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# zbMATH — the first resource for mathematics

Spinor bundles on quadrics. (English) Zbl 0657.14006
By means of the spinor varieties (parametrising the families of k-planes in a smooth quadric $$Q_{2k}$$ or $$Q_{2k+1}$$ in $${\mathbb{C}}P^{n+1})$$, we can embed the quadric in a Grassmannian. A spinor bundle on Q is what the author calls the pull back of the universal bundle of the Grassmannian under this embedding. This gives a stable vectorbundle S of degree $$2^{k-1}$$ $$(n=2k$$ or 2k-1). - Actually, in the case of n even, there are two spinor bundles corresponding to the two different families of linear spaces of maximal dimension in the quadric. These generalise the universal bundle and the dual of the quotient bundle on $$Q_ 4\simeq Gr(1,3).$$
Work concerning these two bundles on Q by R. Hernandez and I. Sols [J. Reine Angew. Math. 360, 124-135 (1985)] is extended to $$Q_ 5$$ and $$Q_ 6$$ using the new spinor bundles. In particular, every rank 3 stable vectorbundle on $$Q_ 5$$, whose Chern classes are all 2, arises as the quotient E of the dual of the $$Q_ 5$$-spinor bundle S in a sequence $$0\to {\mathcal O}\to S^*\to E\to 0.$$
Finally $$CP^ 7\setminus Q_ 6$$ is the fine moduli space for all such stable bundles on $$Q_ 5$$ (as well as for stable bundles on $$Q_ 6$$ with $$c_ 1=2=c_ 2$$, $$c_ 3=(2,0)$$ or (0,2)).
Reviewer: P.Bryant

##### MSC:
 14D20 Algebraic moduli problems, moduli of vector bundles 14M15 Grassmannians, Schubert varieties, flag manifolds 14F05 Sheaves, derived categories of sheaves, etc. (MSC2010) 14D22 Fine and coarse moduli spaces
##### Keywords:
Grassmannian; spinor bundle; fine moduli space
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##### References:
 [1] M. Atiyah, On the Krull-Schmidt theorem with application to sheaves, Bull. Soc. Math. France 84 (1956), 307 – 317. · Zbl 0072.18101 [2] Edoardo Ballico, Uniform vector bundles on quadrics, Ann. Univ. Ferrara Sez. VII (N.S.) 27 (1981), 135 – 146 (1982) (English, with Italian summary). · Zbl 0495.14008 [3] I. N. Bernšteĭn, I. M. Gel$$^{\prime}$$fand, and S. I. Gel$$^{\prime}$$fand, Algebraic vector bundles on \?$$^{n}$$ and problems of linear algebra, Funktsional. Anal. i Prilozhen. 12 (1978), no. 3, 66 – 67 (Russian). A. A. Beĭlinson, Coherent sheaves on \?$$^{n}$$ and problems in linear algebra, Funktsional. Anal. i Prilozhen. 12 (1978), no. 3, 68 – 69 (Russian). [4] Klaus Fritzsche, Linear-uniforme Bündel auf Quadriken, Ann. Scuola Norm. Sup. Pisa Cl. Sci. (4) 10 (1983), no. 2, 313 – 339 (German). · Zbl 0527.14017 [5] Phillip Griffiths and Joseph Harris, Principles of algebraic geometry, Wiley-Interscience [John Wiley & Sons], New York, 1978. Pure and Applied Mathematics. · Zbl 0408.14001 [6] Rafael Hernández and Ignacio Sols, On a family of rank 3 bundles on \?\?(1,3), J. Reine Angew. Math. 360 (1985), 124 – 135. · doi:10.1515/crll.1985.360.124 · doi.org [7] R. Lazarsfeld and A. Van de Ven, Topics in the geometry of projective space, DMV Seminar, vol. 4, Birkhäuser Verlag, Basel, 1984. Recent work of F. L. Zak; With an addendum by Zak. · Zbl 0564.14007 [8] Masaki Maruyama, Moduli of stable sheaves. I, J. Math. Kyoto Univ. 17 (1977), no. 1, 91 – 126. · Zbl 0374.14002 [9] Masaki Maruyama, Boundedness of semistable sheaves of small ranks, Nagoya Math. J. 78 (1980), 65 – 94. · Zbl 0456.14011 [10] Christian Okonek, Michael Schneider, and Heinz Spindler, Vector bundles on complex projective spaces, Progress in Mathematics, vol. 3, Birkhäuser, Boston, Mass., 1980. · Zbl 0438.32016 [11] Giorgio Ottaviani, A class of \?-bundles on \?\?(\?,\?), J. Reine Angew. Math. 379 (1987), 182 – 208. · Zbl 0611.14015 · doi:10.1515/crll.1987.379.182 · doi.org [12] -, Some extensions of Horrocks criterion to vector bundles on Grassmannians and quadrics, preprint. · Zbl 0718.14010 [13] Manfred Steinsiek, Transformation groups on homogeneous-rational manifolds, Math. Ann. 260 (1982), no. 4, 423 – 435. , https://doi.org/10.1007/BF01457022 Manfred Steinsiek, Über homogen-rationale Mannigfaltigkeiten, Schriftenreihe des Mathematischen Instituts der Universität Münster, Ser. 2 [Series of the Mathematical Institute of the University of Münster, Ser. 2], vol. 23, Universität Münster, Mathematisches Institut, Münster, 1982 (German). · Zbl 0491.32025 [14] Jacques Tits, Sur la trialité et certains groupes qui s’en déduisent, Inst. Hautes Études Sci. Publ. Math. 2 (1959), 13 – 60 (French). · Zbl 0088.37204 [15] Hiroshi Umemura, On a theorem of Ramanan, Nagoya Math. J. 69 (1978), 131 – 138. · Zbl 0345.14017
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.


