Mathematics - 2018-06-01 (page 3 of 3)

Ted Shifrin

19:00

With just $c_1$ in there, the result is false.

So your proof isn't going to do it, as you're using just the usual Jacobian matrix.

Mary Star

Oh you're right. It's a typo. It should be:

\begin{equation*}\det \begin{pmatrix}\frac{\partial{f_1}}{\partial{x_1}}(c_1) & \ldots & \frac{\partial{f_1}}{\partial{x_n}}(c_1)\\ \vdots & \vdots & \vdots \\ \frac{\partial{f_n}}{\partial{x_1}}(c_n) & \ldots & \frac{\partial{f_n}}{\partial{x_n}}(c_n)\end{pmatrix}\neq 0 \ \text{ for all } c_1, c_2, \ldots , c_n\in G\end{equation*}

@TedShifrin

MatheinBoulomenos

@TedShifrin can't we translate the orientation argument to LA? Take the volume form $\Omega^{n/2}$, then if $f$ satisfies $f^*(\Omega)=\Omega$, then we will also have $f^*(\Omega^{n/2})=\Omega^{n/2}$ since pullbacks commute with wedge products. But $f^*(\Omega^{n/2})=\operatorname{det}(f) \Omega^{n/2}$

Ted Shifrin

OK, now I believe it might be right, but your proof sure won't work, @MaryStar.

Sure, symplectic gives volume-preserving, @Mathein, but you won't get complex linear.

Oh, you're trying to prove $\det = 1$

Mary Star

@TedShifrin Ah ok. But why? Which is the mistake?

MatheinBoulomenos

@TedShifrin yes, that's what I want to prove

Mike Miller

19:02

Yeah, I thought that's why you brought up the Pfaffian

Ted Shifrin

Right, but I think Mathein's right that there's an easier proof of that.

Mike Miller

I couldn't think of any at least

MatheinBoulomenos

Is there a simple way to see that $\Omega^{n/2} \neq 0$?

Mike Miller

That is the definition of non-degenerate

Ted Shifrin

@MaryStar: First, it makes no sense to pull $b-a$ out. You have to apply the linear map to the vector $(b-a)$, but there's no contradiction to saying that that integral will be $0$.

MatheinBoulomenos

19:04

okay, then the proof should be pretty simple

$f^*(\Omega^{n/2})=\operatorname{det}(f) \Omega^{n/2}$ is actually the definition of the determinant I was given

(with any non-degenerate $n$-form)

Mike Miller

Yeah I should have just recited that at you immediately

But did not think it

Ted Shifrin

Nor did I. Bad me.

Mike Miller

You are in good company

Ted Shifrin

BTW, @Mathein, the volume form will need a factorial in the denominator. :P

Mike Miller

Or bad company, as it were

MatheinBoulomenos

19:06

@TedShifrin no way, my proof works over $\Bbb F_2$

Ted Shifrin

Yeah, but I'm the differential forms person. So badder me.

No, @Mathein, by my definition of volume form $\Omega^n$ is something like $n!$ times the volume form on $\Bbb R^{2n}$. (You shouldn't have $n/2$'s in there.)

MatheinBoulomenos

I can just omit the term "volume form" from the proof

Ted Shifrin

Try squaring $dx\wedge dy + du\wedge dv$ :P

Well, fine, then.

OK, lunchtime for me. Back later.

Mike Miller

Maybe my notion of non-degenerate is wrong over F_2

I just mean Omega gives an isomorphism to the dual space

MatheinBoulomenos

@MikeMiller I don't think so. The only thing that changes is that alternating is no longer the same as antisymmetric

Mike Miller

19:09

Over the reals at least ((maybe everywhere I have done 0 thought)) this implies the top power is nonzero

MatheinBoulomenos

other than that, exterior powers work as usual

hmm, I have to think about it

Ted Shifrin

No, your proof doesn't work in characteristic $2$.

