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I am studying the basic of representation. Coming back from https://en.wikipedia.org/wiki/Tensor_product_of_representations, I am having trouble filling out details why tensor product of two (finite dimensional) representations $V$ and $W$ over $\mathbb{C}$ of a finite group $G$ is also a representation.

I know Is the tensor product of two representations a representation? has an answer but I cannot understand such "high level" reasoning yet.

From the first link above, the action is defined as $g.(v\otimes w):=(g.v)\otimes (g.w)$, but I am not sure about its well-definedness.

Question 1: This definition cannot be seen as $g$ acting on arbitrary element in the tensor product, right? I mean, isn't arbitrary element in the tensor product a linear combination? Hence, what would the "true" action of $g$ be?

I think if I assume the that $g$ acts on an arbitrary element by linearity (meaning it can pass through the sigma and scalars), then here is my attempt based on my understanding (but I am not sure maybe something is wrong):

We first choose a basis $\{v_i\}$ and $\{w_j\}$ for $V$ and $W$, respectively. Thus, all $\text{dim } V\times \text{dim } W$ elements $v_i \otimes w_j$ form a basis for $V\otimes W$. To show well-definedness, I take elements $a,b\in G$ and two more elements in $V \otimes W$, but the elements in this tensor product is not simply basic tensor, right? Hence, I take $\sum c_{i,j}(v_i\otimes w_j),\sum d_{i,j}(v_i\otimes w_j)\in V\otimes W$ instead. If $a=b$ and $\sum c_{i,j}(v_i\otimes w_j)=\sum d_{i,j}(v_i\otimes w_j)$, then by linear independence, $c_{i,j}=d_{i,j}$ for each $i,j$. Hence, $$\sum c_{i,j}((a.v_i)\otimes (a.w_j)) =\sum d_{i,j}((a.v_i)\otimes (a.w_j))\text{ (by $c_{i,j}=d_{i,j}$)}\\=\sum d_{i,j}((b.v_i)\otimes (b.w_j))\text{ (by well-definedness of each $G$-action on $V$ and $W$)}.$$

Thus, the action map is well-defined. I think showing it satisfies the action property + linearity should be OK.

Question 2: Is the above reasoning correct?

Any suggestion or guidance is really appreciated. Thanks!.

10understanding
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1 Answers1

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A representation $V$ of $G$ is a homomorphism $\rho :G \to GL(V)$, where $GL(V)$ is the group of automorphisms of V.

Then, if $g(v \otimes w)=g(v)\otimes g(w)$ gives a representation $V\otimes W$ of $G$, then it should be an automorphism of $V\otimes W$.

It's not necessary to check arbitrary elements of the tensor product. Indeed, the rule defines the action of $g$ only on elements in the form $v\otimes w$ because, although omitted from the definition sometimes, it's linear, so $$g(\sum_i \alpha_i (v_i\otimes w_i))=\sum_i \alpha_i g(v_i\otimes w_i).$$

Now, the action is also well-defined, because $$g((\alpha_1 v_1+\alpha_2v_2)\otimes w) = g(\alpha_1 v_1+\alpha_2 v_2)\otimes g(w) = (\alpha_1 g(v_1)+ \alpha_2 g(v_2))\otimes g(w)$$ $$= \alpha_1 g(v_1)\otimes g(w) +\alpha_2g(v_2)\otimes g(w) = \alpha_1g(v_1\otimes w) + \alpha_2 g(v_2\otimes w)$$ and in the same way, $$g(v\otimes (\alpha_1 w_1+ \alpha_2w_2)) = \alpha_1g(v\otimes w_1) +\alpha_2g(v\otimes w_2).$$

Also, $ker(g)$ is trivial: $$g(v\otimes w) = 0 \implies g(v)\otimes g(w)=0 \implies g(v)=0\ \lor \ g(w)=0$$ But since $g$ is an automorphism on $V$ and $W$, then $v=0$ or $w=0$ and $v\otimes w=0$.

Since $g$ is linear and $ker(g)$ is trivial, then $g$ is an automorphism of $V\otimes W$. The fact that $g^{-1}$ is inverse of $g$ as an automorphism follows from definition on $V$ and $W$ and can be checked directly, thus the action of $G$ is isomorphic to a subgroup of $GL(V\otimes W)$, and $V\otimes W$ is a representation of $G$ by $g(v\otimes w) = g(v)\otimes g(w)$.

Teodora
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