A simple question on representations and tensor products

[vertices]

I have question, can someone please check whether my answer is correct or not:

1)Let \pi_i be representations of a group G on vector spaces V_i, i = 1, 2. Give a formula for the tensor product representation \pi_1 \otimes \pi_2 on V_1 \otimes V_2

Answer: \pi_1 V_1 \otimes \pi_2 V_2

2)Check that it obeys the representation property.

Answer: A representation is a group homomorphism, ie it satisfies:

\pi(g.h)= \pi(g) . \pi(h)

Now,

<br /> [\pi_1 V_1 \otimes \pi_2 V_2](g.h)<br /> =\pi_1 V_1 (gh) \otimes \pi_2 V_2 (gh)<br />

I am a little stuck here: we know that \pi_i is a representation, can we also say that \pi_i V_i is also a representation? If it is, we can use the homomorphism property and show that

\pi_1 V_1 (gh) \otimes \pi_2 V_2 (gh)=\pi_1 V_1 (g)\pi_1 V_1 (h) \otimes \pi_2 V_2 (g) \pi_2 V_2 (h)=[\pi_1 V_1 \otimes \pi_2 V_2](g)[\pi_1 V_1 \otimes \pi_2 V_2](h)

which I think the question is trying to get at.

 

[rochfor1]

If \pi_i are representations, then they are functions G \to V_i, so how are they eating elements of V_i? You have things somewhat backwards here.

 

[vertices]

If \pi_i are representations, then they are functions G \to V_i, so how are they eating elements of V_i? You have things somewhat backwards here.

I think it's because \pi_i \in End (V_i)

 

[rochfor1]

ah yes, excuse me...fair enough. You still have it a bit backwards though. \pi_i \not \in \operatorname{End}(V_i), \pi_i(g) \in \operatorname{End}(V_i) for all g in G. \pi_i : V \to \operatorname{End}(V_i). Pardon the initial confusion, its been a while since I though about this type of representation.

 

[Fredrik]

What do you mean by this? Answer: \pi_1 V_1 \otimes \pi_2 V_2

Aren't you supposed to find a function \pi:G\rightarrow\mbox{End}(V_1\otimes V_2) and show that it's a representation? The first idea that occurs to me is

\pi(g)=\pi_1(g)\otimes\pi_2(g)

where the right-hand side is defined by

\pi_1(g)\otimes\pi_2(g)(x_1\otimes x_2)=\pi_1(g)(x_1)\otimes\pi_2(g)(x_2)[/itex]<br /> <br /> I haven&#039;t checked if it satisfies the requirements.<br /> <br /> (I think rochfor1&#039;s last LaTeX expression should be \pi_i : G \to \operatorname{End}(V_i) ).

 

[rochfor1]

Indeed it should.

 

[vertices]

Hi Fredrik:

Well, I was mindlessly reeling off the following in my notes:

if A \in end(V)

and B \in end (W)

then

A \otimes B \in end(V \otimes W), and this is defined via:

(A \otimes B)(V \otimes W)= (AV) \otimes (AW)

So I just assumed we can let A=\pi_1 and B=\pi_2,

This clearly is silly because it would imply \pi_1: V_1 \rightarrow V_1 and \pi_2: V_2 \rightarrow V_2, where the maps infact should be \pi_1: G \rightarrow V_1 and \pi_2: G \rightarrow V_2 as you point out.

I think you're first idea is correct.

Thanks.