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5 Fixed incorrect presentation for divided square.

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

Idiv = R*(y*Fxy+z*Fxz)+R*(x*Fxy+z*Fyz)+R*(x*Fxz+y*Fyz)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);print ({Sym,Div} / (s -> hilbertPolynomial(HH^0(s(*)))));print ({Sym,Div} / (s -> poincare(HH^0(s(*)))));print rank HH^1(Sym(-3)),rank HH^1(Div(-3)))Hom(Sym,Div);

and the result is $(3,0)$ so {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 symmetric and divided squares Hilbert polynomials are non-isomorphic. (Note that I put the partial derivatives $\partial/\partial x,\dots$ in degree $0$ instead of degree $-1$ where same (as they belong.should having the same Chern classes)

A perhaps better way of seeing and then show that they their Hilbert functions are not isomorphism is to look at the homomorphisms from Sym to Divdifferent. Macualay gives

rank Hom(Sym,Div);

but Finally, I show that there is only one non-zero homomorphism $\mathrm{Sym}\to\mathrm{Div}$. As we already know one homomorphism $\mathrm{Sym}\to\mathrm{Div}$ 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.)

4 Elaboration of non-isomorphism

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. 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);
Sym = sheaf (F/Isym);
Div = sheaf (F/Idiv);
print (rank HH^1(Sym(-3)),rank HH^1(Div(-3)));


and the result is $(3,0)$ so the symmetric and divided squares are non-isomorphic. (Note that I put the partial derivatives $\partial/\partial x,\dots$ in degree $0$ instead of degree $-1$ where they belong.)

A perhaps better way of seeing that they are not isomorphism is to look at the homomorphisms from Sym to Div. Macualay gives

rank Hom(Sym,Div);
1


but we already know one homomorphism $\mathrm{Sym}\to\mathrm{Div}$ namely the one that is given by multiplication in the divided power algebra and it is not an isomorphism.

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. 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);
Sym = sheaf (F/Isym);
Div = sheaf (F/Idiv);
print (rank HH^1(Sym(-3)),rank HH^1(Div(-3)));


and the result is $(3,0)$ so the symmetric and divided squares are non-isomorphic. (Note that I put the partial derivatives $\partial/\partial x,\dots$ in degree $0$ instead of degree $-1$ where they belong.)

2 Added existence of very natural example.
1