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15:00
I don't know if there is a faster way to determine the dimension of $A^n$ without diagonalisation, though
The rule I use: Since $*$ is arbitrary, we can assume $* \times * = *$ and $* \times 0 = 0$
Linear dependence: $av_1+bv_2+cv_3 = 0$ for $(a,b,c)\neq (0,0,0)$
$v_3$ not linear combination of $v_1,v_2$, $c=0$
$av_1+bv_2=0$ $(a,b)\neq (0,0)$
But $a=0$ or $b=0$ or $a,b \neq 0$
Hence C
@secret mmm.
@Secret it was a question appeared in CSIR.
@ManeeshNarayanan no, the 5 lines above are to silent's, my response to you is in the pictures
15:15
@Secret sorry for not using reply option.
But basically, since $*$ is arbitrary. the number of columns in the matrix determines the dimension
@skull awesome
@Secret mmm
what if columns are linearly dependent
that could be a possibility, but since $*$ is unspecified, we assume we can take the largest dimension possible thus we can assume they are linearly independent
15:21
also, if you only have two nonzero columns, any mismatch in the row of zeros will mean they must be linearly dependent (1A and 2A and 3A for example)
since nothing will help you escape from a 0 entry by multiplying stuff
yes I got one help from you to reduce the calculation. 4 can be eliminated from the calculations.
since columns can be linearly dependent.
right?
@LeakyNun Please let me know why!
@ManeeshNarayanan That will depend on how it is interpreted in the context of the question. If the question want you to find the option with the specified dimension regardless of what A is, then yes 4 will be eliminated since there is a chance it can be linearly dependent.
@Secret okay.
16:23
Any semisimple ring is isomorphic to a direct product of matrix rings over division rings
Whats the name of this theorem? DOnt look it up!!
16:35
@NicholasRoberts Artin-Wedderburn
@MatheinBoulomenos yeeeeee
Such a great theorem :')
yeah, it's really nice
Helped me on my algebra 2 grad school final
wow, cool. I haven't taken grad algebra yet
Fix 4 lines in $\mathbb R^3$. How many lines intersect all of them?
16:37
you will love it if you like algebra! What is your current status as a student?
I'm an undergrad student
but I will get my bachelor's soon
Ah, so its very good you know artin-webberdurn as an undergrad
the plan is to start my thesis in a few months
you need a thesis for undergrad? damn
16:38
@Nûr uncountably many regardless of configurations
@MatheinBoulomenos i will post the question from my final that relied on Webberdurn
please do!
actually no, some configuration can have a unique line passing through it
for example, two lines parallel to the x axis and two lines parallel to the y axis
I have no idea how to prove it though
The 4 lines are supposed to be in "any position" ("quelconque" in french)
16:41
@MatheinBoulomenos it is a tough question, almost no one in my class solved it correctly. As a hint, the professor says to regard the group action as a representation of G on V. From there, using rep. theory, one can deduce the number of irreducible reps of G.
@MatheinBoulomenos Suppose $L/K$ is a finite abelian extension of local fields. Then, denote by $N:L^\times \to K^\times$ the norm map. Then, $K^\times/N(L^\times) \cong \operatorname{Gal}(L/K)$
@LeakyNun yes, that follows from norm compatability of the Artin map
could you tell me what the Artin map is?
it's the isomorphism from $K^\times$ to $W(K^{ab}/K)$
that one...
(you forgot abelian again)
16:50
I wanted to write ab instead of alg
but yeah
have you read tate "number-theoretic background"?
17:22
@NicholasRoberts by Maschke, $\Bbb C[S_3]$ is semisimple. There are three conjugacy classes in $S_3$ (as conjugacy classes in $S_n$ correspond to cycle types), so we get $\Bbb C[S_3] \cong \Bbb C \times \Bbb \times M_{2\times2 }(\Bbb C)$. Let $V_1$, $V_2$, $V_3$ be the three irreducible representations such that $V_1$ and $V_2$ are 1-dimensional and $V_3$ is two-dimensional.
Let $\chi$ be the character of $V$.
Choose a basis $v,w$ of $\Bbb C^2$, then we get a basis for $V$ over $\Bbb C$ by $B= (v \otimes v \otimes v, v \otimes v \otimes w, v \otimes w \otimes v, w \otimes v \otimes v, v \ot
@NicholasRoberts this was a cool exercise, thanks for sharing!
oh, I didn't use the character at all
lol
fixing notation and then never using it
it's not wrong ...
