I am trying to understand the Schur functor $S^{(2,1)}$. Let's try on a vector space $V$ of dimension 3. The general definition is :

$S^{(2,1)}V = V^{\otimes 3} \otimes_{S_n} V^{(2,1)}$

where $V^{(2,1)}$ is the irreductible representation of $S_n$ (the symmetric group) associated to $(2,1)$. I understand that representation, which has the basis $\{ (x_2 - x_1), (x_3 - x_1) \}$ as a subrepresentation of a representation with basis $\{ x_1, x_2, x_3 \}$. However, I have difficulties to understand the tensor power on $S_n$.

In the case $S^{(3)}V$, I easily see that, with $V^{(3)} = \mathbb{C}\{x_1 + x_2 + x_3\}$ as a subrepresentation, the condition of the tensor product is

$ \begin{align} & \ \ \ \ \ v_{\sigma(1)} \otimes v_{\sigma(2)} \otimes v_{\sigma(3)} \otimes_{S_n} (x_1 + x_2 + x_3) \\ & \equiv v_{1} \otimes v_{2} \otimes v_{3} \otimes_{S^n} (x_{\sigma(1)} + x_{\sigma(2)} + x_{\sigma(3)}) \\ &= v_1 \otimes v_2 \otimes v_3 \otimes_{S_n} (x_1 + x_2 + x_3) \end{align} $

So it is equivalent to say that the variables commutes, so $S^{(3)}V = Sym^3V$, the component of the symmetric algebra of $V$.

In the case $S^{(1,1,1)}V$, using the same definition and with $V^{(1,1,1)} = \mathbb{C} \{(x_3 - x_2)(x_3 - x_1)(x_2 - x_1)\}$, I also clearly understand why $S^{(1,1,1)}V = \Lambda^3V$, the component of the exterior algebra of $V$.

But there is something I don't understand when $\dim(V^{\lambda}) \neq 1$. In the case $\lambda = (2,1)$, $\dim(V^{\lambda}) = 2$, so by the tensor product, it creates differents classes of tensors and I don't know what to do with them. I also see that different permutations $\sigma \in S_n$ of $\mathbb{C}\{x_3 - x_1, x_2 - x_1 \}$ is always the span of two of the three subspaces $x_3 - x_1, x_2 - x_1$ and $x_3 - x_2$.

By «different classes of tensors», I mean that, by example, we have:

$ \begin{align} v_{\sigma(1)} \otimes v_{\sigma(2)} \otimes v_{\sigma(3)} \otimes_{S_n} (x_3 - x_1) \equiv \\ &\ \ \ \ \ \ \ \ \ \ v_1 \otimes v_2 \otimes v_3 \otimes_{S_n} (x_3 - x_1) \ \text{if} \ \sigma = \epsilon \\ &\ \ \ \ \ - v_1 \otimes v_2 \otimes v_3 \otimes_{S_n} (x_3 - x_1) \ \text{if} \ \sigma = (13) \\ &\ \ \ \ \ \ \ \ \ \ v_1 \otimes v_2 \otimes v_3 \otimes_{S_n} (x_3 - x_2) \ \text{if} \ \sigma = (12) \\ &\ \ \ \ \ - v_1 \otimes v_2 \otimes v_3 \otimes_{S_n} (x_3 - x_2) \ \text{if} \ \sigma = (132) \\ &\ \ \ \ \ \ \ \ \ \ v_1 \otimes v_2 \otimes v_3 \otimes_{S_n} (x_2 - x_1) \ \text{if} \ \sigma = (23) \\ &\ \ \ \ \ - v_1 \otimes v_2 \otimes v_3 \otimes_{S_n} (x_2 - x_1) \ \text{if} \ \sigma = (123) \\ \end{align} $

What do I do with that? How do I interpret it? Is it helpful in understanding this Schur functor? Can you help me in this case and maybe the general case if not too hard? Thanks in advance!

