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Singular Homology/Cohomology as a derived functor? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T23:32:01Z http://mathoverflow.net/feeds/question/66401 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/66401/singular-homology-cohomology-as-a-derived-functor Singular Homology/Cohomology as a derived functor? Blade 2011-05-29T22:32:52Z 2011-05-30T15:49:49Z
Hello, Learning some Alg.geometry and Sheaf theory, I got used to the notion that cohomology arises naturally as a derived functor of some sort.

This has led me thinking, singular cohomology, from algebraic topology, was never defined (In all books i've checked) as a derived functor, but just by giving cycles and boundaries. I could not figure out by myself any reasonable functor whose derived functors yield singular cohomology, So I pose this question out here.

I hope this might shed some more insight on what singular cohomology actually measures.

Thanks
 http://mathoverflow.net/questions/66401/singular-homology-cohomology-as-a-derived-functor/66406#66406 Answer by David Roberts for Singular Homology/Cohomology as a derived functor? David Roberts 2011-05-30T00:27:37Z 2011-05-30T00:27:37Z
Just some high-level thoughts...

Singular cohomology sits more naturally in the setting of (Quillen) model categories, using the chain of Quillen functors $Top \to sSet \to sMod_R \to Chain_R$. Quillen invented/discovered model categories in order to describe 'non-abelian derived functors', that is homotopical algebra rather than homological algebra. Essentially this means functors on the homotopy category, but described using objects of the original category (think: homotopy type represented by spaces). Reciprocally, we now can think about derived functors as being homotopical in nature.

If I was Urs Schreiber* I would now say that really this was all about $(\infty,1)$-categories and functors between them, but that is probably beyond the scope of the question. Or perhaps not! If so, ask away.

*A good colleague of mine :)
 http://mathoverflow.net/questions/66401/singular-homology-cohomology-as-a-derived-functor/66410#66410 Answer by Greg Friedman for Singular Homology/Cohomology as a derived functor? Greg Friedman 2011-05-30T01:51:57Z 2011-05-30T01:51:57Z
Elaborating a bit on Qiaochu Yuan's comment, if X is "nice enough" and $\mathcal{R}$ is the constant sheaf with stalks in $R$, then the singular cohomology agrees with the derived functor of the global section functor: $H^\ast(X;\mathcal{R})\cong H^\ast(X;R)$. This result is scattered throughout Bredon's book on sheaf theory, though I grant that it's not super-intuitive there. Alternatively, it's not hard to show the sheafification of the presheaf complex of singular cochains $U\to C^\ast(U;R)$ is a resolution of $\mathcal{R}$ and it can be shown to be homotopically fine, by a dual argument to the proof that the sheaf complex of singular chains is homotopically fine. Furthermore the presheaf of cochains is conjunctive and, while it's not a mono-presheaf, the cohomology with zero supports of the cochain presheaf is trival. Putting all those things together, the singular cohomology is isomorphic to the hypercohomology of the cochain sheaf complex, which is a derived functor of the global section functor, though in the "hyper" sense. Similarly things can be done with homology, the sheaf of germs of singular chains being homotopically fine. Swan's book on sheaf theory is a good reference for that.
 http://mathoverflow.net/questions/66401/singular-homology-cohomology-as-a-derived-functor/66441#66441 Answer by Tim Porter for Singular Homology/Cohomology as a derived functor? Tim Porter 2011-05-30T12:00:57Z 2011-05-30T12:00:57Z
On seeing the question, I knew that I had seen something like this. Rinehart had a TAMS paper: 1972 called SINGULAR HOMOLOGY AS A DERIVED FUNCTOR. His interpretation of derived functor is, of course, that of his era but it does give a setting that may be adapted to a more modern treatment. Following up on David's reply, it is also possible to use Quillen's treatment of cohomology from the original Homotopical Algebra SL notes, together with ideas on monadic cohomology (the Triples volume of SLN paper by Applegate and Tierney, plus ideas from Mike Barr) to get another take on the question.... but without the various volumes in front of me I will not attempt to do that here and now. :-)
 http://mathoverflow.net/questions/66401/singular-homology-cohomology-as-a-derived-functor/66467#66467 Answer by Saul Glasman for Singular Homology/Cohomology as a derived functor? Saul Glasman 2011-05-30T15:49:49Z 2011-05-30T15:49:49Z
Here's an elaboration of David Roberts' answer, a perspective that can be found in Quillen's 'Homotopical Algebra'. Quillen thought of homology as "derived abelianisation", and the abelianisation of a space (by which I mean simplicial set) can only be the free simplicial abelian group on that space, seeing as it's left adjoint to the forgetful functor $sAb \to sSet$.

