\documentclass[11pt]{amsart} \usepackage[T1]{fontenc} \usepackage{lmodern} \usepackage{amsmath,amssymb,amsthm,mathtools} \usepackage[letterpaper,margin=1.05in]{geometry} \usepackage{microtype} \usepackage[dvipsnames]{xcolor} \usepackage[colorlinks=true,linkcolor=MidnightBlue,citecolor=MidnightBlue, urlcolor=MidnightBlue,bookmarksnumbered=true]{hyperref} \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{proposition}[theorem]{Proposition} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \DeclareMathOperator{\Sym}{Sym} \DeclareMathOperator{\Pic}{Pic} \DeclareMathOperator{\coker}{coker} \DeclareMathOperator{\rk}{rk} \DeclareMathOperator{\Hilb}{Hilb} \DeclareMathOperator{\Nm}{Nm} \newcommand{\OO}{\mathcal O} \newcommand{\PP}{\mathbb P} \emergencystretch=2em \title{Natural cohomology of general Steiner bundles} \author{Vlado Vojdanovski} \date{} \keywords{Steiner bundles, natural cohomology, elliptic curves, maximal rank} \hypersetup{pdftitle={Natural cohomology of general Steiner bundles}, pdfauthor={Vlado Vojdanovski}} \begin{document} \begin{abstract} We prove that a general Steiner bundle on projective space over a field of characteristic zero has natural cohomology. The proof constructs a matrix of linear forms from multiplication of sections on a curve. When the rank exceeds the dimension of the projective space, a finite \'etale isogeny of elliptic curves splits the bundles controlling the graded kernels and cokernels into line bundles of equal degree. Choosing the auxiliary line bundle outside a finite set gives maximal rank in every degree. A rational curve provides the boundary case. Two open rank conditions then yield the assertion for general matrices and all twists. \end{abstract} \maketitle \section{Introduction} Let $k$ be a field of characteristic zero. A vector bundle $F$ on $\PP^n_k$ is a \emph{Steiner bundle} if it admits a presentation \begin{equation}\label{eq:steiner} 0\longrightarrow \OO_{\PP^n}(-1)^{\oplus t} \longrightarrow \OO_{\PP^n}^{\oplus(t+r)} \longrightarrow F\longrightarrow0. \end{equation} It has \emph{natural cohomology} if, for each integer $a$, at most one of the groups $H^i(\PP^n,F(a))$ is nonzero. Here $t\geq1$ and $r\geq n\geq1$; the parameter $r$ is the rank of $F$. Coskun, Huizenga, and Smith conjecture that a general Steiner bundle has natural cohomology~\cite[Conjecture~1.4]{chs}. They recall the known case of $\PP^2$ and establish the conjecture for rank divisible by $n$ and in an asymptotic range. We prove the assertion for all the parameters above. \begin{theorem}\label{thm:main} Let $k$ have characteristic zero, and let $n\geq1$, $t\geq1$, and $r\geq n$. In the affine space of $t\times(t+r)$ matrices of linear forms in $S=k[x_0,\ldots,x_n]$, there is a nonempty Zariski open subset on which every induced map \begin{equation}\label{eq:mu} \mu_s:S_{s-1}^{\oplus(t+r)}\longrightarrow S_s^{\oplus t}, \qquad s\geq0, \end{equation} has maximal rank, where $S_{-1}=0$. For every matrix in this open subset, the transpose defines a presentation~\eqref{eq:steiner}, and $F$ has natural cohomology. \end{theorem} The proof combines several established constructions. Arrondo obtains Steiner bundles from multiplication of sections in~\cite[Example~1.5]{arrondo} and gives elliptic examples in~\cite[Remark~2.9]{arrondo}. The version using a generating subspace of sections is explicit in Lazarsfeld and Sheridan~\cite[Lemma~1.1]{lazarsfeld-sheridan}. Elliptic evaluation kernels obtained by choosing one more generating section than the rank of a globally generated bundle and dualizing appear in Hein's appendix to Brenner~\cite[Theorem~A.1]{brenner-hein} and in Aprodu, Farkas, and Ortega~\cite[Proposition~1.3]{aprodu-farkas-ortega}. The isogeny pushforward and splitting into translates are classical constructions treated by Oda~\cite[\S1]{oda}. Symmetric contraction, which we use to describe