How to understand the RHS in the explicit example of Schur-Weyl duality for $\mathbb{C}^2\otimes\mathbb{C}^2$?

Question

I try to understand the Schur-Weyl duality. I would be alreday happy with a very rudimentary understanding based on the possibly simplest non trivial instance. So it says for example that

$$ \mathbb{C}^2\otimes\mathbb{C}^2 = Sym^2(\mathbb{C}^2)\oplus Alt^2(\mathbb{C}^2)$$

where $Sym^2(\mathbb{C}^2)$ are the symmetric complex 2x2 tensors and $Alt^2(\mathbb{C}^2)$ the skew-symmetric ones. Do I understand correctly that the left hand side can be represented by

$$ \begin{pmatrix} a & b\end{pmatrix}\otimes \begin{pmatrix} a & b \end{pmatrix} = \begin{pmatrix} a^2 & ab & ab & b^2 \end{pmatrix} $$ where $\begin{pmatrix} a & b\end{pmatrix}$ shall represent the vector space of complex 2-vectors for all $a,b\in\mathbb{C}$.

I am somehow lost as to what concerns the right hand side.

Answer 1

Well one big issue is that you are taking the tensor product of two different matrices. $GL_2(\mathbb{C}) \times GL_2(\mathbb{C})$ acts on $\mathbb{C}^2 \otimes \mathbb{C}^2$ via the matrix you wrote down, but it is an irreducible representation. In Schur-Weyl duality we are decomposing $\mathbb{C}^2 \otimes \mathbb{C}^2$ for the diagonal copy of $GL_2(\mathbb{C})$ inside $GL_2(\mathbb{C}) \times GL_2(\mathbb{C})$.

Answer 2

Okay so $\mathbb{C}^2 \otimes \mathbb{C}^2$ is indeed isomorphic to $\mathbb{C}^4$. This means though that each $2\times 2$ matrix will now act via a $4 \times 4$ matrix. In coordinates this looks like:

$$ \begin{pmatrix} a & b \\ c & d \\ \end{pmatrix} \to \begin{pmatrix} a^2 & ab & ab & b^2 \\ ac & ad & bc & bd \\ ac & bc & ad & db \\ c^2 & cd & dc & d^2 \end{pmatrix} $$

In these coordinates $\Lambda^2(\mathbb{C}^2)$ is the subspace spanned by $(0,1,-1,0)^\top$, which you can easily check just gets scaled via the determinant $ad-bc$. Similarly, $Sym^2(\mathbb{C}^2)$ is the three dimensional subspace where the second and third coordinates are equal to one another.