00:00 - 18:0018:00 - 23:00

12:03 AM
@MatheinBoulomenos oh, they’re all reflections!

@LeakyNun yes
so you integrate over a lot of $\chi(1)$
Since $[O(2):SO(2)]=2$, the set of $-1$-determinant matrices has measure $1/2$ and $\chi(1)=2$, so it all works out

@topologicalmagician How about now?  Given that a function is increasing for some $a,b∈R$, assume $a≠b$. Since the function is strictly increasing, then $a < b$ which would also imply that $f(a) < f(b)$ . Therefore, this shows that $f(a)≠f(b)$ and that the function is injective as needed to be shown.

... so I need 20 rep in each community I want to chat in, that makes meta's challenging. (that's good, I guess?)
there was talk of a 'day of silence' on the 18th - wanted to ask in meta how those who participated could find out how it went ...

12:33 AM
@MatheinBoulomenos so is $\sigma(-)$ isomorphic to $-$?
I'm not sure if it even is a functor yet
$V \to \sigma(V)$, $\sigma(bv) = \sigma(b) \sigma(v)$
$f: V \to W$
$f': \sigma(V) \to \sigma(W)$, $f'(\sigma(v)) := \sigma(f(v))$
$f'(\sigma(b) \sigma(v)) = \sigma(f(bv)) = \sigma(bf(v)) = \sigma(b) \sigma(f(v))$
yay it is a functor

@LeakyNun if you have a ring homomorphism $R \to S$, it induces a functor on the module categories by composing the structure morphism with $R \to S$. You can also do this for automorphisms of course

then why isn't the rep U(1) -> GL(1,C) isomorphic to its conjugate?

I was only talking about vector spaces
not representations

If $E_{ij}$ denotes the matrix with $0$'s in every entry except for a $1$ in the $(i,j)$-th entry, what is $E_{ij} E_{pq}$? Is there a nice formula?

but representations are functors from G to C-Vec

12:41 AM
I know that if $j \neq p$, then the product is $0$.

isomorphic functors compose to isomorphic functors right

so what's going on
@user193319 $\delta_{jp} E_{iq}$

@LeakyNun you have an equivalence of categories from a category to itself. That doesn't mean that it's isomorphic to the identity functor

oh
so $\sigma(-)$ and $-$ are not isomorphic?

12:45 AM
no

weird
$\sigma^{-1}(\sigma(V))$ looks the same as $V$

I mean they're the same set even
they're literally set-theoretically the same set

$\sigma^{-1}(-)$ is an inverse to $\sigma()$ yes

oh what am I doing
hahaha
sorry

12:46 AM
$\sigma^{-1}(\sigma(-))$ is isomorphic to the identity functor yes, but $\sigma(-)$ is not

look
$-$ is an isomorphism of categories
$\sigma(-)$ is an isomorphism of categories
they are two isomorphisms that are not isomorphic
that's what confused me :P
yesterday, by Alessandro Codenotti
$(\infty,1)$-categories are clearly the correct setting for Galois theory
yesterday, by Alessandro Codenotti
I mean, they are obviously the correct setting for all math, which just happens to include Galois theory
wait
you said earlier that End(S-Vec) = Z(S) = S
wait this is irrelevant
what am I looking at

No, End(S-Vec) is not Z(S)
Z(S) is the endomorphisms of the identity functor

what was?
aha

so natural transformations $\mathrm{id} \Rightarrow \mathrm{id}$

now I want to look at End(S-Vec)
S-Vec is an abelian category so End(S-Vec) is an abelian "monoid"

12:51 AM
fun fact: for a Dedekind domain R, Aut(R-Mod) is isomorphic to the class group of R

every field is a Dedekind domain
but why

@HarryBattersby just because $a\neq b$ does not imply $a<b$. You should say without loss of generality.

