Mathematics - 2019-07-19

Ultradark

00:48

See this unit disk with lines in it?

Ultradark

01:01

nevermind

Perturbative

06:52

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

07:32

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

08:50

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

09:06

@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

09:08

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

09:16

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

09:21

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

09:29

$S^1 \to pt \to K(\Bbb Z, 2)$, 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

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

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

@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

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

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

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

@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

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

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

@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

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

13:32

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

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

13:55

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

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

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

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

user193319

14:20

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.

Ryan Unger

hmm

Alessandro Codenotti

functions which are pointwise limit of continuous functions are called Baire one functions, it's not trivial to show but it's well known that if $f$ is Baire one then every closed interval contains a point of continuity of $f$, so in particular it cannot be everywhere discontinuous

functions which are poinwise limit of Baire one functions are called Baire two (and so on), the indicator function of the rationals that you're looking at is Baire two for example

I found a nice reference

user193319

@AlessandroCodenotti Thanks! I appreciate it.

Ryan Unger

14:37

@user193319 notes from my analysis course that get straight to the point math.utk.edu/~freire/teaching/m447f16/BaireCategory.pdf

Alessandro Codenotti

Those are nice notes

(if you want the really hardcore version you should look at chapter 24 of Kechris's classical descriptive set theory)

flowian

14:55

Hello

Can somebody help me with a problem - find all subrings of $\mathbb{Z}_12$ ?

Tobias Kildetoft

@flowian Do you require subrings to contain $1$?

flowian

I thought that since for $n\in \mathbb{N}, n < 12$ $\mathbb{Z}_n\subset \mathbb{Z}_{12}$
then all subrings of $\mathbb{Z}_{12}$ are $\mathbb{Z}_n,\; n < 12$

Tobias Kildetoft

both are incorrect

flowian

@TobiasKildetoft I suppose so, since it is in definition of subring ?

Tobias Kildetoft

whether the definition includes it is a matter of style

flowian

14:59

I see

Tobias Kildetoft

Anyway, if some subset includes $1$ and needs to satisfy the requirements of being a subring, what other elements must it contain?

Ryan Unger

is there any legimately useful example of a ring without 1

flowian

$-1,0$

Tobias Kildetoft

@RyanUnger Mostly (from what I have encountered) stuff like the infinite sum of copies of some ring

But you can of course always just adjoint a $1$, and most things don't change at all

I do recall this being must less straightforward when trying to deal with 2-semicategories

since adjunctions were basically impossible to define

flowian

since subring includes $1$, and according to definition subring is Abelian group under addition, then subring contains $\mathbb{N}$ ?

Tobias Kildetoft

15:03

no

because that is not a subset of our ring

flowian

oh

I don't follow why $\mathbb{Z}_n$ for $ n < 12$ is not subring ?

Tobias Kildetoft

it is not even a subset unless you take a too naive definition of the ring to be useful

flowian

oh yes, indeed !

My understanding was not correct

The subring must contain all elements of $\mathbb{Z}_{12}$ and $1,-1,0$

I still don't understand

Tobias Kildetoft

What are the requirements on a subring?

flowian

if $a,b$ is in subring $R'$, then $a+b\in R'$,
if $a,b \in R'$ then $ab\in R'$,

if $a \in R'$ then $-a \in R'$,

Leaky Nun

15:15

maybe he doesn't understand $\Bbb Z_{12}$

flowian

for any $a,b$ in ring $R'$

Tobias Kildetoft

so let's start with the first one. So since you have $1$ and also you have $1$, then you need...?

flowian

and $1\in R'$

Ryan Unger

"the subring must contain all the elements of $\Bbb Z_{12}$"

Leaky Nun

lmao

flowian

15:17

" since you have 1 and also you have 1" don't quite understand

Leaky Nun

I missed that

Tobias Kildetoft

@flowian It says that whenever you have two elements, you also have their sum. And you have the two elements $1$ and $1$.

So this means it must also contain...?

