On the maximum number of complexes of a given degree containing subvarieties of Grassmannians

Abstract

Let ${\mathbb G}(k,r)$ be the Grassmannian of k-subspaces in ${\mathbb P}^r$ embedded in ${\mathbb P}^{N(k,r)}$, with $N(k,r)={{r+1}\atopwithdelims (){k+1}}-1$, via the Plücker embedding. In this paper, extending some classical results by Gallarati (see Gallarati in Rend Accad Naz Lincei Ser VIII 14(2):213–220, 1953, Rend Accad Naz Lincei Ser VIII 14(3):408–412, 1953), we give a sharp upper bound for the number of independent sections of $H^0({\mathbb G}(k,r), {\mathcal O}_{{\mathbb G}(k,r)}(m))$ vanishing on a subvariety X of ${\mathbb G}(k,r)$ such that the union of the k-subspaces corresponding to the points of X spans ${\mathbb P}^r$.

1 Introduction

Let ${\mathbb G}(k,r)$ be the Grassmannian of k-subspaces (i.e., linear subspaces of dimension k) in ${\mathbb P}^r$ embedded in ${\mathbb P}^{N(k,r)}$, with $N(k,r)={{r+1}\atopwithdelims (){k+1}}-1$, via the Plücker embedding. To avoid trivial cases we will assume $k\ne 0,r-1$.

For every positive integer m we consider $H^0({\mathbb G}(k,r), {\mathcal O}_{{\mathbb G}(k,r)}(m))$ whose dimension we denote by $\varepsilon _{k,r}(m)$. This is a well known number, computed for instance in [7, Thm. III, p. 387], which is not necessary to make explicit here. Given any non-zero section $s\in H^0({\mathbb G}(k,r), {\mathcal O}_{{\mathbb G}(k,r)}(m))$, the zero locus scheme (s) of s is called a m-complex of ${\mathbb G}(k,r)$.

Let now X be an irreducible, projective subvariety of ${\mathbb G}(k,r)$. We can consider the subvariety

$$\begin{aligned} Z(X)=\bigcup _{\pi \in X}\pi \end{aligned}$$

of ${\mathbb P}^r$. We will say that X is Grassmann non-degenerate if Z(X) is non-degenerate in ${\mathbb P}^r$, i.e., if Z(X) spans ${\mathbb P}^r$. We will say that X presents the cone case if the subspaces corresponding to the points of X pass through one and the same linear subspace of dimension $k-1$, in which case Z(X) is a cone with vertex a linear space of dimension $k-1$.

Let $\mathcal I_{X,{\mathbb G}(k,r)}$ be the ideal sheaf of X in ${\mathbb G}(k,r)$. We will set

$$\begin{aligned} \theta _X(m):=h^0({\mathbb G}(k,r),\mathcal I_{X,{\mathbb G}(k,r)}(m)) \end{aligned}$$

that is the maximum number of independent m-complexes of ${\mathbb G}(k,r)$ containing X.

In the two papers [4, 5], D. Gallarati proved the following two results:

Theorem 1

Let X be an irreducible projective Grassmann non-degenerate curve in $\mathbb G(1,r)$. Then for any positive integer m one has

$$\begin{aligned} \theta _X(m) \leqslant \varepsilon _{1,r}(m)-m(r-1)-1. \end{aligned}$$

(1)

Moreover:

• (i) If $m>1$ and X is a rational normal curve of degree $r-1$ (equivalently if Z(X) is a rational normal scroll surface of degree $r-1$), then the equality holds in (1);

• (ii) If $m=1$ equality holds in (1) if and only if either X is a rational normal curve of degree $r-1$ or X presents the cone case.

Theorem 2

Let X be an irreducible projective Grassmann non-degenerate curve in $\mathbb G(k,r)$. Then

$$\begin{aligned} \theta _X(1) \leqslant {{r+1}\atopwithdelims (){k+1}} -r+k-1=\varepsilon _{k,r}(1) -r+k-1. \end{aligned}$$

(2)

Moreover, equality holds in (2) if and only if either X is a rational normal curve of degree $r-k$ (equivalently Z(X) is a $(k+1)$-dimensional rational normal scroll of degree $r-k$) or X presents the cone case.

It is may be the case to notice that Grassmannians do contain many rational normal curves, corresponding to rational normal scrolls.