I am trying to understand the Schur functor $S^{(2,1)}$. Let's try on a vector space $V$ of dimension 3. The general definition is :

$S^{\lambda}V = V^{\otimes n} \otimes_{S_n} V^{\lambda}$

where $V^{\lambda}$ is the irreductible representation of $S_n$ (the symmetric group) associated to $\lambda \vdash n$.

In the case that is of interest for me, i.e. $\lambda = (2,1)$, I understand the representation $V^{(2,1)}$ of $S_3$, which has the basis $\{ (x_2 - x_1), (x_3 - x_1) \}$ (as a subrepresentation of a representation with basis $\{ x_1, x_2, x_3 \}$). However, I have difficulties to understand the tensor product on $S_3$.

In the case $S^{(3)}V$, I easily see that, with $V^{(3)} = \mathbb{C}\{x_1 + x_2 + x_3\}$, the definition of elements of $S^{(3)}V$ with the tensor product is

$ \begin{align} & \ \ \ \ \ v_{\sigma(1)} \otimes v_{\sigma(2)} \otimes v_{\sigma(3)} \otimes_{S_n} (x_1 + x_2 + x_3) \\ & \equiv v_{1} \otimes v_{2} \otimes v_{3} \otimes_{S^n} (x_{\sigma(1)} + x_{\sigma(2)} + x_{\sigma(3)}) \\ &= v_1 \otimes v_2 \otimes v_3 \otimes_{S_n} (x_1 + x_2 + x_3) \end{align} $

So it is equivalent to say that the variables commutes, so $S^{(3)}V = Sym^3V$, the component of the symmetric algebra of $V$.

In the case $S^{(1,1,1)}V$, using the same definition and with $V^{(1,1,1)} = \mathbb{C} \{(x_3 - x_2)(x_3 - x_1)(x_2 - x_1)\}$, I also clearly understand why $S^{(1,1,1)}V = \Lambda^3V$, the component of the exterior algebra of $V$.

But there is something I don't understand when $\dim(V^{\lambda}) \neq 1$. In the case $\lambda = (2,1)$, $\dim(V^{\lambda}) = 2$, so by the tensor product, it creates differents classes of tensors and I don't know what to do with them. I also see that different permutations $\sigma \in S_3$ of $\mathbb{C}\{x_3 - x_1, x_2 - x_1 \}$ is always the span of two of the three subspaces $x_3 - x_1, x_2 - x_1$ and $x_3 - x_2$.

By «different classes of tensors», I mean that, by example, we have:

$ \begin{align} v_{\sigma(1)} \otimes v_{\sigma(2)} \otimes v_{\sigma(3)} \otimes_{S_n} (x_3 - x_1) \equiv \\ &\ \ \ \ \ \ \ \ \ \ v_1 \otimes v_2 \otimes v_3 \otimes_{S_n} (x_3 - x_1) \ \text{if} \ \sigma = \epsilon \\ &\ \ \ \ \ - v_1 \otimes v_2 \otimes v_3 \otimes_{S_n} (x_3 - x_1) \ \text{if} \ \sigma = (13) \\ &\ \ \ \ \ \ \ \ \ \ v_1 \otimes v_2 \otimes v_3 \otimes_{S_n} (x_3 - x_2) \ \text{if} \ \sigma = (12) \\ &\ \ \ \ \ - v_1 \otimes v_2 \otimes v_3 \otimes_{S_n} (x_3 - x_2) \ \text{if} \ \sigma = (132) \\ &\ \ \ \ \ \ \ \ \ \ v_1 \otimes v_2 \otimes v_3 \otimes_{S_n} (x_2 - x_1) \ \text{if} \ \sigma = (23) \\ &\ \ \ \ \ - v_1 \otimes v_2 \otimes v_3 \otimes_{S_n} (x_2 - x_1) \ \text{if} \ \sigma = (123) \\ \end{align} $

What do I do with that? How do I interpret it? Is it helpful in understanding this Schur functor? Can you help me in this case and maybe the general case if not too hard? Thanks in advance!