Ryan Unger

3!

flowian

it must contain $2$ :)

Balarka Sen

@RyanUnger everyday maffs on the bloq

Tobias Kildetoft

15:19

ok, and then...?

flowian

and $3,\ldots,11$

Ryan Unger

quick maffs

Balarka Sen

smoke trees

Ryan Unger

sauce

flowian

it cannot contain integer higher than $11$, since it is not in $\mathbb{Z}_{12}$

and $12=0$ in $\mathbb{Z}_{12}$

I feel slow after week of labour

in construction

Tobias Kildetoft

15:22

@flowian You will be greatly helped if you stop thinking of the elements of this ring as integers at all

flowian

aha

@TobiasKildetoft Is my reasoning uncomplete ?

regarding the problem

Tobias Kildetoft

@flowian Well, it lacks a conclusion

also, it is slightly imprecise, considering the stuff about larger integers

flowian

the subring is $\mathbb{Z}_{12}$

Tobias Kildetoft

i.e. there is only one subring

flowian

really, I was thinking to include $\mathbb{Z}_6$ as well

hmm

MatheinBoulomenos

15:33

@flowian what is $\Bbb Z_n$, as a set?

Ryan Unger

n-adic integers?

flowian

$\Bbb Z_n = \{\bar 0, \bar 1,\ldots, \bar{n-1} \}$, where $\bar a = \{b : a = b (\mod n)\}$

Ryan Unger

@BalarkaSen are you allowed to write in an MSE answer that they shouldn't be reading the paper they're reading based on the questions they're asking

MatheinBoulomenos

right, so $\Bbb Z_n$ is not a subset of $\Bbb Z$, or of $\Bbb Z_m$ for $n \neq m$

@RyanUnger I wanted to write something like that before, but I decided against it

Ultradark

Why say $\Bbb R^{n+1}$ when you can just say $\Bbb R^n$?

Ryan Unger

15:38

because then you don't have to write dimension $n-1$ for hypersurfaces

Ultradark

haha haha ^^^^

Tobias Kildetoft

@RyanUnger I have made comments along that line before, though I do try to put it as nicely as I can

MatheinBoulomenos

It feels slightly concerning to use notes that are no longer available at the author's website which start with a bold "USE AT YOUR OWN RISK" on the first page

Balarka Sen

@RyanUnger Be Nice

Ryan Unger

2

Q: Proposition $3.2$ of Mean curvature flow with surgeries of two–convex hypersurfaces

I'm reading "Mean curvature flow with surgeries of two–convex hypersurfaces" by Gerhard Huisken and Carlo Sinestrari and I didn't understand how to prove the proposition $3.2$ on page $148$. The authors indicate the paper "Four-Manifolds with Positive Isotropic Curvature" by Hamilton for the defi...

Balarka Sen

15:42

Yikes

Ryan Unger

He's asking if the metric is induced by the immersion

Part 4 is an annoying computation

Thorgott

16:14

@RyanUnger maybe this: mathoverflow.net/a/22584

Ryan Unger

> If 𝐴 is a ring (possibly non-unital), its unitalization is defined to be the universal arrow from 𝐴 to the forgetful functor from unital rings to rings.

Balarka Sen

lol

Leaky Nun

@RyanUnger it's true

Ryan Unger

16:53

@BalarkaSen math.stackexchange.com/questions/3294520/… whut

Balarka Sen

Shrug @Ryan

Ryan Unger

If you have the divergence theorem, you don't need to work with differential forms (other than the volume/area forms) at all. You just need to prove a product rule $\mathrm{div}\,\langle F, G \rangle = \langle \nabla F, G \rangle + \langle F ,\mathrm{div}\,G \rangle$ (where the $\langle F,G \rangle$ denotes contraction of $F$ with the first $k$ slots of $G$, producing a $1$-tensor) and apply the divergence theorem to the LHS. — Anthony Carapetis Jul 11 at 10:56

no angry comment from Ted

disappointed

Alessandro Codenotti

17:41

Dumb question: what's an easy example of an element of a C*-algebra whose spectral radius is strictly smaller than its norm?