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# Is 7 days too short for bounty question?

I saw this question today: A Topology such that the continuous functions are exactly the polynomials

Franklin.vp almost used up two thirds of his today reputation to offer this bounty, it is pretty a bummer for him if there is no nice answer popping up in 7 days. For this question, which looks like not an easy problem to me though I am not by any means an expert in algebraic topology, I wonder if there is any mechanism to offer a bounty with much longer period.

Another example came into my mind was this question: Probability that a stick randomly broken in five places can form a tetrahedron

Should a more flexible bounty system be introduced to MSE? Like with 5k rep you can offer a bounty up to a month, and with 10k rep you can offer bounty up to half a year.

EDIT: I did a search and saw Jeff Atwood's answer here: Is the time scale for bounties too short on stack exchange? I certain do not agree with him on point 3, at least bounty makes the question in the featured tab and gets the question more exposure over a long period of time.

I don't believe it is too short. Well, actually one of my bounties just ran through the week without someone writing an answer to it.
In my opinion making the bounties lasting longer there will be (at least I expect it) much more questions in the featured tab, which makes it less attractive for someone to offer a bounty, because there will be less extra attention because of the bounty.
On the other hand a short lasting bounty is maybe a motivation to spend more time in a short interval on the question.

• Well maybe there should be two kinds of bounty - one 7 day "feature me" bounty, and another "until answered" incentive without time limitation – Tobias Kienzler Jul 1 '13 at 14:39

I don't think bounty rules will be changed any time soon, but I don't think that's really a problem:

1. You can just post a comment saying that you offer N rep bounty for answer to this question (and after question gets a useful answer you can technically start a bounty and quickly award it).

2. ...And you can keep a question in featured tab for one month paying 750 rep (less than 20% of that 5k); keeping it there indefinitely is much more expensive (500 rep / week), but I'm not sure that would be a good idea anyway.