symmetric powers of evaluation kernels, appears in Kaid and Kasprowitz~\cite[Proposition~4.1, equation~(4.3)]{kaid-kasprowitz}. Our contribution is the uniform deduction of natural cohomology from these constructions. For $r>n$, symmetric contraction identifies the duals of the graded kernels and cokernels with $H^1$ and $H^0$ of $B\otimes\Sym^sK$. After isogeny pullback, this bundle splits into line bundles of degree $nt-rs$. Avoiding finitely many choices of $B$ handles degree zero, giving maximal rank in every degree. The boundary $r=n$ uses the classical rational Schwarzenberger construction \cite[Example~1.7]{arrondo}. The passage from two critical rank conditions to all degrees follows \cite[Lemma~2.10]{chs}. We include proofs of these steps. All curve constructions through Section~\ref{sec:rational} are over an algebraically closed field of characteristic zero. We descend the resulting open subset to $k$ in Section~\ref{sec:generality}. We write $\PP(V)$ for the projective space of lines in $V$, with homogeneous coordinate ring $\Sym(V^*)$. A map of finite-dimensional vector spaces has \emph{maximal rank} when it is injective or surjective. \section{Multiplication of sections on a curve}\label{sec:sections} Let $C$ be a smooth projective integral curve. Suppose that $L$ is a line bundle and that a base-point-free subspace $V\subset H^0(C,L)$ of dimension $n+1$ gives an exact sequence \begin{equation}\label{eq:evaluation} 0\longrightarrow K\longrightarrow V\otimes\OO_C \xrightarrow{e}L\longrightarrow0. \end{equation} Let $B$ be another line bundle satisfying \begin{equation}\label{eq:B} H^1(C,B)=0,\qquad h^0(C,B)=t,\qquad h^0(C,B\otimes L)=t+r. \end{equation} Multiplication of sections gives \[ m:H^0(C,B)\otimes V\longrightarrow H^0(C,B\otimes L). \] Its dual determines a matrix of linear forms over $S=\Sym(V^*)$: \begin{equation}\label{eq:matrix} M:H^0(C,B\otimes L)^*\otimes S(-1) \longrightarrow H^0(C,B)^*\otimes S. \end{equation} After choosing bases, its degree-$s$ component is the map $\mu_s$ in~\eqref{eq:mu}. \begin{lemma}[The cohomology of the graded maps]\label{lem:sections} For every $s\geq0$, the matrix~\eqref{eq:matrix} satisfies \begin{equation}\label{eq:dual-cohomology} (\coker\mu_s)^*\simeq H^0(C,B\otimes\Sym^s K),\qquad (\ker\mu_s)^*\simeq H^1(C,B\otimes\Sym^s K). \end{equation} Moreover, its transpose has constant rank $t$ on $\PP(V)$, and hence defines a Steiner bundle as in~\eqref{eq:steiner}. \end{lemma} \begin{proof} For $s\geq1$, contraction with the quotient map $e$ gives \begin{equation}\label{eq:contraction} 0\longrightarrow\Sym^s K \longrightarrow\Sym^s V\otimes\OO_C \xrightarrow{\delta_s}\Sym^{s-1}V\otimes L \longrightarrow0, \end{equation} where \[ \delta_s(v_1\cdots v_s)= \sum_{j=1}^s v_1\cdots\widehat{v_j}\cdots v_s\otimes e(v_j). \] To verify exactness, split~\eqref{eq:evaluation} locally. The summand $\Sym^{s-j}K\otimes L^j$ maps to the corresponding summand of the target by multiplication by $j$. These scalars are invertible for $1\leq j\leq s$ in characteristic zero. The summand with $j=0$ is exactly the kernel. Tensor~\eqref{eq:contraction} with $B$ and take cohomology. Since $H^1(C,B)=0$ and $C$ is a curve, the result is \begin{equation}\label{eq:cohomology-sequence} \begin{aligned} 0\longrightarrow H^0(C,B\otimes\Sym^sK) &\longrightarrow H^0(C,B)\otimes\Sym^sV\\ &\xrightarrow{H^0(\delta_s)} H^0(C,B\otimes L)\otimes\Sym^{s-1}V\\ &\longrightarrow H^1(C,B\otimes\Sym^sK) \longrightarrow0. \end{aligned} \end{equation} Here $H^0(\delta_s)$ denotes the map after tensoring with $B$. Choose dual bases $x_0,\ldots,x_n$ and $v_0,\ldots,v_n$. The differential pairing \[ \langle x^\alpha,v^\beta\rangle= \begin{cases} \alpha!,&\alpha=\beta,\\ 0,&\alpha\ne\beta \end{cases} \] identifies $S_s^*$ with $\Sym^sV$. Under this pairing, multiplication by $x_i$ is adjoint to differentiation with