Ryan Unger

@AlessandroCodenotti uh aren't those always the same

or is there some condition for that

Alessandro Codenotti

I know they are the same for self-adjoint elements

Ryan Unger

Oh right

everything is self-adjoint

so there's no example

Alessandro Codenotti

Those notes I'm reading prove that $\rho(a)\leq\|a\|$, with equality if $a$ is a self-adjoint element of a unital C*-algebra (the proof doesn't seem to use the unit of the algebra as far as I can tell however)

@RyanUnger what

Ah I need to leave for a while, sorry

Ryan Unger

@AlessandroCodenotti any useful application of this stuff will be to self adjoint operators :P

Ryan Unger

18:04

@AlessandroCodenotti actually the stability operator for MOTS (marginally outer trapped surfaces) in GR is not self-adjoint

this causes headaches

Ryan Unger

18:15

@BalarkaSen are you here

Alessandro Codenotti

@RyanUnger fair

In a way the *-condition says that all elements are close to a self-adjoint

Ryan Unger

@AlessandroCodenotti it's a nontrivial theorem that the smallest eigenvalue of a non self-adjoint elliptic operator is real

smallest meaning one with the smallest real part, I believe

Alessandro Codenotti

I see

This whole $\rho(a)\leq\|a\|$ thing works in any Banach algebra, but I guess that it's easy to get a strict inequality in a norm which is not the C*-norm (if one exists)

Ryan Unger

the failure of self-adjointness comes from a lower order term

so it's not that crazy

but the variational characterization of eigenvalues fails

and there's a crazy remedy for this

Ultradark

19:02

Speaking of eigenvalues...the set of eigenvalues in let’s say an $n \times n$ Hermitian matrix is called the spectrum, correct?

Semiclassical

19:24

Doesn’t need to be hermitian

Tanner Swett

I'm at 4,958 reputation! Boy am I excited about being able to approve tag wiki edits. \o/

Semiclassical

That’s a convenient case, though

@TannerSwett blows party kazoo

My editing is not on point today. I blame mobile

Ultradark

@Semiclassical have you gone through the proof of the spectral theorem yet? I was thinking about trying to work through it in the next few days.

Semiclassical

Not recently, no

Daminark

@Ultradark so the reason both words are used is this

In finite dimensions, let's say $A$ is an $n\times n$ matrix. If $\lambda I - A$ is not surjective, then it's not injective either, and vice versa, right?

Ultradark

19:32

@Daminark yeah I'd agree with that

Daminark

So answering the question "When is $\lambda I - A$ injective/surjective/bijective" is all the same thing

But if you're in infinite dimensions, it's now possible to be one but not the other

For example, take the space of sequences of real numbers

The shift map sending $(x_1,x_2,\ldots)$ to $(0,x_1,x_2,\ldots)$ is injective but not surjective

So the point is that the spectrum in general asks when $\lambda I - T$ (where $T$ is now a linear operator on some vector space, possibly infinite-dimensional) invertible. Eigenvalues specifically ask when that guy is non-injective

For the shift map, $0$ is not an eigenvalue, but it's in the spectrum

As for the spectral theorem, I remember how it goes so I can help if you'd like

Ultradark

"If one interprets the second derivative of the eigenvalues as being proportional to a "force" on those eigenvalues (in analogy with Newton's second law), (1.74) is asserting that each eigenvalue $\lambda_j$ "repels" the other eigenvalues $\lambda_k$ by exerting a force that is inversely proportional to their separation (and also proportional to the square of the the matrix coefficient of $A$ in the eigenbasis)." I just thing this is such a cool thing even though I don't completely understand it

@Daminark yeah maybe, I have the proof in front of me I'm just gonna try to work through it once or twice

Daminark

Not exactly sure what (1.74) is, but also I'm not good at physics so if this is more than just a cute analogy with no substance then I probably won't get too far interpreting it :P

Ultradark

I can't type it cause i don't know all the symbols in math type lol. But it's called "Hadamard second variation formula"

