On secant varieties of Compact Hermitian Symmetric Spaces

Introduction

If the ideal of a variety $X$ is generated in degree two, the minimal possible degree of generators for the ideal of $\sigma(X)$ is three ( Cor. 3.2), although in general one does not expect generators in degree three (e.g. this almost always fails for complete intersections of quadrics). On the other hand, when $X$ is homogeneous, i.e., $V$ is an irreducible $G$-module where $G$ is a a semisimple algebraic group and $X$ is the orbit of a highest weight line (so in particular, the ideal of $X$ is generated in degree two), in all previously known examples (mostly just the rank two compact Hermitian symmetric spaces), the ideal of $\sigma(X)$ was generated in degree three.

In this paper we determine the generators of the ideals of the secant varieties of rank three compact Hermitian symmetric spaces in their minimal homogeneous embeddings, which we abbreviate CHSS. There is one suprise, the ideal of the secant variety of the $D_{7}$ spinor variety is not generated in degree three, which answers a question posed in , Section 3. Recently, L. Manivel has made significant progress towards determining the generators of the ideals of secant varieties of spinor varieties in general, see .

While determining the generators of the ideals of secant varieties of higher rank CHSS seems out of reach at the moment, we show that for all other CHSS other than spinor varieties, there are indeed generators in degree three. Moreover

For a vector space $A$, let $K_{S}(A)\subset A^{{\mathord{\,\otimes}\,}3}$ denote the kernel of the symmetrization map $S^{2}A{\mathord{\,\otimes}\,}A{\mathord{\;\rightarrow\;}}S^{3}A$, it is a $GL(A)$-module isomorphic to $S_{21}A$. We let $\pi_{S}:(A{\mathord{\,\otimes}\,}B)^{{\mathord{\,\otimes}\,}3}{\mathord{\;\rightarrow\;}}(A{\mathord{\,\otimes}\,}B)^{{\mathord{\,\otimes}\,}3}$ denote the symmetrization map whose image is $S^{3}(A{\mathord{\,\otimes}\,}B)$.

The degree three statement in Theorem 1.4 is a consequence of the more general result:

Theorem 1.4 and Proposition 1.5 are proven in §7.

For higher rank irreducible CHSS we have the following result, which is proved in §3:

In it is shown that $D_{7}/P_{7},D_{8}/P_{8}$ are the only spinor varieties whose secant varieties have empty ideal in degree three, thus these are the only CHSS in not having cubics in the ideal of its secant variety. (Segre products and Veronese re-embeddings of any varieties contain cubics in the ideals of their secant varieties as there are cubics in the ideals of the Segre products and Veronese embeddings of projective spaces.)

We obtain our results using the methods of , as described in Theorem 2.1 below, along with some new results about induced representations. In brief, in each case we obtain a desingularization of $\sigma(X)$, by exploiting that fact that each $X$ has a Legendrian “smaller cousin”, and apply Weyman’s method to this desingularization.

When dealing with $\mathfrak{a}_{n}$-modules we sometimes use partitions to index highest weights, with the dictionary $\pi=(p_{1},...,p_{n+1})$ corresponds to the weight $(p_{1}-p_{2})\omega_{1}+(p_{2}-p_{3})\omega_{2}+\cdots+(p_{n}-p_{n+1})\omega_{n}$. We write $S_{\pi}K^{n}$ for the associated module. Sometimes we abbreviate a partition $(i_{1},...,i_{1},i_{2},...,i_{2},...,i_{k},...,i_{k})=((i_{1})^{a_{1}},(i_{2})^{a_{2}},...,(i_{k})^{a_{k}})$ where $i_{s}$ occurs $a_{s}$ times.

Acknowledgments

We thank W. Kraśkiewicz and C. Robles for significant help with computer calculations and L. Manivel for pointing out and correcting an error in an earlier version of this paper.

Method of proof

If the sheaf cohomology groups $H^{i}({\mathcal{B}},S^{d}\eta)$ are all zero for $i>0$ and $d>0$ and if the linear maps $H^{0}({\mathcal{B}},S^{d}\eta){\mathord{\,\otimes}\,}V^{*}{\mathord{\;\rightarrow\;}}H^{0}({\mathcal{B}},S^{d+1}\eta)$ are surjective for all $d\geq 0$, then

$\hat{Y}$ is normal, with rational singularities

The coordinate ring $K[\hat{Y}]$ satisfies $K[\hat{Y}]_{d}\simeq H^{0}({\mathcal{B}},S^{d}\eta)$.