In this note we will prove an extension of Gallarati’s results. Before stating our result, let us introduce some notation. Given positive integers n, k, r, m we will set

$$\begin{aligned} \theta (n,k,r,m)= \varepsilon _{k,r}(m)-(r-k) \sigma (n,m)- \tau (n,m) \end{aligned}$$

where $\sigma (1,m)=m$, $\sigma (2,m)=\frac{m(m+1)}{2}$, whereas if $n\geqslant 3$ one has

$$\begin{aligned} \sigma (n,m)=\sum _{i_{n-2}=1}^m\sum _{i_{n-3}=1}^{i_{n-2}} \cdots \sum _{i_1=1}^{i_2} \frac{i_1(i_1+1)}{2}, \end{aligned}$$

and

$$\begin{aligned} \tau (1,m)=1,\quad \tau (2,m)=m+1,\quad \tau (3,m)=\frac{m(m+1)}{2}+m+1 \end{aligned}$$

whereas if $n\geqslant 4$ one has

$$\begin{aligned} \begin{aligned}&\tau (n,m)=\sum _{i_{n-3}=1}^m\sum _{i_{n-4}=1}^{i_{n-3}} \cdots \sum _{i_1=1}^{i_2} \frac{i_1(i_1+1)}{2}+\\&+ \sum _{i_{n-4}=1}^m\sum _{i_{n-5}=1}^{i_{n-4}} \cdots \sum _{i_1=1}^{i_2} \frac{i_1(i_1+1)}{2}+\cdots + \frac{m(m+1)}{2}+m+1. \end{aligned} \end{aligned}$$

One notes that $\theta (1,1,r,m)$ coincides with Gallarati’s upper bound in (1) and $\theta (1,k,r,1)$ coincides with the upper bound in (2). In addition we notice that

$$\begin{aligned} \sigma (n,m)=\sum _{i=1}^m\sigma (n-1,i) \quad \text{ and }\quad \tau (n,m)=\sum _{i=1}^m\tau (n-1,i)+1. \end{aligned}$$

(3)

Moreover, given positive integers $n>1,k,r,m$ we set

$$\begin{aligned} {\bar{\theta }}(n,k,r,m)= \varepsilon _{k,r}(m)-{{m+n-1}\atopwithdelims ()n}(r-k-n)-{{m+n}\atopwithdelims ()n}. \end{aligned}$$

Then we can state our result:

Theorem 3

Let X be an irreducible projective Grassmann non-degenerate variety of dimension n in $\mathbb G(k,r)$.

(a) If either $n=1$ or $n>1$ and X does not present the cone case, then

$$\begin{aligned} \theta _X(m) \leqslant \theta (n,k,r,m). \end{aligned}$$

(4)

Moreover:

• (i) For $m=1$ equality holds in (4) if and only if either X is a rational normal variety of degree $r-k$ in a subspace of dimension $r-k+n-1$ in ${\mathbb P}^{N(k,r)}$ or $n=1$ and X presents the cone case;

• (ii) For $n=1$ and $m>1$ equality holds in (4) if X is a rational normal curve of degree $r-k$;

• (iii) For $n>1$ and $m>1$, equality holds in (4) if and only if X is a rational normal variety of degree $r-k$ spanning a subspace of dimension $r-k+n-1$ in ${\mathbb P}^{N(k,r)}$.

(b) If $n>1$ and X presents the cone case then

$$\begin{aligned} \theta _X(m) \leqslant {\bar{\theta }}(n,k,r,m). \end{aligned}$$

(5)

Moreover:

• (iv) Equality holds in (5) for $m=1$;

• (v) If equality holds in (5) for some $m\geqslant 2$, then X is a rational normal variety of degree $r-k-n+1$ in a linear subspace of dimension $r-k$ contained in ${\mathbb G}(k,r)$ whose points correspond to k-subspaces containing a fixed $(k-1)$-subspace, or, equivalently, Z(X) is a cone with vertex a $(k-1)$-subspace over a variety of dimension n and minimal degree $r-k-n+1$ in a $(r-k)$-linear subspace of ${\mathbb P}^r$.

The proof of part (a) works by induction on the dimension n of X. We first treat the curve case in Sect. 1, then the surface and threefold case in Sect. 2. Then, to complete the proof of part (a) in the rest of Sect. 2, we proceed by induction using a classical method going back to Castelnuovo (see [2]), passing from X to a general hyperplane section of it. We will treat case (b) in Sect. 3, again using Castelnuovo’s method. In Sect. 4 we finish with some examples showing that the varieties mentioned in the statement of Theorem 3, for which equality holds in (4) and (5), do really occur.

2 The curve case

In this section we will prove Theorem 3 in the curve case $n=1$. A great part of this proof is already in Gallarati’s papers [4, 5]. However it is the case to review some details of the proof that are a bit too terse in Gallarati’s treatment.

Let us start with the following easy lemma:

Lemma 4

Let X be an irreducible projective Grassmann non-degenerate curve in $\mathbb G(k,r)$. Then given $r-k$ general points $P_1,\ldots , P_{r-k}$ of X, there is a hyperplane section of $\mathbb G(k,r)$ not containing X and containing $P_1,\ldots , P_{r-k}$.