respect to $v_i$. Consequently $H^0(\delta_s)$ is precisely $\mu_s^*$, and \eqref{eq:cohomology-sequence} gives~\eqref{eq:dual-cohomology}. For $s=0$, the source of $\mu_0$ is zero; the same identities follow directly from~\eqref{eq:B}. At a point $[v]\in\PP(V)$, with $v\ne0$, the transpose of the twisted sheaf map associated with $M$ is \[ H^0(C,B)\longrightarrow H^0(C,B\otimes L), \qquad b\longmapsto b\,e(v). \] The section $e(v)$ is nonzero because $V\subset H^0(C,L)$. Multiplication by a nonzero section of a line bundle on an integral curve is injective. The fiber map therefore has rank $t$ at every point, proving the final assertion. \end{proof} Thus it suffices to choose $(C,L,V,B)$ for which the two groups in~\eqref{eq:dual-cohomology} never occur simultaneously. The elliptic construction below achieves this by splitting the relevant bundles after a finite cover. \section{The elliptic construction}\label{sec:elliptic} We first record the cohomological facts used in the construction. \begin{lemma}\label{lem:elliptic-cohomology} Let $D$ be an elliptic curve and let $T$ be a line bundle on $D$. If $\deg T>0$, then $H^1(D,T)=0$ and $h^0(D,T)=\deg T$. If $\deg T<0$, then $H^0(D,T)=0$. If $\deg T=0$ and $T\not\simeq\OO_D$, both cohomology groups vanish. If $q:D\to C$ is a finite \'etale cover of degree $n$ in characteristic zero and $W$ is a vector bundle on $C$, then $H^i(C,W)$ is a direct summand of $H^i(D,q^*W)$ for every $i$. \end{lemma} \begin{proof} The canonical bundle of an elliptic curve is trivial. A line bundle of negative degree has no nonzero section, and a degree-zero line bundle with a nonzero section is trivial. The first assertions follow from these facts, Serre duality, and Riemann--Roch. For the last assertion, the unit and trace satisfy \[ \OO_C\longrightarrow q_*\OO_D\longrightarrow\OO_C, \qquad 1\longmapsto n. \] Dividing the trace by $n$ splits the unit. Tensoring with $W$, applying the projection formula, and taking cohomology gives the claimed direct summand. The trace construction is recalled in~\cite[Tag~0BVH]{stacks}. \end{proof} \begin{proposition}[A matrix for $r>n$]\label{prop:elliptic} Suppose $n\geq1$, $t\geq1$, and $r>n$. There are curve data satisfying~\eqref{eq:evaluation} and~\eqref{eq:B} for which the matrix~\eqref{eq:matrix} has maximal rank in every degree $s\geq0$. More precisely, \begin{equation}\label{eq:rank-directions} \begin{aligned} nt-rs>0&\ \Longrightarrow\ \ker\mu_s=0,\\ nt-rs<0&\ \Longrightarrow\ \coker\mu_s=0,\\ nt-rs=0&\ \Longrightarrow\ \mu_s\text{ is an isomorphism}. \end{aligned} \end{equation} \end{proposition} \begin{proof} Choose an elliptic curve $D$ and a cyclic subgroup of order $n$. The quotient is a cyclic \'etale isogeny \[ q:D\longrightarrow C,\qquad \deg q=n. \] For $n=1$, take the identity. Write $G$ for the group of deck translations, and choose a line bundle $A$ of degree $r$ on $D$. Set $Q=q_*A$. It is a vector bundle of rank $n$, and finite pushforward and Riemann--Roch give \[ \deg Q=\chi(C,Q)=\chi(D,A)=r,\qquad h^0(C,Q)=r. \] Base change gives \begin{equation}\label{eq:split} q^*Q\simeq\bigoplus_{g\in G}g^*A. \end{equation} Indeed, $D\times_C D$ is the disjoint union of the graphs of the translations in $G$, so each component contributes one of these line bundles. The bundle $Q$ is globally generated. For any $p\in C$, finite pushforward and the projection formula give \[ H^1(C,Q(-p))\simeq H^1\bigl(D,A\otimes q^*\OO_C(-p)\bigr)=0, \] because the line bundle on the right has degree $r-n>0$. The evaluation map $H^0(C,Q)\to Q|_p$ is therefore surjective. We next choose $n+1$ sections generating $Q$ at every point. Let $\mathcal T=H^0(C,Q)^{\oplus(n+1)}$. For a fixed $p$, evaluation maps $\mathcal T$ onto the space of $n\times(n+1)$ scalar matrices. The matrices of rank at most $n-1$ have codimension two. The closed incidence of pairs $(p,\mathbf