The vector space of minimal generators of the ideal of $Y$ in degree $d$ is isomorphic to $H^{d-1}({\mathcal{B}},\Lambda^{d}\xi)$ which is also the homology of the complex

More generally, $\oplus_{j}H^{j}({\mathcal{B}},\Lambda^{i+j}\xi)$ is isomorphic to the $i$-th term in the minimal free resolution of $Y$.

If moreover $Y$ is a $G$-variety and the desingularization is $G$-equivariant, then the identifications above are as $G$-modules.

2. The basic theorem applies in our case

Notations as above, if $Y$ is a $G$ variety, ${\mathcal{B}}=G/P$ and $\eta$ is induced from an irreducible $P$-module, then the sheaf cohomology groups $H^{i}({\mathcal{B}},S^{d}\eta)$ are all zero for $i>0$ and the linear maps $H^{0}({\mathcal{B}},S^{d}\eta){\mathord{\,\otimes}\,}V^{*}{\mathord{\;\rightarrow\;}}H^{0}({\mathcal{B}},S^{d+1}\eta)$ are surjective for all $d\geq 0$. In particular all the conclusions of 2.1 apply.

An irreducible homogeneous bundle can have nonzero cohomology in at most one degree, but a quotient bundle of a trivial bundle has nonzero sections, thus $H^{0}(B,\eta)$ is a nonzero irreducible module and all other $H^{j}(B,\eta)$ are zero. Let ${\mathfrak{f}}\subset{\mathfrak{p}}\subset{\mathfrak{g}}$ be a semi-simple Levi factor, so the weight lattice of ${\mathfrak{f}}$ is a sublattice of the weight lattice of ${\mathfrak{g}}$, let ${\mathfrak{t}}^{c}$ denote the complement of ${\mathfrak{t}}_{{\mathfrak{f}}}$ (the torus of ${\mathfrak{f}}$) in ${\mathfrak{t}}_{{\mathfrak{g}}}$ and let ${\mathfrak{g}}_{0}={\mathfrak{f}}+{\mathfrak{t}}^{c}$ denote the Levi factor of ${\mathfrak{p}}$. $\eta$ is induced from an irreducible ${\mathfrak{g}}_{0}$-module $U$ which is a weight space for ${\mathfrak{t}}^{c}$ having non-negative weight, say $(w_{1},...,w_{p})$. The bundle $\eta^{{\mathord{\,\otimes}\,}d}$, corresponds to a module which is $U^{{\mathord{\,\otimes}\,}d}$ as an ${\mathfrak{f}}$-module and is a weight space with weight $(dw_{1},...,dw_{p})$ for the action of ${\mathfrak{t}}^{c}$. Thus $S^{d}\eta$ is completely reducible and each component of $S^{d}\eta$ is very ample and in particular acyclic.

To prove the second assertion, consider the maps $V^{*}{\mathord{\,\otimes}\,}H^{0}(B,S^{r-1}\eta){\mathord{\;\rightarrow\;}}H^{0}(B,S^{r}\eta)$. Note that $H^{0}(B,S^{j}\eta)\subset S^{j}V^{*}$. The proof of Proposition 2.2 will be completed by Lemma 2.3 below applied to $U$ and each irreducible component of $H^{0}(B,S^{r}\eta)$.∎

Let $M^{{\mathfrak{g}}}_{{\mathfrak{g}}_{0}}$ denote the sub-category of the category of ${\mathfrak{g}}_{0}$-modules generated under direct sum by the irreducible ${\mathfrak{g}}_{0}$ modules with highest weight in $\Lambda^{+}_{{\mathfrak{g}}}\subset\Lambda^{+}_{{\mathfrak{g}}_{0}}$ and note that it is closed under tensor product. Let $M_{{\mathfrak{g}}}$ denote the category of ${\mathfrak{g}}$-modules. Define an additive functor $\mathcal{F}:M^{{\mathfrak{g}}}_{{\mathfrak{g}}_{0}}{\mathord{\;\rightarrow\;}}M_{{\mathfrak{g}}}$ which takes an irreducible ${\mathfrak{g}}_{0}$-module with highest weight $\lambda$ to the corresponding irreducible ${\mathfrak{g}}$-module with highest weight $\lambda$.

