On degenerate sections of vector bundles

Abstract

We consider the locus of sections of a vector bundle on a projective scheme that vanish in higher dimension than expected. We will find the largest components of this locus asymptotically, after applying a high enough twist to the vector bundle. We will also give an interpretation in terms of a limit in the Grothendieck ring of varieties.

Appendices

Appendix A: Facts about Hilbert functions

Throughout the appendix, we assume that we are working in a projective scheme X and V is a vector bundle on X.

1.1 Lower bound on Hilbert function given degrees

To give crude bounds on Hilbert function, we will generalize a well-known lemma (see [8, Lemma 3.1]). Recall $h_{Z,E}(n):= \dim (\mathrm{im}(H^0(V(n))\rightarrow H^0(V(n)|_{Z})))$ (Definition 2.6).

Lemma A.1

If $H^1(V(n-1))=0$, $Z\subset X$ is a closed subscheme and H is a hyperplane section of X that is a nonzero divisor when restricted to Z, then

$$\begin{aligned} h_{Z,V}(n)\ge h_{Z,V}(n-1)+h_{Z\cap H,V}(n). \end{aligned}$$

Proof

Consider the diagram

The rows are exact on the left since H is a nonzero divisor on X and Z. Taking the image under the middle vertical map yields a short exact sequence with $\mathrm{im}(H^0(V(n))\rightarrow H^0(V(n)|_{Z}))$ in the middle.

$$\begin{aligned}&0 \rightarrow \mathrm{im}(\mathrm{im}(H^0(V(n))\rightarrow H^0(V(n)|_{Z}))\cap \mathrm{im}(H^0(V(n-1)|_{Z})\xrightarrow {\times H} H^0(V(n)|_{Z})))\rightarrow \\&\quad \mathrm{im}(H^0(V(n))\rightarrow H^0(V(n)|_{Z})) \rightarrow \mathrm{im}(H^0(V(n))\rightarrow H^0(V(n)|_{H\cap Z}))\rightarrow 0 \end{aligned}$$

Then,

$$\begin{aligned} h_{Z,V}(n)&:= \dim (\mathrm{im}(H^0(V(n))\rightarrow H^0(V(n)|_{Z})))\\&=\dim (\mathrm{im}(H^0(V(n))\rightarrow H^0(V(n)|_{Z}))\cap \mathrm{im}(H^0(V(n-1)|_{Z})\rightarrow H^0(V(n)|_{Z})))\\&\quad +\dim (\mathrm{im}(H^0(V(n))\rightarrow H^0(V(n)|_{H\cap Z})))\\&\ge \dim (\mathrm{im}(H^0(V(n-1))\rightarrow H^0(V(n-1)|_{Z})))\\&\quad +\dim (\mathrm{im}(H^0(V(n))\rightarrow H^0(V(n)|_{H\cap Z})))\\&= h_{Z,V}(n-1)+h_{Z\cap H,V}(n). \end{aligned}$$

$\square$

Definition A.1

A vector bundle E on X is said to be k-very ample for k a nonnegative integer if the restriction map $H^0(E)\rightarrow H^0(E|_{Z})$ is surjective for every 0-dimensional length $k+1$ subscheme $Z\subset X$.

The following lemma should be well-known to experts, but the author couldn’t find it in the literature.

Lemma A.2

If V is globally generated and L is a k-very ample line bundle, then $V \otimes L$ is k-very ample.

