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12:48 AM
See this unit disk with lines in it?
 
1:01 AM
nevermind
 
 
6 hours later…
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?
 
Yup
 
Cool cool
 
7:32 AM
Mornin'
 
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!
 
 
1 hour later…
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?
 
@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.
 
9:06 AM
@LeakyNun yes
@BalarkaSen sure!
 
Where should I start? Geometric realization?
 
@BalarkaSen did you look at my SSS?
 
@LeakyNun Yeah I saw. Were you able to carry out the computation?
 
ok let me try again
it's way too messy
and I received guidance from Mathein and Loch
 
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
 
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?
 
@Alessandro Let's move to garbology
 
should I move there as well?
 
@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!
 
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 $E^2$
 
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
 
$K(\Bbb Z,1) \to \Bbb RP^\infty \to K(\Bbb Z,2)$
 
$(p, q) \to (p + 2, q - 1)$ 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
 
it isn't
$d(y) = 2x$
 
Zero kernel, potato pohtato
 
$H^\bullet(\Bbb CP^\infty) = \Bbb Z[x]$, $|x|=2$
ok
 
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 $2$
 
right
 
Because the $(2, 0)$ term survives to $E^\infty$
Cool
 
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?
 
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
 
then how do you prove that $H^\bullet(K(\Bbb Z,2)) = \Bbb Z[x]$, $|x|=2$?
 
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 $E^\infty \equiv 0$
 
I mean, how do you work out the cup product?
 
$d : (0, 1) \to (2, 0)$ 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?
 
hmm
I see
 
These are tricky shit
 
10:14 AM
8
A: Cup product of cohomology in a Serre spectral sequence

Allen HatcherMaybe 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
 
lol nice
 
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...
 
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
 
$H^*(\Bbb{RP}^\infty; \Bbb Z_2)$ 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 on earth
 
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.
 
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?
 
But since $H^*(\Bbb{RP}^\infty; \Bbb Z)$ only has even degrees, this justifies it's $\Bbb Z[y]/(2y)$
 
@northerner You need to copy the text as the address in a bookmark
then click on the bookmark while in chat
 
$\Bbb Z[y]/(2y)$ is a different thing... right
 
Well if I click on the book mark it takes me away from chat and to that site.
 
10:29 AM
@northerner Then you put in the wrong text
 
you have $y^2$ and $2y^2$
 
the text is pure javascript as far as I recall, it should not take you anywhere else
 
no
 
Those are all zero
$2y^2$ is in the ideal $(2y)$
 
10:31 AM
oh what am I thinking
I can't algebra
 
@northerner you need the text in those links (or just drag it or right click it as it says)
 
It is a graded ring which is $\Bbb Z_2$ in even degrees and $0$ in odd degrees (except degree $0$, where it's $\Bbb Z$)
 
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?
 
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
 
what on earth
 
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.
 
to me it feels like you're inventing a new tool to solve every new problem
lol
 
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.
 
$\Bbb RP^\infty$ is $K(\Bbb Z/2\Bbb Z,1)$ right
(why?)
 
and why is $\Bbb CP^\infty$ a $K(\Bbb Z,2)$?
 
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)
 
can the universal cover be not contractible?
 
got it working thanks
 
the path space is always contractible, we proved it before, right
@northerner nice
 
@LeakyNun Universal cover of $\Bbb{RP}^2$ is $S^2$...
 
so the path space quotient homotopy isn't the universal cover?
 
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.
 
hi @ÉricoMeloSilva
 
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?
 
yeah
so quotient of contractible... isn't contractible!
 
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
 
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.
 
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 $\neq2$)
 
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.
 
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}$.
 
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 :)
 
No problem. Good luck with your exam!
 
Thank you :)
 
 
1 hour later…
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 $C_{i-1}^\ast \xrightarrow {\beta^\ast} C_i^\ast \xrightarrow {\alpha^\ast} C_{i+1}^\ast$
 
is that a complex cobordism cochain
 
1:55 PM
$C_i \twoheadrightarrow C_i/B_i \twoheadrightarrow C_i/Z_i \xrightarrow \beta B_{i-1} \hookrightarrow Z_{i-1} \hookrightarrow C_{i-1}$
$0 \to Z_i/B_i \to C_i/B_i \to C_i/Z_i \to 0$
$0 \to H_i \to C_i/B_i \to B_{i-1} \to 0$
$0 \to B_i \to Z_i \to H_i \to 0$
 
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.
 
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)
 
I thought that condition was odd.
 
hmm
 
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
 
@AlessandroCodenotti Thanks! I appreciate it.
 
2:37 PM
@user193319 notes from my analysis course that get straight to the point math.utk.edu/~freire/teaching/m447f16/BaireCategory.pdf
 
Those are nice notes
(if you want the really hardcore version you should look at chapter 24 of Kechris's classical descriptive set theory)
 
2:55 PM
Hello
Can somebody help me with a problem - find all subrings of $\mathbb{Z}_12$ ?
 
@flowian Do you require subrings to contain $1$?
 
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 $\mathbb{Z}_n,\; n < 12$
 
both are incorrect
 
@TobiasKildetoft I suppose so, since it is in definition of subring ?
 
whether the definition includes it is a matter of style
 
2:59 PM
I see
 
Anyway, if some subset includes $1$ and needs to satisfy the requirements of being a subring, what other elements must it contain?
 
is there any legimately useful example of a ring without 1
 
$-1,0$
 
@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
 
since subring includes $1$, and according to definition subring is Abelian group under addition, then subring contains $\mathbb{N}$ ?
 
