« first day (2855 days earlier)      last day (2463 days later) » 

18:00
@loch No, a Lie group is more than just a group object in the category of topological spaces (since that would just be a topological group).
@TobiasKildetoft do you know about algebraic torus?
@LeakyNun I know what it is, sure
wonderful
(and I know about its representation theory)
do you mind if I ask you a lot of questions over the following weeks
18:04
sure
So I know that a torus is a group object in the category of k-schemes
no, that's a split torus
I'll try to tell you what I know, maybe you can correct me
@TobiasKildetoft oops
so a split torus is just Spec(k[X1,X1^-1,...,Xn,Xn^-1])
denote this by Spec(R)
@LeakyNun a group object in the category of $k$-schemes is just an algebraic group over $k$
ok so a split torus is a special kind of group object given by Spec(R)
where the group struction, seen in the opposite category (i.e. k-alg) is given by maps R->R tensor_K R, R->R, and R->K
the first map sends Xi to Xi tensor Xi
the second map sends Xi to Xi^-1
the third map sends Xi to 1
18:08
right
it is just the direct product of some number of copies of the multiplicative group
now I'm lost in the definition of torus
what do they mean by T(E) when T is a torus?
What is $E$?
a finite Galois extension of k, I suppose
I don't really know
@TobiasKildetoft what is the multiplicative group?
So a $k$-algebra?
The multiplicative group is just the case $n=1$ of what you wrote above
since that is identified with the group of units under multiplication
(when we see this as a functor)
which one is a functor?
18:11
Spec
Or rather, Spec(R)
probably relevant - if you have a scheme $X$ youmight think of $X(F)$ as $F$-valued points on $X$
how is that defined?
@LeakyNun Spec(R) is the functor that associates to a $k$-algebra $A$ the set of homomorphisms of $k$-algebras from $R$ to $A$.
As loch says, these are the $A$-valued points of the scheme
@TobiasKildetoft I can see that there is a (covariant?) functor sending A : k-alg to Hom(Spec(A),Spec(k[X,X^-1))
which corresponds to the units in A
in which category?
groups?
from the outset, Spec(R) just takes values in Set
18:13
I think you want an abelian group
But being an algebraic group then precisely means taking values in Groups and with composition by the forgetful functor giving back Spec(R)
But it does not sound like this is the point of view you are learning, so it might be more confusing than enlightening
i think the functor pov is really good for alg groups
yeah, it makes it much easier to work with (I think)
@TobiasKildetoft you said two contradictory things in two messages...
But it does make some of the more geometric things more tricky
18:16
anyway, so a torus is a finite direct product of copies of Spec(k[X,X^-1])?
isn't that a split torus?
Or rather, that seems to be called a split torus for you
I don't recall what additional condition "split" really is
I'll split your torus if you know what I mean ;)
(I don't know what I mean)
Groupprops says:
> A torus (sometimes called an algebraic torus) over a field is an algebraic group that can be described as a direct product of finitely many multiplicative groups of finite extensions of that field. Note that we could use the multiplicative group of the field itself in one or more of the direct factors.

The term split torus is used for a torus that is a direct product of copies of the multiplicative group of the field. When the field is an algebraically closed field, any torus over it is a split torus.
18:18
@LeakyNun Ahh, of course
but that isn't what my professor tells me
Which I why I usually only see the term split mentioned when moving to points over a finite field
What is he telling you?
he tells me that a split torus is one where it becomes split when you tensor by the algebraic closure
maybe that's equivalent
When you tensor what?
18:20
I'll tensor your torus if you know what Imean
@TobiasKildetoft let's say I take the direct product with the Spec of the algebraic closure of the field
(that's the tensor product in the opposite category, i.e. the category of k-algebras)
Ahh, so you tensor the coordinate algebra
right
I.e. you extend scalars to the algebraic closure
bear in mind that I don't know 50% of the words i utter
so
you know
would appreciate if you could help me
18:22
Right, that should be equivalent. One direction is clear, the other might take small argument
which definition is easier?
@Daminark hi
Easier for what?
for understanding
for motivation
etc
Probably the direct product one
But you should work out why they are equivalent
I can convince myself of that...
no actually I can't
let's say I managed to do so
18:26
One direction is just seeing what happens when you tensor by the algebraic closure
In general, if $A$ is an $R$-algebra, what is $R[x]\otimes_R A$?
let's work with $\Bbb R$
wiki says there are two tori over $\Bbb R$ with rank 1
$\Bbb R^\times$ and $SO(2)$
but groupprops says $\Bbb R^*$ and $\Bbb C^\ast$
18:30
Well, $SO(2)$ is just $\mathbb{C}^*$, right?
say what
$\Bbb C^\ast \cong \Bbb R_{>0} \times \Bbb R/\Bbb Z$
I don't see it
What does a special orthogonal $2\times 2$ matrix look like?
