# Global sections of determinant bundle of symmetric powers

Let $E$ be a globally generated vector bundle of rank $r$ on a normal irreducible projective variety $X$. Suppose that $E$ induces a finite map $$X \to \mathbb{G}r (H^0(E), r)$$ to the Grassmannian of rank r quotients. Can we say that a symmetric power of $S^mE$ for $m>>0$ induces an immersion $$X \hookrightarrow \mathbb{G}r (H^0(S^mE), R)$$ where $R = \binom{m+r-1}{m} = \mathrm{rk} S^m E$?

In case $r=1$ it is straightforward, as such a line bundle would be ample, here we ae not supposing $E$ to be ample, the hypothesis imply only that $E$ is nef and $\det E$ is ample.

A positive answer would be equivalent to say, under the same hypothesis for $E$, that for $m >>0$ the image $V$ of the determinant map $$\Lambda^R H^0(S^m E) \twoheadrightarrow V \subseteq H^0 (\det S^m E)$$ separate points and tangent vectors of $X$.

• I think this should be true, at least in characteristic $0$ (in positive characteristic there are issues arising from the difference between the subsheaf of symmetric tensors and the symmetrization quotient). I guess I am also assuming that there is a trace map for the morphism $X\to X'$, where $X'$ is the image of $X$ in the Grassmannian. Denote by $\mathcal{I}$ the kernel of the trace map from the pushforward of $\mathcal{O}_X$ to $\mathcal{O}_{X'}$. The goal is to have enough sections of $S^mE\otimes \mathcal{I}$. This should be possible . . . Nov 13, 2013 at 14:05
• My comment was wrong: see the (negative) answer below. Nov 14, 2013 at 18:13

The suggestion in my comment turns out to be impossible, and this leads to counterexamples to the OP's question, already for $\text{dim}(X)=2$. Let $U$ be a $2$-dimensional vector space over a field $k$. Let $\mathbb{P}(U^\vee)$ denote $\text{Proj} \text{Sym}^\bullet_k U$, i.e., $\mathbb{P}^1_k$. To be very precise, there is a short exact sequence on $\mathbb{P}(U^\vee)$, $$0 \to (\bigwedge^2_k(U^\vee))^\vee\otimes_k \mathcal{O}(-1) \xrightarrow{\alpha^\dagger} U\otimes_k \mathcal{O}_{\mathbb{P}(U^\vee)} \xrightarrow{\beta} \mathcal{O}(+1) \to 0,$$ and the surjection $\beta$ is universal in the usual sense.

Let $V$ be $\text{Syt}_k^2(U)$, the $k$-subspace of $U\otimes_k U$ of symmetric tensors (the same as the symmetric quotient $\text{Sym}_k^2(U)$ except in characteristic $2$). Then $V$ is a $3$-dimensional $k$-vector space. Let $\mathbb{P}(V^\vee)$ be $\text{Proj}\text{Sym}_k^\bullet V$, i.e., $\mathbb{P}^2_k$. To be very precise, on $\mathbb{P}(V^\vee)$ there is a short exact sequence, $$0 \to F^\vee \xrightarrow{\gamma^\dagger} V\otimes_k \mathcal{O}_{\mathbb{P}(V^\vee)} \xrightarrow{\delta} \mathcal{O}(1),$$ and the surjection $\delta$ is universal in the usual sense.

Consider $X=\mathbb{P}(U^\vee) \times_k \mathbb{P}(U^\vee)$, i.e., $\mathbb{P}^1_k\times_k \mathbb{P}^1_k$, with its two projection $\pi_1,\pi_2: X \to \mathbb{P}(U^\vee)$. There is a surjection, $$\pi_1^*\beta\otimes \pi_2^*\beta: (U\otimes_k U)\otimes_k \mathcal{O}_X \twoheadrightarrow \pi_1^*\mathcal{O}(+1)\otimes_{\mathcal{O}_X}\pi_2^*\mathcal{O}(+1).$$ Computing locally, the restriction of $\pi_1^*\beta\otimes\pi_2^*\beta$ to the subsheaf $$\text{Syt}^2_k(U)\otimes_k \mathcal{O}_X\subset (U\otimes_k U)\otimes_k \mathcal{O}_X$$ is still surjective, i.e., there is an induced surjection, $$\text{Syt}^2\beta:\text{Syt}^2_k(U)\otimes_k \mathcal{O}_X \twoheadrightarrow \pi_1^*\mathcal{O}(+1)\otimes_{\mathcal{O}_X}\pi_2^*\mathcal{O}(+1).$$ Thus, there exists a unique morphism $f:X\to \mathbb{P}(V^\vee)$ such that $f^*\delta$ equals $\text{Syt}^2\beta$. Of course this is nothing other than the usual degree $2$ morphism $\mathbb{P}_k^1\times_k \mathbb{P}_k^1 \to \mathbb{P}_k^2$, but with due attention paid in characteristic $2$.