Let $G$ be a group. does some engineering
Hey @TobiasKildetoft
@MatheinBoulomenos Hi
17:28
did you see my solution to the rep theory exercise? the task is to compute $\operatorname{dim}_{\Bbb C}( \operatorname{End}_{\Bbb C[G]}(\Bbb C^2 \otimes \Bbb C^2 \otimes \Bbb C^2))$ where $G=S_3$ acts by permutation of the factors
@TobiasKildetoft my idea was to identify the $G$-module with the $G$-module induced from the permutation representation of $S_3$ acting on the power-set of a three-element set
hmm, how do you do that identification?
@MatheinBoulomenos Ahh, nice
Seems like you do much more than you need to though
17:37
I didn't want to compute scalar produts of characters
You just needed the dimension of the endomorphism space, which is the inner product of the character with itself
ahh
and of course the character is easy, as it is just the number of fixed points
true
I wonder if you can generalize this argument in some way? If $S_k$ acts on $\bigotimes_{i=1}^k \Bbb C^n$ by permuting the factors, then I think a similar argument should identify this with the permutation representation of $S_k$ from the action on $\operatorname{Hom}_{\mathbf{Set}}(\{1,2, \dots, k\}, \{1, 2, \dots, n\})$ induced by the action of $S_k$ on $\{1,2,\dots, k\}$
@Akash.B np, but does it sound reasonable to you?
\o @anon
17:45
@MatheinBoulomenos Does Schur-Weyl help here?
good job @MatheinBoulomenos
I was just thinking of looking at the action on the obvious basis again
So Schur-Weyl says that this endomorphism space is the same as the space generated by all the endomorphisms coming from $GL_2$
Probably it is actually easier to use your idea here instead. Which then says something about $GL_2$
@TobiasKildetoft can you share your solution using characters?
17:53
@TobiasKildetoft okay, to use language more common in combinatorics, we could say that the action of $S_k$ on $\bigotimes_{i=1}^k \Bbb C^n$ is the same as the representation induced from action of $S_k$ on the n-colorings of a $k$-element set
@Secret did not see that! thank you very much
@NicholasRoberts Once we have it as a permutation representation, the character is easy to find, since it is given by the number of fixed points
And the dimension of the Hom-space between two reps is given by the inner product of their characters. So it is just a matter of adding some numbers
o/
yea but in this case, the vector space splits its into a direct sum of irreps. so youre saying you need to find the character of each one?
In order to do the inner product
18:00
No, we just need the one character
@Secret, so only b and c are true there?
or are you saying simply do the inner product of the perm rep with itself
Ok, and with this its possible to get the correct answer of 20?
im surprised my professor didnt use this as a solution
I don't know how to prove if the dimension of V is 2
18:02
I didn't check, but it will indeed give the correct answer (the part about inner products is true for any characters)
\o @danu did you get your question answered in the LaTeX room?
Yea
cool
The computer I have at work is just crap, is essentially the answer.
All the software is outdated
@Secret no it says dimension less than or equal to 2
18:03
@MatheinBoulomenos I finished grading the reexams yesterday. Rather depressing how poorly they did. But not a huge surprise I guess
I still don't how how to prove that
@TobiasKildetoft reexams are always worse than the first exams
Yeah, because these were the weakest students. But at least it was quick to grade, as there were only 4 of them
Suppose $k >2$. Call the $k$-element set $X$ and the set of $n$-colorings of $X$ $\operatorname{Col}_n(X)$. Fix two colors $a,b$ from our $n$ available colors, then we can write down an injection of $X$ to $\operatorname{Col}_n(X)$ given by sending $x \in X$ to the coloring that paints $x$ in $a$ und paints all other elements$b$.
This defines $2^n$ different embeddings of $X$ to $\operatorname{Col}_n(X)$ and it's easy that the associated embeddings of representations have pairwise trivial intersection. Thus in the decomposition of the permutation representation associated to $\operatorname{
They're also frequently harder
18:04
I feel like there is some non-trivial combinatorics to be applied here, but I know zero about that
None of them really did the last exercise which was on representation theory
It was a nice little and quite easy (I thought) one. Very concrete
what was the exercise?
They were given $D_4$ as a subgroup of $S_4$ and considered permutation representation on $4$ elements.