Let's say the first column is $(a,b)$. What is the second column?
$(-b,a)$
Well, up to a scalar
are we in the reals or the complexes?
18:37
reals
then isn't it (-b,a)?
Ahh, right, we need orthonormal, not just orthogonal columns
So now, let's denote this matrix just as $(a,b)$. What will we get when we multiply $(a,b)$ with say $(c,d)$?
$\begin{pmatrix}a&-b\\b&a\end{pmatrix} \begin{pmatrix}c&-d\\d&c\end{pmatrix} = \begin{pmatrix}ac-bd&-ad-bc\\ad+bc&ac-bd\end{pmatrix}$
2
Q: Fourier transform of a sequence and inverse fourier transform

user8469759If $$ h(k) = \begin{cases} \frac{1}{2l+1} & -l \leq k \leq l \\0 & \text{otherwise}\end{cases} $$ Where $l \geq 0$ is some integer. I've done some computation and the summation $$ F[h](\omega)=\sum_{k=-\infty}^{+\infty} h(k)e^{j\omega k} = \frac{1}{2l+1}e^{-j\omega} \frac{\sin(\omega(l+1/2))}...

another update
then I can see that SO(2) = S^1...
18:47
hmm, yeah
18:59
Is this for your first year project? @LeakyNun
Finally finished the combo final
Woot
congrats
19:20
How did it go? @Dami
Hello. Could someome tell me how to know if this $2\sum_{k=1}^n\sum_{j\neq k}^n(\nabla w_k\cdot\nabla w_j)c_kc_j$ is equal to 0?
I think it went pretty well, thanks!
I screwed up one question completely, I think I got at least half the points on another, and on a third one... I'm not sure, probably under half the points
Rest of the test is hopefully good
@Alt. Need more context methinks
the red color text
Great! @Dami do you have more exams to deal with now?
19:34
Next week I have complex (analytic number theory mostly) on Monday and algebra (Galois) on Tuesday
@Alt. But that's not just a dot product, it's an inner product?
@Lozansky yes, it's an inner product, I mean this $(x\cdot y)=x_1y_1+\dots+x_ny_n$
@Alt. Mhm and integrate over $D$
? no. The integration part comes later
Then I guess you can apply Green 1 to cancel the terms where $w_{k \neq 0}$ appear in the surface integral (since they are $0$ on the boundary)
19:42
Guys, say $p$ and $q$ are prime, and we have a finite group $G$ whose non-trivial elements have order $p$ or $q$. Let $N=\{g\in G: g^q=e\}\neq\{e\}$. How can I show that the order of $G/N$ is a power of $p$? I’ve already shown that $N$ is normal, and I know that by definition, $N$ only contains elements of order $q$, and I'm guessing then that #$N=q^k$, if #$G=p^lq^k$, but not sure
@Lozansky I'm in the algebraic part of the solution, nothing about integrals
@Sha do you know Cauchy's theorem on finite groups?
@Alt. Not sure what you mean by "algebraic part". The inner product is defined as a scalar product integrated over $D$
@AlessandroCodenotti o yea, that's it, thanks!
hi, $Fr$ is the frobenius transform($x\to x \ ^p$) i showed that $<Fr> \le Gal(\overline{F_p} / F_p)$ and that it is infinite. also i showed that $\overline{F_p} \ ^{<Fr>} = F_p$. now i need to show that $Gal(\overline{F_p}/F_p)$ is abelian. someone can help?
i thought i solved it but i didn't :/
19:48
@ShaVuklia You can use it to show that $|N|=q^k$
@Lozansky I mean algebraic calculations with the inner product. So Green theorem it's applied to pass from $(\sum_{j=1}^nc_j\nabla w_j\cdot \sum_{j=1}^nc_j\nabla w_j)$ to $\sum_{j=1}^n(\nabla w_j\cdot\nabla w_j)c_j^2+2\sum_{k=1}^n\sum_{j\neq k}^n(\nabla w_k\cdot\nabla w_j)c_kc_j$ ?
@AlessandroCodenotti not sure how? say $\vert N\vert < q^k$. somehow we should find an element not in $N$ that actually does belong in $N$? (ie contradiction)
@Liad do you know what is the definition of $Gal(\overline{\mathbb{F}_p}/\mathbb{F}_p)$?