Let ${\mathfrak{l}}\subset{\mathfrak{g}}$ and ${\mathcal{F}}$ be as above. Let $U,W$ be irreducible ${\mathfrak{l}}$-modules Then

Let $N\subset P$ denote the unipotent radical of $P$. Any $L$-module $W$ may be considered as a $P$-module where $N$ acts trivially. Saying $V=\mathcal{F}(W)$ means that $V$ is the $G$-module parabolically induced from $W$ and $W$ is the set of $N$-invariants of $V$. The $N$-invariants of $\mathcal{F}(U){\mathord{\,\otimes}\,}\mathcal{F}(W)$ contain $U{\mathord{\,\otimes}\,}W$. ∎

Proof of Proposition 1.6

Let $G=SL(n,K)$ and $W=\Lambda^{k}K^{m}=V_{\omega_{k}}$. It follows from the Pieri formulas that the modules $V_{2\omega_{k-2}+\omega_{k+4}}$ and $V_{\omega_{k-4}+2\omega_{k+2}}$ do not occur in $V_{2\omega_{k}}{\mathord{\,\otimes}\,}V_{\omega_{k}}$. To see that they occur in $S^{3}W$, identify $V_{\omega_{k}}$ with $\Lambda^{k}(K^{n})$ where $K^{n}$ has a basis $\{e_{1},\ldots,e_{n}\}$. First observe that $\Lambda^{6}(K^{6})\subset S^{3}(\Lambda^{2}(K^{6}))$, in fact if $(f_{1},...,f_{6})$ is a basis of $K^{6}$, then the inclusion takes it to the Pfaffian

where we sum over all permutations $\sigma\in{\mathfrak{S}}_{6}$ satisfying

Now considering $K^{6}\subset K^{n}$ as the span of $\{e_{k-1},\ldots,e_{k+4}\}$ we can produce a highest weight vector of $V_{2\omega_{k-2}+\omega_{k+4}}$ by wedging each term $f_{i}\wedge f_{j}=e_{i+k-2}\wedge e_{i+k-2}$ in the summation with $e_{1}\wedge\ldots\wedge e_{k-2}$. We leave it to the reader to check the resulting vector has the desired properties. The module $V_{\omega_{k-4}+2\omega_{k+2}}$ occurs in $S^{3}W$ as well by symmetry (or one can define an analogous map).

Desingularizations for secant varieties of Rank 3 CHSS

In our situation the desingularizations are based on the observation that in each case $X$ is swept out by the union of Legendrian varieties $X_{small}$ and $\sigma(X)$ is the union of the $\sigma(X_{small})$’s which are linear spaces.

Here is a table of $X,X_{small},{\mathcal{B}}$ and the desingularizing bundle $E$ over $B$:

Lemma 4.1 combined with Theorem 2.1 and Proposition 2.2 prove Theorem 1.1.

We now proceed with a case by case study.

Case of X=G​(3,W)𝑋𝐺3𝑊X=G(3,W)

$\hat{\sigma}(G(3,W^{*}))=R_{6}(\Lambda^{3}W^{*})$.

A general point of $\hat{\sigma}(G(3,W^{*}))$ is of the form $v_{1}\wedge v_{2}\wedge v_{3}+v_{4}\wedge v_{5}\wedge v_{6}$, so $\hat{\sigma}(G(3,W^{*}))\subseteq R_{6}(\Lambda^{3}W^{*})$ because the latter is compact and the former connected. But both varieties are of the same dimension $6(\text{dim}\,W)-17$ and are reduced and irreducible so they must be equal. ∎

The last module corresponds to a partition of length seven and thus by the remark above, it is among the generators of $I(R_{6}(\Lambda^{3}W^{*}))$ (because the ideal in degree two of any secant variety is empty), and the rest are not as their partitions have length at most six.

To show there are no generators in degree greater than three, we need to prove exactness in the middle step of (1) which in this case is:

The largest partition that can show up in the middle has length nine, so once we have solved the problem for $G(3,K^{9})$ we are done.

Thus one could proceed calculate $H^{d}(G(6,K^{9}),\Lambda^{d+1}\xi)$ with the aid of a computer to conclude (although the passage from the cohomology of $\Lambda^{d+1}gr(\xi)$ to $\Lambda^{d+1}\xi$ might require some effort). We will proceed differently, resolving the cases of $\text{dim}\,W=7,8,9$ iteratively using rank varieties with $p=\text{dim}\,W-1$.