MatheinBoulomenos

ah, too bad

Mike Miller

And the way I would do that off the top of my head is showing the existence of a symplectic basis

MatheinBoulomenos

the thing with determinants also holds in characteristic

Ted Shifrin

19:11

See my comment about squaring that $2$-form. You get $0$ in characteristic $2$. Things go crazy with skew-symmetric matrices in characteristic $2$.

The Pfaffian stuff probably doesn't work in weird characteristic, either.

Off to lunch.

MatheinBoulomenos

@TedShifrin what if I lift to $\Bbb Z$, then normalize, then reduce the normalized form?

Mike Miller

That's what I figured but I didn't want to think about whether it was non degenerate

@Mathei How can you normalize? You hella divide by 2

MatheinBoulomenos

hmm

Mary Star

@TedShifrin So, do you mean that I have to make something completely different? But what?

MatheinBoulomenos

Algebraic De Rham cohomology also doesn't work in char p

maybe there's something pathological with alternating forms in finite characteristic

ah, I think the exterior algebra exists, but it's no longer a quotient of the tensor algebra in char p

Mike Miller

19:17

Oh I see now why det Sp(2n) = 1 in characteristic p lol

@Mathei I think the keyword is divided power algebra

MatheinBoulomenos

@MikeMiller good point

do these actually come up in topology?

Abcd

How do I split $2^{2r+1}$ into x and y such that $x- y = 2^ r$ and $xy = 2^{2r+1}$

Is it possible? (need this for an inverse trig series)

Mike Miller

I have never seen them @MatheinBoulomenos

Maybe when talking about Chern classes, there are some arguments where n! appears

MatheinBoulomenos

19:35

@MikeMiller do you know if there are classifications of finite subgroups of $SO(n)$ for $n>3$? The case $n=3$ is pretty well known. The thing with periodic Tate cohomology has made that question a lot more interesting for me

Abcd

(never mind, that question^)

MatheinBoulomenos

Also are there more conditions than just acting freely and orientation-preservingly? Certainly cyclic groups also act in such a way on a $2$-sphere, but the Tate-cohomology is not 3-periodic

Vrouvrou

Hello, what is the definition of compact set using closed covering

?

GFauxPas

every set of closed sets covering the space admits a finite subset of closed sets that still covers the space

I would guess

MatheinBoulomenos

are you sure that is equivalent to being compact?

I'm quite sure it isn't

and I don't think there's a definition in terms of closed coverings

GFauxPas

19:48

no I'm not sure at all, it was just a guess /shrug

Leaky Nun

rip guess

MatheinBoulomenos

Take the cofinite topology on an infinite set, where a set is open iff it's empty or the complement is finite

this is compact

but you can cover it with infinitely many closed singletons which doesn't have a finite subcover

GFauxPas

math.stackexchange.com/questions/2804546/… what is Aortiz getting at with his comment on my question

Vrouvrou

I found this $A$ is compact iff for any family of closed set from E $\cap F_i\subset E\setminus A$ there exists a finite sub family $\cap_{I\in J}F_i\subset E\setminus A$

MatheinBoulomenos

yeah, but that's not a "closed covering"

Vrouvrou

19:54

how?

@MatheinBoulomenos i want to prove that the intersection of compact sets is compact using closed covering

MatheinBoulomenos

a covering is something such that the union contains the subspace you consider (or is equal to the whole space if you consider coverings of the whole space)

Ted Shifrin

@MaryStar Here's a hint: What does the mean value theorem (as you were doing it) tell you for each component function $f_i$ separately?

Albas

I don't get this, what is so nice about a diffeomorphism.
I mean that two smooth Manifolds being diffeomorphic to each other depends upon what kind of a smooth structure is imposed. But I can give those Manifolds some absurd smooth structure and they will no longer be diffeomorphic to each other.

