Retract of symmetric algebra If we consider a vector space over an algebraically closed field $k$ of characteristic $0$, we have the following exact sequence:
$$0\rightarrow \wedge^{2}V\rightarrow \text{T}^{2}V\rightarrow\text{S}^{2}(V)\rightarrow 0\text{,}$$
where $\wedge^{2}, \text{T}^{2}$ and $\text{S}^{2}$ denotes the second degree component of the exterior algebra (resp. tensor and symmetric algebra). In fact, we can define explicitly the retract
$$\text{S}^{2}(V)\rightarrow\text{T}^{2}(V)$$
which is just:
$$v_{1}\cdot v_{2}\mapsto \frac{1}{2!}\sum_{\sigma\in S_{2}}v_{\sigma(1)}\otimes v_{\sigma_{2}}\text{.}$$
My question is the following one: can we describe explicitly the kernel of the map
$$\text{T}^{n}(V)\rightarrow\text{S}^{n}(V)$$
as before? It is easy to define the retract for the general case, we just can replace $2$ by $n$, but how we describe the kernel? What happens in the case $n=3$ and $n=4$ (which should be the easiest ones)?
Thank you for your time.
 A: (Below I need $k$ to have characteristic zero, which maybe can be weakened to characteristic bigger than $n$, but I don't need it to be algebraically closed.)
It depends on what you mean by "explicitly." Abstractly Schur-Weyl duality implies that the tensor power $V^{\otimes n}$ breaks up, under the commuting actions of $GL(V)$ and $S_n$, into a direct sum of Schur functors
$$V^{\otimes n} \cong \bigoplus_{\lambda \vdash n} S^{\lambda}(V) \boxtimes M^{\lambda}$$
corresponding to partitions of $n$. This is precisely the decomposition into irreducible representations of $GL(V)$, and when $n \ge 3$ there are strictly more of them than just the symmetric and exterior power.
For example, when $n = 3$ there are three Schur functors which can be thought of as corresponding to the three irreducible representations of $S_3$:

*

*The symmetric cube $S^3(V)$, corresponding to the trivial representation of $S_3$,

*The exterior cube $\wedge^3(V)$, corresponding to the sign representation of $S_3$, and

*The Schur functor $S^{(2, 1)}(V)$, which doesn't have a name that I'm aware of, corresponding to the nontrivial $2$-dimensional irreducible representation of $S_3$.

So the kernel of the symmetrization map $V^{\otimes 3} \to S^3(V)$ is isomorphic, as a $GL(V)$-representation, to $\wedge^3(V) \oplus 2 S^{(2, 1)}(V)$.
What I mean by "corresponds" is the following. If $M^{\lambda}$ is the Specht module (irreducible representation) of $S_n$ labeled by $\lambda$, then $S^{\lambda}(V)$ is constructed as $V^{\otimes n} \otimes_{S_n} M^{\lambda}$. The isotypic component of $S^{\lambda}(V)$ in $V^{\otimes n}$ can be isolated as the image of the idempotent
$$\frac{\dim M^{\lambda}}{n!} \sum_{\pi \in S_n} \chi_{\lambda}(\pi^{-1}) \pi$$
(where $\chi_{\lambda}$ is the character of the Specht module $M^{\lambda}$), which reproduces the familiar symmetrization idempotent when applied to the trivial representation and the familiar antisymmetrization idempotent when applied to the sign representation, as expected. So this can all be done as explicitly as you can write down the character table of $S_n$.