For $\text{dim}\,W=7$, the method in , §7.3 shows that the ideal of $R_{6}(\Lambda^{3}W)$ is generated by $S_{31^{6}}W$ and we are done. For the next two cases we proceed indirectly, calculating the ideal of $R_{7}(\Lambda^{3}K^{8})$ (resp. $R_{8}(\Lambda^{3}K^{9})$), and show these are in the ideal generated by $S_{31^{6}}W$ to complete the proof.

The ideal of the rank variety $R_{6}(\Lambda^{3}K^{7*})$ is generated in degree three by $S_{3,1^{6}}K^{7}$ included in $S^{3}(\Lambda^{3}K^{7})$ as described in the recipe in the proof.

The ideal of the rank variety $R_{7}(\Lambda^{3}K^{8*})$ is generated in degree five by $S_{3,2^{2},1^{5}}K^{8}$ included in $S^{5}(\Lambda^{3}K^{8})$ as described in the recipe in the proof.

The ideal of the rank variety $R_{8}(\Lambda^{3}K^{9*})$ is generated in degrees four and five by $S_{4,1^{8}}K^{9}$ and $S_{3^{2},2^{2},1^{5}}K^{9}$ respectively included in $S^{4}(\Lambda^{3}K^{9})$ and $S^{5}(\Lambda^{3}K^{9})$ as described in the recipe in the proof.

Thanks to the irreducibility of $\xi$ and its exterior powers, determination of modules generating the ideal is a straightforward application of the methods of and is left to the reader. It remains to show the above modules are all in the ideal generated by $S_{31^{6}}W$. To do this we give explicit descriptions of the modules as spaces of polynomials.

We will encode the representations occurring in the $d$-th symmetric powers of $\Lambda^{3}W$ by Young tableaux of shape $\pi=\lambda(D)$ with $3d$ boxes, filled with the numbers $1,\ldots,d$ with each number occurring three times. These tableaux are also assumed to be weakly increasing in rows and strictly increasing in columns. We associate to such tableau $D$ the map $\rho(D)$

where $\pi^{\prime}=(d_{1}^{\prime},...,d_{r}^{\prime})$ is the conjugate partition to $\pi$.

The map $\rho(D)$ is defined as the composition of the following maps:

a) assuming there are $e_{i,s}$ boxes filled with $s$ in the $i$-th row, apply the embedding

b) Noting that for each $s$, $e_{1,s}+\cdots+e_{r,s}=3$, wedge the factors coming from different rows corresponding to the same number $s$ in $D$, i.e., after rearranging the factors define the projection to $(\Lambda^{3}W)^{{\mathord{\,\otimes}\,}d}$ by sending, for each $s$,

c) Project $(\Lambda^{3}W)^{{\mathord{\,\otimes}\,}d}{\mathord{\;\rightarrow\;}}S^{d}(\Lambda^{3}W)$ by symmetrizing.

We call the Young diagram $D$ the numbering scheme associated to the map $\rho(D)$. Write $\lambda(D)$ for the weight of the numbering scheme whose $i$-th entry is the length of the $i$-th column of the Young diagram of $D$ - this will be the highest weight of the associated module.

The four Schur functors mentioned in the statement of the lemma correspond to four numbering schemes.

It is clear that the images of the corresponding maps are in the ideals of corresponding rank varieties because of the length of the first row of each numbering scheme.

Decomposing the domain and range of $\rho(D_{i})$ into irreducible representations, in all four cases $S_{\lambda(D_{i})}W$ is the only Schur functor occurring in both the domain and range of $\rho(D_{i})$, and it occurs there with multiplicity one. Thus it only remains to see the maps $\rho(D_{j})$ are nonzero, which is the purpose of the following lemma.

The numbering schemes $D_{1},D_{2},D_{3},D_{4}$ all yield nonzero modules.

is nonzero in $S_{4}(\Lambda^{3}W)$. The other cases are similar.