Another way to phrase this will be, what does a diffeomorphism say about the smooth structure that has been imposed

Tobias Kildetoft

@Albas That makes no sense to me

A diffeomorphism says that the two smooth structures are "the same". It does not say anything about them individually

Also, who says you can find "absurd" smooth structures?

Albas

20:11

@TobiasKildetoft what I meant to say is that there might be other smooth structures for which the Manifolds are not diffeomorphic to each other.

Tobias Kildetoft

Yes, of course

Albas

So a diffeomorphism basically shows us which two smooth structures are same in some sense

Tobias Kildetoft

just like there might be other topologies for which they are not homeomorphic

Alessandro Codenotti

@Albas it's kinda like saying rings are uninteresting because I can make isomorphic groups nonisomorphic by imposing different ring structures

Albas

@AlessandroCodenotti Ahh I see.

Mary Star

20:19

@TedShifrin Do you mean from $f(b)-f(a)=(b-a)\int_0^1J_f(a+t(b-a))dt $ ?

Albas

This question is more motivated from a physics perspective but are infinite dimensional manifolds interesting structures to study?
Like can we do things like define cotangent spaces and subsequently differential forms and other things which are done for finite dimensional Manifolds on them as well or do we fall into some trouble?

Faust

good morning

Leaky Nun

20:44

@MatheinBoulomenos are you here?

MatheinBoulomenos

yeah

Leaky Nun

Does Galois descent fail if your fields have characterstic $p$ and Galois group being a $p$-group?

@loch hi

loch

Hey

MatheinBoulomenos

@LeakyNun I don't think so

Leaky Nun

@MatheinBoulomenos I feel like surjectivity of $B^\Gamma \otimes_F K \to B$ relies on taking average

@loch did you see my statement of Galois descent?

Albas

20:46

Hey@loch

Leaky Nun

20 hours ago, by Leaky Nun

[Galois Descent (for Algebras)] [statement only]
Let $F/E$ be a Galois extension with Galois group $\Gamma := \operatorname{Gal}(F/E)$.
For an $E$-algebra $B$, a semilinear action by $\Gamma$ is a map $\varphi : \Gamma \times B \to B$ satisfying $\varphi(\sigma, eb) = \sigma(e) \varphi(\sigma, b)$ for every $e \in E$ and $b \in B$.
Writing $\sigma(b)$ for $\varphi(\sigma,b)$, the conditoin becomes $\sigma(eb) = \sigma(e) \sigma(b)$.
Then, the category of ($F$-algebras) is equivalent with the category of ($E$-algebras with semilinear action by $\Gamma$)

MatheinBoulomenos

do you mean $(B \otimes_F K)^\Gamma \to B$? $\Gamma$ doesn't act on the first component

Leaky Nun

oh I changed variables

eh

$K$ is the bigger field now

$F$ is the smaller field

$B$ is an algebra over $K$ with a semilinear $\Gamma$-action

Then $B^\Gamma$ is an algebra over $F$

now we want to show that $B^\Gamma \otimes_F K = B$

loch

@LeakyNun uh yeah but I don't think I have time to think about it now lol

MatheinBoulomenos

oh okay

@LeakyNun it's enough to prove this for vector spaces, that's probably easier

Leaky Nun

20:53

@MatheinBoulomenos right, but I'm saying that surjectivity requires taking average

MatheinBoulomenos

the proof in Keith Conrad's notes doesn't take averages

loch

@Albas I know nothing about these things - but I think they show up in the context of things like moduli space of complex structures etc - which one might want to look at eg if they want to do gromov witten theory via differential (rather than algebraic) geometry - I can't guarantee what I'm saying here is accurate though

By differential I guess I probably meant symplectic

Leaky Nun

@MatheinBoulomenos could you point me to the page number?

