2 deleted 2 characters in body

$\newcommand{\SbV}{\mathrm{Sym}^2 V}$ $\newcommand{\ScV}{\mathrm{Sym}^3 V}$ $\newcommand{\quotA}{\left< v\otimes v\otimes v \mid v\in V\right>}$ $\newcommand{\quotB}{\left< vvv \mid v\in V\right>}$

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

$\left(xxx,yyy,zzz,yxx+xyx+xxy\text{ and 5 similar terms},xyz+xzy+yzx+yxz+zxy+zyx\right)$sums},xyz+xzy+yzx+yxz+zxy+zyx\right)$. Here, we are suppressing the$\otimes$signs for the sake of clarity. It is thus easily seen that$C$has basis$\left(xxy,xxz,yyz,yyx,zzx,zzy,xyz,yzx\right)$(again, the$\otimes$signs are being suppressed). On the other hand, the subspace$\quotB$of$\ScV$has basis$\left(xxx,yyy,zzz\right)$(because, when projecting$\quotA$onto$\ScV$, the basis elements$yxx+xyx+xxy$(along with the$5$similar termssums) and$xyz+xzy+yzx+yxz+zxy+zyx$are mapped to$0$). Hence,$B$has basis$\left(x^2y,x^2z,y^2z,y^2x,z^2x,z^2y,xyz\right)$. We now know an$8$-element basis of$C$and a$7$-element basis of$B$. Thus, our map$f:B\to C$can be represented by a$8\times 7$-matrix. Now, our map$f$, being$\mathrm{GL}\left(V\right)$-equivariant, must commute with the actions of all diagonal matrices in$\mathrm{GL}\left(V\right)$. In other words, it should not matter whether we first multiply$x$,$y$,$z$with any three nonzero elements$\alpha$,$\beta$,$\gamma$of$K$, and then apply$f$, or if we do that the other way round. As a consequence, we clearly have (1)$f\left(x^2y\right)=axxy$for some$a\in K$; (2)$f\left(x^2z\right)=bxxz$for some$b\in K$; (3)$f\left(y^2z\right)=cyyz$for some$c\in K$; (4)$f\left(y^2x\right)=dyyx$for some$d\in K$; (5)$f\left(z^2x\right)=ezzx$for some$e\in K$; (6)$f\left(z^2y\right)=izzy$for some$i\in K$(sorry, couldn't call it$f$); (7)$f\left(xyz\right)=gxyz+hyzx$for some$g,h\in K$. Let me explain why these equations are indeed clear: For example, we know that (8)$f\left(x^2y\right)=a_1xxy+a_2xxz+a_3yyz+a_4yyx+a_5zzx+a_6zzy+a_7xyz+a_8yzx$for some$a_1,a_2,...,a_8\in K$. But$f$, being$\mathrm{GL}\left(V\right)$-equivariant, must commute with the action of all diagonal matrices in$\mathrm{GL}\left(V\right)$. Thus, for every nonzero$\alpha,\beta,\gamma\in K$, we have (9)$\alpha^2\beta f\left(x^2y\right) = a_1\alpha^2\beta xxy+a_2\alpha^2\gamma xxz+a_3\beta^2\gamma yyz+a_4\beta^2\alpha yyx+a_5\gamma^2\alpha zzx+a_6\gamma^2\beta zzy+a_7\alpha\beta\gamma xyz+a_8\alpha\beta\gamma yzx$(by applying the action of the diagonal matrix$\mathrm{diag}\left(\alpha,\beta,\gamma\right)$to both sides of (8)). Since$K$is infinite, we can forget that$\alpha,\beta,\gamma\in K$were nonzero elements of$K$, but rather consider (9) as a polynomial identity, and conclude that it is an identity coefficient-wise. Thus,$f\left(x^2y\right)=a_1xxy$,$0=a_2xxz$,$0=a_3yyz$,$0=a_4yyx$,$0=a_5zzx$,$0=a_6zzy$,$0=a_7xyz+a_8yzx$, so that$a_2=a_3=a_4=a_5=a_6=a_7=a_8=0$. This proves (1). Similarly, (2), (3), ..., (7) are proven. Since$f$also commutes with permutation matrices in$\mathrm{GL}\left(V\right)$, it does not matter whether we first permute$x$,$y$,$z$, and then apply$f$, or if we do that the other way round. As a consequence,$a=b=c=d=e=i$, by looking at what happens to the basis elements$xxy$,$xxz$,$yyz$,$yyx$,$zzx$,$zzy$. But also, by looking at what happens to the basis element$xyz$, we get$g=h=0$. Finishing move: The map$V\to V$given by$x\mapsto x+z$,$y\mapsto y$,$z\mapsto z$is an element of$\mathrm{GL}\left(V\right)$and maps$xxy$to$xxy+xzy+zxy+zzy$. Thus, we must have$f\left(x^2y+xzy+zxy+z^2y\right)=a\left(xxy+xzy+zxy+zzy\right)$. This becomes$axxy+azzy=a\left(xxy+xzy+zxy+zzy\right)$, quickly resulting