Proof

Let $Z\subset X$ be a length $k+1$ subscheme and $Z=\mathrm{Spec}(A)$ for an Artinian ring A. Let $A=\oplus _{i=1}^{\ell }{A_\ell }$, where each $A_i$ is a local Artinian ring. Consider

To show $\phi$ is surjective, it suffices to show that $\psi$ hits every element $e_{ij}\in A^{\mathrm{rank}(V)}\cong \oplus _{i=1}^{\ell }{A_{\ell }}$, where $e_{ij}$ is the element that projects to 0 under each projection $A^{\mathrm{rank}(V)}\rightarrow A_{p}^{\mathrm{rank}(V)}$ for $p\ne i$ and maps to the element

$$\begin{aligned} (0,0,\ldots , \underbrace{1}_{\text {index}\ j}, \ldots , 0)\in A_i^{\mathrm{rank}(V)} \end{aligned}$$

for $p=i$. For convenience of notation, assume $j=1$. Let $m_i\subset A_i$ be the maximal ideal. Since V is globally generated, we can find $s\in H^0(V)$ such that s maps to $(1,0,\ldots ,0)$ under the composition

$$\begin{aligned} H^0(V)\rightarrow H^0(V|_Z)\xrightarrow {\sim } A^{\mathrm{rank}(V)}\rightarrow A_i^{\mathrm{rank}(V)}\rightarrow (A_i/m_i)^{\mathrm{rank}(V)}. \end{aligned}$$

Let $a\in A$ be the image of s in $A_i$ under the composition

$$\begin{aligned} H^0(V)\rightarrow H^0(V|_Z)\xrightarrow {\sim } A^{\mathrm{rank}(V)}\rightarrow A_i^{\mathrm{rank}(V)}\rightarrow A_i, \end{aligned}$$

where the last map is projection onto the first factor. In particular, a is invertible in $A_i$. Since L is k-very ample, the composition

$$\begin{aligned} H^0(L)\rightarrow H^0(L|_Z)\xrightarrow {\sim } A \end{aligned}$$

is surjective, so we can find $t\in H^0(L)$ such that t projects to 0 in each factor $A_p$ for $p\ne i$ and t projects to $a^{-1}$ in the factor $A_i$. Then, the element $s\otimes t\in H^0(V)\otimes H^0(L)$ maps to a section of $H^0(V\otimes L)$ that maps to $e_{ij}$ under $\psi$ above. $\square$

Lemma A.3

If V 0-regular in the sense of Castelnuovo-Mumford regularity, $Z\subset X$ a subscheme of degree at least d and dimension m, then for $n\ge 1$

$$\begin{aligned} h_{Z,V}(n-1)\ge {\left\{ \begin{array}{ll} \mathrm{rank}(V)\left( {\begin{array}{c}n+m\\ m+1\end{array}}\right) &{}\text { if } n\le d \\ \mathrm{rank}(V)\sum _{i=0}^{m}{\left( {\begin{array}{c}n-d+i-1\\ i\end{array}}\right) \left( {\begin{array}{c}d+m-i\\ m-i+1\end{array}}\right) }&\text { if n> d}. \end{array}\right. } \end{aligned}$$

Proof

We will use induction and Lemma A.1. First, we show the base case $m=0$. Let $Z=\mathrm{Spec}(A)$ for an Artinian K-algebra A and fix an trivialization $V|_{Z}\cong {\mathscr {O}}_{\mathrm{Spec}(A)}^{\mathrm{rank}(V)}$. We want to show $h_{Z,V}(n-1)\ge \mathrm{rank}(V)\min \{n,d\}$. Since V is globally generated [12, Theorem 1.8.3] and ${\mathscr {O}}(n-1)$ is n-very ample [10, Theorem 1.1], we can apply Lemma A.2 to see $V(n-1)$ is n-very ample.

Without loss of generality, we can assume $d\le n$, or else we can just restrict Z to a subscheme of length n. (Recall A has quotients of any length less than A. To show this, it suffices to see that if A is local, then there is an ideal of A of length 1. To exhibit such an ideal, we can pick any element $a\in A$ that is anniliated by the maximal ideal of A and take the ideal generated by a.) Then, by n-very ampleness, we see $H^0(V(n-1))\rightarrow H^0(V(n-1)|_{Z})$ is surjective, so $h_{Z,V}(n-1)=n$. The reason why we only prove an inequality is that we restricted Z above.

