Mathematics

Ultradark

12:48 AM

See this unit disk with lines in it?

Ultradark

1:01 AM

nevermind

Perturbative

6:52 AM

Ordinary homology/cohomology is the same as homology/cohomology with a trivial coefficient system which is the same as homology/cohomology with untwisted coefficients right?

Balarka Sen

Yup

Perturbative

Cool cool

ÍgjøgnumMeg

7:32 AM

Mornin'

Alessandro Codenotti

Morning everyone

@Balarka remember that I asked you about simplicial sets and geometric realization? I'm done with exams now so whenever you have time to talk about them I'd be very happy to listen!

Leaky Nun

8:50 AM

Jun 30 at 14:05, by Alessandro Codenotti

@Balarka I have some confusion concerning what precisely is the relationship between (small) $\infty$-groupoids and topological spaces

@AlessandroCodenotti this?

Balarka Sen

@Alessandro Neat! I can explain now if you want.

We can read Goerss-Jardine in more detail after 24th as well. I only read the first couple chapters.

Alessandro Codenotti

9:06 AM

@LeakyNun yes

@BalarkaSen sure!

Balarka Sen

Where should I start? Geometric realization?

Leaky Nun

@BalarkaSen did you look at my SSS?

Balarka Sen

@LeakyNun Yeah I saw. Were you able to carry out the computation?

Leaky Nun

ok let me try again

it's way too messy

and I received guidance from Mathein and Loch

Alessandro Codenotti

Wait let me get my laptop, I'll write down the definition of simplicial sets I'm familiar with to see if we agree :P

Leaky Nun

9:08 AM

I needed to compute $H^1(\Bbb RP^\infty)$ and $H^2(\Bbb RP^\infty)$ separately (i.e. not using SSS)

is this normal?

Balarka Sen

@Alessandro Let's move to garbology

Leaky Nun

should I move there as well?

Alessandro Codenotti

ok

Balarka Sen

@LeakyNun You should do it here, otherwise there'd be too much clutter in two simultaneous discussions!

@LeakyNun It doesn't seem necessary at first glance, but let me do the computation quickly myself.

You immediately get $H^1(\Bbb{RP}^\infty) = 0$ by thinking about the $E^2$ page. It's the $(1, 0)$-term right? No differential hits or comes out of it. It can't survive to $E^\infty$, otherwise $H^1(\Bbb{CP}^\infty)$ would be nonzero!

Leaky Nun

I don't think so, since $E^2_{0,1} = H^0(\Bbb CP^\infty; H^1(S^1))$ points to $E^2_{2,0} = H^2(\Bbb CP^\infty; H^0(S^1))$ on $E^2$

Balarka Sen

9:16 AM

Oh yeah I have the fiber bundle backwards, you're right.

$S^1 \to \Bbb{RP}^\infty \to \Bbb{CP}^\infty$. OK

$E^2_{p, q} = H^p(\Bbb{CP}^\infty; H^q(S^1))$.

There are two columns $q = 0, 1$ and only even rows

Leaky Nun

$K(\Bbb Z,1) \to \Bbb RP^\infty \to K(\Bbb Z,2)$

Balarka Sen

$(p, q) \to (p + 2, q - 1)$ because cohomological indexing

So goes two rows up and one column back

Right, so you need to prove $d^2_{(0, 1)}$ is an isomorphism to start off with

Leaky Nun

it isn't

$d(y) = 2x$

Balarka Sen

Zero kernel, potato pohtato

Leaky Nun

$H^\bullet(\Bbb CP^\infty) = \Bbb Z[x]$, $|x|=2$

ok

Balarka Sen

9:21 AM

OK, so yeah you need to know $H^1 = 0$ and $H^2 = \Bbb Z/2$ to show it's multiplication by $2$

Leaky Nun

right

Balarka Sen

Because the $(2, 0)$ term survives to $E^\infty$

Cool

Leaky Nun

then it's Leibniz rule

$d(y)=2x$, so $d(x^{2n}y) = 2nx^{2n-1}d(x)y + x^{2n} d(y) = x^{2n} d(y) = 2x^{2n+1}$

so $H^{2n+2} = \Bbb Z/2\Bbb Z$

$H^{2n+1} = 0$

and how do I obtain the cup product?

Balarka Sen

I'd write it as $d(1 \otimes y) = 2x \otimes 1$, so $d(x^k \otimes y) = kx^{k-1} \otimes 0 + 2x (x^k \otimes 1) = 2x^{k+1} \otimes 1$, just to avoid confusion.