Now consider the dual of the short exact sequence on $\mathbb{P}(V^\vee)$, i.e., $$0 \to \mathcal{O}(-1) \xrightarrow{\delta^\dagger} V^\vee\otimes_k \mathcal{O}_{\mathbb{P}(V^\vee)} \xrightarrow{\gamma} F \to 0.$$ The surjection $\gamma$ is also universal, thus identifying $\mathbb{P}(V^\vee)$ with the Grassmannian, $\mathbb{G}r(2,V)$. Define $E$ to be $f^*F$ on $X$. Consider the restriction of $F$ to points and lines in $\mathbb{P}(V^\vee)$.

For a point of $\mathbb{P}(V^\vee)$ corresponding to a rank $1$ quotient, $V\twoheadrightarrow Q_1$, the fiber $F|_{[Q_1]}$ is simply the dual of the $2$-dimensional kernel of the quotient. For a pointed line in $\mathbb{P}(V^\vee)$ corresponding to a flag of quotients, $V\twoheadrightarrow Q_2\twoheadrightarrow Q_1$, the restriction of $F$ to the line is isomorphic to $\mathcal{O}\oplus \mathcal{O}(1)$, but where the fiber of the positive summand at the marked point $[Q_1]$ corresponds to the annihilator of the kernel of the surjection to $Q_2$. In particular, this depends on the choice of the line containing the specified point $[Q_1]$. As we hold $Q_1$ fixed and vary the line, this image in $F|_{[Q_1]}$ of the positive summand varies over all $1$-dimensional subspaces of the $2$-dimensional fiber.

Claim. For every integer $m\geq 0$, the locally free sheaf $F_m := \text{Sym}^m_{\mathcal{O}}(F)\otimes_{\mathcal{O}}\mathcal{O}(-1)$ on $\mathbb{P}(V^\vee)$ has only the zero section.

Proof of Claim. Since the restriction of $F$ to a pointed line is isomorphic to $\mathcal{O}\oplus \mathcal{O}(1)$, also the restriction of $F_m$ to every pointed line $(L,[Q_1])$ is isomorphic to a direct sum, $$\mathcal{O}(-1)\oplus \mathcal{O}\oplus \mathcal{O}(+1)\dots \oplus \mathcal{O}(m).$$ In particular, the subsheaf spanned by global sections is a locally free subsheaf whose quotient is an invertible sheaf. For a specified point $[Q_1]$ of the line, the fiber at $[Q_1]$ of the "spanned subsheaf" is a codimension $1$ linear subspace $S_L$ of the fiber $F_m|_{[Q_1]}$. By the same analysis as in the previous paragraph, for fixed $[Q_1]$, for every nonzero element $v$ of $F_m|_{[Q_1]}$, there is a choice of line $L$ containing $[Q_1]$ such that $v$ is not contained in $S_L$. Since the fiber $v$ of a global section of $F_m$ at $[Q_1]$ is contained in $S_L$ for every choice of $L$, it follows that every global section of $F_m$ vanishes at $[Q_1]$. Since this holds for every point, and since $\mathbb{P}(V^\vee)$ is reduced, etc., it follows that $F^m$ has only the zero global section.

QED Claim.

Now consider the locally free sheaf on $X$, $f^*\text{Sym}^m_{\mathcal{O}}(F) \cong \text{Sym}^m_{\mathcal{O}}(E)$. The global sections on $X$ of $f^*\text{Sym}^m_{\mathcal{O}}(F)$ are the same as the global sections on $\mathbb{P}(V^\vee)$ of $f_*(f^*\text{Sym}^m_{\mathcal{O}}(F))$.
By the projection formula, this is canonically isomorphic to $\text{Sym}^m_{\mathcal{O}}(F)\otimes_{\mathcal{O}}f_*\mathcal{O}_X$. However, $f_*\mathcal{O}_X$ is canonically isomorphic to $\mathcal{O}_{\mathbb{P}(V^\vee)}\oplus \mathcal{O}(-1)$, where the summand $\mathcal{O}_{\mathbb{P}(V^\vee)}$ is just the image of $f^\#$. Since, as proved in the claim, $\text{Sym}^m_{\mathcal{O}}(F)\otimes_{\mathcal{O}}\mathcal{O}(-1)$ has only the zero global section, it follows that the pullback map on global sections, $$f^*: H^0(\mathbb{P}(V^\vee),\text{Sym}^m_{\mathcal{O}}(E)) \to H^0(\mathbb{P}(U^\vee)\times_k \mathbb{P}(U^\vee), \text{Sym}^m_{\mathcal{O}}(f^*E)),$$ is an isomorphism. Therefore, for every induced morphism as in the OP's question, the morphism factors through $f$. So the induced morphism can never be a closed immersion.

• Thanks a lot for your useful answer. This is what I was looking for. I see the example like this: take a double cover $\pi \colon X \to \mathbb{P}^2$ ramified over a smooth conic, and call $Q$ the universal rank 2 quotient bundle on $\mathbb{P}^2$. Then $\pi_*(\pi^* S^m Q) \cong S^m Q \oplus S^m Q(-1)$ so for the claim you prove the map induced on $X$ by the global sections $H^0 (X, \pi^* S^m Q)$ is the same as the one defined on $H^0 (\mathbb{P}^2, S^m Q)$ i.e. it will always factor thru the degree 2 map and will never be injective. Nov 18, 2013 at 17:03
• @OscarAmalfitano: Yes, that is the idea of the example. I just wanted to be careful about characteristic 2. Nov 18, 2013 at 18:09