MatheinBoulomenos

page 5

Leaky Nun

interesting

MatheinBoulomenos

20:58

I think this should also follow from the fact that $H^1(\Gamma,\operatorname{GL}_n(K))=\{1\}$ which should also give a profinite version of thi if $K/F$ is not finite, but I'm too lazy to work out the details right now

(that's non-abelian Galois cohomology)

Leaky Nun

ok

MatheinBoulomenos

@LeakyNun the thing is, the condition $\varphi(\sigma,eb)=\sigma(e)\varphi(\sigma,b)$ looks a lot like a cocycle

do you know group cohomology?

@LeakyNun I've searched for a connection between non-abeilan Galois cohomology and Galois descent and I found this: asving.com/2016/11/26/descent-on-vector-spaces-and-cohomology

Leaky Nun

21:15

@MatheinBoulomenos some

MatheinBoulomenos

that proof is more elegant if you know the result that $H^1(\Gamma, \operatorname{GL}_n(K))$ vanishes

Leaky Nun

hmm

MatheinBoulomenos

that's a form of Hilbert's Satz 90, by the way

Leaky Nun

I see

Evinda

21:27

Hello

I want to prove that each continuous function $f$ in a closed and bounded interval $[a,b]$ can be approximated uniformly with polynomials, as good as we want, i.e. for a given positive $\epsilon$, there is a polynomial $p$ such that

$$\max_{a \leq x \leq b} |f(x)-p(x)|< \epsilon.$$

Firstly, we should make sure that we can assume without loss of generality that the interval $[a,b]$ is contained in the open interval $(-\pi,\pi)$.

But why can we assume this?

mercio

who puts bright blue text on bright red background in 2018

Evinda

Hey @LeakyNun
Do you maybe have an idea?

Maximus

21:43

solve $0 \leq (-3 + x log(2))/(2 x)$ with x>0

wolfram alpha says $3/log(2) \leq x$

but when I do it by hand I get $3/(log(2)-2) \leq x$

GFauxPas

where's the $-2$ coming from

Maximus

What silly mistake and I making

GFauxPas

/log

Maximus

ah nvm

GFauxPas

:)

user193319

22:10

This document defines the operator norm on $B(H)$ as $||T|| = \sup_{||x||=1} ||Tx||$, and the proceeds with the following calculation: $||T^*T|| = \sup_{||x||=1=||y||} |\langle T^*Tx,y \rangle | = \sup_{||x||=1=||y||} | \langle Tx,Ty \rangle| = ||T||^2$

Why are there two variables ($x$ and $y$) involved in that calculation? That doesn't follow the given definition. I don't understand the calculation.

By the way, I am referring to page 16 (or 18) in that document.

From my understanding, $||T^*T|| = \sup_{||x||=1} \sqrt{\langle T^* Tx, T^* Tx \rangle}$, which doesn't even remotely agree with the calculation.

Mikhail

Yo, does this integral have a name $\int_{-\text{ko}}^{\text{ko}} e^{-i z \sqrt{\text{ko}^2-k^2}} \, dk$ ? Or evaluate to something?

Semiclassical

22:38

I think it’s a Bessel function?

Mike Miller

@MatheinBoulomenos Cyclic groups do not act freely on S^2. The only map without fixed points is (conjugate to) the antipodal map

MatheinBoulomenos

ahh I see

Mike Miller

So Z/2 gets that 3-periodicitg and only in Z/2 homology

And of course it has the 1-periodicity from S^0 ;)

I don't know about SO(n), of course A_n sits inside that so any group of order n-1, right?

But there are classifications of which groups act freely on spheres

MatheinBoulomenos

So suppose a finite group $G$ acts freely on S^n, can we write down a period complex $P_\bullet$ of $G$-modules such that for any $G$-module $A$, the Tate cohomology is the homology of the complex $Hom_G(P_\bullet,A)$?

it works for cyclic groups

Mike Miller

13

Q: Which groups act freely on $S^n$?

