okay this is the question =p

Hmm I feel like we talked a bit about this yesterday

that's pure group theory

oh really :D

do you know what conjugation looks like in $S_n$?

well hmm

i remember that one plug in s(a_1) s(a_2 ) ...s(a_n)

did not understand it well tbh in Group theory

sts'() = s(a_1) s(a_2 ) ...s(a_n)

soemthing of that nature

right, if t=(a_1, ... a_n)

that should allow you to do a)

yes =p

let me try :D

damn it :(

the notation is what makes me get lots

what about the notation?

so we need to show that p_1 and p_2 are conjugates right?

for all elements of S_6

no

for ever a in Z/2xZ/2 you have to show that p_1(a) and p_2(a) are conjugate

p_1(a) = sigma p_2(a) sigma'

sigma in S_6 is that right?

yes

for every a you have to find a sigma that works

since we have 4 elements in (Z/2)^2

dont we just do this by computation ?

or there is a better way?

yes, you just compute this

hmm let me see :D

so basiclly the Point of this exerice is what?

that knowing p (1,0)

and we need to figure out the rest?

knowning that we have a representation

hmm

for summer break, ill go thruought all algebra in a detailed way :D many gaps i got!

the point of this exercise is to show that just because for every a you cand find that f(a) and g(a) are conjugate, this doesn't mean there is some s such that sf(a)s'=g(a) for all a

where f,g:G -> H are group homomorphisms

this is important because in representation theory you often are interested if you can find a s such that sf(a)s'=g(a) for all a

okay that make sense =p

okay, am gonna continue this exercice from here, we were given 6 exercices to do exept those others from serre, there are 2 i did not understand well, ill post the other one, in case you have to leave so i can work on them,

that one and then I will work on the rest alone :D

and show you my solutions :D

thanks alot mathein ! you are the best ! :D

hmm, that's an interesting exercise

he gave us some stuff from AA and others from rep theory

Are you working over $\Bbb C$?

general field K

but he said mostly wil be like serre

using C

I hate when the HW problems are needed to be solved in time in other to be able to pass ><

anyway I have a week to solve these and to reviste some linear algebra and Group theory ( only small parts )

so we have an isomorphism between the representation and its dual representation, so we have an invertible matrix $A$ such that $A^{-1}r(g)A=(r(g)^{-1})^T$, we can rewrite this as $r(g)Ar(g)^T=A$, so we have to show that we can choose $A$ to be either orthogonal or symplectic

hmm

I don't see how Schur's lemma is useful when we're not over an algebraically closed field, to be honest

am not really that far yet :D but let me check again if we working over C or any field

ye over Any field K , but we gonna mainly be using Complex

I have an extra note that might be usefull on that question

He put this like on the top of the question

t^A <--- its his way to write A^t ( transpose )

i meant the t, is on the left side

My problem is that I don't see how Schur's lemma says anything useful if we're not over an algebraically closed field

sorry, I'm stumped

better ask Tobias

its fine =p

It might be also that he made a mistake ><

i mean if you look at first problem i sent you, second line p_2 should have been p_1

he wrote p_2 = ... , 3 times

@MatheinBoulomenos still here? =p

yeah

am still working on first problem =p

I just want to show you what i got so far

p_1(1,0) = (12) (34)

p_1 (0,1) = (13)(24)

p_1(1,1) = (14) (23)

p_1(0,0) = (1)

p_2 (1,0) = (12) (34)

p_2(0,1) = (12)(56)

p_2(1,1) = (34) (56)

p_2(0,0) = (1)

so to answer this question , one just needs to find the particualr element that makes p_1(a) and p_2(a) conjugate

eg (13) (24) = sigma (12) (56) sigma '

yeah

okay so far so good :D

so what sigmas do you choose?

still trying to figure out that method

i mean i know that the elements of same cycle type are conjugates

in S_n

that also solves the exercise

if you can use that

you should be able to give the sigmas explicitely though

okay but all of them are of type [2,2]

exept for the identity

could it be that what he wants us to do :D

you need to understand how you can conjugate one to get to the other one, or you won't be able to solve b)

okay hmm

sigma (12) (56) sigma ' = (13) (24) , can you remind me how to find it in that systematic way ?

i did this Before but forgot the algoritm

what if you send 1->1, 2 ->3, 5->2,6->4 and do something with the rest

it does not matter what I do with rest right?

no

I mean, yes it doesn't matter

haha ><

got me a bit confused for a sec there =p

no idea how to answer negative question affirmatively

yes it is better to say the full thing back ><

to avoid comfusion

(235)(46) should do for that part

I think I solved the second exercise, but I must have done something wrong

I haven't used that the representation is irreducible

and I showed that it's always orthogonal

Hmm, if you said that something is not right am pretty sure it is :D

I can send email to my teacher if you can give me the question =p

he often makes mistakes so its fine =p

Okay, my solution goes like this: we know that because he representation is self-dual, we can find and invertible matrix such that $A^{-1}r(g)A=r(g^{-1})^T$ for all $g \in G$

Now we define a non-degenerate bilinear form on the vector space by setting $\langle v,w \rangle =v^T A^TAw$

then for $g$ in $G$, we have $\langle r(g) v, r(g) w \rangle= v^T r(g)^T A^T A r(g) w$

so it suffices to show that $r(g)^T A^T A r(g)= A^T A$

hmm okay :D ill let you know asap =p

Ah, no I start with $A^{-1}r(g^{-1})^TA=r(g)$

I'm not finished, yet!