@LeakyNun the isomorphism $\mathrm{Cl}(R) \to \mathrm{Aut}(R-Mod)$ is given by $I \mapsto (M \mapsto I \otimes_R M)$

nice
maybe this can be generalized to Picard stuff

12:55 AM
but what can we say about End

@topologic I am not sure how to do it :/ Could you perhaps show me an example of an acceptable proof for this question? I feel like I am going nowhere with my proof. I am new to proofs, so I don't really understand how a formal proof should look like. Do we we always suppose something and then arrive at a conclusion based on that assumption that we made?

to be clear it isn't Aut(R-Mod)
rather Aut(R-Mod)/~
where ~ is isomorphism of isomorphisms

yeah true

which begs the question: what is an isomorphism of isomorphisms of isomorphisms?

12:57 AM
infinity categories intensify

@HarryBattersby The last proof you wrote would be correct if you wrote suppose without loss of generality $a<b$, this is because if you assumed $a\neq b$ then $a<b$ or $a>b$, but because $a,b$ are arbitrary, proving it for one case suffices.
If you want to show $p \implies q$ you assume p and then deduce true statements given p is true until you reach q

@topologicalmagician Oh, I see! Thanks a lot! Do you perhaps recommend any resources for learning proofs and strengthening my abilities in constructing them? I find that even though I understand concepts, constructing proofs about those concepts is sometimes hard because I'm not sure how to start building up the proof using the given information in the question.

@HarryBattersby just keep on solving more and more and you'll eventually get it

@MatheinBoulomenos what time is it

time is an illusion

1:06 AM
@topologicalmagician But the problem is, how would I check if I'm right? I am mostly unsure if I am even on the right track.

@HarryBattersby that's why this site is useful
@LeakyNun why sleep when you can do math

because if you sleep you can live longer to do more math

@LeakyNun makes sense

@topologicalmagician Gotcha, thanks.

1:16 AM
Is a Baire Space Lindelof? I don't think it is because in my ebook it says that Baire Spaces are either compact Hausdorff or a complete metric.space :V

it's not true that Baire spaces are either compact Hausdorff or a complete metric space
and also Baire spaces are not necessarily Lindelöf, just take the discrete topology on an uncountable set

@MatheinBoulomenos if Frac(R) is module-finite over R then R is a field
simply because that is an integral extension and being a field is preserved under integral extension
however the same cannot be said about algebra-finite, since R=Z_(2) is a counter-example

1:37 AM
Baire category theorem mentioned compact Hausdorff or complete metric space

4 hours later…
5:23 AM
@MatheinBoulomenos show that for an irreducible character $\chi$ we have $\frac1{|G|} \sum_{g \in G} \chi(g^2) \in \{-1,0,1\}$

5:34 AM
complex character btw
generalize that howsoever you will

6:23 AM
@LeakyNun do you know the characters of $\mathrm{Sym}^2(V)$ and $\mathrm{Alt}^2(V)$?

not really
also, Galois descent is a special case of Morita equivalence of matrix algebra!

yes
one can compute that if $\chi$ is the character of $V$, then $\chi(g)^2+\chi(g^2)$ is the character of $\mathrm{Sym}^2(V)$ and $\chi(g)^2-\chi(g^2)$ is the character of $\mathrm{Alt}^2(V)$

no wait
both miss a factor of $\frac{1}{2}$

ok
what is $\Bbb C \otimes_\Bbb R \Bbb H$?

6:29 AM
$M_{2 \times 2}(\Bbb C)$

why?
apart from, you know, $\Bbb H$ corresponds to $\frac12$

it's a CSA over $\Bbb C$ since being a CSA is preserved by a base change under field extensions but by Artin-Wedderburn the only CSA over an algebraically closed field are matrix algebras

I guess
C isn't central over R btw

yes
but if $L/K$ is a field extension and $A$ is CSA over $K$, then $L \otimes_K A$ is CSA over $L$

hmm
cool

6:32 AM
that's how you get functoriality for the Brauer group
apart from the group cohomology formalism

oh
I think $\Bbb C$ in $\Bbb C \otimes_\Bbb R V$ should be $a+ib$
instead of $a+bi$

about the Frobenius-Schur indicator question, we make the classic distinction between three cases:
- No non-degenerate $G$-invariant bilinear form exist
- There's a symmetric bilinear $G$-invariant form
- There's a skew-symmetric bilinear $G$-invariant form