Proof

Given a linear subspace $\Pi$ of dimension $r-k-1$, one can consider the Schubert cycle of all linear k-subspaces of ${\mathbb P}^r$ intersecting $\Pi$. This is a linear complex, i.e., a hyperplane section $H_\Pi$ of ${\mathbb G}(k,r)$. Given $P_1,\ldots , P_{r-k}$ general points of X, these correspond to $r-k$ general subspaces sweeping out Z(X), as the points move. Let us fix a general point $p_i\in P_i$ for any $1\leqslant i\leqslant r-k$, so that $p_1,\ldots , p_{r-k}$ are general points of Z(X). The Grassmann non-degeneracy of X is equivalent to Z(X) being non-degenerate, so $p_1,\ldots , p_{r-k}$ are linearly independent in ${\mathbb P}^r$. Then $\Pi =\langle p_1,\ldots , p_{r-k}\rangle$ is a $(r-k-1)$-subspace of ${\mathbb P}^r$, and actually it is a general such subspace, so that it intersects Z(X), that has dimension $k+1$, in finitely many points. Then $\Pi$ intersects the subspaces $P_1,\ldots , P_{r-k}$ but does not intersect all the subspaces $P\in X$. Hence $H_\Pi$ is a hyperplane section of $\mathbb G(k,r)$ not containing X and containing $P_1,\ldots , P_{r-k}$, as required. $\square$

Let $X\subset {\mathbb P}^r$ be an irreducible non-degenerate projective variety. For all non-negative integers m we let $h_X(m)$ be the Hilbert function of X, i.e., $h_X(m)$ is the dimension of the image of the restriction map

$$\begin{aligned} H^0({\mathbb P}^r, \mathcal O_{{\mathbb P}^r}(m))\longrightarrow H^0(X, {\mathcal O}_X(m)). \end{aligned}$$

So $h_X(m)-1$ is the dimension of the linear series cut out on X by the hypersurfaces of degree m of ${\mathbb P}^r$.

Now suppose that X is an irreducible projective Grassmann non-degenerate variety of dimension n in $\mathbb G(k,r)$. For any non-negative integer m, we have the exact sequence

$$\begin{aligned} 0\longrightarrow \mathcal I_{X,{\mathbb G}(k,r)}(m)\longrightarrow {\mathcal O}_{{\mathbb G}(k,r)}(m)\longrightarrow {\mathcal O}_{X}(m)\longrightarrow 0 \end{aligned}$$

which gives

$$\begin{aligned} 0\longrightarrow & {} H^0({\mathbb G}(k,r),\mathcal I_{X,{\mathbb G}(k,r)}(m))\longrightarrow H^0({\mathbb G}(k,r),{\mathcal O}_{{\mathbb G}(k,r)}(m))\nonumber \\\longrightarrow & {} H^0(X,{\mathcal O}_{X}(m)) \end{aligned}$$

(6)

The image of

$$\begin{aligned} \rho _m: H^0({\mathbb G}(k,r),{\mathcal O}_{{\mathbb G}(k,r)}(m))\longrightarrow H^0(X,{\mathcal O}_{X}(m)) \end{aligned}$$

(7)

coincides with the image of

$$\begin{aligned} H^0({\mathbb P}^r,{\mathcal O}_{{\mathbb P}^r}(m))\longrightarrow H^0(X,{\mathcal O}_{X}(m)) \end{aligned}$$

because ${\mathbb G}(k,r)$ is projectively normal and therefore

$$\begin{aligned} H^0({\mathbb P}^r,{\mathcal O}_{{\mathbb P}^r}(m))\longrightarrow H^0({\mathbb G}(k,r),{\mathcal O}_{{\mathbb G}(k,r)}(m)) \end{aligned}$$

(8)

is surjective for all non-negative integers m. So the dimension of the image of $\rho _m$ is $h_X(m)$.

By (6) we have

$$\begin{aligned} \theta _X(m) =\varepsilon _{k,r}(m)-h_X(m) \end{aligned}$$

(9)

Now we are ready for the:

Proof of Theorem 3 for $n=1$

By (9), to bound $\theta _X(m)$ from above we have to bound $h_X(m)$ from below. To do so, notice that given $m(r-k)$ general points of X, there is some m-complex of ${\mathbb G}(k,r)$ containing those points and not containing X. It suffices to divide the $m(r-k)$ general points of X in m subsets of $r-k$ general points, and then apply Lemma 7, thus finding the required m-complex splitting in m linear complexes each containing one of the m subsets of $r-k$ points. This proves that

$$\begin{aligned} h_X(m)-1\geqslant m(r-k). \end{aligned}$$

(10)

Then by (9) we get

$$\begin{aligned} \theta _X(m) =\varepsilon _{k,r}(m)-h_X(m) \leqslant \varepsilon _{k,r}(m)-m(r-k)-1=\theta (1,k,r,m) \end{aligned}$$

proving (4) in this case.