Consider the contribution to the monomial $(e_{1}\wedge e_{2}\wedge e_{3})(e_{1}\wedge e_{2}\wedge e_{3})(e_{1}\wedge e_{4}\wedge e_{5})(e_{6}\wedge e_{7}\wedge e_{8})$ in the image of highest weight vector $\rho(D_{2})(e_{1}\wedge e_{2}\wedge e_{3}\wedge e_{4}\wedge e_{5}\wedge e_{6}\wedge e_{7}\wedge e_{8}){\mathord{\,\otimes}\,}(e_{1}\wedge e_{2}\wedge e_{3}){\mathord{\,\otimes}\,}(e_{1})$. All occurrences of this monomial can be divided to 24 classes (corresponding to permutations of $\{1,2,3,4\}$) according to the order in which the factors appear in $(\Lambda^{3}W)^{{\mathord{\,\otimes}\,}d}$ after applying parts a) and b) of the definition of $\rho(D_{2})$. In fact only two classes out of 24 are non-empty. The factor $e_{6}\wedge e_{7}\wedge e_{8}$ has to come from the first factor, and one of the factors $e_{1}\wedge e_{2}\wedge e_{3}$ has to come from the fourth factor. The factor $e_{1}\wedge e_{4}\wedge e_{5}$ can come from the third factor (and this gives contribution $3$ to the coefficient) or from the second factor (and this gives contribution $1$ to the coefficient). Thus the coefficient is nonzero and and therefore $\rho(D_{2})\neq 0$.

To finish the proof of Theorem 1.2 we need to show that the ideal generated by the first module contains the other modules. But this is clear by the definition of maps $\rho(D)$ and by the last part of the proof of Proposition 5.2 as the other numbering schemes all contain the first.

Note that $V_{\omega_{7}}^{D_{7}}$ decomposes to $V_{\omega_{6}}^{D_{6}}{\mathord{\,\oplus}\,}V_{\omega_{5}}^{D_{6}}$ as a $D_{6}$ module, and this splitting gives rise to the the bundles $\xi$ and $\eta$ over $Q^{12}$. Thus they are both irreducible and dual to one another. In this case it is straightforward to calculate $H^{j}(\Lambda^{j+1}\xi)$ if one knows the decomposition of $\Lambda^{j+1}V_{\omega_{6}}^{D_{6}}$. In fact we calculated the entire minimal free resolution which is available at http://www.math.neu.edu/$\sim$weyman/mathindex.html for the interested reader. In particular the only generator of the ideal is the module $V_{\omega_{4}}$ as stated in the theorem.

All spaces discussed in this section are to be considered as linear subspaces of $(A{\mathord{\,\otimes}\,}B)^{{\mathord{\,\otimes}\,}3}=A^{{\mathord{\,\otimes}\,}3}{\mathord{\,\otimes}\,}B^{{\mathord{\,\otimes}\,}3}$ and all evaluations are as multi-linear forms. In particular, the symmetrization map

realizes $S^{3}(A{\mathord{\,\otimes}\,}B)\subset(A{\mathord{\,\otimes}\,}B)^{{\mathord{\,\otimes}\,}3}$ as $\pi_{S}((A{\mathord{\,\otimes}\,}B)^{{\mathord{\,\otimes}\,}3})$. Similarly we regard $S^{2}A{\mathord{\,\otimes}\,}A\subset A^{{\mathord{\,\otimes}\,}3}$ as the image of the symmetrization map $x{\mathord{\,\otimes}\,}y{\mathord{\,\otimes}\,}z\mapsto\frac{1}{2}(x{\mathord{\,\otimes}\,}y{\mathord{\,\otimes}\,}z+y{\mathord{\,\otimes}\,}z{\mathord{\,\otimes}\,}z)$ and likewise for $S^{2}B{\mathord{\,\otimes}\,}B\subset B^{{\mathord{\,\otimes}\,}3}$.

where $K_{S}(A)$ is the kernel of the map $S^{2}A{\mathord{\,\otimes}\,}A{\mathord{\;\rightarrow\;}}S^{3}A$, which is a $GL(A)$-module isomorphic to $S_{21}A$. In particular, if $R\in K_{S}(A)$, we have $R(u,v,w)=R(v,u,w)$ for all $u,v,w\in A^{*}$ and

which holds because $R(u,v,w)+R(u,w,v)+R(v,w,u)+R(v,w,u)+R(w,u,v)+R(w,v,u)=0$, and setting $w=u$ gives $2R(u,v,u)+2R(u,u,v)+2R(v,u,u)=4R(u,v,u)+2R(u,u,v)=0$.

By definition $S_{1}(Y)=\{T\in I_{2}(Y){\mathord{\,\otimes}\,}A\mid\pi_{S,A}(T)=0\}$, where

Elements of $S_{1}(Y)\subset K_{S}(A)$ have the property that as tri-linear forms they vanish on any triple of the form $(v,v,w)$ with $[v]\in Y$ and $w$ arbitrary.