In mathematics, and especially the discipline of representation theory, the Schur indicator, named after Issai Schur, or Frobenius–Schur indicator describes what invariant bilinear forms a given irreducible representation of a compact group on a complex vector space has. It can be used to classify the irreducible representations of compact groups on real vector spaces. == Definition == If a finite-dimensional continuous complex representation of a compact group G has character χ its Frobenius–Schur indicator is defined to be ∫ g...
ok now let's generalize it to L/K using our Brauer group technology
does it work? is it just a coincidence for C/R?

note that a symmetric bilinear $G$-invariant form is a non-zero element of $\mathrm{Hom}_{\Bbb C[G]}(\mathrm{Sym}^2(V),1)$ and an anti-symmetric bilinear $G$-invariant form is a non-zero element of $\mathrm{Hom}_{\Bbb C[G]}(\mathrm{Alt}^2(V),1)$
where $1$ is the trivial 1-d representation

I know why you wrote $1$ ( ͡° ͜ʖ ͡°)

6:41 AM
Okay so one thing you have to check is that $\mathrm{dim}_{\Bbb C}(V \otimes_{\Bbb C} V)^G \leq 1$

and why is that true?

but that's not that hard: $V \otimes_{\Bbb C} V \cong \mathrm{Hom}_{\Bbb C}(V,V^*)$, so $(V \otimes_{\Bbb C} V)^G \cong \mathrm{Hom}_{\Bbb C[G]}(V,V^*)$ and that is at most one-dimensional by Schur's lemma
I'm assuming $V$ irreducible ofc

oh no

if $V$ is not irreducible, then $\frac{1}{G}\sum \chi(g^2)$ can take other values than $0,1,-1$

that's neat

6:47 AM
okay, so knowing that $\mathrm{dim}_{\Bbb C}(V \otimes_{\Bbb C} V)^G \leq 1$ and by definition $V \otimes_{\Bbb C} V = \mathrm{Sym}^2(V) \oplus \mathrm{Alt}^2(V)$, we immediately get that exactly one of three cases must occur:
- $\mathrm{Sym}^2(V)^G=\mathrm{Alt}^2(V)^G=0$
- $\mathrm{dim}_{\Bbb C} \mathrm{Sym}^2(V)^G=1$ and $\mathrm{Alt}^2(V)^G=0$
- $\mathrm{Sym}^2(V)^G=0$ and $\mathrm{dim}_{\Bbb C}\mathrm{Alt}^2(V)^G=1$
For any representation $W$, we have $\mathrm{dim}_{\Bbb C}W^G = \langle \chi_W,1\rangle=\frac{1}{|G|} \sum_{g \in G} \chi_W(g)$
so we just apply that here: $\chi_{\mathrm{Sym}^2(V)}(g)=\frac{1}{2}(\chi_V(g)^2+\chi_V(g^2)$ and $\chi_{\mathrm{Alt}^2(V)}=\frac{1}{2}(\chi_V(g)^2-\chi_V(g^2)$, so $\frac{1}{|G|}\sum_{g \in G}\chi(g^2)=\langle \chi_{\mathrm{Sym}^2(V)},1 \rangle - \langle \chi_{\mathrm{Alt}^2(V)},1\rangle = \mathrm{dim}_{\Bbb C} (\mathrm{Sym}^2(V)^G)- \mathrm{dim}_{\Bbb C}(\mathrm{Alt}^2(V)^G)$

so the fact that it characterises the Brauer group structure is a coincidence?

by the above three cases, you know that this difference of dimensions is always $\in \{0,1,-1\}$ and by the previous discussion you know when each case occurs
@LeakyNun yeah kind of

no way

I mean, the Frobenius-Schur indicator over an algebraically closed field will still tell you if there's a $G$-invariant bilinear form and if it is symmetric or skew-symmetric
but that doesn't neccarily tell you the whole story about the Brauer group

how can one accept coincidences

6:59 AM
begrudgingly
I mean, you can still view the coincidence as a combination of two non-coincidental facts:
1) the generalized Galois descent for elements of the Brauer group
2) the fact that the Frobenius-Schur indicator (over an algebraically closed field of chararacteristic not 2) tells you about the existence of G-invariant bilinear forms and if it's symmetric or skew-symmetric

hmm
C is overrated anyways
let's move to C((x))
what's its Brauer group?