• Regarding 2, the price is higher: it will quickly reach 500/week. If a user already offered a bounty of $x$ points on a question, their next bountry of that question must be for at least $\min(2x,500)$ points. Source – ˈjuː.zɚ79365 Jun 26 '13 at 23:18
• @ˈjuː.zɚ79365 Oh, you're right. – Grigory M Jun 26 '13 at 23:20

format: plain


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# Math Help - Summation Problem

1. ## Summation Problem

Show that

Ʃn i2 = n(n+1)(2n+1)/6, n>=1
i=1

Should I use the formula
Ʃn i = n(n+1)/2
i=1

and if so, how would I go about doing this?

2. Use induction.

3. Originally Posted by jaclyn91

Should I use the formula
Ʃn i = n(n+1)/2
i=1
Yes, this works.

Now, use this simple fact: $i(i+1)=\frac{1}{3}\left( i(i+1)(i+2)-i(i-1)(i+1) \right).$

So the sum becomes $\sum\limits_{i=1}^{n}{i(i+1)}=\frac{1}{3}\sum\limi ts_{i=1}^{n}{\left( i(i+1)(i+2)-i(i-1)(i+1) \right)},$ now this is

$\underbrace{\sum\limits_{i=1}^{n}{i^{2}}}_{\text{Y ou need this}}+\underbrace{\sum\limits_{i=1}^{n}{i}}_{\tex t{You know this}}=\frac{1}{3}\underbrace{\sum\limits_{i=1}^{n }{\left( i(i+1)(i+2)-i(i-1)(i+1) \right)}}_{n(n+1)(n+2),\text{ because the above telescopes}}.$

I think you can take it from there.


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# Symbols:I/Imaginary Part

## Imaginary Part

$\map \Im z$ or $\map {\operatorname {Im} } z$

The imaginary part of a complex number $z$.

The $\LaTeX$ code for $\map \Im z$ is \map \Im z .

The $\LaTeX$ code for $\map {\operatorname {Im} } z$ is \map {\operatorname {Im} } z .

## Variants

Variants of $\map {\operatorname {Im} } z$ and $\map \Im z$ that can often be found are:

$\map {\operatorname {\mathscr I} } z$
$\map {\operatorname {im} } z$

The $\LaTeX$ code for $\map {\operatorname {\mathscr I} } z$ is \map {\operatorname {\mathscr I} } z .

The $\LaTeX$ code for $\map {\operatorname {im} } z$ is \map {\operatorname {im} } z .


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# Finding recursive relations in Mathematica

1. Aug 13, 2011

### mnb96

Hello,
I have a sequence of polynomials defined in the following way:
$$P_k(x) = \frac{\partial^k}{\partial x^k}e^{s(x)}\vert_{x=0}$$

Essentially the polynomial Pk is the k-th derivative of $\exp(s(x))$ evaluated at x=0. The function s(x) is a polynomial of 2nd degree in x.

In mathematica I define the polynomials with the following code:

D[Exp[s[x]], {x, k}]

For k=1..n one obtains a list of n polynomials P1(x),...,Pn(x).
My question is: is it possible to ask Mathematica to find a recurrent relation that expresses any Pk as a function of the previous polynomials Pk-1, Pk-2... ? (assuming it exists).

A similar well-known problem exists for http://en.wikipedia.org/wiki/Hermite_polynomials#Definition"

Thanks.

Last edited by a moderator: Apr 26, 2017
2. Aug 13, 2011

### Simon_Tyler

Try

Code (Text):
In[1]:= s[x_] := a x^2+b x+c

In[2]:= poly[k_] = Simplify[SeriesCoefficient[Exp[s[x]], {x, 0, k}], k>0]

Out[2]= DifferenceRoot[Function[{y,n},
{-2 a y[n] - b y[1+n] + (2+n) y[2+n]==0, y[0]==E^c,  y[1]==b E^c} ]]

The above shows the difference equation (recurrence relation) and the initial conditions that the polynomials must satisfy for any given quadratic s[x]. We can now use the recurrence relation to calculate the sequence. Below we compare calculating using poly[n] (the DifferenceRoot object) and Recurrence table

Code (Text):

In[11]:= dr = Table[poly[n], {n, 0, 20}]; // Timing
drExp = Expand[dr]; // Timing

