Can the projection (tensor algebra) -> (symmetric algebra) be forced to split in char. p by factoring out p-th powers? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T05:07:37Z http://mathoverflow.net/feeds/question/69940 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/69940/can-the-projection-tensor-algebra-symmetric-algebra-be-forced-to-split-in Can the projection (tensor algebra) -> (symmetric algebra) be forced to split in char. p by factoring out p-th powers? darij grinberg 2011-07-10T14:54:00Z 2011-07-12T09:46:20Z <p><strong>Question 1 (the weak and simple statement, which, I think, already is wrong):</strong> Let $p$ be a prime. Let $k$ be a field with characteristic $p$.</p> <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</p> <p>$V^{\otimes p} / \left&lt; v\otimes v\otimes ...\otimes v \text{ (}p\text{ times)} \mid v\in V\right> \to \mathrm{Sym}^p V / \left&lt; vv...v \text{ (}p\text{ times)} \mid v\in V\right>$</p> <p>(where the angular brackets mean "$k$-linear span"). Does this projection canonically split?</p> <p>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 <a href="http://mathoverflow.net/questions/66925/restricted-universal-enveloping-algebra-of-abelian-p-lie-algebra/66934#66934" rel="nofollow">Torsten Ekedahl's disproof of my previous guess</a>, I should learn some modular representation theory of $\mathrm{GL}$; is there a good source for it?</p> <p>If Question 1 actually happens to get a positive answer, then here is the question I am coming from:</p> <p><strong>Question 2 (generalization of Question 1; ignore if Question 1 is answered No):</strong> Much of the following is copied over from <a href="http://mathoverflow.net/questions/66925/restricted-universal-enveloping-algebra-of-abelian-p-lie-algebra" rel="nofollow">http://mathoverflow.net/questions/66925/restricted-universal-enveloping-algebra-of-abelian-p-lie-algebra</a> .</p> <p>Let $p$ be a prime. Let $k$ be a commutative ring such that $p=0$ in $k$.</p> <p>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 $v\in\mathfrak g$.</p> <p>Let $U^{[p]}\left(\mathfrak g\right)$ be the <a href="http://en.wikipedia.org/wiki/Restricted_Lie_algebra#Restricted_universal_enveloping_algebra" rel="nofollow">restricted universal enveloping algebra</a> 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.</p> <p>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 $x\in\mathfrak g$.</p> <p>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</p> <p>$\otimes^{[p]} \mathfrak g\to\mathrm{Sym}\mathfrak g \to U^{[p]}\left(\mathfrak g\right)$.</p> <p>Does this projection split canonically?</p> http://mathoverflow.net/questions/69940/can-the-projection-tensor-algebra-symmetric-algebra-be-forced-to-split-in/70102#70102 Answer by darij grinberg for Can the projection (tensor algebra) -> (symmetric algebra) be forced to split in char. p by factoring out p-th powers? darij grinberg 2011-07-12T09:40:29Z 2011-07-12T09:46:20Z <p>$\newcommand{\SbV}{\mathrm{Sym}^2 V}$ $\newcommand{\ScV}{\mathrm{Sym}^3 V}$ $\newcommand{\quotA}{\left&lt; v\otimes v\otimes v \mid v\in V\right>}$ $\newcommand{\quotB}{\left&lt; vvv \mid v\in V\right>}$</p> <p>I think I have solved this, with the help of mt and Tom Goodwillie.</p> <p>Question 1 is wrong (and thus Question 2 is wrong as well).</p> <p><em>Proof.</em> 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</p> <p>$A=V^{\otimes 3} / \quotA$</p> <p>and</p> <p>$B=\ScV / \quotB$.</p> <p>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.</p> <p>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</p> <p>$A=\left(\left(V\otimes \SbV\right)/\quotA\right) \oplus V\otimes \wedge^2 V$.</p> <p>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</p> <p>$C:=\left(V\otimes \SbV\right)/\quotA$.</p> <p>(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.)</p> <p>First we notice that the subspace $\quotA$ of $V^{\otimes 3}$ is $10$-dimensional and has basis</p> <p>$\left(xxx,yyy,zzz,yxx+xyx+xxy\text{ and 5 similar sums},xyz+xzy+yzx+yxz+zxy+zyx\right)$.</p> <p>Here, we are suppressing the $\otimes$ signs for the sake of clarity. It is thus easily seen that $C$ has basis</p> <p>$\left(xxy,xxz,yyz,yyx,zzx,zzy,xyz,yzx\right)$</p> <p>(again, the $\otimes$ signs are being suppressed).</p> <p>On the other hand, the subspace $\quotB$ of $\ScV$ has basis</p> <p>$\left(xxx,yyy,zzz\right)$</p> <p>(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</p> <p>$\left(x^2y,x^2z,y^2z,y^2x,z^2x,z^2y,xyz\right)$.</p> <p>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.</p> <p>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</p> <p>(1) $f\left(x^2y\right)=axxy$ for some $a\in K$;</p> <p>(2) $f\left(x^2z\right)=bxxz$ for some $b\in K$;</p> <p>(3) $f\left(y^2z\right)=cyyz$ for some $c\in K$;</p> <p>(4) $f\left(y^2x\right)=dyyx$ for some $d\in K$;</p> <p>(5) $f\left(z^2x\right)=ezzx$ for some $e\in K$;</p> <p>(6) $f\left(z^2y\right)=izzy$ for some $i\in K$ (sorry, couldn't call it $f$);</p> <p>(7) $f\left(xyz\right)=gxyz+hyzx$ for some $g,h\in K$.</p> <p>Let me explain why these equations are indeed clear: For example, we know that</p> <p>(8) $f\left(x^2y\right)=a_1xxy+a_2xxz+a_3yyz+a_4yyx+a_5zzx+a_6zzy+a_7xyz+a_8yzx$ for some $a_1,a_2,...,a_8\in K$.</p> <p>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</p> <p>(9) $\alpha^2\beta f\left(x^2y\right) = a_1\alpha^2\beta xxy+a_2\alpha^2\gamma xxz+a_3\beta^2\gamma yyz+a_4\beta^2\alpha yyx+a_5\gamma^2\alpha zzx+a_6\gamma^2\beta zzy+a_7\alpha\beta\gamma xyz+a_8\alpha\beta\gamma yzx$</p> <p>(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,</p> <p>$f\left(x^2y\right)=a_1xxy$, $0=a_2xxz$, $0=a_3yyz$, $0=a_4yyx$, $0=a_5zzx$, $0=a_6zzy$, $0=a_7xyz+a_8yzx$,</p> <p>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.</p> <p>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 $g=h=0$.</p> <p>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 $a=0$.</p> <p>Our map $f$ is thus the zero map, qed.</p>