Any element of $K_{S}(A){\mathord{\,\otimes}\,}K_{S}(B)\subset(A{\mathord{\,\otimes}\,}B)^{{\mathord{\,\otimes}\,}3}$ may be written as a sum $\sum R_{i}{\mathord{\,\otimes}\,}T_{i}$ with $R_{i}\in K_{S}(A)$ and $T_{i}\in K_{S}(B)$, thus any element of $\pi_{S}(K_{S}(A){\mathord{\,\otimes}\,}K_{S}(B))$ is of the form $P=\pi_{S}(\sum R_{i}{\mathord{\,\otimes}\,}T_{i})$ with $R_{i}\in K_{S}(A)$ and $T_{i}\in K_{S}(B)$. We compute

The first equality holds because of the six permutations in ${\mathfrak{S}}_{3}$, only three yield different elements, the second because $R_{i}\in S^{2}A{\mathord{\,\otimes}\,}A$ and $T_{i}\in S^{2}B{\mathord{\,\otimes}\,}B$, and the third by (2).

which is zero because $R_{i,\alpha}\in I_{2}(Y)$ for all $i,\alpha$. Similarly $\pi_{S}(K_{S}(A){\mathord{\,\otimes}\,}S_{1}(Z))\subset I_{3}(\sigma(Seg(Y\times Z))$.

Now say $P=\pi_{S}(\sum R_{i}{\mathord{\,\otimes}\,}T_{i})\in(K_{S}(A){\mathord{\,\otimes}\,}K_{S}(B))\cap I_{3}(\sigma(Seg(Y\times Z))$. Without loss of generality we assume the $R_{i}$ are linearly independent modulo $S_{1}(Y)$ and the $T_{i}$ are linearly independent modulo $S_{1}(Z)$. Fix $(y,z)\in U$, we obtain a linear equation

where $c_{i,y,z}=T_{i}(y,y,z)$ and if $y,z$ are chosen generically all the coefficients are nonzero because we are working mod $S_{1}(Z)$. Note that the index range for $i$ is at most from $1$ to $\text{ min }\{\text{dim}\,K_{S}(A),\text{dim}\,K_{S}(B)\}$. We will show each $R_{i}(v,v,w)$ must be zero for all $v,w\in\hat{Y}$. Since $\hat{Y}$ spans $A$ and the expression is linear in $w$, we can have this hold for all $v\in\hat{Y}$ and $w\in A$, but this in turn implies that each $R_{i}\in S_{1}(Y)$.

To obtain the desired vanishing, fix $v,w$ and consider the $R_{i}(v,v,w)=r_{i}$ as constants. We have an equation

As remarked above, since $Z$ is linearly non-degenerate, we may choose $\text{dim}\,B$ elements $z_{s}\in\hat{Z}$ that give a basis of $B^{*}$. Similarly, we may choose $\binom{\text{dim}\,B+1}{2}-\text{dim}\,I_{2}(Z)$ elements $y_{t}\in\hat{Z}$ such that the vectors $y_{t}^{2}$ span $I_{2}(Z){}^{\perp}\subset S^{2}B^{*}$. Thus the vectors $y_{t}^{2}{\mathord{\,\otimes}\,}z_{s}$ give a basis of $I_{2}(Z){}^{\perp}{\mathord{\,\otimes}\,}B^{*}$. Thus the pairing with elements of $K_{S}(B)/S_{1}(Z)$ is perfect, which implies that the matrix given by pairing the $y_{t}^{2}{\mathord{\,\otimes}\,}z_{s}$ with the $T_{i}$ has a one-sided inverse, so we have enough independent equations to force all the $r_{i}$ to vanish.

The argument for the $S^{3}A{\mathord{\,\otimes}\,}S^{3}B$ factor is similar, but easier, as there is no need to symmetrize. Write $P=\sum R_{i}{\mathord{\,\otimes}\,}T_{i}$ with $R_{i}\in S^{3}A$ and $T_{i}\in S^{3}B$.

Now $R_{i}\in I_{3}(\sigma(Y))$ iff $R_{i}(v,v,w)=0$ for all $v,w\in\hat{Y}$ and one concludes as above. ∎

We now show there are no new generators in degrees greater than three.