Next, for the induction step, suppose we know Lemma A.3 in dimension $m-1$. From the long exact sequence, we have the exactness of

$$\begin{aligned} H^i(V(-i))\rightarrow H^i(V(-i)|_{H})\rightarrow H^{i+1}(V(-i-1)) \end{aligned}$$

so $V|_{H}$ is still 0-regular. Also, $H^0(V)\rightarrow H^0(V|_{H})$ is surjective as $H^1(V(-1))=0$. Therefore, we can apply the induction hypothesis and Lemma A.1.

If Z is dimension m and degree d, we have for $n\le d$

$$\begin{aligned} h_{Z,V}(n-1)&\ge h_{Z,V}(-1)+h_{Z\cap H,V}(0)+h_{Z\cap H,V}(1)+\cdots + h_{Z\cap H,V}(n-1)\\&\ge 0+\mathrm{rank}(V)\left( {\begin{array}{c}m\\ m\end{array}}\right) +\cdots +\mathrm{rank}(V)\left( {\begin{array}{c}m+n-1\\ m\end{array}}\right) =\mathrm{rank}(V)\left( {\begin{array}{c}n+m\\ m+1\end{array}}\right) . \end{aligned}$$

If $n>d$,

$$\begin{aligned} h_{Z,V}(n-1)&\ge h_{Z,V}(d-1)+\left( h_{Z\cap H,V}(d)+\cdots +h_{Z\cap H,V}(n-1)\right) \\&= \mathrm{rank}(V)\left( {\begin{array}{c}d+m\\ m+1\end{array}}\right) +\sum _{j=d+1}^{n}{\mathrm{rank}(V)\sum _{i=0}^{m-1}{\left( {\begin{array}{c}j-d+i-1\\ i\end{array}}\right) \left( {\begin{array}{c}d+m-1-i\\ m-1-i+1\end{array}}\right) }}\\&= \mathrm{rank}(V)\left( {\begin{array}{c}d+m\\ m+1\end{array}}\right) +\mathrm{rank}(V)\sum _{i=0}^{m-1}{\left( {\begin{array}{c}d+m-1-i\\ m-1-i+1\end{array}}\right) \sum _{j=d+1}^{n}{\left( {\begin{array}{c}j-d+i-1\\ i\end{array}}\right) }}\\&= \mathrm{rank}(V)\left( {\begin{array}{c}d+m\\ m+1\end{array}}\right) +\mathrm{rank}(V)\sum _{i=0}^{m-1}{\left( {\begin{array}{c}d+m-1-i\\ m-1-i+1\end{array}}\right) \left( {\begin{array}{c}n-d+i\\ i+1\end{array}}\right) }\\&= \mathrm{rank}(V)\left( {\begin{array}{c}d+m\\ m+1\end{array}}\right) +\mathrm{rank}(V)\sum _{i=1}^{m}{\left( {\begin{array}{c}d+m-i\\ m-i+1\end{array}}\right) \left( {\begin{array}{c}n-d+i-1\\ i\end{array}}\right) }. \end{aligned}$$

$\square$

Corollary A.4

If $Z\subset X$ is a subscheme of degree at least d and dimension m, then for $n\ge 0$,

$$\begin{aligned} h_{Z,V}(n)\ge d\cdot \mathrm{rank}(V)\left( {\begin{array}{c}n-d+m\\ m\end{array}}\right) \end{aligned}$$

for a 0-regular vector bundle V on X.

Proof

We bound $h_{Z,V}(n)$ from below by the $i=m$ term in summation in the statement of Lemma A.3. $\square$

Appendix B: Scheme theoretic facts

Lemma B.1

Suppose X is an integral projective scheme and $Z\subset X$ a closed subset. Let S be a finite type K-scheme, $Y\subset S\times (X\backslash Z)$ be a closed subset and $\pi :Y\rightarrow S$ be the projection. Then,

$$\begin{aligned} \{s\in S: \dim (\pi ^{-1}(s))\ge d\} \end{aligned}$$

is a closed subset of S for all $d>\dim (Z)$.