@LeakyNun I don't think you can compute the cohomology ring from this, actually. It should be a fiber or a base space to figure out the differentials using specseq formalism

i might be wrong

Leaky Nun

then how do you prove that $H^\bullet(K(\Bbb Z,2)) = \Bbb Z[x]$, $|x|=2$?

Balarka Sen

9:29 AM

$S^1 \to pt \to K(\Bbb Z, 2)$, no? That's the pathspace fibration

The base is a $K(\Bbb Z, 2)$

You're using the double cover of this fibration, $S^1 \to S^\infty \to \Bbb{CP}^\infty$

$E^2_{p, q} = H^p(\Bbb{CP}^\infty, H^q(S^1))$. Again two columns. Converges to $E^\infty \equiv 0$

Leaky Nun

I mean, how do you work out the cup product?

Balarka Sen

$d : (0, 1) \to (2, 0)$ is an isomorphism, because this times $E^\infty$ is full zero

So $d(1 \otimes y) = x \otimes 1$

$d(x^k \otimes y) = x^{k+1} \otimes 1$. These all have to be isomorphisms so that none survives to $E^\infty$. So cup product with $x$ is an isomorphism.

That means $H^*(\Bbb{CP}^\infty) = \Bbb Z[x]$

Correct?

Leaky Nun

hmm

I see

Balarka Sen

These are tricky shit

Leaky Nun

10:14 AM

8

A: Cup product of cohomology in a Serre spectral sequence

Maybe the simplest example is the following. There are two fiber bundles with base and fiber both $S^2$. One is the product $S^2\times S^2$ and the other consists of two copies of the mapping cylinder of the Hopf map $S^3 \to S^2$ glued together via the identity map between their "source" ends $S...

God himself hath spoken

@BalarkaSen

Balarka Sen

lol nice

Secret

Dehn invariant

In geometry, the Dehn invariant of a polyhedron is a value used to determine whether polyhedra can be dissected into each other or whether they can tile space. It is named after Max Dehn, who used it to solve Hilbert's third problem on whether all polyhedra with equal volume could be dissected into each other. Two polyhedra have a dissection into polyhedral pieces that can be reassembled into either one, if and only if their volumes and Dehn invariants are equal. A polyhedron can be cut up and reassembled to tile space if and only if its Dehn invariant is zero, so having Dehn invariant zero is...

hmm...

Leaky Nun

numberphile much?

@BalarkaSen so how to work up the cup product in $\Bbb RP^\infty$?

can I just use simplicial cohomology?

it doesn't even exist

Balarka Sen

$H^*(\Bbb{RP}^\infty; \Bbb Z_2)$ is $\Bbb Z_2[x]$. Now you use the change of coefficients map $\Bbb Z \to \Bbb Z_2$ to get $H^*(\Bbb{RP}^\infty; \Bbb Z) \to H^*(\Bbb{RP}^\infty; \Bbb Z_2)$ and naturality to argue

Leaky Nun

what on earth

Balarka Sen

10:25 AM

This is an isomorphism in degree 2 and injective in all degrees, so the generator in degree $2$, say $y$ with $|y| = 2$ maps to $x^2$

Let's see. By naturality, $y^k$ maps to $x^{2k}$, which generates degree $2k$ there. So it's an isomorphism in even degrees, actually.

northerner

Guys I've been trying to get latex to work here. I know it's something to do with chat.stackexchange.com/rooms/36/mathematics but what exactly am I supposed to do? Does it work with firefox?

Balarka Sen

But since $H^*(\Bbb{RP}^\infty; \Bbb Z)$ only has even degrees, this justifies it's $\Bbb Z[y]/(2y)$

Tobias Kildetoft

@northerner You need to copy the text as the address in a bookmark

then click on the bookmark while in chat

Leaky Nun

$\Bbb Z[y]/(2y)$ is a different thing... right

northerner

Well if I click on the book mark it takes me away from chat and to that site.

