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,

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

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

[tex]\pi_1(g)\otimes\pi_2(g)(x_1\otimes x_2)=\pi_1(g)(x_1)\otimes\pi_2(g)(x_2)[/itex]

I haven't checked if it satisfies the requirements.

(I think rochfor1'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.