Proof

Let ${\overline{Y}}$ be the closure of Y in $S\times X$ and $\pi :{\overline{Y}}\rightarrow S$ be the projection. Let $Y_d\subset Y$ be the closed subset $\{y\in Y: \dim (\pi ^{-1}(\pi (y))\cap Y)\ge d\}$. Let ${\overline{Y}}_d\subset {\overline{Y}}$ be the closed subset $\{y\in {\overline{Y}}: \dim (\pi ^{-1}(\pi (y)))\ge d\}$.

Since $Y_d\subset {\overline{Y}}_d$, $\pi (Y_d)\subset \pi ({\overline{Y}}_d)$. We want to show $\pi (Y_d)= \pi ({\overline{Y}}_d)$.

To see the other inclusion, we first claim $Y_d={\overline{Y}}_d\backslash (S\times Z)$. Given this, $\pi (Y_d)\supset \pi ({\overline{Y}}_d)$ also holds, as if $s\in \pi ({\overline{Y}}_d)$ is a closed point, then $\pi ^{-1}(s)$ has dimension at least d in X, so $\pi ^{-1}(s)\backslash Z$ still has dimension at least d as $\dim (Z)<d$.

To see $Y_d={\overline{Y}}_d\backslash (S\times Z)$, suppose $\xi \in {\overline{Y}}$ is a scheme theoretic point with $\dim (\overline{\{\xi \}})\ge d$ that maps to a closed point of S and $y\in \overline{\{\xi \}}\backslash Z$. Equivalently, $y\in {\overline{Y}}_d\backslash (S\times Z)$. But then $\overline{\{\xi \}}$ cannot be contained in Z since $\dim (Z)<\dim (\overline{\{\xi \}})$, so $\xi \in Y$ and $y\in Y_d$. $\square$

Lemma B.2

Suppose $Z\subset \mathbb {P}^r$ is a integral projective scheme, and $\mathcal {Z}\subset \mathbb {P}^r_{\mathrm{Spec}(K[\epsilon ]/(\epsilon ^2))}$ is a nontrivial infinitesimal deformation of the embedding $Z\subset \mathbb {P}^r$. Let $Z'\subset \mathbb {P}^r$ be the scheme theoretic image of $\mathcal {Z}\rightarrow \mathbb {P}^r$. Then, $\deg (Z')=2\deg (Z)$.

Proof

Assume Z is not contained in hyperplane $\{X_0=0\}$. In the chart $\{X_0\}$, the data of the embedding $Z\subset \mathbb {P}^r$ is the surjection $K[x_1,\ldots ,x_r]\rightarrow A$ for $\mathrm{Spec}(A)=Z\backslash \{X_0=0\}$ and similarly for $\mathcal {Z}\subset \mathbb {P}^r_{\mathrm{Spec}(K[\epsilon ]/(\epsilon ^2))}\rightarrow \mathcal {A}$ for $\mathrm{Spec}(\mathcal {A})=\mathcal {Z}\backslash \{X_0=0\}$.

The scheme theoretic image of $\mathcal {Z}\rightarrow \mathbb {P}^r$ restricted to $\{X_0=0\}$ can be computed affine locally [15, Tag 01R8] and is $\mathrm{Spec}(B)$ for $B=\mathrm{im}(K[x_1,\ldots ,x_r]\rightarrow \mathcal {A})$. Equivalently, it is defined by the ideal $I\subset K[x_1,\ldots ,x_r]$, where $I=\ker (K[x_1,\ldots ,x_r]\rightarrow \mathcal {A})$. Let $\eta \in \mathrm{Spec}(A)$ be the generic point. It suffices to show the multiplicity of B at $\eta$ is 2, where multiplicity is defined to be the length of $B_{\eta }$ [7, Section 1.5]. In the argument below, we will use A is integral to see $A\rightarrow A_{\eta }$ is injective.

