Mathematics - 2019-10-22 (page 1 of 2)

Leaky Nun

12:03 AM

@MatheinBoulomenos oh, they’re all reflections!

MatheinBoulomenos

@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

Harry Battersby

@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.

maybeso

... 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 ...

Leaky Nun

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

MatheinBoulomenos

@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

Leaky Nun

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

MatheinBoulomenos

I was only talking about vector spaces

not representations

user193319

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?

Leaky Nun

but representations are functors from G to C-Vec

user193319

12:41 AM

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

Leaky Nun

isomorphic functors compose to isomorphic functors right

MatheinBoulomenos

yes

Leaky Nun

so what's going on

@user193319 $\delta_{jp} E_{iq}$

MatheinBoulomenos

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

Leaky Nun

oh

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

MatheinBoulomenos

12:45 AM

no

Leaky Nun

weird

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

MatheinBoulomenos

...

Leaky Nun

I mean they're the same set even

they're literally set-theoretically the same set

MatheinBoulomenos

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

Leaky Nun

oh what am I doing

hahaha

sorry

MatheinBoulomenos

12:46 AM

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

Leaky Nun

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

MatheinBoulomenos

No, End(S-Vec) is not Z(S)

Z(S) is the endomorphisms of the identity functor

Leaky Nun

what was?

aha

MatheinBoulomenos

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

Leaky Nun

now I want to look at End(S-Vec)

S-Vec is an abelian category so End(S-Vec) is an abelian "monoid"

MatheinBoulomenos

12:51 AM

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

Leaky Nun

every field is a Dedekind domain

but why

topologicalmagician

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

MatheinBoulomenos

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

Leaky Nun

nice

maybe this can be generalized to Picard stuff

MatheinBoulomenos

yes

Leaky Nun

12:55 AM

but what can we say about End

MatheinBoulomenos

idk

Harry Battersby

@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?

Leaky Nun

to be clear it isn't Aut(R-Mod)

rather Aut(R-Mod)/~

where ~ is isomorphism of isomorphisms

MatheinBoulomenos

yeah true

Leaky Nun

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

MatheinBoulomenos

12:57 AM

infinity categories intensify

topologicalmagician

@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

Harry Battersby

@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.

topologicalmagician

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

Leaky Nun

@MatheinBoulomenos what time is it

MatheinBoulomenos

time is an illusion

Harry Battersby

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.

Leaky Nun

your health isn't

topologicalmagician

@HarryBattersby that's why this site is useful

@LeakyNun why sleep when you can do math

Leaky Nun

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

topologicalmagician

@LeakyNun makes sense

Harry Battersby

@topologicalmagician Gotcha, thanks.

usukidoll

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

MatheinBoulomenos

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

Leaky Nun

@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

usukidoll

1:37 AM

Baire category theorem mentioned compact Hausdorff or complete metric space

Leaky Nun

5:23 AM

@MatheinBoulomenos show that for an irreducible character $\chi$ we have $\frac1{|G|} \sum_{g \in G} \chi(g^2) \in \{-1,0,1\}$

Leaky Nun

5:34 AM

complex character btw

generalize that howsoever you will

MatheinBoulomenos

6:23 AM

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

Leaky Nun

not really

also, Galois descent is a special case of Morita equivalence of matrix algebra!

MatheinBoulomenos

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 $\mathrm{Alt}^2(V)$

Leaky Nun

ok

MatheinBoulomenos

no wait

both miss a factor of $\frac{1}{2}$

Leaky Nun

ok

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

MatheinBoulomenos

6:29 AM

$M_{2 \times 2}(\Bbb C)$

Leaky Nun

why?

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

MatheinBoulomenos

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

Leaky Nun

I guess

C isn't central over R btw

MatheinBoulomenos

yes

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

Leaky Nun

hmm

cool

MatheinBoulomenos

6:32 AM

that's how you get functoriality for the Brauer group

apart from the group cohomology formalism

Leaky Nun

oh

I think $\Bbb C$ in $\Bbb C \otimes_\Bbb R V$ should be $a+ib$

instead of $a+bi$

MatheinBoulomenos

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

Leaky Nun

Frobenius–Schur indicator

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?

MatheinBoulomenos

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

Leaky Nun

I know why you wrote $1$ ( ͡° ͜ʖ ͡°)

MatheinBoulomenos

6:41 AM

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

Leaky Nun

and why is that true?

MatheinBoulomenos

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

Leaky Nun

oh no

MatheinBoulomenos

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

Leaky Nun

that's neat

MatheinBoulomenos

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 $\frac{1}{|G|}\sum_{g \in G}\chi(g^2)=\langle \chi_{\mathrm{Sym}^2(V)},1 \rangle - \langle \chi_{\mathrm{Alt}^2(V)},1\rangle = \mathrm{dim}_{\Bbb C} (\mathrm{Sym}^2(V)^G)- \mathrm{dim}_{\Bbb C}(\mathrm{Alt}^2(V)^G)$

Leaky Nun

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

MatheinBoulomenos

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

Leaky Nun

no way

MatheinBoulomenos

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

Leaky Nun

how can one accept coincidences

MatheinBoulomenos

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

Leaky Nun

hmm

C is overrated anyways

let's move to C((x))

what's its Brauer group?

MatheinBoulomenos

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

Leaky Nun

again?