Out[11]= {0.21, Null}
Out[12]= {1.57, Null}

In[13]:= rt = RecurrenceTable[{-2 a y[n] - b y[1 + n] + (2 + n) y[2 + n] == 0,
y[0] == E^c, y[1] == b E^c}, y, {n, 0, 20}]; // Timing
rtExp = Expand[rt]; // Timing

Out[13]= {0.01, Null}
Out[14]= {1.66, Null}

In[15]:= dr == rt
Out[15]= True

In[16]:= drExp == rtExp
Out[16]= True

The problem is that both evaluating the DifferenceRoot poly[k] and using RecurrenceTable, whilst great for purely numeric functions, are horrible for symbolic functions such as the one above. the reason for this is that they don't simplify y[n] after every step and end up with a terrible mess, e.g.

Code (Text):

In[17]:= dr[[6]]
drExp[[6]]
Out[17]= 1/5 (2/3 a (2 a b E^c+1/2 b (2 a E^c+b^2 E^c))+1/4 b (a (2 a E^c+b^2 E^c)+1/3 b (2 a b E^c+1/2 b (2 a E^c+b^2 E^c))))
Out[18]= 1/2 a^2 b E^c+1/6 a b^3 E^c+(b^5 E^c)/120
Finally, compare calculating by taking derivatives with calculating using http://reference.wolfram.com/mathematica/tutorial/FunctionsThatRememberValuesTheyHaveFound.html" [Broken] (in the latter, it's the Expand that makes the difference compared to the DifferenceRoot and RecurrenceTable options given above)

Code (Text):

In[25]:= deriv=Table[1/k! D[Exp[s[x]], {x,k}] /. x->0, {k, 0, 100}]//Expand;//Timing
Out[25]= {0.36, Null}

In[26]:= Clear[y]
y[n_Integer?Positive] := y[n] = 1/n (2 a y[-2+n] + b y[-1+n]) //Expand
y[0] = E^c;
y[1] = b E^c;

In[30]:= mem = Table[y[n], {n, 0, 100}]//Expand;//Timing
Out[30]= {0.3, Null}

In[31]:= deriv==mem
Out[31]= True

Last edited by a moderator: May 5, 2017
3. Aug 13, 2011

### Simon_Tyler

Also, if you want the recurrence relation for the polynomials not evaluated at x=0, try

Code (Text):

s[x_] := a x^2 + b x + c
poly[k_] = Simplify[SeriesCoefficient[Exp[s[x]], {x, x, k}], k > 0]
which returns

Code (Text):
DifferenceRoot[Function[{y,n},
{-2 a y[n] - (b + 2 a x) y[1+n] + (2+n) y[2+n]==0,
y[0]==E^(c+b x+a x^2),  y[1]==(b+2 a x) E^(c+b x+a x^2)}
]][k]

4. Aug 14, 2011

### Simon_Tyler

I must have had my brain turned off this morning, since I didn't mention that what you supplied was an exponential generating function. Just google "generating function recursion relation" to find stuff on how to approach these problems more generally. In particular the free book "generatingfunctionology" is worth reading (not that I can remember much of it)

5. Aug 16, 2011

### mnb96

Thanks a lot for the answers!


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# Convergent Summation Proof

I'll preface and apologize for the somewhat lengthy intro.

Alright, so recently I've been self-studying summation and found interest in convergent series. I was looking at some fairly common series and read that the reciprocals of triangular numbers summed indefinitely converge to $2$.

Immediately I thought of the reciprocals of powers of $2$ starting at $0$ and summed indefinitely, which also converge to $2$.