@ShaVuklia Think about $G/N$
I don't know man:l
19:56
Hmm, quick sanity check: if $Gal(E/F) = \mathbb{Z}/6$ and $E$ is the splitting field of some irreducible $f$, then $f$ should have degree $6$, right? My argument is that $\mathbb{Z}/6$ can't act transitively on more than 6 things by orbit-stabilizer, 1 and 2 are obviously impossible since they'll split in smaller fields, and the thing is that if you take a root of an irreducible degree 3/4/5 which splits in $E$, it generates a proper subfield which should be Galois since stuff is abelian
@ShaVuklia How many elements does it have?
apparently a power of $p$, but that's what I want to show
@ShaVuklia Well suppose that's not the case and argue for a contradiction
@loch yes
@LeakyNun i see - what is your topic on?
19:59
I don't know man, I should be going to bed really, my test is tomorrow
I can't think right now
@loch local langlands for abelian
I assume they mean something
lol i have no idea
@Alt. Ah. No, that is just grouping together terms with equal indices and terms with non-equal indices.
back then i just did some stuff with $p$-adics
@Sha good luck
20:02
oh i think i see it now
yeah, having a good sleep before an exam is important
(but, rarely possible)
But how are you supposed to cram an entire course in one day if you sleep?
true dat
@loch yes
@Daminark $Gal(E/F) \hookrightarrow S_{\deg f}$ right
you know that Z/6Z doesn't go into S3
caveat, it does go into S5
but that embedding isn't transitive
so S6 it is
@loch if T is a torus, what does $T_K$ mean?
field is $F$, $K$ is finite Galois extension over which $T$ splits
so $T_K$ means $T \times \operatorname{Spec}(K)$?
it is said that $L := \operatorname{Hom}(T_K,\operatorname{Spec}(K[X,X^{-1}]))$ determines $T_K$ up to isomorphism
but that is just the units of the coordinate algebra of $T_K$?
20:21
@Daminark @LeakyNun generally if $\operatorname{Gal}(E/F)$ is cyclic of order $n$ then $E/F$ is the splitting field of some irreducible polynomial $f$ of degree $n$.
The reason is that the embedding into $S_{\operatorname{deg}(f)}$ given by the action on the roots must be generated by a single $n$-cycle (since having multiple cycles would give different orbits and a non-transitive action) and the only way the subgroup generated by a cycle acts transitively is if the length of the cycle is equal to the number of elements it acts on
and it's also abelian!
I see
@MatheinBoulomenos do you know about algebraic torus?
I probably should, but I don't
Also I talked to someone who read Kronecker-Weber in D&F and tried to invoke that
But I was just like uhhhhh
Like, that's only over $\mathbb{Q}$, and also even if we were only thinking of $\mathbb{Q}$ it was probably way too strong of a result to be acceptable to use given that we haven't proven it
for sure
I also don't see how it helps, honestly
@MatheinBoulomenos So let $R=k[X_1, X_1^{-1}, \cdots, X_n, X_n^{-1}]$
20:26
Lol I didn't see how he concluded, I just saw him say Kronecker-Weber and I was just like "uh fam?"
form the canonical maps $m : R \to R \otimes_k R$, $i : R \to R$, and $e : R \to k$
So you're talking a product of some copies of $k^\times$ as an algebraic group?
right
in the opposite category, i.e. that of schemes over $k$, you have $m^* : G \times G \to G$, $i^\ast : G \to G$, and $e^\ast : \{pt\} \to G$
so you have a group object
where $G = \operatorname{Spec}(R)$
I'm familiar with a bit of algebraic groups
(so you don't have to explain how Hopf algebras correspond to affine group objects)
but can you explain that to me?
Anyway, a torus is just $\prod_{i=1}^n \operatorname{Spec}(E_i[X,X^{-1}])$, where each $E_i$ is a finite Galois extension of $k$, and the product is taking place in the category of affine $k$-schemes
20:32
Is Kronecker-Weber in D&F?