u)$ for which the sections $\mathbf u$ do not span $Q|_p$ thus has dimension at most $\dim\mathcal T-1$. Its image in $\mathcal T$ is closed, since $C$ is projective. A tuple outside that image generates $Q$ everywhere. For the resulting map from $U\otimes\OO_C$, with $\dim U=n+1$, let $L^{-1}$ denote its kernel. We obtain \begin{equation}\label{eq:Q-evaluation} 0\longrightarrow L^{-1}\longrightarrow U\otimes\OO_C \longrightarrow Q\longrightarrow0,\qquad \deg L=r. \end{equation} Dualizing gives~\eqref{eq:evaluation} with $K=Q^*$ and $V=U^*$. By~\eqref{eq:split}, $q^*K$ is a direct sum of line bundles of degree $-r$, so $H^0(C,K)=0$ by Lemma~\ref{lem:elliptic-cohomology}. The map $V\to H^0(C,L)$ is consequently injective. Thus $V$ is the required base-point-free subspace. Choose $B\in\Pic^t(C)$, with a restriction to be imposed below. Both $B$ and $B\otimes L$ have positive degree, so Lemma~\ref{lem:elliptic-cohomology} gives~\eqref{eq:B}. For \[ T_s=B\otimes\Sym^sK \] the splitting~\eqref{eq:split} yields \begin{equation}\label{eq:symmetric-split} q^*T_s\simeq \bigoplus_{\substack{\alpha_g\geq0\\\sum_{g\in G}\alpha_g=s}} q^*B\otimes\bigotimes_{g\in G}(g^*A)^{-\alpha_g}. \end{equation} Every line bundle in this sum has degree \begin{equation}\label{eq:common-degree} nt-rs. \end{equation} If this number is positive, Lemma~\ref{lem:elliptic-cohomology} gives $H^1(C,T_s)=0$; if it is negative, it gives $H^0(C,T_s)=0$. There is at most one integer $s$ for which $nt-rs=0$. At that degree, choose $B$ so that none of the finitely many summands in~\eqref{eq:symmetric-split} is trivial. For each multi-index $\alpha$, the forbidden condition is \begin{equation}\label{eq:forbidden} q^*B\simeq\bigotimes_{g\in G}(g^*A)^{\alpha_g}. \end{equation} It excludes only finitely many points of $\Pic^t(C)$. In fact, two solutions differ by an element of the kernel of $q^*:\Pic^0(C)\to\Pic^0(D)$. The norm identity $\Nm_q(q^*P)\simeq P^{\otimes n}$ puts this kernel in the finite $n$-torsion subgroup of $\Pic^0(C)$. Since $\Pic^t(C)$ is an elliptic curve over an infinite algebraically closed field, it has points outside the finite union of forbidden sets. For this choice of $B$, every degree-zero summand is nontrivial. Both cohomology groups of $T_s$ then vanish, again by Lemma~\ref{lem:elliptic-cohomology}. The one chosen line bundle $B$ therefore provides the required vanishings for all $s\geq0$. Lemma~\ref{lem:sections} now proves~\eqref{eq:rank-directions}. Finally, the difference between the target and source dimensions of $\mu_s$ is \begin{equation}\label{eq:dimension-difference} t\binom{n+s}{n}-(t+r)\binom{n+s-1}{n} =\binom{n+s-1}{n-1}\left(t-\frac{rs}{n}\right), \end{equation} where the second term on the left is zero at $s=0$. Its sign is the sign of~\eqref{eq:common-degree}, so these vanishings give maximal rank in the required direction. \end{proof} \section{The rational boundary case}\label{sec:rational} \begin{proposition}[A matrix for $r=n$]\label{prop:rational} For $n,t\geq1$ and $r=n$, there are curve data satisfying~\eqref{eq:evaluation} and~\eqref{eq:B} whose matrix has maximal rank in every degree. It is injective for $st$. \end{proposition} \begin{proof} Take \[ C=\PP^1,\qquad L=\OO_{\PP^1}(n),\qquad V=H^0(\PP^1,L),\qquad B=\OO_{\PP^1}(t-1). \] For the degree-$n$ monomial basis $x^{n-j}y^j$, $0\leq j\leq n$, the adjacent relations give a map \[ \OO_{\PP^1}(-1)^{\oplus n}\longrightarrow \OO_{\PP^1}^{\oplus(n+1)} \] whose column indexed by $0\leq jt$. The assertions follow from Lemma~\ref{lem:sections}. \end{proof} \section{Generality and cohomology in every twist}\label{sec:generality} \begin{proof}[Proof of Theorem~\ref{thm:main}] Let $\mathcal M$ be the affine space of $t\times(t+r)$ matrices with entries in $S_1$, and put \[ \beta=\left\lfloor\frac{nt}{r}\right\rfloor,\qquad \alpha=\left\lceil\frac{nt}{r}\right\rceil. \] Impose