From here, I got the idea to do a proof that the two series' sums are equal. For reciprocals of powers of $2$, I didn't have any trouble getting the following:

$$\sum_{i=0}^{\infty}2^{-i} =\,\,2$$

However, writing the reciprocals of triangular numbers in summation notation was a bit more difficult. I started off with the following basic summation to simply generate triangular numbers:

$$\sum_{i=1}^n i\,\,\,|\,\,\,n\in\Bbb N_1$$ However, I needed to generate these terms some other way to function as my summand, so I used the following identity with a binomial coefficient to do so.

$$\sum_{i=1}^{n-1} i = \begin{pmatrix}n\\2\\ \end{pmatrix}$$

The difference of $1$ is negligible since we are summing indefinitely. Thus, we finally reach summation notation describing the reciprocals of triangular numbers indefinitely summed, which is the following:

$$\sum_{i=2}^{\infty} \begin{pmatrix}i\\2\\ \end{pmatrix}^{-1} =\,\,2$$

Now we get to our equation, which is:

$$\sum_{i=0}^{\infty}2^{-i} = \sum_{i=2}^{\infty} \begin{pmatrix}i\\2\\ \end{pmatrix}^{-1}$$

This is where the proof begins. Gauss' method obviously won't work with either part since we're working with indefinite values, not to mention exponents and reciprocals. I've tried researching for infinite summation properties, but I've only found some where the summand base is less than $1$. I have no idea how to deal with a summand that's a binomial coefficient.

I'm not asking for users to do the proof for me (In fact I don't want someone to do that), but if you could give me some applicable summation identities that would help me generate a proof from here, that would be fantastic.

• It's not clear what you're asking. Are you asking for a proof of that last equality without first knowing that they both sum to $2$? – Casteels Aug 30 '15 at 18:45
• Note that the reciprocal of $\binom{i}{2}$ is $\frac{2}{i(i-1)}$. This is equal to $\frac{2}{i-1}-\frac{2}{i}$. Now use telescoping. – André Nicolas Aug 30 '15 at 18:45
• @Casteels Not asking for someone to give me a proof, just asking for help in proving that that equation is true. – Lanier Freeman Aug 30 '15 at 18:58
• @AndréNicolas Fantastic, this is precisely what I needed to solve the rightmost summation! – Lanier Freeman Aug 30 '15 at 19:02
• Good, I am pleased that given a start you were able to finish things. – André Nicolas Aug 30 '15 at 19:09

We have the following: $$2^{-i} = \Bigg(\frac{1}{2}\Bigg)^i$$

And thus: $$\sum_{i=0}^\infty 2^{-i} =\sum_{i=0}^\infty \Bigg(\frac{1}{2}\Bigg)^i$$

Now you can use your properties for infinite sums that you have, if your base is $<1$

Infact, what you have here, is called geometric series by mathematicans. Look it up on wikipedia.

• Exactly what I needed. Now the left side of the equation is done. Thank you so much! – Lanier Freeman Aug 30 '15 at 18:53

Hint: For $T(n)=\frac{n(n-1)}{2}$ we have $\frac{1}{T(n)}=2\left(\frac{1}{n-1}-\frac{1}{n}\right)$. And please correct the statement "the two series are equal." You mean to say the two series' sums are equal.

• Grammar has been fixed, thanks for the catch. Could you write that first part in $LaTeX$? – Lanier Freeman Aug 30 '15 at 18:54
• @L.A.F.2. The formatting has been done. – callculus Aug 30 '15 at 19:18
• Fantastic, this will resolve the right side of the equation. Thanks! – Lanier Freeman Aug 30 '15 at 19:20


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# How many points are there on an elliptic curve reduced at a bad prime?

Given an elliptic curve $E$ defined over $\mathbb{Z}$, and a prime $p$, I know that Hasse's theorem gives, when $p$ is a good prime, a relation between the number of solutions over $\mathbb{F}_{p^n}$ and the number of solutions over $\mathbb{F}_p$ (for this, the coefficients of the equation are reduced mod $p$).

Is there such a relation also at the bad primes?