@Symposium yeah, I'm kinda surprised, too. Maybe it's mentioned, but I don't think there's a proof
@LeakyNun so basically you just take the morphisms that define a group object in the category of affine schemes over $k$ (note that finite products of affine schemes are affine again, so that's just saying that $G$ is affine) and apply your duality to the cateogory of $k$-algebras, which turns all morphisms around and makes a tensor product out of the product of schemes
proofwiki.org/wiki/User:GFauxPas/Sandbox I'm stuck on the set arithmetic here, can anyone give me a hint to prove disjointness?
and then these commutative diagrams define a "cogroup object" in $k-\mathbf{Alg}$ which are basically by definition commutative (but not necessarily cocommutative) Hopf algebras over $k$
trying to prove that a $\sigma$-algebra is closed under countable unions iff closed under countable disjoint unions
@GFauxPas everything is intersection
20:37
okay
and
WLOG m < n
sure
if $x \in F_n$, then $x \in E_n \setminus E_m$
if $x \in F_m$, then $x \in E_m$
done
oh lol I dont need all those \bigcups
thanks Leaky, that's obvious in retrospect
just draw a diagram ^^
20:39
okay Ted
(If I call you "Ted" it's a compliment)
@LeakyNun there's another way to define group schemes (affine or not) which I actually prefer. So basically by a Yoneda argument, we can view any scheme over $k$ as a functor $F:k-\mathbf{Alg} \to \mathbf{Set}$ by looking at some Hom-functor.
One can show, by a Yoneda exercise I won't spoil for you that having a group object is equivalent to having a functor $\tilde{F}: k-\mathbf{Alg} \to \mathbf{Grp}$ such that $\tilde{F}$ composed with the forgetful functor $\mathbf{Grp} \to \mathbf{Set}$ is $F$
in practise, one often goes the other way around: you start with a functor to the category of groups and then show it's representable
For algebraic tori, this functors are simple to describe
@GFauxPas just use dominated convergence
so $k^\times$ as a group scheme over $k$ is just the functor $k-\mathbf{Alg} \to \mathbf{Grp}$ given by $A \to A^\times$
invoking the axiom of choice, there exists a set that is disjoint from another set
20:45
i thought we were just using big theorems
use LDC here
oh thats from a long time ago from undergrad
and if $E_i$ is a finite extension of $k$ (Galois if you want), then $E_i^\times$ is the functor $k-\mathbf{Alg} \to \mathbf{Grp}$ is given by $A \mapsto (E_i \otimes_k A)^\times$
And if you take some products of that, you take just product of groups
if $T$ is a torus, what do you think $T_K$ and $T(K)$ mean?
people who don't do notations...
he also wrote $T_K(K)$
I assume $K$ is some extension of $K$?
20:47
and said that $T_K(K) = T(K)$
$K$ is some extension of $F$
we're using $F$ as our base field now
$T$ splits over $K$
$T(K)$ are the $K$-rational points of the scheme
i.e. $T \times \operatorname{Spec}(K)$ is a split torus
@MatheinBoulomenos how is that defined?
so if we regard it as a functor, it's just evaluated at $K$
20:49
hmm
so if $T = \prod_{i=1}^n \operatorname{Spec}(E_i[X,X^{-1}])$
where each $E_i$ is a finite extension of $F$
this means it's the set of homomorphisms (of schemes over $F$) from $K$ to $T$
because $T$ is a group scheme, this has a group structure
then $T(K) = \prod_{i=1}^n (E_i \otimes_F K)^\times$?
then what do you suppose $T_K$ means?
not 100% sure on that, but probably the base change to $K$
20:51
can you write it down concretely?
aha
So $T$ is a $F$-algebra
and we tensor over $F$ with $K$ and get a $K$-algebra
it should be $T_K(-) = \prod_{i=1}^n ((E_i \otimes_F K) \otimes_K -)^\times$, right
you don't have to call me gott, Mathein is alright
20:53
god why do people not do notations
like in front of every paper should be one page of pure notation
I think this is standard stuff for people in alg geo
fair enough
oh and in front of every statement I made in this ten minutes, adjoin "Spec"
@loch hi
@LeakyNun you don't have to always write Spec if you regard things as functors
oh nvm remove the Spec
someone who knows more alg geo than us noobs! @loch
20:56
$T_K(-) = \operatorname{Hom}_{F-Sch}(\operatorname{Spec}(-), T \times \operatorname{Spec}(K))$?
I'd write $\times_F$ instead of $\times$ but yeah
two contravariants make a covariant
So $T_K = T \times_F \operatorname{Spec}(K)$ as an affine scheme
no wait, you take homomorphisms of schemes over $K$
Now $L := \operatorname{Hom}(T_K, \operatorname{Spec}(K[X,X^{-1}]))$
@MatheinBoulomenos right
I suppose $L$ has some meaning

« first day (2855 days earlier)      last day (2463 days later) »