# Example of tensor product of two representations

This is similar to what Serre wrote in his book on linear representations of finite groups by Serre:

There is also the tensor product which has the properties of "multiplication". Let $$V_1$$ and $$V_2$$ be two vector spaces. The tensor product $$\otimes: V_1 \times V_2 \rightarrow V_1 \otimes V_2$$ constructs a new vector space $$V_1 \otimes V_2$$ with the map $$(\xi_1,\xi_2) \rightarrow \xi_1 \cdot \xi_2$$ which can be defined as the set of formal linear combinations $$\xi_1 \otimes \xi_2$$ subject to the conditions:

$$(i)$$ $$\xi_1 \cdot \xi_2$$ is linear in both variables $$\xi_1$$ and $$\xi_2$$.

$$(ii)$$ If $$\lbrace e_{i_1} \rbrace$$ and $$\lbrace e_{i_2} \rbrace$$ is a basis for $$V_1$$ and $$V_2$$ respectively then $$e_{i_1} \otimes e_{i_2}$$ is a basis for $$V_1 \otimes V_2$$.

We can show that this space exits and is unique. By condition $$(ii)$$ we have: $$\dim(V_1 \otimes V_2) = \dim(V_1)\cdot \dim(V_2)$$

Now let $$\rho^1\colon G \rightarrow GL(V_1)$$ and $$\rho^2\colon G \rightarrow GL(V_2)$$ be two linear representations of a group $$G$$. For $$s \in G$$, define an element $$\rho_s$$ of $$GL(V_1 \otimes V_2)$$ by the condition: $$\rho_s(\xi_1 \cdot \xi_2) = \rho_s^1(\xi_1)\cdot \rho_s^2(\xi_2) \quad \text{for } \xi_1 \in V,\; \xi_2 \in V_2.$$

We write: $$\rho_s = \rho_s^1 \otimes \rho_s^2$$

The $$\rho_s$$ above defines a linear representation of $$G$$ in $$V_1 \otimes V_2$$ which is called the tensor product of the given representations.

Defining the above using matrix notation:

let $$\lbrace e_{i_1} \rbrace$$ be a basis for $$V_1$$ and let $$r_{i_1j_1}(s)$$ be the matrix of $$\rho_s^1$$ with respect to this basis, define $$\lbrace e_{i_2} \rbrace$$ and $$r_{i_2j_2}(s)$$ in a similar manner. Now we have: $$\rho_s^1(e_{i_1}) = \sum_{i_1} r_{i_1j_1}(s) \cdot e_{i_1}, \qquad \rho_s^2(e_{i_2}) = \sum_{i_2} r_{i_2j_2}(s) \cdot e_{i_2}$$

which implies: $$\rho_s(e_{i_1}\cdot e_{i_2}) = \sum_{i_1,i_2} r_{i_1j_1}(s) \cdot r_{i_2j_2}(s)\cdot e_{i_1} \cdot e_{i_2}$$

Consequently the matrix of $$\rho_s$$ is $$(r_{i_1j_1}(s) \cdot r_{i_2j_2}(s))$$ which is the tensor product of the matrices $$\rho_s^1$$ and $$\rho_s^2$$.

We must add, the tensor product of two irreducible representations is not in general irreducible. It decomposes into a direct sum of irreducible representations which can be determined by means of character theory, which we shall discuss in the next chapter.

My question therefore is, can someone construct a concrete example to make me understand this

• Have you seen tensor products of vector spaces (without a group acting) before? – Nate Apr 17 '19 at 13:41

Let $$\rho: \langle g: g^4=e \rangle \to GL(\mathbb{R}^2)$$ be the representation of the cyclic group with four elements (i.e. $$\mathbb{Z}/ 4\mathbb{Z}$$) into $$\mathbb{R}^2$$ such that $$\rho_g = \begin{bmatrix} 0 & -1 \\ 1 &0 \end{bmatrix} = M$$ and $$\rho_{g^k}=M^k$$. Note that there are no non-trivial $$g$$-invariant subspaces (one can verify this by showing $$M$$ does not have 1 as an eigenvalue) and so $$\rho$$ is irreducible.
However, let's look at what happens when we tensor $$\rho$$ with itself to get a representation of $$\mathbb{Z}/ 4\mathbb{Z}$$ into $$\mathbb{R}^2 \otimes\mathbb{R}^2$$. Letting $$\rho' = \rho \otimes \rho$$ we get that for $$v,w\in \mathbb{R}^2$$: $$\rho'_g(v\otimes w) = \rho(v) \otimes \rho(w)$$ So your intuition should be that $$\rho'$$ mimics $$\rho$$ on each "copy" of $$\mathbb{R}^2$$. This is made clearer by actually computing the matrix of $$\rho'_g$$.
As you wrote, if $$\{e_1,e_2 \}$$ is the standard basis of $$\mathbb{R}^2$$ then the set $$\{e_1\otimes e_1, e_1\otimes e_2, e_2\otimes e_1,e_2\otimes e_2\}$$ forms a basis of $$\mathbb{R}^2 \otimes \mathbb{R}^2$$. Observe: $$\rho'_g(e_1\otimes e_1) = Me_1 \otimes Me_1 = e_2 \otimes e_2$$ $$\rho'_g(e_1\otimes e_2) = Me_1 \otimes Me_2 = e_2 \otimes -e_1 = -e_2 \otimes e_1$$ $$\rho'_g(e_2\otimes e_1) = Me_2 \otimes Me_1 = -e_1 \otimes e_2$$ $$\rho'_g(e_2\otimes e_2) = Me_2 \otimes Me_2 = -e_1 \otimes -e_1 = e_1 \otimes e_1$$ So the matrix of $$\rho'_g$$ is: $$\rho'_g = \begin{bmatrix} 0 & 0 & 0 & 1 \\ 0 & 0 & -1 & 0 \\ 0 & -1 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{bmatrix}.$$ This actually shows us something mildly interesting. While the original $$\rho$$ was an irreducible representation our new $$\rho'$$ is not since the subspace of $$\mathbb{R}^2\otimes \mathbb{R}^2$$ spanned by $$\{e_1\otimes e_1, e_2\otimes e_2 \}$$ is a proper invariant subspace. This confirms the statement at the end of your post.
Remark: I'm not sure if you know this but $$\mathbb{R}^2\otimes \mathbb{R}^2 \cong \mathbb{R}^4$$ as both are four dimensionals vector spaces over $$\mathbb{R}$$. So there really is nothing particularly fancy going on here, but I tried to keep everything in terms of tensors to help you see how things would work in a more general case.
• A representation is just a group homomorphism from a group into $GL(V)$ for some vector space $V$. So I wouldn't say that I "came up" with a matrix $\rho_g$, but rather that this is how I defined the representation to begin with. I guess you could say that I came up with the idea for such a representation by trying to find a $2\times 2$ matrix that satisfied $M^4 =I$, since that would be enough for me to define my homomorphism. – Edgar Jaramillo Rodriguez Apr 24 '19 at 20:49