-
For bad primes it is much easier as there are essentially three cases to consider: additive reduction, split multiplicative reduction and non-split multiplicative reduction. It is a good exercise to try to count the number of points in each case yourself. –  Daniel Loughran May 31 '12 at 7:38

If the reduction is additive, there are $p+1$ points including one singular point. If it is split multiplicative it is $p$ and if non-split multiplicative, then it is $p+2$. See Washington "Elliptic curves, Number Theory and Cryptography ", section 2.10 on page 59.
... and see François comment below for $n>1$.
The way to remember it is that when you remove the singularity, the rest has a group structure, and it is isomorphic to the additive group (order $p$) in the case of additive reduction, to the multiplicative group of the base field (order $p-1$) or to the kernel of the norm map from the quadratic extension of the base (order $p+1$) in the two cases of multiplicative reduction. –  Chandan Singh Dalawat May 31 '12 at 8:05
So when one considers the reduction of $E$ as a curve defined over $\mathbf{F}_{p^n}$, then the number of points is $p^n+1$, $p^n$ in the case of additive resp. split multiplicative reduction. In the non-split multiplicative case it will depend whether $n$ is odd or even. –  François Brunault May 31 '12 at 8:14
The question was specifically when $n>1$, so Francois answered that. As a consequence of all this, the relation $|E(\mathbb{F}_p)| = p+1-a_p$ remains true for primes of bad reduction, where $a_p$ are coefficients of the Hasse-Weil L-function. It fails however when $p$ is replaced by $q=p^n$, except for primes where the reduction is additive. –  Anonymous May 31 '12 at 8:55
Sorry I did not see the $n$ in the unedited version. –  Chris Wuthrich May 31 '12 at 9:45
Careful. This answer is correct when the elliptic curve is given in Weierstrass form but not in general. For instance, $xy(x-y)=p$ defines an elliptic curve whose reduction modulo $p$ has $3p+1$ points. –  Felipe Voloch May 31 '12 at 17:25


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# Back-testing Stock Selection - Data Analysis Help Needed

 Sci Advisor P: 3,313 In my opinion, you can't think clearly about problems involving statistics unless you have a probability model for the phenomenon you are studying. If you do the traditional type of statistics without formulating such a model, the methods you use are actually assuming a particular probability model, so you haven't escaped the requirement - you've only succeeded in pushing it into the background or remaining ignorant of it. If the primary goal is do a conceptually clear analysis, you should formulate a probability model for how the data is generated. On the other hand, if the primary goal is to write an acceptable academic document, then focus you attention on which people are going to approve it. The use of statistics is subjective and different people may have very strong opinions on statistical methods. Certain methods are established traditions in certain fields of study. The simplest course of action is to ask the people who are going to review the document for suggestions and to follow those suggestions. Your question of how extract information from the data is a natural question, but it puts people who answer in the position of doing a mind-reading exercise as to goals of the analysis. My mind reading attempt is this: There is some sort of "utility" function that defines how well your valuation method works. I can't follow exactly how you compute it, but I'll imagine it as something like this: We assume that at time t = n, investor A invests, say, $10,000, in stocks that your method reccomends, dividing his money equally among those stocks. At the same time, investor B invests$10,000 in the same manner in stocks your method recommends against. We assume that at time t = n + 1, both investors sell their stocks. The utiity of your method from time t = n to t = n + 1 is (profit of investor A - profit of investor B). Let's assume your goal is to use the traditional (non-Bayesian) sort of statisitcs to "prove" that your method "works". The simplest formulation of this is to a "hypothesis test". For this, we need a "null hypothesis", which will express the general idea "There is no difference in the performance of investor A and investor B". However, the null hypothesis must say more than this. It must say enough to let us compute the probability of observing some statistic ( like the t-statistic ). For example, if you assume that each "step" in the observed data from time t = n to time t = n+1 is an independent draw from a random variable representing the distribution of utility then you can compute the proability of various statistics. Since your are dealing with time series data, I'm dubious of such a simplified assumption. It seems to me that what happens in successive steps of the process aren't independent events if you believe in "trends" in the stockmarket. Again, I emphasize that if you know the people who will review your document, then you should try to divine their opinions on such matters. If you can't speak to them directly, then try to look at work that they have written and see what they did.