Tobias Kildetoft

10:29 AM

@northerner Then you put in the wrong text

Leaky Nun

you have $y^2$ and $2y^2$

Tobias Kildetoft

the text is pure javascript as far as I recall, it should not take you anywhere else

northerner

Is this the right text? math.ucla.edu/~robjohn/math/mathjax.html

Leaky Nun

no

Balarka Sen

Those are all zero

$2y^2$ is in the ideal $(2y)$

Leaky Nun

10:31 AM

oh what am I thinking

I can't algebra

Tobias Kildetoft

@northerner you need the text in those links (or just drag it or right click it as it says)

Balarka Sen

It is a graded ring which is $\Bbb Z_2$ in even degrees and $0$ in odd degrees (except degree $0$, where it's $\Bbb Z$)

Leaky Nun

oh right

10 mins ago, by Balarka Sen

$H^*(\Bbb{RP}^\infty; \Bbb Z_2)$ is $\Bbb Z_2[x]$. Now you use the change of coefficients map $\Bbb Z \to \Bbb Z_2$ to get $H^*(\Bbb{RP}^\infty; \Bbb Z) \to H^*(\Bbb{RP}^\infty; \Bbb Z_2)$ and naturality to argue

what is the naturality, and why is it an isomorphism?

Balarka Sen

Like always think about polynomial rings $R[x]$ being a graded ring which is $R$ in $k$-th degree, generated by $x^k$

@LeakyNun Naturality says this map commutes with cup product

Leaky Nun

what on earth

Balarka Sen

10:33 AM

It's an isomorphism in degree 2 because... uh.

It should be clear, but I don't know how to say it.

Leaky Nun

to me it feels like you're inventing a new tool to solve every new problem

lol

Balarka Sen

I certainly didn't invent them. But these are very tricky, yeah

I can't do shit if you give me a complicated enough cohomology ring.

Oh yeah do cellular cohomology. That should make it clear why it's injective in every degree.

Leaky Nun

$\Bbb RP^\infty$ is $K(\Bbb Z/2\Bbb Z,1)$ right

(why?)

Balarka Sen

Ya

Leaky Nun

and why is $\Bbb CP^\infty$ a $K(\Bbb Z,2)$?

Balarka Sen

10:36 AM

Because it's double covered by $S^\infty$, contractible

$K(G, 1)$ space = $\pi_1 = G$ and universal cover contractible (this implies all higher homotopy vanishes)

Leaky Nun

can the universal cover be not contractible?

northerner

got it working thanks

Leaky Nun

the path space is always contractible, we proved it before, right

@northerner nice

Balarka Sen

@LeakyNun Universal cover of $\Bbb{RP}^2$ is $S^2$...

Leaky Nun

so the path space quotient homotopy isn't the universal cover?

Balarka Sen

10:38 AM

@LeakyNun This is harder. Let me get a cigarette to see if I have a simple explanation which doesn't go into classifying space bullshit.

Leaky Nun

hi @ÉricoMeloSilva

Balarka Sen

Oh alternatively you can brute force. Run homotopy fiber sequence on $S^1 \to S^\infty \to \Bbb{CP}^\infty$

That computes $\pi_n(\Bbb{CP}^\infty) = \Bbb Z$ if $n = 2$ and $0$ otherwise.

@LeakyNun It is, it just changes the homotopy type badly

In fact that's a very interesting point

The path-space fibration is $\Omega X \to P X \to X$, yeah?

Leaky Nun

yeah

so quotient of contractible... isn't contractible!

Balarka Sen

Note that $\pi_0 \Omega X = \pi_1 X$. That means the path components of $\Omega X$ is in bijection with $\pi_1 X$

Collapse each path component to a point

You get universal cover $\pi_1 X \to \widetilde{X} \to X$

So universal cover is path-space fibration with the connected components of the fibers collapsed to points

It's the 1-connected type of the path-space fibration, or something, as some people would say

A'ight gotta get a cigarette

jublikon

10:53 AM

Hi,

Every quadratic matrix can be rewritten as the sum of a symmetric and a skew-symmetric matrix.
I have a task, where I have to show for two linear mappings that they are diagonalizable.
I am able to create a generator tuple consisting of symmetric and skew symmetric matrices (these are the eigenvectors of the linear mappings). Naive way: Take standard bases and rewrite every vector in there as the sum of symmetric and skew symmetric matrix -> add the corresponding symmetric and skew symmetric matrix to the new generator tuple.

Thorgott

11:19 AM

Not sure if I'm correctly understanding the issue, but by your first sentence you have $M_n(F)=\mathrm{Sym}_n(F)+\mathrm{Skew}_n(F)$ and thus you can obtain a basis of $M_n(F)$ only containing symmetric and skew-symmetric matrices. You can even show that that sum is direct. It requires $2\in F^{\times}$ though.