When $n$ is even, it is easy to classify groups which act freely on $S^n$ using degree theory: if $G$ acts on $S^n$, then associating to each element $g \in G$ the degree of the map obtained from multiplication by $g$, one gets a map $d : G \to \{\pm 1\}$. It is easy to verify this is a homomorph...

@Mathei Allow me to not think about this, I am too scared of saying something wrong

Mikhail

22:52

@Semiclassical I cant' tell.

Current game plan is to Series expand in Mathematica and then integrate those terms :-/

Then be like "high order terms are negligible" behold my exact solution to a well known, and much studied physics problem.

Semiclassical

Ehh, I’d start by subbing $k=k_0 \sin x$

Mikhail

Integrate[ Series[Exp[-I*z*Sqrt[ko^2 - k^2]], {k, 0, ducks_given}], {k, -ko, +ko}]

Semiclassical

Also, since the integrand is an even function you can restrict to positive values at the cost of a factor of 2

MatheinBoulomenos

@MikeMiller another question, about covering spaces. Since covering spaces of $X$ are equivalent to $\pi_1(X)$-Sets (or locally constant set-valued sheaves), we see that the category of them is complete and cocomplete. How do we compute limits and colimits? Just like in the slice category $\mathbf{Top}/X$?

Jake Rose

$$
\begin{matrix}
1-\lambda & 2 \\
2 & 1-\lambda
\end{matrix}$$

Is there an easy way to get the eigenvalues for this matrix?

*EIGENVECTORS

*eigenvectors

Mikhail

22:59

@Semiclassical $\frac{\text{ko} e^{-i \text{ko} z} (3604299271372800+\text{ko} z (\text{ko} z (-162886633273665+\text{ko} z (\text{ko} z (3847551062985+\text{ko} z (415261645785 i-11 \text{ko} z (3174578001+13 \text{ko} z (\text{ko} z (-720450+17 \text{ko} z (19 \text{ko} z-1275 i))+15601572 i))))-28107268226100 i))+763603178502465 i))}{1802149635686400}$ (20 terms for fun)

GFauxPas

what in the world

Semiclassical

That seems proof more of the futility of that approach

GFauxPas

what is $\text{ko}$

Mikhail

around eight : -)

GFauxPas

some number?

ah

Mikhail

23:15

mmmm, seems to diverge a lot when I keep adding other terms

like: Column[N[Table[
Integrate[
Series[Exp[-I*z*Sqrt[ko^2 - k^2]], {k, 0,
terms}], {k, -r, +r}] /. {r -> 7, ko -> 7, z -> 2}, {terms,
15}], 2]]

> 1.9-13.9 I
34.-9. I
34.-9. I
30.+59. I
30.+59. I
-85.+68. I
-85.+68. I
-1.3*10^2-8.*10^1 I
-1.3*10^2-8.*10^1 I
2.*10^1-1.8*10^2 I
2.*10^1-1.8*10^2 I
1.7*10^2-8.*10^1 I
1.7*10^2-8.*10^1 I
1.4*10^2+8.*10^1 I
1.4*10^2+8.*10^1 I

Ted Shifrin

@JakeRose: Your question confuses me. Do you mean the matrix $\begin{bmatrix} 1 & 2 \\ 2 & 1\end{bmatrix}$?

Jake Rose

No the matrix was correct as written

Ive included the eigenvalues

Ted Shifrin

No, the matrix isn't correct as written.

You then want the nullspace of that matrix.

For the particular values of $\lambda$ which are eigenvalues of the original.

Abcd

Hello @Ted

Ted Shifrin

Hi @Abcd

So what are the eigenvalues, numerically?

@Mikhail: Please don't enter stuff like that. It messes up all our screens for hours.

Abcd

23:22

If $\alpha$ is the only real root of the equation $x^3+ bx^2 +cx +1=0$ $(b<c)$, then what can we say about $\sin \alpha$? I know that $\alpha<0$ (leakynun explained it)

Jake Rose

Could somebody help me work out $div(ar-b)$ where a = $(x,y,z)$, b is a constant vector and a is a scalar multiple

Abcd

I mean is $\sin \alpha>0$ or <0

@Ted Do you have any idea about it?