First exercise was to show that two suitable vectors spanned subrepresentations
Secone was to show that these were not isomorphic
that sounds rather easy
Third was to show that the character was given by a certain formula
18:07
whatever happened with those high schoolers you were showing university math to @TobiasKildetoft?
ok. @Secret, $v_3=(0,1,0)$ is not a linear combination of $v_1=(1,0,0),v_2=(2,0,0)$, and $v_1,v_2,v_3$ linearly dependent. So, how did u conclude that $ v_3 $ has to be zero vector?
and last one as to show that the inner product of the character with itself was $3$ and that there was an irreducible $2$-dimensional rep such that the full rep was the sum of it plus the two from part 1
@skull They were never heard from again
by anyone
:(
no interest
Actually, last time I did anything with non-university students they were not high schoolers, but students about to enter high school
right
18:09
And it went fine. They found it hard to understand, but also interesting
@TobiasKildetoft what do think about the generalization of the approach to $\bigotimes_{i=1}^k \Bbb C^n$? Is this something that might lead to some interesting stuff?
cool
@MatheinBoulomenos It does look interesting
yeah, it feels quite combinatorical
@Silent ok so that means only B is the one not ruled out but why dim <= 2 ?
18:10
But it is too hot here to do any combinatorics
which I guess isn't surprising for representations of $S_k$
Or really to think for that matter
@Secret oh, right, there is no info that these three vectors span $V$.
@Secret, why a and d don't hold?
@TobiasKildetoft so if we write down the obvious basis for $\bigotimes_{i=1}^k \Bbb C^n$, then so far, I analysed the $k2^n$-dimensional subrepresentation spanned by those elementary tensors with exactly two different basis components
no, not even that
the subrepresentation spanned by those elementary tensors where all but one basis components are the same
@Silent v3 is not a linear combination of v1, v2, thus by rearrangement, v1 cannot be a linear combination of v2, v3
unless something like v1=av2+0v3 count as a valid linear combintion
18:18
@Secret oh! this generally doesn't count?
It is also possible for v1, v2 to be lienarly independent, but either v1 or v2 can be linearly dependent with v3
@Silent I don't know
18:35
@MatheinBoulomenos, ring homomorphism between commutative rings with unity has to preserve multiplicative identity, ie, if $f:R\to S$ is ring homomorphism, then $f(1_R)=f(1_S)$, right?
@Silent that's the usual convention nowadays, yes
ok, thanks!
I think you have one $f$ too much
@MatheinBoulomenos i am sorry, but i can't understand this sentence
you wrote $f(1_R)=f(1_s)$, but it's supposed to be $f(1_R)=1_S$
18:41
oh sorry
@MatheinBoulomenos, is this a ring homomorphism: $f: Z_4\to R$ defined as $f(x)=xe$ where $R$ is a ring with unity. I am confused here, because, if $R=\Bbb Z$ then $f(2)*f(3)=2e*3e=6*e$ while $f(2*3)=f(2)=2*e$.
it's not necessarily well-defined, even as a map
if your ring satisfies $4=0$, then it works and it will be a ring homomorphism
hi @MatheinBoulomenos @Silent
hi
:)
$tr(A^p) \equiv tr(A) \mod p$
@Nûr it's just the generalization of $(x+y)^p = x^p + y^p$ ^^
18:50
Yes :)
oh I was gone when Tobias came
@MatheinBoulomenos who do you think would be familiar with Weil group in this chat?
don't know
I only worked with the abelianization
18:53
@Silent you firstly need to check whether they are well-defined
@LeakyNun I checked: a,b,c are well defined, but none of them preserves multiplicative identity. d does, but that may not be well defined!
@TobiasKildetoft consider any way of writing $k=(a_1+a_2+\dots+a_l)$ with $a_i \in \Bbb N$, consider the set $\operatorname{Part}(X,(a_1,a_2, \dots, a_l))$ of all partitions of $X$ into a partition consisting of an $a_1$ element set, an $a_2$ element set etc. $\operatorname{Part}(X,(a_1,a_2, \dots, a_l))$ carries a natural $S_k$ action.
