Mathematics

2:02 PM

After blundering the queen back the computer gives the evaluation at +8

Which is converted 99% even by people at my level

It was so winning that the lichess analysis doesn't count giving the queen back as a mistake at all. Not even as an innacuracy

Balarka Sen

2:21 PM

@AlessandroCodenotti truth

Alessandro Codenotti

I'm reading an example which I feel I don't understand at all

Leaky Nun

@AlessandroCodenotti what is it

Alessandro Codenotti

Consider $H=\ell^2$ and $K=\prod_{n=1}^\infty [0,1/n]$ embedded into $\ell^2$ in the obvious way. Let $P$ be the subspace of $K$ of sequences with only rational coordinates. Let $Q$ be subspace of $H$ of sequences with only rational coordinates. Then $\dim P=0$ and $\dim Q=1$ but how

Don't they look the same locally

Balarka Sen

Weird

Alessandro Codenotti

Yeah apparently that $Q$ is the standard example to show that the inequality in the product theorem $\dim (X\times Y)\leq\dim X+\dim Y$ can be strict also for nice spaces

(Because $Q\times Q\cong Q$)

Leaky Nun

2:30 PM

"nice spaces"

Alessandro Codenotti

Separable metrizable

Doesn't get much better than that

Balarka Sen

Yeah this is a reasonable space. Surprising

Alessandro Codenotti

I'm going through the proof that $\dim Q=1$ now. $\dim P=0$ was very reasonable, let's see if I understand what's going on

Balarka Sen

You have to see some sort of connectedness of $Q$, or what?

Leaky Nun

what is dim here?

Balarka Sen

2:34 PM

Covering dimension surely

Alessandro Codenotti

Yes

The proof is actually for the small inductivee dimension because that's easier to work with, but they all agree for separable metric spaces anyway

@BalarkaSen Not sure, the point is that in $P$ points have a local basis of open sets with empty boundary, while in $Q$ that does not happen apparently

Balarka Sen

Ah

Strange, beats me

Alessandro Codenotti

I will inaugurate the blog I forgot to open with a write up of this stuff if I manage to understand it first lol

Balarka Sen

Excellent, I will read it

Knight

@Semiclassical What is the Eqaution of a wave on a string in the n th normal mode?

I want to know if it involves $kx$ like in these equations of waves $$ y(x,t) = A \cos (kx -\omega t)$$

Knight

2:53 PM

Hello Astyx

Astyx

Hi

Balarka Sen

If $X$ is an H-space, then $X_{\Bbb Q}$ is a product of $K(\Bbb Q, n)$'s which is either $S^{2k+1}_{\Bbb Q}$ or $\Omega S^{2k}_{\Bbb Q}$ by investigating the Postnikov tower. Therefore $H^*(X; \Bbb Q) = \Lambda_{\Bbb Q}(x_1, x_2, \cdots) \otimes \Bbb Q[y_1, y_2, \cdots]$

Knight

Why amplitude of a wave decreases even in non-absorbing medium?

Balarka Sen

If I demand $X$ is finite (finite cohomological dimension and finite dimension on each grade), then those generators are finite in number, and the polynomial algebra fucks off

Thus, $H^*(X; \Bbb Q) = \Lambda_{\Bbb Q}(x_1, \cdots, x_n)$

Knight

You see, intensity decreases with increase in distance and intensity is directly proportional to the square of the amplitude

Balarka Sen

3:08 PM

Well, OK, $X$ is rationally a product of odd-dimensional spheres. In particular $X$ is a $\Bbb Q$-PD space

Apparently it's true that an odd dimensional H-space is in fact homotopy equivalent to a manifold. Can we see this from here somehow?