Ted Shifrin

Not offhand, @Abcd.

Jake Rose

Is it 3a?

Ted Shifrin

@JakeRose: I'm not inclined to answer your questions if you ignore me when I try to.

Jake Rose

23:25

@TedShifrin OH my bad sorry I didnt see your previous message

I calculated the eigenvalues as -1 and 3

Ted Shifrin

Right. So now put one of those in and find the nullspace (and maybe you call it kernel) of the resulting matrix. Both cases will be very easy to do.

Abcd

@TedShifrin Specifically the question is: Find the value of $2\arctan (\csc \alpha)+ \arctan(2\sin\alpha \sec^2 \alpha)$ which simplifies to $2(\arctan \sin(\alpha)+ \arctan (\csc \alpha))$ . Then it eventually boils down to finding the sign of sin alpha because that decides whether the answer is $-\pi$ or $\pi$

Ted Shifrin

@Abcd: You come up with the most arcane questions ... and I never find them very interesting (to me).

Jake Rose

Thats what I did

I was just wondering if there was a simple trick to doing it faster

Abcd

Okay :/

Ted Shifrin

23:29

No, @JakeRose. But for $2\times 2$ matrices if you have the correct eigenvalues, it's very easy, because the rows will be multiples of one another, so you just need a vector orthogonal to one of the rows. How do you give me a vector orthogonal to $(a,b)$?

@Abcd: Oh, I see. If you know that equation has only one real root (so everything else is irrelevant), the fact that the polynomial $f(x)=x^3+bx^2+cx+1$ has value $1$ at $0$ means the graph must cross the $x$-axis before you get to $x=0$. So the root must be negative.

Jake Rose

axb?

Ted Shifrin

We're in 2D, @JakeRose. No.

Very easily, what's a vector orthogonal to $(3,4)$?

Jake Rose

How do you know that the rows will be multiples of eachother btw?

Abcd

@TedShifrin Yes, I know that the root is negative ...but how much negative? That doesn't help us find the sign of sin alpha

Ted Shifrin

Because having an eigenvalue means that the matrix you wrote down ($A-\lambda I$) is singular.

Abcd

23:32

Because sin can be positive for negative arguments too

Jake Rose

Because the det(...) =0?

Ted Shifrin

So you are asking if $-\pi<\alpha<0$, @Abcd.

Abcd

Yes

Ted Shifrin

Right, @JakeRose. That makes the rows linearly dependent (scalar multiples).

Abcd

Putting -1 gives:

b-c

and b<c

So for -1 its negative

Ted Shifrin

23:34

Oh, so that does it.

Abcd

\implies root is between -1 and 0

Done :D

Ted Shifrin

Right. You have it.

See ... you didn't need me at all.

Abcd

No :P . You gave the idea of "putting the values to check"

Ted Shifrin

@JakeRose: The hint is to switch the two numbers and make one negative.

Jake Rose

Is the linear dependence resulting in the determinant = 0 a general property for any nxn matrix?

Ted Shifrin

23:37

Yes.

Jake Rose

How do you prove it?

Not asking for you to do so but just the general direction

Ted Shifrin

Depends how much you know.

Do you know properties of determinant if you do row operations?

Leaky Nun

just use the wedge power man

Ted Shifrin

smacks Leaky

Besides, that doesn't help.

Jake Rose

Yes I think so

Ted Shifrin

23:38

So if I can take row $n$ and subtract a bunch of multiples of earlier rows and get $0$ what does that tell me about the determinant?

Leaky Nun

@TedShifrin if it's 0, then it's in the ideal generated by the ei wedge ei guys, so linear dependent

Ted Shifrin

@Leaky. Shaddup.