Fix a $l$ possible colors $c_1, \dots, c_l$
Define a map $\operatorname{Part}(X,(a_1,a_2, \dots, a_l)) \to \operatorname{Col}_n(X)$ by sending a partition $P=\{P_1, \dots, P_l\}$ to the coloring that colors all elements from $P_1$ in $a_1$, all elements in $
@Silent the analysis is wrong
@MatheinBoulomenos did you read any books / notes?
this description also shows that the images of all those maps give a direction sum decomposition of $\otimes_{i=1}^k \Bbb C^n$
@LeakyNun an Annals papers about the proof of local class field theory via formal groups
do you still have the link?
18:59
@LeakyNun in which part? in part d?
a,b,c
oh!
oh lol, it's not even the "Annals of mathematics"
thanks
@TobiasKildetoft are you still there?
Since there are partitions involved now, it seems that maybe something could be said by using Young diagrams
19:02
@LeakyNun i m sorry! i can't find out where i am wrong in concluding well defined-ness
let's take B
consider f([0])
f([0]) = 5x0 = 0
f([5]) = 5x5 = 25
[0] = [5]
but 0 != 25
oh!
So, a, c are well defined right?
hm. is the set of n x n positive semidefinite matrices that have diagonal bounded above by 1 the convex hull of something?
@Silent show that they are
19:21
a: $5(4n+i)=20n+5i\equiv 5i \pmod {10}$ and c: $3(4n+i)=12n+3i\equiv 3i\pmod {12}$
@LeakyNun
So, which one is ring homomorphism?
@LeakyNun, none of those two preserves multiplicative identity!
then identity is not preserved
it's just convention
ok, so both a and c are ring homomorphisms?
thank you!
you haven't checked multiplication
19:33
can somebody confirm if
$$\int_{-1}^{1} dsgn = 2?$$
 
1 hour later…
20:37
Hi @Mathei
20:47
If f:N-->N, then is $x^{2} + {x} + 1$ onto?
I think it is. My test analysis doesn't.
What gets mapped to $0$?
Hi @AlessandroCodenotti
@AlessandroCodenotti Not a natural right?
A function is onto if Range = Codomain.
Range and codomain both are $N$.
@SwapnilDas what gets mapped to 1 then?
20:52
N includes 0 for me but that's not universally agreed upon
buonasera @AlessandroCodenotti
I'm reading some cool algebra about formally real fields @Mathei
Yes, such an idiot I am. Thanks.
@AlessandroCodenotti ah, that's really cool
from a model theoretic perspective?
@MatheinBoulomenos if we assume that $W_K^{ab} \cong K^\times$, is there a short proof of $K^\times/N(L^\times) \cong \operatorname{Gal}(L/K)$?
20:54
Hi @Leaky
@LeakyNun just having any isomorphism of abstract groups doesn't give you much
in that direction
@MatheinBoulomenos what do we need?
@MatheinBoulomenos kinda, I need a reference for some algebraic facts about formally real fields, because I want to talk about real closed fields without proving them
So I found this cool book which contains everything I need (and much much more) and I'm reading a bit of it
Well, there is a special isomorphism $Art_K:K^\times \to W_K^{ab}$ (it's actually uniquely characterized by some properties) this has a certain compatability. If $L/K$ is finite separable, then we have that $Art_K \circ N_{L/K} = (Art_L)_{\mid K^{ab}}$
this is called norm-compatability of the Artin map
20:57
yes
It's corollary 5.16, but there's always Lubin-Tate extensions instead of abelian extensions. Theorem 6.15 gives you that the maximal Lubin-Tate extension is equal to the maximal abelian extension
So corollary 5.16 + Thm 6.15 imply the statement
once you have norm compatability, proving the isomorphism is just elementary Galois theory
@AlessandroCodenotti there's actually a theory of formally p-adic fields and p-adically closed fields that's very analogous to formally real fields and real closed fields
you can prove that p-adically closed fields are model complete
Given that the model completeness of algebraically closed fields of a fixed characteristic implies the Nullstellensatz, I'm willing to believe that this could be pretty useful
(since you expressed interest in p-adics lately)
That was Daminark I think
I don't know much about p-adics, but it does sound cool
21:03
oh, but I thought you wanted to read Neukirch chapter 2, too?