Composing the free simplicial abelian group functor with the Dold-Kan correspondence - an equivalence of categories between simplicial abelian groups and chain complexes - gives a functor $sSet \to Ch$, which happens to coincide (more or less) with the singular complex functor. If we endow $Ch$ with the injective model structure this is a left Quillen functor. By Quillen's philosophy, homology of spaces should be the total left derived functor of this functor. But the word 'derived' is sort of vacuous here, because all simplicial sets are cofibrant, and this negates the need to explicitly use any model category language. This is why we think of the singular complex as the "homology object" of a space.


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# Crossnumber winners

Are you a winner?

Before we get on to winners, here is the solution of the Chalkdust prize crossnumber #2:

 6 7 1 2 5 1 1 6 5 5 3 7 6 7 7 4 9 3 9 4 1 0 0 6 2 5 4 8 9 2 1 4 0 9 1 2 2 3 7 1 8 8 4 3 5 9 5 3 8 9 7 3 4 0 1 6 3 1 1 1 8 2 7 2 7 7 7 7 7 7 8 8 8 8 8 8 9 9 7 2 1 4 5 0 8 2 1 8 9 0 1 9 1 8 9 0 8 8 8 8 8 8 8 6 7 6 0 9 4 1 0 9 6 1 2 5 7 6 1 4 5 5 9 0 2 1 2 4 2 1 9 6 0 0 0 1 6 1 0 3 8

The sum of the across clues was $277777800764962$.

### Picking the Winners

We received 50 entries of which 38 were correct.

This time, we put all the correct entries in a Santa hat and got our glorious leader, Rafael Prieto Curiel, to pick the winners:

Thank you to everyone who entered the crossnumber competition. The competition will return in issue 3 of Chalkdust in March.

Matthew Scroggs is a PhD student at UCL working on finite and boundary element methods. His website, mscroggs.co.uk, is full of maths and now features a video of him completing a level of Pac-Man optimally.
@mscroggs      mscroggs.co.uk    + More articles by Matthew

• ### Power-up review

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Learn how to model your heart one beat at a time ❤
• ### Math Blaster

Belgin plays a classic mathsy game from her childhood...in 16-bit graphics! Here's her review...
• ### Why Knot?

You can un-knot a knot, by cutting it not?
• ### Surname Extinction

Why do surnames die out? We take a look at the Galton-Watson process for modelling the extinction of surnames to answer the question: 'When will we all be Smiths?'
• ### Crossnumber winners, Issue 06

Did you solve it?

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# The Unapologetic Mathematician

## Proving the Classification Theorem IV

We continue proving the classification theorem. The first three parts are here, here, and here.

1. Any connected graph $\Gamma$ takes one of the four following forms: a simple chain, the $G_2$ graph, three simple chains joined at a central vertex, or a chain with exactly one double edge.
2. This step largely consolidates what we’ve done to this point. Here are the four possible graphs:

The labels will help with later steps.

Step 5 told us that there’s only one connected graph that contains a triple edge. Similarly, if we had more than one double edge or triple vertex, then we must be able to find two of them connected by a simple chain. But that will violate step 7, and so we can only have one of these features either.

3. The only possible Coxeter graphs with a double edge are those underlying the Dynkin diagrams $B_n$, $C_n$, and $F_4$.
4. Here we’ll use the labels on the above graph. We define

$\displaystyle\epsilon=\sum\limits_{i=1}^pi\epsilon_i$

$\displaystyle\eta=\sum\limits_{j=1}^qj\eta_j$

As in step 6, we find that $2\langle\epsilon_i,\epsilon_{i+1}\rangle=-1=2\langle\eta_j,\eta_{j+1}\rangle$ and all other pairs of vectors are orthogonal. And so we calculate

\displaystyle\begin{aligned}\langle\epsilon,\epsilon\rangle&=\sum\limits_{i=1}^pi^2\langle\epsilon_i,\epsilon_i\rangle+2\sum\limits_{i

And similarly, $\langle\eta,\eta\rangle=\frac{q(q+1)}{2}$. We also know that $4\langle\epsilon_p,\eta_q\rangle^2=2$, and so we find