haha ok good i did not hit send

haha

From that we get $r(g^{-1})^TA=Ar(g)$

so we get $(Ar(g))^T=(r(g^{-1})^TA)^T=A^Tr(g^{-1})$

but also $(Ar(g))^T=r(g)^TA^T$

So we get $r(g)^TA^T=A^Tr(g^{-1})$

If we plug that into $r(g)^TA^TAr(g)$ we get $r(g)^TA^TAr(g)=A^Tr(g^{-1})Ar(g)$

hmm, I think I made an mistake :D

we don't have $r(g^{-1})Ar(g)=A$

Okay nevermind

Hmm :D

I know how to solve it if the field is algebraically closed though

Then that is the Point of ti

i mean if he said in the hint to use schur so that makes sense =p

The thing is

Schur says something even if the field is not algebraically closed

it's just not that useful

Anyway, I'm going to bed now

Good night!

okay Ill ask the teacher and let you know :D

thanks alot mathein as allways :D

Goodnight ! :D

hi

hi

Oups you are here :)

So Im(p_2) = { (12) (34) , (12) (56) , (34) (56) , (1) }

Im ( p_1) = (12) (34) , (13)(24) , (14)(23), (1)

We can call those K and H respectivly

So since K fixes nothing

and H fixes the Points 5,6

in the set X= {1,2,3,4,5,6}

We have if sigma p_1(a) sigma' = p_2(a)

this means that

Sigma H sigma' = K

But if H fixed a Point x, so does K

contradiction :D

@MatheinBoulomenos well ? :D

good work

haha thanks _

the last line was not very sure about it

if H fixes a point

so does K or sigma(K)

i need to Review the lecture on conjugation in S_n properly

it was very index stuff so i skipped it

You have to think about conjugation in S_n as just relabelling of the letters you use

Can you explain more please? :)

example if you have

So what does conjugating with (a,b) do?

It's always switching a and b

thanks :D

Okay, i have Another question that am thinking of

G acts transtlivly on X ,we need to show that

Summation _g in G Fix (g) = ord (G)

transitive actions means that there exist g in G

such that gx =y , for all x ,y in X right?

ie, by multipling with element of G, we can get from one Point of the set to another

right

okay ill keep thinking about it meanwhile :)

Hint: x in Fix(g) <=> gx=x <=> g in Stab(x)

hmm

well we do know that G/ stab(x) is in bijective correspondance with orb(x)

but since we have one orbit because our action is transitive

|G| = |Stab(x) |

@MatheinBoulomenos so far so good?

No, you don't have that. That's true if X has only one element

grrr ok

so we gonna sum up what elements of G fixes

and that has to add up to the order of G

Okay so you know that for every x in X, you have |Stab(x)||Orb(x)|=|G|

yes

But since the action is transitive, |Orb(x)|=|X| for every x

yes so far so good

So we can say that |Stab(x)|=|G|/|X| for all x

Yes =p

So we have |G|=|X|*(|G|/|X|)= sum_{x in X} |Stab(x)|

Because we sum over |X| terms and each has size |G|/|X|

Now here's a little trick

define a function from GxX -> {0,1} by setting f(g,x)=1 if gx=x and f(g,x)=0 if gx /= x

Then if we fix x, we can say that |Stab(x)|=sum_{g in G} f(g,x)

okay hmm

And if we fix g, we have |Fix(g)|=sum_{x in X} f(g,x)

Do you see how this helps us?

can you please fix the latex typing

could not read it

Oh I thought you weren't using latex :P

I copy and past it so i can see it haha

in the "ask a question " on this site

$|G|=\sum_{x \in X} |\operatorname{Stab}(x)|$

We define $f:G\times X \to \{0,1\}$ by $f(g,x)=1$ if $gx=x$ and $f(g,x)=0$ if $gx \neq x$

For any fixed $x \in X$, we have that $|\operatorname{Stab}(x)|=\sum_{g \in G} f(g,x)$

Are you following so far?

Yes so far so good

Okay, for a fixed $g \in G$ we have $|\operatorname{Fix}(g)|= \sum_{x \in X} f(g,x)$

So now I'm asking you to put these pieces together

okay that idea of introducing that function was very neat

let me put the pieces together and see how well it goes :D

i might need some time :D

@MatheinBoulomenos arent those quantities equal ?

what quantities?

x being in the Fix(g) is equivalent to g being in Stab(x)

Yes, but we're summing over different things

ahh true

we're fixing x or we're fixing g

damn it mathein :(

Its hard for me to work with these stuff :(

grrrr never invented new function to solve a problem

Okay I gave you 3 equations

i would have double sum

$|G|=\sum_{x \in X} |\operatorname{Stab}(x)|$, $|\operatorname{Stab}(x)|=\sum_{g \in G} f(g,x)$ and $|\operatorname{Fix}(g)|= \sum_{x \in X} f(g,x)$

yes, the double sum is good!

but hmm

what happens if you change the order of summation

let me see how I can make sense of that :D

i want to solve it :D

give me a min =p

ahaaa

if we interchange the order

the inside sum will be the fix(g) :D

exactly!

You got it

very NEAT solution mathein :D im gonna copy this and try to re-use it :D

But first of all, I need to repeat step by step

this is genious :D you are the best mathein :D

Yes, I think the introduction of this functions allows a very elegant way of writing this down

you can do the same idea without the function, but the notation will be more cumbersome

the thing is they never showed us such methods

Yeah I can see how it would be bad Writing if we define it using words

btw if you get the solution of that symplectic/orthogonal representation exercise, I'd be interested to see it

@MatheinBoulomenos yes sir! I will tell you as soon as I got one :D