in$a=0$. Our map$f$is thus the zero map, qed. 1$\newcommand{\SbV}{\mathrm{Sym}^2 V}\newcommand{\ScV}{\mathrm{Sym}^3 V}\newcommand{\quotA}{\left< v\otimes v\otimes v \mid v\in V\right>}\newcommand{\quotB}{\left< vvv \mid v\in V\right>}$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$\left(xxx,yyy,zzz,yxx+xyx+xxy\text{ and 5 similar terms},xyz+xzy+yzx+yxz+zxy+zyx\right)$. Here, we are suppressing the$\otimes$signs for the sake of clarity. It is thus easily seen that$C$has basis$\left(xxy,xxz,yyz,yyx,zzx,zzy,xyz,yzx\right)$(again, the$\otimes$signs are being suppressed). On the other hand, the subspace$\quotB$of$\ScV$has basis$\left(xxx,yyy,zzz\right)$(because, when projecting$\quotA$onto$\ScV$, the basis elements$yxx+xyx+xxy$(along with the$5$similar terms) and$xyz+xzy+yzx+yxz+zxy+zyx$are mapped to$0$). Hence,$B$has basis$\left(x^2y,x^2z,y^2z,y^2x,z^2x,z^2y,xyz\right)$. We now know an$8$-element basis of$C$and a$7$-element basis of$B$. Thus, our map$f:B\to C$can be represented by a$8\times 7$-matrix. Now, our map$f$, being$\mathrm{GL}\left(V\right)$-equivariant, must commute with the actions of all diagonal matrices in$\mathrm{GL}\left(V\right)$. In other words, it should not matter whether we first multiply$x$,$y$,$z$with any three nonzero elements$\alpha$,$\beta$,$\gamma$of$K$, and then apply$f$, or if we do that the other way round. As a consequence, we clearly have (1)$f\left(x^2y\right)=axxy$for some$a\in K$; (2)$f\left(x^2z\right)=bxxz$for some$b\in K$; (3)$f\left(y^2z\right)=cyyz$for some$c\in K$; (4)$f\left(y^2x\right)=dyyx$for some$d\in K$; (5)$f\left(z^2x\right)=ezzx$for some$e\in K$; (6)$f\left(z^2y\right)=izzy$for some$i\in K$(sorry, couldn't call it$f$); (7)$f\left(xyz\right)=gxyz+hyzx$for some$g,h\in K$. Let me explain why these equations are indeed clear: For example, we know that (8)$f\left(x^2y\right)=a_1xxy+a_2xxz+a_3yyz+a_4yyx+a_5zzx+a_6zzy+a_7xyz+a_8yzx$for some$a_1,a_2,...,a_8\in K$. But$f$, being$\mathrm{GL}\left(V\right)$-equivariant, must commute with the action of all diagonal matrices in$\mathrm{GL}\left(V\right)$. Thus, for every nonzero$\alpha,\beta,\gamma\in K$, we have (9)$\alpha^2\beta f\left(x^2y\right) = a_1\alpha^2\beta xxy+a_2\alpha^2\gamma xxz+a_3\beta^2\gamma yyz+a_4\beta^2\alpha yyx+a_5\gamma^2\alpha zzx+a_6\gamma^2\beta zzy+a_7\alpha\beta\gamma xyz+a_8\alpha\beta\gamma yzx$(by applying the action of the diagonal matrix$\mathrm{diag}\left(\alpha,\beta,\gamma\right)$to both sides of (8)). Since$K$is infinite, we can forget that$\alpha,\beta,\gamma\in K$were nonzero elements of$K$, but rather consider (9) as a polynomial identity, and conclude that it is an identity coefficient-wise. Thus,$f\left(x^2y\right)=a_1xxy$,$0=a_2xxz$,$0=a_3yyz$,$0=a_4yyx$,$0=a_5zzx$,$0=a_6zzy$,$0=a_7xyz+a_8yzx$, so that$a_2=a_3=a_4=a_5=a_6=a_7=a_8=0$. This proves (1). Similarly, (2), (3), ..., (7) are proven. Since$f$also commutes with permutation matrices in$\mathrm{GL}\left(V\right)$, it does not matter whether we first permute$x$,$y$,$z$, and then apply$f$, or if we do that the other way round. As a consequence,$a=b=c=d=e=i$, by looking at what happens to the basis elements$xxy$,$xxz$,$yyz$,$yyx$,$zzx$,$zzy$. But also, by looking at what happens to the basis element$xyz$, we get$g=h=0$. Finishing move: The map$V\to V$given by$x\mapsto x+z$,$y\mapsto y$,$z\mapsto z$is an element of$\mathrm{GL}\left(V\right)$and maps$xxy$to$xxy+xzy+zxy+zzy$. Thus, we must have$f\left(x^2y+xzy+zxy+z^2y\right)=a\left(xxy+xzy+zxy+zzy\right)$. This becomes$axxy+azzy=a\left(xxy+xzy+zxy+zzy\right)$, quickly resulting in$a=0$. Our map$f\$ is thus the zero map, qed.