# Can the projection (tensor algebra) -> (symmetric algebra) be forced to split in char. p by factoring out p-th powers?

Question 1 (the weak and simple statement, which, I think, already is wrong): Let $p$ be a prime. Let $k$ be a field with characteristic $p$.

For any $k$-vector space $V$, consider the canonical projection $V^{\otimes p}\to \mathrm{Sym}^p V$ from the $p$-th tensor power of $V$ to the $p$-th symmetric power of $V$. This projection does not canonically split. But it induces a projection

$V^{\otimes p} / \left< v\otimes v\otimes ...\otimes v \text{ (}p\text{ times)} \mid v\in V\right> \to \mathrm{Sym}^p V / \left< vv...v \text{ (}p\text{ times)} \mid v\in V\right>$

(where the angular brackets mean "$k$-linear span"). Does this projection canonically split?

It trivially does for $p=2$, because it is the identity map in this case. But I highly suspect that it fails even for $p=3$. However I cannot prove it. Judging from Torsten Ekedahl's disproof of my previous guess, I should learn some modular representation theory of $\mathrm{GL}$; is there a good source for it?

If Question 1 actually happens to get a positive answer, then here is the question I am coming from:

Question 2 (generalization of Question 1; ignore if Question 1 is answered No): Much of the following is copied over from Restricted universal enveloping algebra of Abelian p-Lie algebra .

Let $p$ be a prime. Let $k$ be a commutative ring such that $p=0$ in $k$.

Let $\mathfrak g$ be an abelian $p$-restricted Lie algebra over $k$. In other words, let $\mathfrak g$ be a $k$-module along with a $\mathbb Z$-linear map ${}^{[p]}:\mathfrak g\to\mathfrak g$ (written postfix) that satisfies $\left(\lambda v\right)^{[p]}=\lambda^p v^{[p]}$ for all $\lambda\in k$ and $v\in\mathfrak g$.

Let $U^{[p]}\left(\mathfrak g\right)$ be the restricted universal enveloping algebra of $\mathfrak g$. In other words, let $U^{[p]}\left(\mathfrak g\right)$ be the factor algebra of the symmetric algebra of $\mathfrak g$ modulo the ideal generated by elements of the form $x^p-x^{[p]}$ with $x\in\mathfrak g$. Note that $U^{[p]}\left(\mathfrak g\right)$ is not a graded algebra, but a filtered one.

Let $\otimes^{[p]}\left(\mathfrak g\right)$ be the factor algebra of the algebra $\otimes \mathfrak g$ (this is the tensor algebra of the $k$-module $\mathfrak g$) modulo the ideal generated by elements of the form $\underbrace{x\otimes x\otimes ...\otimes x}_{p\text{ times}}-x^{[p]}$ with $x\in\mathfrak g$.

The canonical projection $\otimes \mathfrak g\to\mathrm{Sym}\mathfrak g \to U^{[p]}\left(\mathfrak g\right)$ (where $\otimes \mathfrak g$ means the tensor algebra of $\mathfrak g$) induces a canonical projection

$\otimes^{[p]} \mathfrak g\to\mathrm{Sym}\mathfrak g \to U^{[p]}\left(\mathfrak g\right)$.

Does this projection split canonically?

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$V\otimes V$ is canonically $S^2(V)\oplus\Lambda^2(V)$ for $p>2$, so for $p=3$ you get $V^{\otimes 3} \cong V\otimes S^2(V) \oplus V\otimes \Lambda^2(V)$, with the first summand in the kernel of the map to $S^3(V)$. That might help you work out if the map splits as a map of GL-modules for $p=3$ (I think it will). Sandy Green's LNM (Polynomial representations of $\mathsf{GL}_n$) could be a good start for modular rep thy of GL. – M T Jul 10 '11 at 16:24
I think the second summand, not the first, is in the kernel. Thanks a lot for the reference. – darij grinberg Jul 10 '11 at 16:40
oops, you're right about the summand. J.A. Green is James Alexander, I imagine that's where the Sandy is from. – M T Jul 10 '11 at 16:44
I believe that the only natural map from $Sym^3(V)/\langle vvv\rangle$ to $V^{\otimes 3}/\langle v\otimes v\otimes v\rangle$ is zero. – Tom Goodwillie Jul 11 '11 at 4:04

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

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

$\left(xxx,yyy,zzz,yxx+xyx+xxy\text{ and 5 similar sums},xyz+xzy+yzx+yxz+zxy+zyx\right)$.

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 sums) 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.

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