Semiclassical

19:51

Waiting to hear back after a (phone) interview is as stressful as the rest of the process

Probably won’t hear back until Monday? I dunno what impression I made

Tobias Kildetoft

@Semiclassical Yeah, waiting after an interview can be terrible

Semiclassical

Yeah

Ultradark

As $n\to \infty$ the amount of stress will drop and tend to zero

$n$ being the number of interviews

but there's many other variations to consider with this

factor that probably carries most of the "weight:" One needs to sculpt an appropriate metric that captures the "desire of the interviewee to obtain the position." One can stick the interviewee with a "patch" which measures the emotional state of the individual in the following hours after the interview.

ncbi.nlm.nih.gov/pmc/articles/PMC4804278

ÍgjøgnumMeg

20:33

Hey y'all

J. Doe

Hola

Ryan Unger

@BalarkaSen I'm looking at a claim that the regular neighborhoods of any two triangulations of $\Bbb R^3$ are ambiently isotopic. But I know that there are triangulations of $\Bbb R^3$ with non-homeomorphic 1-skeleta. This seems at odds with the claim. How can you have non-homeomorphic graphs with homeomorphic regular neighborhoods?

It's easy to see that if the regular neighborhoods are homeomorphic, then the graphs (viewing the 1-skeleta as graphs) have to be homotopy equivalent. Does this not determine the graph up to homeomorphism already?

anakhro

21:09

What is a regular neighbourhood?

Ryan Unger

@anakhro essentially a PL tubular neighborhood

anakhro

21:24

Tubular with respect to the triangulation?

Ryan Unger

@anakhro Uh well the precise definition is in Hempel. It's a bit hard to write. It's a simplicial complex containing your original one that can be obtained by a sequence of (at least one) 'elementary collapsing'

anakhro

Get's complicated quickly, :P

Ryan Unger

yeah this PL stuff is super complicated

anakhro

PL is piecewise linear, right?

Ryan Unger

yeah

there's a book where they prove that the category of PL 3-manifolds is equivalent to smooth 3-manifolds

I imagine its a hellish read

the definition of a PL 3-manifold takes a while to even write down

anakhro

21:29

I had done two lemmas for something which had a deformation in them. I did one with a smooth bump function, and the other with a piecewise linear deformation. The latter was very much nicer to imagine, read, and understand.

skillpatrol

21:41

Beauty Is Not Simplicity: An Analysis of Mathematicians' Proof Appraisals

2

J. Doe

22:21

Aren't beauty and elegance the same thing

skillpatrol

i think "beauty" is a more general term

J. Doe

They are saying elegance is simplicity. So beauty is not elegance, according to the title.

skillpatrol

yeah, maybe the title is worded as "click bait"

J. Doe

"For example, some participants marked down the theorem ‘Every prime number of the form 4n+1 is the sum of two integral squares in exactly one way’ because it lacked a simple proof..."

I mean the proofs I've seen are not complicated. Perhaps simplicity means something else.

J. Doe

23:40

So are any of you gonna be storming Area 51?

skullpetrol

why?

J. Doe

Because apparently 2 million people are on September 20th.

A further 1,300,000 people said they were interested in the event, scheduled for 20 September 2019, and billed as "Storm Area 51, They Can't Stop All of Us", an attempt to "see them aliens." -- From Wiki.

Funniest thing I've read all week lol.

skullpetrol

not me, because I never follow the herd

J. Doe

So you plan to go earlier than then? Nice!

skullpetrol

nah, not interested in "them aliens"

J. Doe

23:45

I visited the event page. People are sharing what they are going to teach "their alien."

Ryan Unger

@skullpetrol wot

skullpetrol

them aliens voted for trump

J. Doe

lool

skullpetrol

Make them aliens great again!

J. Doe

Apparently the U.S. Air Force warned people. But they have no answer for this genius: "What a bunch of morons. We're going to storm Area 51 by land, not air. The air force can't do anything to us."

skullpetrol

23:56

:-/

Transcript for

$Mathematics$ Mathematics