You can also give such a basis explicitly: Take $e_{\alpha,\alpha}$ with $1\le\alpha\le n$, $e_{\alpha,\beta}+e_{\beta,\alpha}$ with $1\le\alpha<\beta\le n$ and $e_{\alpha,\beta}-e_{\beta,\alpha}$ with $1\le\alpha<\beta\le n$ (where $e_{\alpha,\beta}$ is the matrix that has $1$ in the position $(\alpha,\beta)$ and $0$ elsewhere). (this also only works in characteristic $\neq2$)

jublikon

11:48 AM

@Thorgott Thank you for your answer. It directly addresses to my issue.
Can you explain the explicit basis more precisely please? I think I do not understand the formal definition completely.

Thorgott

In that case it might be helpful to look at some examples. In two dimensions, the basis is $\begin{pmatrix}1&0\\0&0\end{pmatrix}$, $\begin{pmatrix}0&0\\0&1\end{pmatrix}$, $\begin{pmatrix}0&1\\1&0\end{pmatrix}$ and $\begin{pmatrix}0&1\\-1&0\end{pmatrix}$.

In three dimensions, it is $\begin{pmatrix}
1&0&0\\0&0&0\\0&0&0
\end{pmatrix},
\begin{pmatrix}
0&0&0\\0&1&0\\0&0&0
\end{pmatrix},
\begin{pmatrix}
0&0&0\\0&0&0\\0&0&1
\end{pmatrix},
\begin{pmatrix}
0&1&0\\1&0&0\\0&0&0
\end{pmatrix},
\begin{pmatrix}
0&0&1\\0&0&0\\1&0&0
\end{pmatrix},
\begin{pmatrix}
0&0&0\\0&0&1\\0&1&0
\end{pmatrix},
\begin{pmatrix}
0&1&0\\-1&0&0\\0&0&0
\end{pmatrix},
\begin{pmatrix}
0&0&1\\0&0&0\\-1&0&0
\end{pmatrix}$ and $
\begin{pmatrix}
0&0&0\\0&0&1\\0&-1&0
\end{pmatrix}$.

jublikon

12:04 PM

very, very nice @Thorgott . Your have made it clear to me now.
This understanding will be very helpful in my exam next week.
Thanks! Have nice day :)

Thorgott

No problem. Good luck with your exam!

jublikon

Thank you :)

Leaky Nun

1:32 PM

$0 \to \operatorname{Ext}_R^1(\operatorname{H}_{i-1}(X; R), G) \to H^i(X; G) \, \overset{h} \to \, \operatorname{Hom}_R(H_i(X; R), G)\to 0$

complex $C_{i-1} \xrightarrow \alpha C_i \xrightarrow \beta C_{i+1}$

$C_{i+1}^\ast \xrightarrow \gamma C_i^\ast \xrightarrow \delta C_{i-1}^\ast$

$\ker \delta / \operatorname{im} \gamma$

$\operatorname{Ext}^1_R(M,G) \to \operatorname{Ext}^1_R(H_{i-1}(X;R),G) \to \operatorname{Hom}_R(P,G) \to \operatorname{Hom}_R(M,G) \to \operatorname{Hom}_R(H_{i-1}(X;R),G)\to 0$

complex $C_{i+1} \xrightarrow \alpha C_i \xrightarrow \beta C_{i-1}$ and $C_{i-1}^\ast \xrightarrow {\beta^\ast} C_i^\ast \xrightarrow {\alpha^\ast} C_{i+1}^\ast$

Ultradark

is that a complex cobordism cochain

Leaky Nun

1:55 PM

$C_i \twoheadrightarrow C_i/B_i \twoheadrightarrow C_i/Z_i \xrightarrow \beta B_{i-1} \hookrightarrow Z_{i-1} \hookrightarrow C_{i-1}$

$0 \to Z_i/B_i \to C_i/B_i \to C_i/Z_i \to 0$

$0 \to H_i \to C_i/B_i \to B_{i-1} \to 0$

$0 \to B_i \to Z_i \to H_i \to 0$

user193319

2:20 PM

Problem:Show that there cannot be a sequence of continuous functions $f_n : [0,1] \to \Bbb{R}$ such that $\{f_n(x)\}$ is decreasing for every $x \in [0,1]$, $\lim_{n \to \infty} f_n(x) = 1$ if $x \in [0,1[ \cap \Bbb{Q}$, and $\lim_{n \to \infty} f_n(x) = 0$ if $x \in [0,1] \setminus \Bbb{Q}$.

I could use a hint.

Alessandro Codenotti

The fact that $\{f_n(x)\}$ is decreasing for every $x$ is actually not needed

(presumably it makes the problem easier even though I don't really see how to use this fact right now)

user193319

I thought that condition was odd.

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$Mathematics$ Mathematics