can reduce 29% of intrahepatic lipid. Huang et al. [40] further noted that a combination of dietary modification and aerobic exercise at 70% of the target heart rate can be effective for improving the liver histology of adults with non-alcoholic steatohepatitis (BMI $>$ 25 kg/m${}^{2}$). In addition, Baba et al. [17] discovered that 60% to 70% of maximal heart rate exercise combined with dietary modification can help decrease ALT concentrations among those with non-alcoholic steatohepatitis. The study also discovered that the normalisation of ALT is independent of weight loss. Similarly, this study showed that a combination of 50-70% of maximal heart rate exercise and dietary modification caused significantly improved liver function. In particular, the intervention can be used for the improvement of liver dysfunction and diseases that are triggered by being overweight or obese.

GGT is produced by the hepatocyte mitochondria and confined to the cytoplasm and intrahepatic bile duct epithelium [41]. Previous studies have demonstrated that GGT is significantly and negatively associated with an increase in physical activity [42,43]. GGT concentrations greater than 109 U/L and exercise times of less than 60 minutes per week are considered to be risk factors for diabetes [44]. Therefore, increased physical activity or an exercise intervention can help reduce GGT concentrations. Chen et al. [45] noted that a 10-week exercise programme in combination with diet improvement can significantly reduce GGT among those with fatty liver disease. Ohno et al. [46] observed that GGT values decreased by 50% following a 10-week exercise programme for sedentary individuals. In this study, it was also discovered that GGT values demonstrate a significant downward trend following 12 weeks of intervention. Further analysis showed that a reduction in GGT is significantly associated with a decrease in weight, waist circumference, BMI and percentage fat (r = 0.56 to 0.79, p $<$ 0.05). To summarise, exercise alone or exercise combined with dietary improvement can improve GGT and the improvement of GGT is associated with a reduction in body weight.

5. Conclusions

It is incredibly important for overweight and obese individuals to reduce their weight and improve their liver function. The results of this study suggest that 12 weeks of a combination of aerobic exercise and dietary modification significantly reduce weight, waist circumference and body fat percentage among overweight and obese men, while significantly improving their liver function. The findings can provide a scientific reference for the improvement of liver function and prevention of liver diseases in overweight and obese people.

Author contributions

Xiao-Guang Zhao is contributed in study design, manuscript writing, and submission; Hui-Ming Huang is contributed in study design and data analysis; Chen-Ya Du is contributed in manuscript writing and data collecting.

Ethics approval and consent to participate

The study protocol was reviewed and approved by the Human Ethics Board of Ningbo University (No: RAGH20190715).

Acknowledgment

The authors thank numerous subjects participated in this study. The authors express their gratitude to all the editors and peer reviewers for their comments and suggestions.

Funding

This research was supported by the Fundamental Research Funds for the Provincial Universities of Zhejiang (SJWY2020005), the Zhejiang Philosophical and Social Science Programme (21NDJC004Z), and the National Social Science Foundation in China (18BTY100).

Conflict of interest

The authors declare no competing interests.

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