The proof of (i) in the case $n=1$ is contained in [4, 5] and we do not dwell on this here. As for (ii), if X is a rational normal curve of degree $r-k$, then for all positive integers m we have $h_X(m)=m(r-k)+1$ and therefore from (9) we find that $\theta _X(m)=\theta (1,k,r,m)$ as wanted. $\square$

3 The general case

Before treating the general case it is necessary to work out first the surface and the threefold cases. We start with some lemmata.

Lemma 5

Let X be an irreducible projective Grassmann non-degenerate variety in $\mathbb G(k,r)$ not presenting the cone case. Let $P_1,\ldots , P_{r-k}$ be general points in X, which correspond to k-subspaces of ${\mathbb P}^r$ that we denote by the same symbols. Then

$$\begin{aligned} \langle P_1,\ldots , P_{r-k}\rangle ={\mathbb P}^r. \end{aligned}$$

Proof

One has $\dim (\langle P_1, P_2\rangle )=k+a_1$, with $a_1\geqslant 1$. If $a_1=1$, two k-subspaces corresponding to general points of X span a linear space of dimension $k+1$, so they intersect in a linear subspace of dimension $k-1$. Then either they all lie in a linear space of dimension $k+1$ or they all pass through a linear space of dimension $k-1$ (this is an elementary fact, see [1, Chapt. I, Sect. 18, p. 16]), but neither case can occur by the non-degeneracy of X and the fact that X does not present the cone case. Hence $a_1\geqslant 2$.

Next, if $\langle P_1, P_2\rangle ={\mathbb P}^r$ the claim clearly holds. So suppose that $\langle P_1, P_2\rangle$ is a proper subspace of ${\mathbb P}^r$, hence $k+a_1<r$. Then $P_3$ cannot lie in $\langle P_1, P_2\rangle$ by the non-degeneracy of X, so $\dim (\langle P_1, P_2, P_3 \rangle )=k+a_1+a_2$, with $a_2\geqslant 1$. If $\langle P_1, P_2, P_3 \rangle ={\mathbb P}^r$, the claim is clearly true, otherwise we iterate the above argument. The upshot is that either a proper subset of $P_1,\ldots , P_{r-k}$ spans ${\mathbb P}^r$ and we are done, or

$$\begin{aligned} \dim ( \langle P_1,\ldots , P_{r-k}\rangle )=k+a_1+\cdots +a_{r-k-1}\geqslant k+r-k=r \end{aligned}$$

proving the lemma. $\square$

Lemma 6

Let X be an irreducible projective Grassmann non-degenerate variety in $\mathbb G(k,r)$. Then

$$\begin{aligned} h_X(1)\geqslant r-k+1. \end{aligned}$$

Proof

If X is a curve, the assertion follows from (10). Assume that X has dimension $n\geqslant 2$. Fix $P_1,\ldots , P_{r-k}$ general points in X. Let C be the complete intersection curve of X with $n-1$ general hypersurfaces of degree $d\gg 0$ containing $P_1,\ldots , P_{r-k}$. Since the successive hypersurface sections of X, up to C, are connected one immediately has $\theta _C(1)=\theta _X(1)$. By Lemma 5, C is Grassmann non-degenerate, and therefore, by (4) one has

$$\begin{aligned} \theta _X(1)=\theta _C(1)\leqslant \theta (1, k, r, 1)={{r+1}\atopwithdelims (){k+1}} -r+k-1, \end{aligned}$$

whence the assertion immediately follows. $\square$

Lemma 7

Let X be an irreducible projective Grassmann non-degenerate variety of dimension $n\geqslant 2$ in $\mathbb G(k,r)$ not presenting the cone case. If Y is a general hyperplane section of X, then Y is also Grassmann non-degenerate and does not present the cone case.

Proof

Let $P_1,\ldots , P_{r-k}$ be general points in X. By Lemma 6, there is a hyperplane section Y of X containing $P_1,\ldots , P_{r-k}$, and Y can be considered to be a general hyperplane section of X. By Lemma 5, Y is Grassmann non-degenerate.

To finish the proof we have to show that for a general hyperplane section Y of X, Y does not present the cone case. Suppose, by contradiction, that this is the case. Then given two general points $P_1,P_2\in X$, there is a hyperplane section of X containing them, and therefore the k-subspaces $P_1,P_2$ intersect along a $(k-1)$-subspace. Then either all subspaces corresponding to points of X pass through the same $(k-1)$-subspace or they all lie in the same $(k+1)$-subspace. Both possibilites lead to contradictions since X is Grassmann non-degenerate and does not present the cone case. $\square$

Next let $X\subset {\mathbb P}^r$ be an irreducible non-degenerate projective variety. Let Y be a general hyperplane section of X. For all $m\geqslant 1$ one has the inequality

$$\begin{aligned} h_X(m)-h_X(m-1)\geqslant h_Y(m) \end{aligned}$$

(11)

and equality holds for all m if and only if X is projectively normal (see [6, Lemma (3.1)]).