$\Bbb Q/\Bbb Z$ I guess?

again?

I'm note sure how to compute $\mathrm{N}_{\Bbb C((x^{1/n})),\Bbb C((x))}(\Bbb C((x^{1/n}))^\times)$
maybe that norm is even surjective
that would imply that the Brauer group is $0$

but that's impossible, obviously we have division algebras over C((x))
by link.springer.com/article/10.1134/S0001434607050227 there is a nondegenerate pairing $K^\times \times \operatorname{Br}(K) \to \Bbb Q/\Bbb Z$ for $K = \Bbb F_q((u))((t))$

7:10 AM
okay

not that anyone asked

$\mathrm{Gal}(\Bbb C((x)) / \Bbb C((x^2)) )=\Bbb Z/2\Bbb Z$ generated $x \mapsto -x$. $\Bbb C((x))^\times=x^{\Bbb Z} \times \Bbb C[[x]]^\times$

$\operatorname{Gal}(\Bbb C((x))/\Bbb C((x^n))) = \Bbb Z/n\Bbb Z$ generated by $x \mapsto \zeta_n x$, $N(\sum a_j x^j) = \prod_i \sum a_j \zeta_n^{ij} x^j$. let $p$ be a polynomial, then $p = \prod_j (a_j-x)$, and let $q = \prod_j (a_j^{1/n}-x)$, then $N(q) = \prod_j \prod_i (a_j^{1/n} - \zeta_n^i x) = \prod_j (a_j - x^n) = N(p(x^n))$
so it is surjective (!?)

yeah seems like it

Fourier theory yay

7:20 AM
that implies that the Brauer group of $\Bbb C((x))$ vanishes

Fourier everywhere
oh, that means there are no central division algebras
14 mins ago, by Leaky Nun
but that's impossible, obviously we have division algebras over C((x))

(One day, I aspire to be able to follow these conversations)

@LeakyNun yeah

cool
@Rithaniel you can learn about the Brauer group in Cassels--Fröhlich

or since every finite extension of $\Bbb C((x))$ is again isomorphic to $\Bbb C((x))$ we have no division algebras over $\Bbb C((x))$ such that the center is finite dimensional over $\Bbb C((x))$

7:23 AM
I do not follow

So, is this a re-verified result or are you guys talking about entirely new topics?

(surely $\Bbb C((x^{1/2}))$ is such a division algebra)

oh
I meant non-commutative
@Rithaniel I'm sure other people have thought about this before

@Rithaniel we're computing the Brauer group of a field that we haven't computed before

Gotcha

7:24 AM
@MatheinBoulomenos $M(2,\Bbb C((x^{1/2})))$

that's not a division algebra

oh
ah I understand what you mean now

so we have an analog of Wedderburn's little theorem I guess

so every finite division algebra over $\Bbb C((x))$ is... a field!

7:25 AM
lol
we said it at the same time

btw the proof of Wedderburn's little theorem by group cohomoogy is really smooth
but I'm sure I've shown it to you before

I'm sure I forgot it
it's in wiki though
In mathematics, Wedderburn's little theorem states that every finite domain is a field. In other words, for finite rings, there is no distinction between domains, skew-fields and fields. The Artin–Zorn theorem generalizes the theorem to alternative rings: every finite alternative division ring is a field. == History == The original proof was given by Joseph Wedderburn in 1905, who went on to prove it two other ways. Another proof was given by Leonard Eugene Dickson shortly after Wedderburn's original proof, and Dickson acknowledged Wedderburn's priority. However, as noted in (Parshall 198...
neat proof

you don't need Herbrand quotients
you can just show that the norm maps are surjective by elementary group theory

hmm?
actually I can't find the message
so, how do you prove it?

