Suppose that $X$ is a smooth projective variety (eg $P^n$) and $E$ is a vector bundle (eg the tangent bundle). If the characteristic is zero, then taking symmetric powers "commutes" with taking duals: $ Sym_m(E)^* $ and $Sym_m(E^*)$ are canonically isomorphic. This is not true in characteristic $p>0$ (one has a canonical isomorphism with a divided power, instead.) But even in characteristic $p$, if the bundle is trivial, then there are non-canonical isomorphisms, and it is not hard to show that the Chern classes are the same.

Question: Is $Sym_m(E)^*$ isomorphic to $Sym_m(E^*)$ (non-canonically) in general?

  • $\begingroup$ I've added the "Algebraic geometry" tag. $\endgroup$
    – algori
    Aug 4, 2010 at 0:37
  • $\begingroup$ Hi David! The examples to try would be tautological bundles on grassmannians (if the isomorphim exists for those, then it should always exist, right?). Have you tried with tangent bundles to projective spaces? $\endgroup$
    – Angelo
    Aug 4, 2010 at 6:41
  • $\begingroup$ Angelo -- could you please explain why: as opposed to the topological case, not all algebraic vector bundles are induced from the tautological bundles over Grassmannians. $\endgroup$
    – algori
    Aug 4, 2010 at 7:18
  • $\begingroup$ @algori: It is almost true, up to a twist by a line bundle every vector bundle (on some quasi-projective variety) is the pullback of the tautological bundle on some Grassmannian. The problem is invariant under twisting by line bundles. So yes the problem is equivalent to the same problem for the tautological bundles. $\endgroup$ Aug 4, 2010 at 7:48
  • $\begingroup$ Torsten -- yes, it is invariant under twisting. My bad. $\endgroup$
    – algori
    Aug 4, 2010 at 8:37

2 Answers 2


Here's another look at the case $m=2$. Maybe it can be used to shed more light on Torsten's nice example. $S^2E$ is part of an exact sequence $0\to \Lambda^2E\to E\otimes E\to S^2E\to 0$. $\Gamma^2E$ is part of an exact sequence $0\to \Gamma^2E\to E\otimes E\to \Lambda^2E\to 0$. The composition of $E\otimes E\to \Lambda^2E \to E\otimes E$ is $1-T$ where $T$ is the involution $x\otimes y\mapsto y\otimes x$. Its kernel $\Gamma^2E$ may be considered as the symmetric bilinear forms on $E^*$, while its cokernel $S^2E$ is the quadratic forms. If $2=0$ then the equation $(1-T)(1+T)=0$ says that the image of $1-T$ is contained in the kernel of $1-T$; we have $\Lambda^2E$ injecting into $\Gamma^2E$. We have in fact an exact sequence $0\to \Lambda^2E\to\Gamma^2E\to E'\to 0$, and another one $0\to E'\to S^2E\to \Lambda^2E\to 0$, where I am writing $E'$ for "$E$ twisted by Frobenius".

If you're not in characteristic $2$ then there's no reason for $S^2E$ (or $\Gamma^2E$) to have a proper nontrivial subbundle.

Added: In the case when $E$ is the tangent bundle of $P^2$, or alternatively the rank $2$ quotient bundle of a trivial rank $3$ bundle which, as Torsten mentions, is the tangent bundle twisted by a line bundle, I believe it is not hard to work out by hand that the only global maps $E\otimes E\to E\otimes E$ are the linear combinations of the identity and the involution $v\otimes w\mapsto w\otimes v$. Of these, the only ones that kill the image of $\Lambda^2 E$ and so give a map $S^2E=coker(\Lambda^2E\to E\otimes E)\to E\otimes E$ are the multiples of $v\otimes w\mapsto v\otimes w+w\otimes v$, so that the only maps $S^2E\to \Gamma^2E=ker(E\otimes E\to\Lambda E)$ are the multiples of the usual one. This argument works over any ground ring, and shows that the two bundles are isomorphic only if $2$ is invertible.

  • $\begingroup$ This is a nice way of looking at it. The generalisation would be that for the rank $n$ tautological bundle $\mathbb E$ on $\mathbb P^n$ (over a ring $R$) we should have that the natural map $R[\Sigma_n]\to\mathrm{End}(\mathbb E^{\oplus n})$ should be an isomorphism for all $R$. For this to be true it is enough that $\mathrm{End}(\mathbb E^{\oplus n})$ have dimension $n!$ for any field. This would be a kind of Weyl theorem (the one where $\mathrm{End}(\mathbb E^{\oplus n})$ is replaced by $R^n$ instead of $\mathbb E$ and morphisms are $\mathrm{GL}_n$-maps). $\endgroup$ Aug 6, 2010 at 6:40

I shall show that the answer is no when $p=2$ (and it seems to me that a somewhat more involved calculation will work for any $p$). We shall show that there exists a vector bundle $\mathcal E$ such that $S^2\mathcal E$ is not isomorphic to $\Gamma^2\mathcal E$ ($=(S^2\mathcal E^\ast)^\ast$).