Proof of part (a) of Theorem 3 for surfaces

Let X be an irreducible projective Grassmann non-degenerate surface in $\mathbb G(k,r)$ not presenting the cone case and let Y be a general hyperplane section of X. By Lemma 7, Y is Grassmann non-degenerate. By (11), for all positive integers m one has the inequalities

$$\begin{aligned}&h_X(m)-h_X(m-1)\geqslant h_Y(m)\\ \nonumber&\ldots \\ \nonumber&h_X(1)-h_X(0)\geqslant h_Y(1) \end{aligned}$$

(12)

and summing up we get

$$\begin{aligned} h_X(n)\geqslant \sum _{i=1}^mh_Y(i)+h_X(0)=\sum _{i=1}^mh_Y(i)+1. \end{aligned}$$

(13)

By (9) we have

$$\begin{aligned} \varepsilon _{k,r}(m)-\theta _X(m)=h_X(n)\geqslant \sum _{i=1}^mh_Y(i)+1=\sum _{i=1}^m(\varepsilon _{k,r}(i)-\theta _Y(i))+1 \end{aligned}$$

(14)

whence, using Theorem 3 for the curve Y (that we can do because Y is Grassmann non-degenerate), we have

$$\begin{aligned} \theta _X(m)&\leqslant \varepsilon _{k,r}(m)-\sum _{i=1}^m(\varepsilon _{k,r}(i)-\theta _Y(i))-1=\\ \nonumber&=\varepsilon _{k,r}(m)-\sum _{i=1}^m\varepsilon _{k,r}(i)+\sum _{i=1}^m\theta _Y(i)-1=\\ \nonumber&=\sum _{i=1}^m\theta _Y(i)-\sum _{i=1}^{m-1}\varepsilon _{k,r}(i)-1\leqslant \\ \nonumber&\leqslant \sum _{i=1}^m \Big (\varepsilon _{k,r}(i)-i(r-k)-1 \Big )- \sum _{i=1}^{m-1}\varepsilon _{k,r}(i)-1=\\ \nonumber&=\varepsilon _{k,r}(m)-\frac{m(m+1)}{2}(r-k)-m-1=\theta (2,k,r,m) \end{aligned}$$

(15)

proving (4) in this case.

Next let us prove (i). If $\theta _X(1)=\theta (2,k,r,1)$ then the above argument shows that $\theta _Y(1)= \theta (1,k,r,1)$. By Theorem 3(i) for Y and by Lemma 7, Y is a rational normal curve of degree $r-k$, because Y cannot present the cone case. Hence X is a rational normal surface of degree $r-k$. Conversely, if X is a rational normal surface of degree $r-k$, then $h_X(1)=r-k+2$ and therefore

$$\begin{aligned} \theta _X(1)=\varepsilon _{k,r}(1)-(r-k+2)=\theta (2,k,r,1) \end{aligned}$$

as wanted.

As for (iii), suppose that for some $m>1$ one has $\theta _X(m)=\theta (2,k,r,m)$. Then the above argument implies that $\theta _Y(i)=\theta (1,k,r,i)$, for $1\leqslant i\leqslant m$. By Theorem 3(ii) for Y, Y is a rational normal curve of degree $r-k$ and therefore X is a rational normal surface of degree $r-k$. Conversely, if X is a rational normal surface of degree $r-k$, then X is projectively normal and we have equalities in (12), (13) and (14). Moreover $\theta _Y(i)=\theta (1,k,r,i)$ for all positive integers i by Theorem 3(ii) for Y, and therefore we have equalities in (15). This implies that $\theta _X(m)=\theta (2,k,r,m)$. $\square$

The proof in the case of threefolds is similar so we will be brief.