3:03 PM
no
because that is not a subset of our ring
 
oh
I don't follow why $\mathbb{Z}_n$ for $ n < 12$ is not subring ?
 
it is not even a subset unless you take a too naive definition of the ring to be useful
 
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
 
What are the requirements on a subring?
 
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 $-a \in R'$,
 
3:15 PM
maybe he doesn't understand $\Bbb Z_{12}$
 
for any $a,b$ in ring $R'$
 
so let's start with the first one. So since you have $1$ and also you have $1$, then you need...?
 
and $1\in R'$
 
"the subring must contain all the elements of $\Bbb Z_{12}$"
 
lmao
 
3:17 PM
" since you have 1 and also you have 1" don't quite understand
 
I missed that
 
@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...?
 
it must contain $2$ :)
 
@RyanUnger everyday maffs on the bloq
 
3:19 PM
ok, and then...?
 
and $3,\ldots,11$
 
quick maffs
 
smoke trees
 
sauce
 
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
 
3:22 PM
@flowian You will be greatly helped if you stop thinking of the elements of this ring as integers at all
 
aha
@TobiasKildetoft Is my reasoning uncomplete ?
regarding the problem
 
@flowian Well, it lacks a conclusion
also, it is slightly imprecise, considering the stuff about larger integers
 
the subring is $\mathbb{Z}_{12}$
 
i.e. there is only one subring
 
really, I was thinking to include $\mathbb{Z}_6$ as well
hmm
 
3:33 PM
@flowian what is $\Bbb Z_n$, as a set?
 
n-adic integers?
 
$\Bbb Z_n = \{\bar 0, \bar 1,\ldots, \bar{n-1} \}$, where $\bar a = \{b : a = b (\mod n)\}$
 
@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
 
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
 
Why say $\Bbb R^{n+1}$ when you can just say $\Bbb R^n$?
 
3:38 PM
because then you don't have to write dimension $n-1$ for hypersurfaces
 
haha haha ^^^^
 
@RyanUnger I have made comments along that line before, though I do try to put it as nicely as I can
 
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
 
@RyanUnger Be Nice
 
2
Q: Proposition $3.2$ of Mean curvature flow with surgeries of two–convex hypersurfaces

GeorgeI'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...

 
3:42 PM
Yikes
 
He's asking if the metric is induced by the immersion
Part 4 is an annoying computation
 
4:14 PM
@RyanUnger maybe this: mathoverflow.net/a/22584
 
> 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.
 
lol
 
@RyanUnger it's true
 
Shrug @Ryan
 
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
 
5:41 PM
Dumb question: what's an easy example of an element of a C*-algebra whose spectral radius is strictly smaller than its norm?
 
@AlessandroCodenotti uh aren't those always the same
or is there some condition for that
 
I know they are the same for self-adjoint elements
 
Oh right
everything is self-adjoint
so there's no example
 
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
 
@AlessandroCodenotti any useful application of this stuff will be to self adjoint operators :P
 
6:04 PM
@AlessandroCodenotti actually the stability operator for MOTS (marginally outer trapped surfaces) in GR is not self-adjoint
this causes headaches
 
6:15 PM
@BalarkaSen are you here
 
@RyanUnger fair
In a way the *-condition says that all elements are close to a self-adjoint
 
@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
 
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)
 
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
 
7:02 PM
Speaking of eigenvalues...the set of eigenvalues in let’s say an $n \times n$ Hermitian matrix is called the spectrum, correct?
 
7:24 PM
Doesn’t need to be hermitian
 
I'm at 4,958 reputation! Boy am I excited about being able to approve tag wiki edits. \o/
 
That’s a convenient case, though
@TannerSwett blows party kazoo
My editing is not on point today. I blame mobile
 
@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.
 
Not recently, no
 
@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?
 
7:32 PM
@Daminark yeah I'd agree with that
 
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
 
"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
 
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
 
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"
 
7:51 PM
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
 
@Semiclassical Yeah, waiting after an interview can be terrible
 
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.
 
8:33 PM
Hey y'all
 
Hola
 
@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?
 
9:09 PM
What is a regular neighbourhood?
 
@anakhro essentially a PL tubular neighborhood
 
9:24 PM
Tubular with respect to the triangulation?
 
@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'
 
Get's complicated quickly, :P
 
yeah this PL stuff is super complicated
 
PL is piecewise linear, right?
 
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
 
9:29 PM
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.
 
10:21 PM
Aren't beauty and elegance the same thing
 
i think "beauty" is a more general term
 
They are saying elegance is simplicity. So beauty is not elegance, according to the title.
 
yeah, maybe the title is worded as "click bait"
 
"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.
 
 
1 hour later…
11:40 PM
So are any of you gonna be storming Area 51?
 
why?
 
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.
 
not me, because I never follow the herd
 
So you plan to go earlier than then? Nice!
 
nah, not interested in "them aliens"
 
11:45 PM
I visited the event page. People are sharing what they are going to teach "their alien."
 
@skullpetrol wot
 
them aliens voted for trump
 
lool
 
Make them aliens great again!
 
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."
 
11:56 PM
:-/
 

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