$\displaystyle\langle\epsilon,\eta\rangle^2=p^2q^2\langle\epsilon_p,\eta_q\rangle^2=\frac{p^2q^2}{2}$

Now we can use the Cauchy-Schwarz inequality to conclude that

$\frac{p^2q^2}{2}=\langle\epsilon,\eta\rangle^2<\langle\epsilon,\epsilon\rangle\langle\eta,\eta\rangle=\frac{p(p+1)q(q+1)}{4}$

where the inequality is strict, since $\epsilon$ and $\eta$ are linearly independent. And so we find

\displaystyle\begin{aligned}\frac{p^2q^2}{2}&<\frac{p(p+1)q(q+1)}{4}\\2p^2q^2&

We thus must have either $p=q=2$, which gives us the $F_4$ diagram, or $p=1$ or $q=1$ with the other arbitary, which give rise the the $B_n$ and $C_n$ Coxeter graphs.

February 25, 2010 Posted by | Geometry, Root Systems | 5 Comments

## Proving the Classification Theorem III

Sorry, but what with errands today and work on some long-term plans, I forgot to put this post up this afternoon.

We continue with the proof of the classification theorem. The first two parts are here and here.

1. If $\{\epsilon_1,\dots,\epsilon_k\}\subseteq\mathfrak{D}$ is a simple chain in the graph $\Gamma$, then we can remove these vectors and replace them by their sum. The resulting set is still admissible, and its graph is obtained from $\Gamma$ by contracting the chain to a single vertex.
2. A simple chain in $\Gamma$ is a subgraph like

We define

$\displaystyle\epsilon=\sum\limits_{i=1}^k\epsilon_i$

I say that if we replace $\{\epsilon_1,\dots,\epsilon_k\}$ by $\{\epsilon\}$ in $\mathfrak{D}$ then we still have an admissible set $\mathfrak{D}'$. Any edges connecting to any of the vertices $\epsilon_i$ will then connect to the vertex $\epsilon$.

It should be clear that $\mathfrak{D}'$ is still linearly independent. If there’s no way to write zero as a nontrivial linear combination of the vectors in $\mathfrak{D}$, then of course there’s no way to do it with the additional condition that the coefficients of all the $\epsilon_i$ are the same.

Because we have a simple chain, we find $2\langle\epsilon_i,\epsilon_{i+1}\rangle=-1$ for $1\leq i\leq k-1$ and $\langle\epsilon_i,\epsilon_j\rangle=0$ otherwise. Thus we calculate (compare to step 2)

\displaystyle\begin{aligned}\langle\epsilon,\epsilon\rangle&=k+\sum\limits_{i

and so $\epsilon$ is a unit vector.

Any $\eta\in\Gamma$ can be connected by an edge to at most one of the $\epsilon_i$, or else $\Gamma$ would contain a cycle (violating step 3). Thus we have either $\langle\eta,\epsilon\rangle=0$ if $\eta$ is connected to no vertex of the chain, or $\langle\eta,\epsilon\rangle=\langle\eta,\epsilon_i\rangle$ if $\eta$ is connected to $\epsilon_i$ by some number of edges. In either case we still find that $4\langle\eta,\epsilon\rangle^2$ is $0$, $1$, $2$, or $3$, and so $\mathfrak{D}'$ is admissible.

Further, the previous paragraph shows that the edges connecting $\epsilon$ to any other vertex $\eta$ of the graph $\Gamma$ are exactly the edges connecting the chain to $\eta$. Thus we obtain the graph $\Gamma'$ of $\mathfrak{D}'$ by contracting the chain in $\Gamma$ to a single vertex.

3. The graph $\Gamma$ contains no subgraph of any of the following forms:
4. Indeed, if one of these graphs occurred in $\Gamma$ it would be (by step 1) the graph of an admissible set itself. However, if this were the case we could contract the central chain to a single vertex by step 6. We would then find one of the graphs

But each of these graphs has four edges incident on the central vertex, which violates step 4!

February 25, 2010 Posted by | Geometry, Root Systems | 4 Comments


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## Are these linear transformations?