Proof of part (a) of Theorem 3 for threefolds

Let X be an irreducible projective Grassmann non-degenerate threefold in $\mathbb G(k,r)$ not presenting the cone case and let Y be a general hyperplane section of X, which, by Lemma 7, is Grassmann non-degenerate and does not present the cone case. Arguing as in the proof of the surface case, and applying part (a) of Theorem 3 for Y, we have

$$\begin{aligned} \theta _X(m)&\leqslant \sum _{i=1}^m\theta _Y(i)-\sum _{i=1}^{m-1}\varepsilon _{k,r}(i)-1\leqslant \\&\leqslant \sum _{i=1}^m \Big (\varepsilon _{k,r}(i)-\frac{i(i+1)}{2} (r-k)-i-1 \Big )- \sum _{i=1}^{m-1}\varepsilon _{k,r}(i)-1=\\&=\varepsilon _{k,r}(m)- \Big (\sum _{i=1}^m\frac{i(i+1)}{2}\Big ) (r-k)-\frac{m(m+1)}{2}-m-1=\theta (3,k,r,m) \end{aligned}$$

proving (4) in this case. The proof of (i) and (ii) proceed exactly in the same way as in the surface case, so we leave it to the reader. $\square$

We can now finish the proof of part (a) of Theorem 3. The proof is similar to the surface and threefold case so we will again be brief.

Proof of part (a) of Theorem 3, the general case

We will work by induction, since we have proved the theorem for $n=1,2,3$. Let X be an irreducible projective Grassmann non-degenerate variety of dimension $n\geqslant 4$ in $\mathbb G(k,r)$ not presenting the cone case. Let Y be its general hyperplane section, which by Lemma 7, is Grassmann non-degenerate and does not present the cone case. So we can apply induction on Y. Arguing as in the surface and threefold case and applying induction and (3), we have

$$\begin{aligned} \theta _X(m)&\leqslant \sum _{i=1}^m\theta _Y(i)-\sum _{i=1}^{m-1}\varepsilon _{k,r}(i)-1\leqslant \\ \nonumber&\leqslant \sum _{i=1}^m (\varepsilon _{k,r}(i)-(r-k)\sigma (n-1,k,r,i)- \tau (n-1,k,r,i)) - \sum _{i=1}^{m-1}\varepsilon _{k,r}(i)-1=\\ \nonumber&=\varepsilon _{k,r}(m)- \Big (\sum _{i=1}^m\sigma (n-1,k,r,i) \Big ) (r-k)- \Big (\sum _{i=1}^m \tau (n-1,k,r,i)\Big )-1=\\&=\theta (n,k,r,m) \end{aligned}$$

proving (4) in this case. The proof of (i) and (ii) proceeds exactly in the same way as in the surface case, so we leave it to the reader. $\square$

4 The cone case

Next we come to the proof of part (b) of Theorem 3. For this we need a preliminary. Let X be an irreducible, non degenerate variety of degree d and dimension $n\geqslant 1$ in ${\mathbb P}^r$. We will denote by $X_ i$ the i-dimensional section of X with a general subspace of dimension $r-n+i$ of ${\mathbb P}^r$. One has $X=X_n$ and $X_0$ is a reduced finite set. We note that

$$\begin{aligned} h_{X_0}(m)\geqslant \min \{d,m(r-n)+1\} \end{aligned}$$

(16)

(see [6, Corollary (3.5)]).

Proposition 8

Let X be an irreducible, non degenerate variety of degree d and dimension $n\geqslant 1$ in ${\mathbb P}^r$, with $r>n$. For all positive integers m, one has

$$\begin{aligned} h_X(m)\geqslant {{m+n-1}\atopwithdelims ()n}(r-n)+{{m+n}\atopwithdelims ()n}. \end{aligned}$$

(17)

Moreover:

• (i) Equality holds in (17) for $m=1$;

• (ii) Equality holds in (17) for some $m\geqslant 2$ if and only if X is a variety of minimal degree $d=r-n+1$.

Proof

This proposition is essentially contained in Theorem 6.1 in [3]. However we give here a proof for completeness.

First we note that trivially equality holds in (17) for $m=1$, i.e., (i) holds. Next we prove (17) by double induction, first on the dimension n and then on m.

In the curve case, we proceed by induction on m, assuming $m\geqslant 2$. By (11) we have

$$\begin{aligned} h_X(m)\geqslant h_X(m-1)+h_{X_0}(m) \end{aligned}$$

and by (16), by the induction and the fact that $d\geqslant r$ since X is non-degenerate, we have

$$\begin{aligned} h_X(m)\geqslant (m-1)(r-1)+m+r=m(r-1)+m+1 \end{aligned}$$

(18)

proving (17) in this case. If equality holds in (18) in particular we have

$$\begin{aligned} \min \{d,m(r-1)+1\}=r \end{aligned}$$

and since $m(r-1)+1\geqslant 2r-1>r$ (because $r>1$), we deduce $d=r$, hence X is a rational normal curve. Conversely, if X is a rational normal curve, one has $h_X(m)=mr+1$ and the equality holds in (17).