Consider a vector bundle $\mathcal E$ which is an extension $0\rightarrow\mathcal O_X\rightarrow\mathcal E\rightarrow\mathcal O_X\rightarrow0$. We shall consider the structure group for such extensions. Hence we assume $\mathcal E$ has a basis $e_1=1\in\mathcal O_X$ and $e_2$ (mapping to $1\in\mathcal O_X$). Adapted base changes have the form $e_1\mapsto e_1$ and $e_2\mapsto e_2+he_1$. On the basis $e_1^2,e_1e_2,e_2^2$ of $S^2\mathcal E$ we get $e_1^2\mapsto e_1^2$, $e_1e_2\mapsto e_1e_2+he_1^2$ and $e_2^2\mapsto e_2^2+h^2e_1^2$. This gives us (globally) an exact sequence $$ 0\rightarrow S^2\mathcal O_Xe_1^2\rightarrow S^2\mathcal E\rightarrow\mathcal O_Xe_1e_2\bigoplus\mathcal O_Xe_2^2\rightarrow0 $$ and if $g$ is the extension class for $\mathcal E$ we get that the extension class for this extension is $ge_1e_2+F(g)e_2^2$ (with $F$ the Frobenius). Hence if $g,F(g)\in H^1(X,\mathcal O_X)$ are linearly independent over the base field $k$, then $S^2\mathcal O_X\rightarrow S^2\mathcal E$ induces an isomorphism on global sections so that $H^0(X,S^2\mathcal E)=k$.

On the other hand, we have a basis $e_1^{(2)},e_1e_2,e_2^{(2)}$ for $\Gamma^2\mathcal E$ ($(-)^{(2)}$ denoting the divided power). This gives $e_1^{(2)}\mapsto e_1^{(2)}$, $e_1e_2\mapsto e_1e_2$ and $e_2^{(2)}\mapsto e_2^{(2)}+he_1e_2+h^2e_1^{(2)}$ so that we get an exact sequence $$ 0\rightarrow \mathcal O_Xe_1^{(2)}\bigoplus\mathcal O_Xe_1e_2\rightarrow \Gamma^2\mathcal E\rightarrow\mathcal O_Xe_2^{(2)}\rightarrow0. $$ (This can also be seen by dualising the argument for $S^2$.) Hence $H^0(X,\Gamma^2\mathcal E)$ is at least $2$-dimensional.

It is easy enough to get examples where $g\in H^1(X,\mathcal O_X)$ is linearly independent from $F(g)$. One may for instance take an ordinary genus $2$ curve for which there is a basis $a,b\in H^1(X,\mathcal O_X)$ with $F(a)=a$ and $F(b)=b$ and let $g=a+\lambda b$ with $\lambda\notin\mathbb F_p$ or supersingular but not superspecial genus $2$ curve which has $a\in H^1(X,\mathcal O_X)$ with $a,Fa$ a basis and $F^2a=0$.

Comment: The example looks very special but in some sense it is exactly the beviour under extensions that is the problem. Indeed, given a vector bundle $\mathcal E$ we may consider the complete flag space of $\mathcal E$ over $X$. If $S^m\mathcal E$ and $\Gamma^m \mathcal E$ were isomorphic over the flag space, then we can push down such an isomorphism and get an isomorphism over $X$. Hence, we may assume that $\mathcal E$ has a complete flag. As was noticed in the question, if $\mathcal E$ is actually a direct sum of line bundles we have an isomorphism.

Addendum: The example above is on a curve and the problem is invariant under twisting (by a line bundle). Every rank $2$ vector bundle on a curve can up to twisting be generated by three sections. Hence the vector bundle (again up to twisting) is the pullback by a map from the curve to $\mathbb P^2$ of the tautological rank $2$-vector quotient bundle (dual of the tautological sub-bundle). Hence for that bundle on $\mathbb P^2$ we have that the symmetric and divided squares are non-isomorphic. This bundle is, up to a twist, the tangent bundle of $\mathbb P^2$ and hence the same is true for the tangent bundle. As graded modules over the polynomial ring $k[x,y,z]$ we have explicit presentations of these two modules. In principle we should be able to directly show that they are non-isomorphic and in practice I guess Macaulay should be able to do it.

Addendum 1: In fact it can. (Previous code had incorrect presentation for divided sqaure..) Here is the code:

R = ZZ/2[x,y,z];
F = R^6;
Fzz = F_0;
Fxx = F_1;
Fxy = F_2;
Fxz = F_3;
Fyy = F_4;
Fyz = F_5;
Isym = R*(x*Fxx+y*Fxy+z*Fxz)+R*(x*Fxy+y*Fyy+z*Fyz)+R*(x*Fxz+y*Fyz+z*Fzz);
Idiv = R*(y*Fxy+z*Fxz)+R*(x*Fxy+z*Fyz)+R*(x*Fxz+y*Fyz)+R*(x^2*Fxx+y^2*Fyy+z^2*Fzz+x*y*Fxy+x*z*Fxz+y*z*Fyz);
Sym = sheaf (F/Isym);
Div = sheaf (F/Idiv);
print ({Sym,Div} / (s -> hilbertPolynomial(HH^0(s(*)))));
print ({Sym,Div} / (s -> poincare(HH^0(s(*)))));
print rank Hom(Sym,Div);

{3*P  + 3*P , 3*P  + 3*P }
      1         2      1         2

              -1          2
{6 - 3T, T   + 3 - T }


Here I first verify that the Hilbert polynomials are the same (as they should having the same Chern classes) and then show that their Hilbert functions are different. Finally, I show that there is only one non-zero homomorphism $\mathrm{Sym}\to\mathrm{Div}$. As we already know one such, namely the one that is given by multiplication in the divided power algebra, and it is not an isomorphism we conclude that they are not isomorphic. (Note that I put the partial derivatives $\partial/\partial x,\dots$ in degree $0$ instead of degree $-1$ where they belong.)

  • $\begingroup$ This can be worked out by hand, too. I added to my answer (but did not give details). $\endgroup$ Aug 6, 2010 at 1:14

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