$H^2(\Bbb F_{q^n},\Bbb F_q,\Bbb F_{p^n}^\times) \cong \Bbb F_p^\times/N(\Bbb F_{q^n}^\times)$ by Tate-Nakayama (as the Galois group is cyclic), so we just need that the norm is surjective. The Galois group is generated by $x \mapsto x^q$, so the norm is just $x \mapsto \prod_{k=0}^{n-1} x^{q^k}= x^{\sum_{k=0}^{n-1}q^k}= x^{\frac{q^n-1}{q-1}}$

7:32 AM
oh

Thus the kernel of the norm is the set of all $a \in \Bbb F_{q^n}^\times$ such that $a^{\frac{q^n-1}{q-1}}-1=0$, that set has at most $\frac{q^n-1}{q-1}$ elements
$\Bbb F_{q^n}^\times$ has $q^n-1$ elements, so the image has at least $q-1$ elements
done

1

 \begin{array}{l}{\text { 2.) Consider the linear transformation } T: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2},\left[\begin{array}{l}{x} \\ {y}\end{array}\right] \mapsto\left[\begin{array}{c}{-2 x-5 y} \\ {2 x+4 y}\end{array}\right]} \\ {\text { Find a basis } \mathcal{B} \text { of } \mathbb{...

@MatheinBoulomenos not only is the Galois group cyclic
but the units group also
which should shorten your proof further

I don't see how that shortens the proof

well "it is surjective by elementary cyclic group theory"

7:35 AM
how?

I mean at this point it's really elementary group theory

okay yeah I see it

hii everyone. i am new here

@EpsilonDelta welcome
@MatheinBoulomenos I don't see it, but I know that the proof will be elementary :P

so what's going now, question solving?

7:37 AM
$\Bbb F_q^\times$ embeds in $\Bbb F_{q^n}^\times$ as the $\frac{q^n-1}{q-1}$ powers
so it's really "by classification of finite fields"

yeah
but if you try to prove Wedderburn's little theorem without the group cohomology machinery it's not that elementary

heh
that's the point of the machinery

I mean there are some elementary proofs
but they require some tricks

to make non-trivial things trivial

@LeakyNun did you know that there's a "tautological" CFT for any field with absolute Galois group $\widehat{\Bbb Z}$?

7:43 AM
what is it?

just consider the $G$-module $\Bbb Z$ with the trivial action and assign that to every finite extension of your field
that satisfies the axioms for a class formation

so you get a reciprocity isomorphism
it's really boring though

I don't really understand this table: the top row and bottom row seem arbitrary
I mean ok the middle row is accepted
but why in the top row and bottom row are some of them restrictions and some of them extensions?

I don't even understand the notation

7:53 AM
Irr(G,R)_R, _C, _H is a partition of Irr(G,R)

oh so these are defined by their endomorphism ring?
no

the top row turns out to be so
the middle row is by F.S. indicator

the columns look more regular than the rows

8:08 AM
@MatheinBoulomenos so upon further reading: (R,R) and (H,H) stands out as that when they are transported to C, it becomes irreducible
and otherwise we must have the other case
arghhhh
I know why it's confusing
it's because they use e and r
whereas the actual difference is whether they are out of C or into C
!!!!!
it's the L[g,f]-Mod "Galois descent" again!
look
let L[g,f] = Mat(something,A) where A is a skew field
then Irr(G,A)_A = { W | the L-rep W is irreducible }
again we have Brauer descent being the equivalence of:
1. L-Vec with G-semilinear f-twisted action
2. L[G,f]-Mod
3. A-Mod
so Irr(G,A)_A is the things in 3 such that the 1-version without the G-semilinear f-twisted action is irreducible

2 hours later…
10:31 AM
let w_n^k be the kth power of the nth root of unity
is it true that -w_n^k = +w_n^{k+n/2} ?
yes :)

11:21 AM
Man, this website and chatroom is a godsend. Thanks to everyone for all the help over the years.

@user193319 what happened?

@LeakyNun Nothing in particular. But thanks for helping me yesterday and all the other times, bro.

Thanks @MatheinBoulomenos and @TedShifrin for all the help over the months/years

3 hours later…
2:07 PM
Does anyone know how to draw contour lines on a surface like this youtu.be/WsZj5Rb6do8?t=120 either in MATLAB or some free online or Windows program? I believe the video was created with a Mac only program.