Next we treat the case of varieties X of dimension $n>1$ and we assume by induction that the proposition holds for varieties of dimension smaller than n. Moreover we proceed by induction on m. By (11), by induction and by (16), we have

$$\begin{aligned} h_X(m)&\geqslant h_X(m-1)+h_{X_{n-1}}(m)\geqslant \nonumber \\&\geqslant {{m+n-2}\atopwithdelims ()n}(r-n)+{{m+n-1}\atopwithdelims ()n}+\nonumber \\&+{{m+n-2}\atopwithdelims (){n-1}}(r-n)+{{m+n-1}\atopwithdelims (){n-1}}=\nonumber \\&={{m+n-1}\atopwithdelims ()n}(r-n)+{{m+n}\atopwithdelims ()n} \end{aligned}$$

(19)

as wanted. Moreover, if the equality holds in (19) for some $m>1$, then the equality holds in (17) for some $m>1$ for $X_{n-1}$. Then by induction $X_{n-1}$ is of minimal degree and therefore also X is of minimal degree. Conversely, if X is of minimal degree, then also $X_{n-1}$ is of minimal degree. Moreover X is projectively normal. Because of this and by induction we have that equalities hold in (19) and therefore equality holds in (17). $\square$

Now we are in position to finish the:

Proof of part (b) of Theorem 3

Let X be an irreducible projective Grassmann non-degenerate variety of dimension n in $\mathbb G(k,r)$ presenting the cone case. Let $\Pi$ be the vertex of Z(X), that is a $(k-1)$-dimensional subspace of ${\mathbb P}^r$. Then X lies in the $(r-k)$-subspace $\Pi ^\perp$ contained in $\mathbb G(k,r)$, whose points correspond to linear spaces of dimension k containing $\Pi$. Since each map $\rho _m$ as in (7) factors through the maps

$$\begin{aligned} H^0({\mathbb G}(k,r),{\mathcal O}_{{\mathbb G}(k,r)}(m))\longrightarrow H^0(\Pi ^\perp ,{\mathcal O}_{\Pi ^\perp }(m)) \longrightarrow H^0(X,{\mathcal O}_{X}(m)) \end{aligned}$$

and since $H^0({\mathbb G}(k,r),{\mathcal O}_{{\mathbb G}(k,r)}(m))\longrightarrow H^0(\Pi ^\perp ,{\mathcal O}_{\Pi ^\perp }(m))$ is clearly surjective, to compute $h_X(m)$ it suffices to compute it as a subvariety of $\Pi ^\perp$. We can then apply Proposition 8. By (9) we have

$$\begin{aligned} \theta _X(m)&=\varepsilon _{k,r}(m)-h_X(m) \leqslant \\&\leqslant \varepsilon _{k,r}(m)-{{m+n-1}\atopwithdelims ()n}(r-k-n)-{{m+n}\atopwithdelims ()n}={\bar{\theta }}(n,k,r,m) \end{aligned}$$

as wanted. If the equality holds for some $m\geqslant 2$, then in particular

$$\begin{aligned} h_X(m)={{m+n-1}\atopwithdelims ()n}(r-k-n)+{{m+n}\atopwithdelims ()n} \end{aligned}$$

and, by Proposition 8, X is a variety of minimal degree in $\Pi ^\perp$, proving the assertion. $\square$

5 Examples

In the statement of Theorem 3 it is mentioned the possibility that either X does not present the cone case and is a rational normal variety of degree $r-k$ in a subspace of dimension $r-k+n-1$ in ${\mathbb P}^{N(k,r)}$, or X presents the cone case and X is a rational normal variety of degree $r-k-n+1$ in a linear subspace of dimension $r-k+n-1$ contained in ${\mathbb G}(k,r)$. The latter case can clearly occur, whereas it is not a priori clear that the former case can occur for all values of $n\geqslant 2$ and k (for $n=1$ also the former case occurs when Z(X) is a rational normal $(k+1)$-dimensional variety of degree $r-k$ in ${\mathbb P}^r$). In this section we provide two examples showing that for $n\geqslant 2$ also the former case can occur. We do not claim that these examples exhaust all the possibilities.

Example 9

Fix positive integers n, k and set $r=n+k-1$. In ${\mathbb P}^r$, fix a k-linear space $\Pi$ and inside it a $(k-2)$-linear space $\pi$. Then let X be the subvariety of ${\mathbb G}(k,r)$ described by all points corresponding to k-subspaces of ${\mathbb P}^r$ intersecting $\Pi$ in a subspace of dimension at least $k-1$ containing $\pi$. It is easy to check that $\dim (X)=n$, that it is Grassmann non-degenerate and that it does not present the cone case. Moreover X is a cone with vertex the point corresponding to $\Pi$ (indeed for any k-subspace $\Pi '$ of ${\mathbb P}^r$ intersecting $\Pi$ in a subspace of dimension $k-1$ containing $\pi$, the span of $\Pi$ and $\Pi '$ is a subspace of dimension $k+1$ and the elements of the pencil inside it spanned by $\Pi$ and $\Pi '$ form a line entirely contained in X). Let us intersect X with a hyperplane H not containing $\Pi$. To do this, let us fix a general linear space $\Pi '$ of dimension $r-k-1=n-2$. Then $\Pi '$ does not intersect $\pi$ and we can consider the hyperplane section $H_{\Pi '}$ of ${\mathbb G}(k,n)$ consisting of all points corresponding to linear spaces of dimension k intersecting $\Pi '$. The points in the intersection $X\cap H_{\Pi '}$ can be obtained in the following way. Take any $(k-1)$–space P in $\Pi$ containing $\pi$, take a point $p\in \Pi '$, take $\Pi _{P,p}:=\langle P,p\rangle$ and consider it as a point of ${\mathbb G}(k,r)$. One has