I guess it's pretty rare for rational homotopy type to determine the full homotopy type

Alessandro Codenotti

Ah maybe I see it now, I don't think that $K$ and $Q$ look the same at all locally, I think that $K$ has empty interior has a subspace of $Q$ even

Balarka Sen

Yeah I just don't see what the natural manifold which will homotopy-model $X$ is

Alessandro Codenotti

By the way I noticed only now all the stars on my chess study group message, if there's actual interest in such a thing we could try going through Silman's endgames book or maybe the first of Yusupov's books

Balarka Sen

Honestly I would love to but I also have n things to do

Alessandro Codenotti

Fair

Balarka Sen

3:24 PM

I will hazard a guess that there is some map $X \to \prod S^{2k+1}$ which is a rational equivalence. So its homotopy fiber is some rationally contractible space. Maybe that is going to be a manifold

Maybe this is backwards and what one really should look at is the rational equivalence $\prod S^{2k+1} \to X$, and surger the domain so that this becomes a homotopy equivalence

@MikeMiller On a scale of 1-10, how bullshit is it that interesting surgery obstructions only lie in rational homology

Throw in simple connectedness of $X$

Lol, this is starting to sound like a super plausible idea to me. Look at the effect of $\prod S^{2k+1} \to X$ on homotopy groups; suppose you miss some torsion class in $\pi_i(X)$, attach an $i$-handle or something to $\prod S^{2k+1}$ so that this class is now in the image. Suppose it has some kernel which is a torsion class, represented by maybe an $i$-dimensional embedded sphere

Surgery along it to make it zero

The effect on $H_*(-; \Bbb Q)$ is remaining the same because you're doing surgery, which doesn't change the cobordism class of the map. So after each step you can do the process again

Mike Miller

3:39 PM

What do you mean "interesting surgery obstructions"?

Balarka Sen

I have no clue. If you have time maybe you can read what I am trying to do

It starts a few messages up

Mike Miller

Think of eg SU(3). This is a 3-sphere bundle over the 5-sphere, but not a trivial one. The bundle is classified by a non-trivial element of $\pi_4 SO(4)$, which is isomorphic to (Z/2)^2. I don't know if you get non-diffeo total spaces for the class (1,1) and (1,0) though, or which one SU(3) is.

I think you should assume simply connected to start.

Balarka Sen

That's a good example

Mike Miller

Even worse there are Lie groups with torsion in homology which are still rationally products of spheres

simply connected!

Balarka Sen

Why is that a bad thing

Mike Miller

3:50 PM

Your extensions are just very interesting is all

Balarka Sen

Ah OK

Mike Miller

So I object to the claim that the interesting surgery stuff is all rational

Balarka Sen

Got it, that makes sense

Yeah I should start by understanding and playing with one example

The SU(3) thing is a really good one

Mike Miller

Seems like Browder proved that an H-space with finitely generated homology has Poincare duality --- somehow by first showing this at each prime $p$ and that therefore it has it for everything at once, and somehow he shows it at each prime using the Bockstein spectral sequence somehow

Balarka Sen

I saw some computation of the mod p cohomology ring somewhere

Mike Miller

3:53 PM

Once you know that it's a PD complex (and you are expecting to produce parallelizable manifolds out of this) you have the input for surgery theory, a PD complex with normal data (given by the trivial bundle)

forgot that it's very easy to show that an s.c. complex with finitely generated homology is equivalent to a finite complex

Balarka Sen

So $\Bbb F_p$ cohomology is a direct sum of $\Lambda(x_1, \cdots, x_n; \Bbb F_p)$ and a truncated polynomial algebra over $\Bbb F_p$

That must be a PD ring

@MikeMiller What is this normal data?

Mike Miller

A vector bundle you want to be the stable normal bundle to your manifold

@BalarkaSen The hard part is ensuring that the top dimension is the same for different $p$

Balarka Sen

Ah makes sense

kauray

4:18 PM

Is a sigmoid function a linear function?

Astyx

4:31 PM

No

Thorgott

4:50 PM

can there be a set of more than rank many linearly independent elements in a free module?

Alessandro Codenotti

@Thorgott The keyword is invariant basis number I think

Thorgott

I only care about commutative rings, so that's a given, but this statement appears to be subtler

this is equivalent to saying that a free module can't have a free submodule of larger rank

Alessandro Codenotti

10

Q: Ranks of free submodules of free modules

Possible Duplicate: Atiyah-MacDonald, exercise 2.11 The following question came up during tea today. Let $R$ be a commutative ring with an identity and let $M \subset R^n$ be a submodule. Assume that $M \cong R^k$ for some $k$. Question : Must $k \leq n$? If $R$ is a domain, then thi...