I shoud show whether $A: V \rightarrow \mathbb{F}^m$ and for all $A(v)=\begin{bmatrix} f_1(v)\\ f_2(v)\\ \vdots\\ f_m(v)\end{bmatrix}$ and $f_i$ is a linear functional $f_i \in L(V, \mathbb{F}^1), i=1,2,\dots ,m$. is a linear transformation. I think no, because the image of the zero vector is not 0. To be honest I am not really sure about these lienar functional thing.

Second. $C \in L(V), \{v_1,v_2,v_3\}$ v1 v2 v3 is a basis of V.
$C(\alpha v_1+\alpha v_2+\alpha v_2)=(\alpha-\beta+2\gamma)v_1 + (2\alpha + \gamma)v_2 + (3\alpha + \beta)v_3$. About this I know it is a linear transformation. Is it invertible. I know it would be enough to show its injectivity but how can I do this in this case?

Thank you!


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# Solving For X on both sides

1. Mar 24, 2009

### Stratosphere

1. The problem statement, all variables and given/known data
A basic example is $$\sqrt{x-1}$$=x-7

2. Relevant equations

3. The attempt at a solution
I end up with an x$$^{2}$$ and a normal x I dont know to to solve it.

2. Mar 24, 2009

### Integral

Staff Emeritus
Please show us your work. You say you have and x2. ok, you should have an x2 term, now show us exactly what you have and how you got it.

3. Mar 24, 2009

### HallsofIvy

Staff Emeritus
So that's a quadratic equation, isn't it? You can solve quadratic equations by factoring, completing the square, or using the quadratic formula.

4. Mar 24, 2009

### Stratosphere

I thought this was a radical equation? Not a quadric one. If this is a quadric one I dont have to worry abought that yet then.
How do you solve somthing with a root on both side then? Like $$\sqrt{3x+2}$$ + $$\sqrt{2x}$$= 35, I keep getting an answer thats too high.

Last edited: Mar 24, 2009
5. Mar 24, 2009

### Office_Shredder

Staff Emeritus
Turn it into

$$\sqrt{3x+2} = 35 - \sqrt{2x}$$

square both sides

$$3x+2 = 35^2 + 2x - 70 \sqrt{2x}$$

and now you only have one square root

6. Mar 24, 2009

### symbolipoint

Stratosphere, your first example, $$\sqrt{x-1}$$ = x - 7, should give you no complications when you square both sides. It WILL give you a quadratic equation. Either you can (after suitable algebraic steps) factor and solve, or you can complete the square/use solution to quadratic equation to solve.

7. Mar 24, 2009

### Stratosphere

were does the -70 come from?

8. Mar 24, 2009

### General_Sax

$$(35-\sqrt{2x})^2$$

Expand it out.

9. Mar 24, 2009

### Chaos2009

Don't forget, when you square the radical, you can end up finding solutions for x that do not work. You have to go through and check these solutions with your original equation to make sure that each value works.

10. Mar 25, 2009

### Stratosphere

Oh and how do you know what is a,b and c when using the quadratic formula?

11. Mar 26, 2009

### yeongil

The general form of a quadratic equation is $$ax^{2} + bx + c = 0$$. Hopefully that answers your question. ;)

01


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Home Net Core Inline Razor Markup Null Model

# Net Core Inline Razor Markup Null Model

Tango Down
1#
Tango Down Published in 2018-01-11 18:12:43Z
 My first go with Razor Pages inline markup. Running into this weird issue after passing a ViewModel to a PartialView. Of course in my parent page I pass the ViewModel to the PartialView: @{Html.RenderPartial("Partial/_RequestView", Model.NewRequest);} public class IndexModel : PageModel { private readonly IActiveDirectoryClient _activeDirectoryClient; private readonly ITravelClient _travelClient; public IEnumerable Requests { get; set; }  In the partial view, I have no issue referencing the model in a lambda expression @Html.HiddenFor(model => model.RequestId)  However when I attempt to reference the Model in razor markup inline the Model is null. Any ideas?

@Model.Name

 The NewRequest property is set within the OnGetAsync() method in the parent page  public async Task OnGetAsync() { NewRequest = BuildNewRequest(); if (NewRequest == null) throw new NullReferenceException("Unable to build new travel request"); return await Task.FromResult(Page()); }
Tango Down
2#
Tango Down Reply to 2018-01-16 16:45:26Z
 Answered my own question. MUST remove the @page directive in order for the partial view to work and for the @Model to be recognized. Kinda bizarre that VS 2017 templates adds this directive for a partial view, I can only assume it's a bug.
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