$$\begin{aligned} X\cap H_{\Pi '}=\bigcup _{\pi \subset P\subset \Pi , p\in \Pi '}\Pi _{P,p}. \end{aligned}$$

Now the subspaces P such that $\pi \subset P\subset \Pi$ vary in a ${\mathbb P}^1$, and $p\in \Pi '$ varies in a ${\mathbb P}^{n-2}$. An easy explicit computation shows that $X\cap H_{\Pi '}$ is isomorphic to the Segre variety $\textrm{Seg}(1,n-2)\cong {\mathbb P}^1\times {\mathbb P}^{n-2}$ which has degree $n-1$ and spans a linear space of dimension $2n-3$. Hence X has degree $n-1$ and spans a linear space of dimension $2n-2$. Thus X is a rational normal variety of dimension n and degree $n-1$ contained in ${\mathbb G}(k,n)$.

Example 10

Fix positive integers n, k and set $r=k+4$. First consider the case $k=2h+1$ is odd. Then fix two skew subspaces $\Pi ,\Pi '$ of dimension $h+2$ in ${\mathbb P}^r$ and inside $\Pi ,\Pi '$ fix two $(h-1)$-subspaces $\pi , \pi '$ respectively. For any h-linear space P [resp. $P'$] contained in $\Pi$ [resp. contained in $\Pi '$] and containing $\pi$ [resp. containing $\pi '$], consider $\Pi _{P,P'}=\langle P,P'\rangle$, that has dimension $2\,h+1=k$, so we can consider it as a point in ${\mathbb G}(k,r)$. The subspaces P [resp. $P'$] contained in $\Pi$ [resp. contained in $\Pi '$] and containing $\pi$ [resp. containing $\pi '$] vary in a ${\mathbb P}^2$, hence

$$\begin{aligned} V=\bigcup _{\pi \subset P\subset \Pi , \pi '\subset P'\subset \Pi '}\Pi _{P,P'} \end{aligned}$$

is a subvariety of ${\mathbb G}(k,r)$ which is easy to check to be isomorphic to the Segre variety $\textrm{Seg}(2,2)\cong {\mathbb P}^2\times {\mathbb P}^2$ embedded in ${\mathbb P}^8$. As well known, $\textrm{Seg}(2,2)$ contains a Veronese surface $X\subset {\mathbb P}^5$, as the image of the diagonal of ${\mathbb P}^2\times {\mathbb P}^2$. The surface X is a rational normal surface contained in ${\mathbb G}(k,n)$ which is Grassman non-degenerate and not presenting the cone case.

Consider now the case $k=2h+2$ (and still $r=k+4$). Now fix two skew subspaces $\Pi ,\Pi '$ of dimension $h+2$ in ${\mathbb P}^r$ and a point p off the hyperplane of ${\mathbb P}^r$ spanned by $\Pi$ and $\Pi '$. As in the odd case, inside $\Pi ,\Pi '$ we fix two $(h-1)$-subspaces $\pi , \pi '$ respectively. For any h-linear space P [resp. $P'$] contained in $\Pi$ [resp. contained in $\Pi '$] and containing $\pi$ [resp. containing $\pi '$], consider $\Pi _{P,P'}=\langle P,P',p\rangle$, that has dimension k, so we can consider it as a point in ${\mathbb G}(k,r)$. The subspaces P [resp. $P'$] contained in $\Pi$ [resp. contained in $\Pi '$] and containing $\pi$ [resp. containing $\pi '$] vary in a ${\mathbb P}^2$, hence

$$\begin{aligned} V=\bigcup _{\pi \subset P\subset \Pi , \pi '\subset P'\subset \Pi '}\Pi _{P,P'} \end{aligned}$$

is a subvariety of ${\mathbb G}(k,r)$ which is isomorphic to the Segre variety $\textrm{Seg}(2,2)$. As in the odd case we see that ${\mathbb G}(k,r)$ contains a Veronese surface $X\subset {\mathbb P}^5$, that is a rational normal surface which is Grassman non-degenerate and not presenting the cone case.