Does anyone know a good book (other than Atiyah-Macdonald and Eisenbud) to study dimension theory of rings? In particular, I'm looking for a book on dimension theory for Noetherian semi-local ring. Thanks.

@AlessandroCodenotti Thanks for all the help over the months/years.

2:26 PM
2

I arrived at an idea of considering rational linear combinations of square roots of non-square naturals and of the irrationality of those combinations. So suppose that we have some $n$-tuple $(\sqrt{a_1},...,\sqrt{a_n})$ where $a_1,...,a_n$ are natural numbers which are not squares so that $\sqr... @JyrkiLahtonen: Do you mind having a look at the above question, if it's in your specialty? 2:38 PM 0 I want to show that$f:(0,2\pi]\to S^1$defined by$t\to (\sin t, \cos t)$is not a homeomorphism. I will do this by showing that its inverse is not continuous. The inverse is defined by$(\sin t, \cos t)\to t$. So at$(0,1)$we want to have$\lvert 2(1-\cos t)\rvert <\delta \Rightarrow \lv...

@user193319 np

3:04 PM
@BalarkaSen Do you happen to be here?

2 hours later…
4:52 PM
Sanity check, what is the smallest $n$ such that $\Bbb R\Bbb P^3$ embeds into $\Bbb R^n$?

5:07 PM
5

thanks
But wait, which manifolds are used to show that the bound from Whitney's embedding theorem is sharp then?

@Alessandro $\Bbb{RP}^2$ does not embed in $\Bbb R^3$, I mean.

hi
yesterday, by feynhat
What is the reason behind the coboundary sign convention in Bredon or May or Milnor&Stasheff?
^ Is anyone familiar with any of these texts?

5:23 PM
@BalarkaSen sure, but is the bound sharp for all$n$ or only some?

@Alessandro Only some. $S^n$ embeds in $\Bbb R^{n+1}$, much smaller than $2n$!

I mean do all $3$ dimensional varieties embed into $\Bbb R^5$?

I think it's true that every (orientable?) 3-manifold can be embedded in $\Bbb R^5$.
I don't know a slick proof

I see
Anyway I pinged you earlier to talk about the Higson thing
But now I'm busy with the dinner

Every $3$-manifold is a surgery on a link so maybe something along the lines
@AlessandroCodenotti Yeah I am swamped with work as well
We can converse later this week maybe

5:31 PM
@BalarkaSen Such is life for students :/

I think I'll drop a course, I'm doing too much stuff now

Hello, when i prove Cauchy Schwarz inequality i found $| \sum a_k b_k |\leq( \sum a_k^2)^(1/2) (\sum b_k^2)^(1/2)$
but there is some references I found $\sum|a_k| |b_k|$
how to do , to go from the first to the second one

Try applying the first inequality to the vectors $(|a_1|,...,|a_n|)$ and $(|b_1|,...,|b_n|)$.

Can I get any feed back from you on this:
0

Let $X$ be a finite, commutative, boolean monoid with a $0$ element. Let $A = \{ x \subset X : 0 \in x$ and $x^2 \subset x\}$ where $x^2 = \{ab: a,b \in x\}$. Define for $x, y \in A$, $x + y = x \Delta y \cup \{0\}$. Then since $x^2 \subset x$ for $x \in A$, we actually have $x \subset x^2$ as...

5:38 PM
@PolineSandra the second one is just triangle inequality

@LeakyNun the first is correct?

yes

but in the triangle inequality we have |x+y|<|x|+|y| not the inverse
@LeakyNun

that isn't the inverse
oh
I misunderstood

@Thorgott yes but it is not the case here no ?
or I can apply it for |a_i| and |b_i|

5:43 PM
that's what he said

I don't understand
what is the right expression

Both versions of the inequality are correct
The first one is the Cauchy-Schwarz inequality. By doing what I suggested above, you can derive the other version.

and you can derive the first one from the second one by using triangle inequality

ok
I understand

5:57 PM
@LeakyNun have you presented the Brauer descent thing in your Lie groups course?

@MatheinBoulomenos nah
btw I have a logic test in half an hour

good luck

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