Oh, right, but I just generally want to learn more algebraic number theory, I'm not specifically interested in p-adics over other topics
chapter 2 is mostly p-adics (well, technically some stuff is more general, but p-adics and their extensions are the main example)
Ah, I see, I didn't know that
Well I guess I'll learn about p-adics eventually then :D
you absolutely should if you're interested in ANT
@LeakyNun fun fact: in characteristic p, the Weil group is not dense as a subset of the absolute Galois group
@MatheinBoulomenos all my fields are local
21:08
From what I've seen of ANT (very little, more or less Neukirch chapter one restricted to the quadratic number fields case) it's a really cool subject so I'd like to see more of it
@LeakyNun $k((T))$ is a local field of characteristic p
but Tate said it had to be dense...
Ah, the mememaster has arrived
Hi @Dami
hi danmark
Hello, $(E,d)$ is a compact space and $f:E\to E$ such that $\forall x,y\in E, x\neq y, d(f(x).f(y))<d(x,y)$how to prove that $f$ has a unique fixed point
21:22
Pick $x_0\in E$ and construct a sequence $x_n=f(x_{n-1})$, show that it converges to a fixed point of $f$ (you're gonne need that a compact metric space is complete at some point)
i don't have compact implies complete !
I think you want to show that the map $E \times E \to \Bbb R, (x,y) \mapsto d(f(x),f(y)) - d(x,y)$ has a maximum as a first step
i know that is a compact space each sequence has a convergent subsequence
@PolineSandra then it's complete!
complete means that any cauchy sequence is converegent
21:25
but any cauchy sequence has a convergent subsequence
and you can prove that the limit is the limit of the whole sequence using Cauchy
How do you solve the differential equation $\frac{d^2y}{dx^2}=y^2=0$?
y^2=0 so y=0 ^^
You don't need to show this for this exercise
just using a subsequence of the sequence that Alessandro mentioned is enough
Ive been given the answer of ax+b?
The quesiton was phrased let $y^{(n)}=\frac{d^ny}{dx^n}$
That means the n:th derivative
21:28
ahhh
bugger
They did not make that explicity clear
21:42
@LeakyNun okay i'll take it back, the Weil group probably is dense even in the function field case
I was just confused that one source only stated denseness for p-adic fields
Hi all, can someone give me a hint on how to find homology of $S^1 \times X$ if $H_*(X)$ a free abelian group
@B.Mehta Künneth?
I don't have that as a tool - only Mayer Vietoris
@B.Mehta take two semicircles $U_1$ and $U_2$ on $S^1$ such that their interiors cover $S^1$, then apply Mayer-Vietoris to $U_1 \times X$ and $U_2 \times X$
prove Kuenneth then
21:48
$U_1$ and $U_2$ are contractible, so $U_1 \times X$ and $U_2 \times X$ are homotopy equivalent to $X$
Can somebody explain to me why $A+Be^{x}+Ce^{ix}+De^{−x}+Ee^{−ix}$ is equivelant to $a+bsinx+ccosx+dsinhx+ecoshx$?
euler's formula
What about the hyperbolics
?
$U_1 \cap U_2$ is homotopy equivlanet to two discrete points, so $(U_1 \times X) \cap (U_2 \times X) = (U_1 \cap U_2) \times X$ is homotopy equivalent to $X \coprod X$
you can pull out a dijsoint union out of the homology as a direct sum (another Mayer-Vietoris argument)
@MatheinBoulomenos is that MV?
I thought that's proved early on
21:51
you can prove it separately if you want, it's not hard
but you can also prove MV early on
ok everything is early
so freeness of $H_*(X)$ gives you that the exact sequence from MV splits which gives you the homology of $H_*(S^1 \times X)$
@LeakyNun just a random thought: maybe the reason why we look at the Weil group instead the absolute Galois group is because its (complex) representations are more interesting
14
A: Why Weil group and not Absolute Galois group?

Johnson JiaOne reason we prefer the Weil group over the Galois group (at least in the local case) is that the Weil group is locally compact, thus it has "more" representations (over $\bf C$). In fact, all $\bf C$-valued characters of $Gal(\bar{\bf Q_p} / \bf Q_p)$ have finite image, where as that of $W_{\bf...

okay, someone else said that too
r e k t
21:55
I can prove to you that continuous homomorphisms $\rho: G \to H$ where $G$ is infinite profinite and $H$ is a (real or complex) Lie group have finite image, if you want
please do
btw this article you linked was received 05/06/2006 and accepted 07/07/2008 :o
zwei jahre :o
@MatheinBoulomenos Thanks, I'll give it a go!

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