on a matrix $M\in\mathcal M$ the conditions that $\mu_\beta$ be injective and $\mu_\alpha$ be surjective. They are open rank conditions. If $\beta=0$, the first condition is vacuous. If $\alpha=\beta$, they say that one square map is invertible. Propositions~\ref{prop:elliptic} and~\ref{prop:rational} give a matrix satisfying both, so their intersection is nonempty. These two conditions imply maximal rank in every degree. Injectivity propagates downward: a nonzero homogeneous vector in $\ker\mu_s$ for $s<\beta$ could be multiplied by a nonzero monomial to give a nonzero vector in $\ker\mu_\beta$. Surjectivity propagates upward: the graded module \[ N=\coker\bigl(M:S(-1)^{\oplus(t+r)}\to S^{\oplus t}\bigr) \] is generated in degree zero. Hence $N_\alpha=0$ implies $N_s=0$ for all $s\geq\alpha$. Since $\alpha$ and $\beta$ are equal or consecutive, every nonnegative degree is covered. The dimension comparison~\eqref{eq:dimension-difference} shows that the injective and surjective ranges are the ones required for maximal rank. In particular, $N$ has finite length. Sheafifying $M$ and twisting by $\OO_{\PP^n}(1)$ gives a surjection \[ \OO_{\PP^n}^{\oplus(t+r)} \longrightarrow\OO_{\PP^n}(1)^{\oplus t}. \] Its kernel $E$ is locally free. Dualizing gives \eqref{eq:steiner} with $F=E^*$. For a twist $a=s-1\geq-1$, the exact sequence \begin{equation}\label{eq:kernel-twist} 0\longrightarrow E(a)\longrightarrow \OO_{\PP^n}(a)^{\oplus(t+r)} \longrightarrow\OO_{\PP^n}(a+1)^{\oplus t} \longrightarrow0 \end{equation} and the cohomology of line bundles on projective space yield \begin{equation}\label{eq:E-cohomology} H^0(\PP^n,E(s-1))=\ker\mu_s,\qquad H^1(\PP^n,E(s-1))=\coker\mu_s. \end{equation} All higher groups vanish, and maximal rank ensures that these two groups are not simultaneously nonzero. For $a\leq-2$, both line bundles in~\eqref{eq:kernel-twist} have no global sections and no intermediate cohomology. The long exact sequence therefore leaves only $H^n(\PP^n,E(a))$ as a possible nonzero group. When $n=1$, the same conclusion follows directly from the vanishing of $H^0$ of the two line bundles. These line-bundle cohomology statements are recorded in~\cite[Tag~01XT]{stacks}. Thus $E$ has natural cohomology in every twist. Serre duality gives \[ H^i(\PP^n,F(a))\simeq H^{n-i}(\PP^n,E(-a-n-1))^*, \] so $F$ has natural cohomology as well. It remains to justify the assertion over the original field $k$. The two critical maps are universal matrices whose entries are integer polynomials in the coefficients of $M$. The construction over an algebraic closure gives maximal minors of the two maps with nonzero product. This product is a nonzero polynomial over $\mathbb Q$, hence also over $k$. Its nonvanishing defines a nonempty open subset of $\mathcal M_k$. An infinite field has rational points in any such open subset of affine space. The same rank and cohomology arguments apply to these matrices over $k$, completing the proof. \end{proof} We record the resulting dimensions explicitly. \begin{corollary}\label{cor:dimensions} For a matrix in the open subset of Theorem~\ref{thm:main}, let $E=F^*$ and set, for $s\geq0$, \[ D_s=t\binom{n+s}{n}-(t+r)\binom{n+s-1}{n}, \] with the second binomial interpreted as zero at $s=0$. Then \[ h^0(\PP^n,E(s-1))=\max\{-D_s,0\},\qquad h^1(\PP^n,E(s-1))=\max\{D_s,0\}. \] All higher cohomology groups in these twists vanish. The graded module $N=\coker M$ has Hilbert series \[ \Hilb_N(z)= \left[\frac{t-(t+r)z}{(1-z)^{n+1}}\right]_+, \] where positive truncation keeps the coefficients preceding the first nonpositive coefficient and sets that coefficient and all later coefficients to zero. \end{corollary} \begin{proof} The formulas for $E$ follow from~\eqref{eq:E-cohomology}, maximal rank, and the difference between the two vector-space dimensions. 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