MatheinBoulomenos

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 $0$

Leaky Nun

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 $K = \Bbb F_q((u))((t))$

MatheinBoulomenos

7:10 AM

okay

Leaky Nun

not that anyone asked

MatheinBoulomenos

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

Leaky Nun

$\operatorname{Gal}(\Bbb C((x))/\Bbb C((x^n))) = \Bbb Z/n\Bbb Z$ 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 (!?)

MatheinBoulomenos

yeah seems like it

Leaky Nun

Fourier theory yay

MatheinBoulomenos

7:20 AM

that implies that the Brauer group of $\Bbb C((x))$ vanishes

Leaky Nun

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))

Rithaniel

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

MatheinBoulomenos

@LeakyNun yeah

Leaky Nun

cool

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

MatheinBoulomenos

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 $\Bbb C((x))$

Leaky Nun

7:23 AM

I do not follow

Rithaniel

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

Leaky Nun

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

MatheinBoulomenos

oh

I meant non-commutative

@Rithaniel I'm sure other people have thought about this before

Leaky Nun

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

Rithaniel

Gotcha

Leaky Nun

7:24 AM

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

MatheinBoulomenos

that's not a division algebra

Leaky Nun

oh

ah I understand what you mean now

MatheinBoulomenos

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

Leaky Nun

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

MatheinBoulomenos

yeah

Leaky Nun

7:25 AM

lol

we said it at the same time

MatheinBoulomenos

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

Leaky Nun

I'm sure I forgot it

it's in wiki though

Wedderburn's little theorem

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

MatheinBoulomenos

you don't need Herbrand quotients

you can just show that the norm maps are surjective by elementary group theory

Leaky Nun

hmm?

actually I can't find the message

so, how do you prove it?

MatheinBoulomenos

$H^2(\Bbb F_{q^n},\Bbb F_q,\Bbb F_{p^n}^\times) \cong \Bbb F_p^\times/N(\Bbb F_{q^n}^\times)$ 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 $x \mapsto \prod_{k=0}^{n-1} x^{q^k}= x^{\sum_{k=0}^{n-1}q^k}= x^{\frac{q^n-1}{q-1}}$

Leaky Nun

7:32 AM

oh

MatheinBoulomenos

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

maths student

1

Q: Linear Transformation and Basis

$$ \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{...

Leaky Nun

@MatheinBoulomenos not only is the Galois group cyclic

but the units group also

which should shorten your proof further

MatheinBoulomenos

I don't see how that shortens the proof

Leaky Nun

well "it is surjective by elementary cyclic group theory"

MatheinBoulomenos

7:35 AM

how?

Leaky Nun

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

MatheinBoulomenos

okay yeah I see it

Epsilon Delta

hii everyone. i am new here

Leaky Nun

@EpsilonDelta welcome

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

Epsilon Delta

so what's going now, question solving?

Leaky Nun

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"

MatheinBoulomenos

yeah

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

Leaky Nun

heh

that's the point of the machinery

MatheinBoulomenos

I mean there are some elementary proofs

but they require some tricks

Leaky Nun

to make non-trivial things trivial

MatheinBoulomenos

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

Leaky Nun

7:43 AM

what is it?

MatheinBoulomenos

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

Leaky Nun

MatheinBoulomenos

so you get a reciprocity isomorphism

it's really boring though

Leaky Nun

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?

MatheinBoulomenos

I don't even understand the notation

Leaky Nun

7:53 AM

Irr(G,R)_R, _C, _H is a partition of Irr(G,R)

MatheinBoulomenos

oh so these are defined by their endomorphism ring?

no

Leaky Nun

the top row turns out to be so

the middle row is by F.S. indicator

MatheinBoulomenos

the columns look more regular than the rows

Leaky Nun

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

Anush

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 :)

user193319

11:21 AM

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

Leaky Nun

@user193319 what happened?

user193319

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

Leaky Nun

np

user193319

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

user10478

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.

Thomas Shelby

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.

user193319

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

user21820

2:26 PM

2

Q: Rational (non-negative) linear combinations of square roots of non-square naturals cannot change the status of irrationality, can they?

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...

irrational-numbers

@JyrkiLahtonen: Do you mind having a look at the above question, if it's in your specialty?

maths student

2:38 PM

0

Q: Homeomorphism to circle.

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...

real-analysis calculus general-topology

MatheinBoulomenos

@user193319 np

Alessandro Codenotti

3:04 PM

@BalarkaSen Do you happen to be here?

Alessandro Codenotti

4:52 PM

Sanity check, what is the smallest $n$ such that $\Bbb R\Bbb P^3$ embeds into $\Bbb R^n$?

Mike Miller

5:07 PM

5

Alessandro Codenotti

thanks

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

Balarka Sen

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

feynhat

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?

Alessandro Codenotti

5:23 PM

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

Balarka Sen

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

Alessandro Codenotti

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

Balarka Sen

I think it's true that every (orientable?) 3-manifold can be embedded in $\Bbb R^5$.

I don't know a slick proof

Alessandro Codenotti

I see

Anyway I pinged you earlier to talk about the Higson thing

But now I'm busy with the dinner

Balarka Sen

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

Alessandro Codenotti

5:31 PM

@BalarkaSen Such is life for students :/

Balarka Sen

:S

Alessandro Codenotti

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

Poline Sandra

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

Thorgott

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

Shine On You Crazy Diamond

Can I get any feed back from you on this:

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Q: Size of a boolean ring induced by a finite, commutative, boolean monoid?

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...