From exactness of localization, it suffices to show that the induced map $B_{\eta }\rightarrow A_{\eta }$ is not an isomorphism. We will assume $B_{\eta }\rightarrow A_{\eta }$ is an isomorphism, and we will show that the deformation $\mathrm{Spec}(\mathcal {A})\rightarrow \mathrm{Spec}(K[\epsilon ]/(\epsilon ^2))$ must be trivial.

The middle row is exact by flatness [9, Proposition 2.2]. From the diagram, we see $\mathcal {A}\rightarrow \mathcal {A}_{\eta }$ is an injection.

Since $B\rightarrow \mathcal {A}_{\eta }$ is injective, we know $B\rightarrow B_\eta$ is injective. Then, $B\rightarrow A_\eta$ is injective, so $B\rightarrow A$ is injective. But $B\rightarrow A$ is surjective, so $B\rightarrow A$ is an isomorphism. This gives a splitting $A\rightarrow \mathcal {A}$ compatible with the splitting $K[x_1,\ldots ,x_r]\rightarrow K[\epsilon , x_1,\ldots ,x_r]/(\epsilon ^2)$. This shows that $\mathrm{Spec}(\mathcal {A})\rightarrow \mathrm{Spec}(K[\epsilon ]/(\epsilon ^2))$ is the trivial deformation. $\square$

Appendix C: Isomorphism in the Grothendieck ring

We will need a criterion for isomorphism in the Grothendieck ring.

Proposition C.1

Suppose $f: Y\rightarrow Z$ is a morphism of finite type K-schemes and $A\subset Y$ and $B\subset Z$ are constructible sets such that $f(A)=B$ and $f^{-1}(p)$ is a single reduced point for each closed $p\in B$. Then $[A]=[B]$ in the Grothendieck ring $\mathcal {M}$.

Remark 2

In characteristic 0, the conclusion of Proposition C.1 holds if $f:A\rightarrow B$ is only a bijection of closed points [1, Lemma 2.1]. As a note of caution, we have contacted the author and Lemma 2.1 in [1] is correct in characteristic zero, but in characteristic p there is a gap in the proof, since separable maps do not necessarily remain separable when restricted to closed subsets. In our case this is guaranteed by the assumption that all of our fibers are reduced points.

Proof

By assumption f induces a bijection $A\rightarrow B$. If we write $B=U_1\cup \cdots U_{\ell }$ as a finite union of locally closed sets, it suffices to show $[f^{-1}(U_i)]=[U_i]$. Therefore, it suffices to show the case where B is locally closed. Similarly, we can break B up further, so that B is irreducible and locally closed.

Since B is locally closed, A is also locally closed. We can equip A and B with the reduced subscheme structure of an open subscheme. Then, $f: A\rightarrow B$ is a map of irreducible finite type K-schemes such that each fiber over a closed point $p\in B$ is a single reduced point. Pick an affine open $\mathrm{Spec}(R_B)\subset B$ and an affine open $\mathrm{Spec}(R_A)\subset f^{-1}(\mathrm{Spec}(R_B))$.

By Grothendieck’s generic freeness lemma [6, Theorem 14.4], we can restrict $R_B$ so that $R_A$ is free over $R_B$. By the assumption on the fibers of $f: A\rightarrow B$, $R_A$ must be rank at most 1 over $R_B$, so the map $R_B\rightarrow R_A$ is an isomorphism. Then, $\mathrm{Spec}(R_A)\rightarrow \mathrm{Spec}(R_B)$ is an isomorphism and we use Noetherian induction on the complement $f: A\backslash \mathrm{Spec}(R_A)\rightarrow B\backslash \mathrm{Spec}(R_B)$. $\square$