2 deleted 2 characters in body


I think I have solved this, with the help of mt and Tom Goodwillie.

Question 1 is wrong (and thus Question 2 is wrong as well).

Proof. Let $V$ be a three-dimensional vector space over an infinite field $K$ of characteristic $3$. Let $\left(x,y,z\right)$ be a basis of $V$. Let

$A=V^{\otimes 3} / \quotA$

and

$B=\ScV / \quotB$.

If Question 1 would have a positive answer, there would be a $\mathrm{GL}\left(V\right)$-equivariant map $B\to A$ splitting the canonical projection $A\to B$ (because canonical morphisms between Schur functors are, in particular, $\mathrm{GL}\left(V\right)$-equivariant maps on each object). We will show that this is not the case.

First, we know that $V\otimes V\cong \SbV\oplus \wedge^2 V$ canonically (since the characteristic of our field is $\neq 2$), so that $V^{\otimes 3}\cong V\otimes \SbV \oplus V\otimes \wedge^2 V$ canonically. We thus identify $V^{\otimes 3}$ with $V\otimes \SbV \oplus V\otimes \wedge^2 V$. Then, clearly, the subspace $\quotA$ of $V^{\otimes 3}$ lies completely inside the direct addend $V\otimes \SbV$, so that $A=V^{\otimes 3} / \quotA$ becomes

$A=\left(\left(V\otimes \SbV\right)/\quotA\right) \oplus V\otimes \wedge^2 V$.

The projection $A\to B$ has the direct addend $V\otimes \wedge^2 V$ in its kernel, and thus it factors through the $\mathrm{GL}\left(V\right)$-module

$C:=\left(V\otimes \SbV\right)/\quotA$.

(Thanks to mt for this idea.) Now, assume that we have a $\mathrm{GL}\left(V\right)$-equivariant map $B\to A$ splitting the canonical projection $A\to B$. Then, this map gives rise to a $\mathrm{GL}\left(V\right)$-equivariant map $f:B\to C$ splitting the canonical projection $C\to B$ (in fact, just compose the map $B\to A$ with the projection $A\to C$ to obtain this map $f$). This map $f:B\to C$ must be injective (since it splits a projection). We will now show that this is impossible by proving that $f=0$. (Tom's idea.)

First we notice that the subspace $\quotA$ of $V^{\otimes 3}$ is $10$-dimensional and has basis