Leaky Nun

For any non-archimedean local field $K$ (ring of integers $\mathcal O_k$ and residue field $k$), the following three categories are canonically equivalent:
1. $L/K$ finite unramified extension
2. $R/\mathcal O_K$ unramified ring extension
3. $l/k$ finite separable extension

MatheinBoulomenos

4. finite sets with a continuous transitive $\widehat{\Bbb Z}$-action

Don't you want some finiteness continuous condition in 2., too, though?

@LeakyNun

Leaky Nun

Gruß am Sylowmeister

@MatheinBoulomenos maybe my professor forgot to write that

@MatheinBoulomenos how does that work?

MatheinBoulomenos

23:50

Galois theory, basically

Ted Shifrin

@JakeRose: Did you figure out what I asked?

Leaky Nun

@MatheinBoulomenos what is the map?

Jake Rose

Sorry Im bordeline falling asleep its rather late here

A vector orthogonal to 3,4?

Ted Shifrin

Well, that was two questions ago. And I gave a hint for that one. Then you asked about determinant.

Maybe you should get some sleep :)

MatheinBoulomenos

If $l/k$ is a finite extension, then this is the splitting field of some irreducible polynomial. $\operatorname{Gal}(k^{alg}/k)$ acts on the roots of that polynomial

Jake Rose

23:51

Is it that the determinant is =0?

MatheinBoulomenos

the residue fields are perfect

Leaky Nun

so I realized

Ted Shifrin

Yes, to that. @JakeRose

Jake Rose

Also I dont know if my brain isnt functioning right but how do you get a vector orthogonal to that?

Ted Shifrin

I told you what to do, but I don't ping on everything I write @JakeRose. Switch the numbers and make one negative.

Jake Rose

23:52

Do you have to have every row satisfy that property also?

MatheinBoulomenos

Well, $\operatorname{Gal}(K^{ur}/K)= \widehat{\Bbb Z}$, so you can do that, too

Ted Shifrin

Yes, and it's true, along with other properties, @JakeRose.

Jake Rose

@TedShifrin Really sorry, not at my usual standard today

MatheinBoulomenos

$K^{ur}$ is the maximally unramified extension

Leaky Nun

@MatheinBoulomenos you just took the external limit of my equivalent categories :P

direct limit

MatheinBoulomenos

23:53

that doesn't make sense

Jake Rose

Oh yeah thats a cool little trick

Leaky Nun

@MatheinBoulomenos Gal(K^ur/K) = Gal(k^alg/k) because of the equivalence of 1 and 3

MatheinBoulomenos

yes

Ted Shifrin

Congruent triangles, if you draw the picture, shows you why those vectors are orthogonal, @JakeRose.

Leaky Nun

@MatheinBoulomenos why is it continuous?

Jake Rose

23:54

I was thinking more in terms of scalar product

MatheinBoulomenos

factors over a discrete quotient

Jake Rose

But that was is also nice

Leaky Nun

I'm not very familiar with the profinite topology

Ted Shifrin

Well, sure, @JakeRose. But I use this example to motivate the definition of scalar product in my book/lectures.

MatheinBoulomenos

@LeakyNun if $l/k$ is finite, then the topology is constructed such that $\operatorname{Gal}(k^{alg}/k) \to \operatorname{Gal}(l/k)$ is continuous with the discrete topology on the RHS

note that this is also equivalent to the category of finite coverings of $S^1$

Jake Rose

23:57

iNTERESTING

*interesting

Leaky Nun

which is?

Jake Rose

Im definitely gonna check out your lectures more when I have some free time over summer

Ted Shifrin

OK, sleep well :)

MatheinBoulomenos

@LeakyNun those 4 categories we mentioned are equivalent to the category of finite connected (forgot that) coverings of $S^1$

and also, up to equivalence, they don't depend on $K$

which is weird

Leaky Nun

interesantissimo

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$Mathematics$ Mathematics