# Young Symmetrizer and Exterior Products, such as $S_{(2,1)}V = Ker(\Lambda^2V \otimes V \to \Lambda^3V )$

Background: Young symmetrizer $c_\lambda$ gives an explicit description of Schur module $S_\lambda V$, which is also the kernel of maps between exterior products (as in Fulton & Harris).

Example: Consider the Young symmetrizer $c_{(21)} = () + (12) - (13) - (132)$, constructed from:

$1 2\\3$

Then its image in $\otimes^3 V$ is spanned by:

$v_1 \otimes v_2 \otimes v_3 + v_2 \otimes v_1 \otimes v_3 - v_3 \otimes v_2 \otimes v_1 - v_3 \otimes v_1 \otimes v_2$

which also gives $S_{(2,1)}V$. And we have:

$S_{(2,1)}V = Ker(\Lambda^2V \otimes V \to \Lambda^3V )$.

=======================

Question 1: The dimension of $S_{(2,1)}V$ shall be $\frac{(n+1)n(n-1)}{3}$, where $n = dim(V)$. However, I am not sure how to directly see the expression:

$v_1 \otimes v_2 \otimes v_3 + v_2 \otimes v_1 \otimes v_3 - v_3 \otimes v_2 \otimes v_1 - v_3 \otimes v_1 \otimes v_2$

has dimension $\frac{(n+1)n(n-1)}{3}$ in $\otimes^3 V$.

That is, without using $\dim( S_{(2,1)}) = \dim(\Lambda^2V \otimes V) - \dim(\Lambda^3V)$.

=======================

Question 2: Consider the map $\Lambda^2V \otimes V \to \Lambda^3V$ again, which is defined by:

$(v_1 \wedge v_2) \otimes v_3 \mapsto v_1 \wedge v_2 \wedge v_3$

It seems there are multiple ways to write down its kernel, such as:

$(v_1 \wedge v_2) \otimes v_3 + (v_1 \wedge v_3) \otimes v_2$, or,

$(v_1 \wedge v_3) \otimes v_2 + (v_2 \wedge v_3) \otimes v_1$, etc.

And there are multiple ways to construct a map $\Lambda^2V \otimes V \to \otimes^3 V$, such as:

$(v_1 \wedge v_2) \otimes v_3 \mapsto v_1\otimes v_2 \otimes v_3 - v_2 \otimes v_1 \otimes v_3$, or,

$(v_1 \wedge v_2) \otimes v_3 \mapsto v_1\otimes v_3 \otimes v_2 - v_2 \otimes v_3 \otimes v_1$, etc.

So there are actually multiple elements in $\otimes^3 V$ corresponding to $Ker(\Lambda^2V \otimes V \to \Lambda^3V )$.

I have computed some of them, and it seems they are either:

• Young symmetrizer $c^\prime_{(21)}$ constructed from all ways of filling the Young tableaux, such as $() + (23) - (12) - (132)$ constructed from:

$2 3\\1$

• $g \cdot c^\prime_{(21)}$ where $g \in S_3$. Example: $(23) + (123) - (132) - (13) = (23) \cdot \Big(() + (13) - (12) - (123)\Big)$. And $() + (13) - (12) - (123)$ can be constructed from:

$1 3\\2$

So my second questions is, is the above true for other partitions $\lambda$?

Question 1: The general construction of the Schur module (see for instance Fulton's book on Young tableaux, Chapter 8; the example $S_{2,1}(V)$ appears in the introduction of Part II) yields
$S_{2,1}(V) = \mathrm{coker}(\omega: \Lambda^3(V) \to \Lambda^2(V) \otimes V),$
where $\omega$ is defined by $$\omega(a \wedge b \wedge c) = (b \wedge c) \otimes a - (a \wedge c) \otimes b + (a \wedge b) \otimes c,$$ or equivalently by the more symmetric definition $$\omega(a \wedge b \wedge c) = (a \wedge b) \otimes c + (b \wedge c) \otimes a + (c \wedge a) \otimes b.$$ Notice that $\omega$ is part of the comultiplication of the graded Hopf algebra $\Lambda^*(V)$. You can also write $S_{2,1}(V)$ as a kernel, but let me work with this cokernel description.
Notice that $\omega$ has a left inverse $s : \Lambda^2(V) \otimes V \to \Lambda^3(V)$ given by mapping $(a \wedge b) \otimes c$ to $\frac{1}{3} (a \wedge b \wedge c)$. In particular, $\omega$ is injective. Hence, $$\dim(S_{2,1}(V)) = \binom{n}{2} \cdot n - \binom{n}{3} = \frac{(n+1)n(n-1)}{3}.$$
We can also write down a basis: If $e_1,\dotsc,e_m$ is a basis of $V$, then the images of the vectors $(e_i \wedge e_j) \otimes e_k$ with $i<j$ and $i \leq k$ form a basis of $S_{2,1}(V)$. For this one checks (a) that they generate $S_{2,1}(V)$, (b) that their number is $\dim(S_{2,1}(V))$.
Here is a sketch of a completely different approach: Using the decomposition $$\Lambda^n(V \oplus W) = \bigoplus_{p=0}^{n} \Lambda^p(V) \otimes \Lambda^{n-p}(W),$$ one finds a decomposition $$S_{2,1}(V \oplus W) \cong S_{2,1}(V) \oplus S_{2,1}(W) \oplus (V^{\otimes 2} \oplus W) \oplus (V \otimes W^{\otimes 2}).$$ In particular, we have $$S_{2,1}(V \oplus \mathbb{K}) \cong S_{2,1}(V) \oplus V^{\otimes 2} \oplus V.$$ From here, it is easy to determine the dimension recursively.