1. Case Y=G​(2,B)𝑌𝐺2𝐵Y=G(2,B)

We need to study the exactness of the middle step of

Here by $\mid_{|\pi|\leq 4}$, we mean the components of $S_{a,b}(\Lambda^{2}B)$ that occur in the decomposition of $S^{p}(A{\mathord{\,\otimes}\,}\Lambda^{2}B)=\oplus_{a+b=p}S_{a,b}A{\mathord{\,\otimes}\,}S_{a,b}(S_{1,1}B)$ that as partitions $S_{\pi}B$ have length at most four.

Let $B$ have dimension $n$ and consider the rank variety

Over ${\mathcal{B}}:=A^{*}{\mathord{\,\otimes}\,}G(n-1,B^{*})$ we have the bundle with fiber $A^{*}{\mathord{\,\otimes}\,}\Lambda^{2}{\mathcal{S}}$ which provides a desingularization of $R_{n-1}(A^{*}{\mathord{\,\otimes}\,}\Lambda^{2}B^{*})$. Our corresponding bundles are $\eta=A{\mathord{\,\otimes}\,}\Lambda^{2}{\mathcal{S}}^{*}$ and $\xi=A{\mathord{\,\otimes}\,}(\Lambda^{2}B/\Lambda^{2}{\mathcal{S}})=A{\mathord{\,\otimes}\,}{\mathcal{S}}^{*}{\mathord{\,\otimes}\,}{\mathcal{Q}}^{*}$. Note that $\text{rank}\,\xi=2(n-1)$.

where $\pi^{\prime}$ denotes the conjugate partition to $\pi$, so if $\pi=(a,b)$, then $\pi^{\prime}=(2^{a},1^{b})$.

We apply the Bott algorithm (see e.g. , §4.1.5) to the weight $\mu=\omega_{a}+\omega_{a+b}-(2a+b)\omega_{n}$ The only potential way to have non-zero cohomology is at step $(n-1)-(a+b)$, or at step $(n-1)-a$ or step $n-1$.

To obtain nonzero $H^{(n-1)-(a+b)}({\mathcal{B}},\Lambda^{n-(a+b)}\xi)$, $-(2a+b)+(n-1-(a+b))$ must be negative and $-(2a+b)+(n-1-(a+b))+2$ must be non-negative, so we must have $3a+2b=n-1$. Write $i=2a+b$, so $H^{i}({\mathcal{B}},\Lambda^{i+1}\xi)=S_{n-i,2i-n}A{\mathord{\,\otimes}\,}S_{2^{2i-n},1^{2n-2i}}B$.

To obtain nonzero $H^{(n-1)-a}({\mathcal{B}},\Lambda^{n-a}\xi)$, $-(2a+b)+(n-1-a)$ must be negative and $-(2a+b)+(n-1-a)+2$ must be non-negative, implying $3a+b=n+1$, contradicting $a+b\leq n-1$.

To obtain nonzero $H^{n-1}({\mathcal{B}},\Lambda^{n}\xi)$, we would have to have $-(2a+b)+(n-1)<-1$ and $-(2a+b)+(n-1)+2$ non-negative, implying $2a+b=n+2$ contradicting $a+b\leq n-1$.

Let $B,A$ be vector spaces respectively of dimensions $n,2$ and consider the rank variety

Then the ideal of $R_{n-1}$ is generated in degrees $\lceil\frac{n}{2}\rceil\leq d\leq\lfloor\frac{2n}{3}\rfloor$ by the modules

Write ${\mathcal{B}}=G/P_{i_{0}}$. Following the conventions of , $i_{0}=4,6$ respectively. We write the Levi factor of ${\mathfrak{p}}$ as ${\mathfrak{g}}_{0}={\mathfrak{f}}+\langle Z_{i_{0}}\rangle$, where ${\mathfrak{f}}$ is semi-simple (respectively $\mathfrak{a}_{3}+\mathfrak{a}_{1}$ and $\mathfrak{d}_{5}$) and $\langle Z_{i_{0}}\rangle$ is the center of ${\mathfrak{g}}_{0}$.

We will obtain the result by computing $H^{i}({\mathcal{B}},\Lambda^{i+1}\xi)$ via $H^{i}({\mathcal{B}},\Lambda^{i+1}gr(\xi))$ and applying a result of Ottaviani and Rubei.

where $E_{\lambda}$ denotes the irreducible bundle corresponding to the ${\mathfrak{g}}_{0}$-module of highest weight $\lambda$.

We first compute the decomposition of the exterior powers of the ${\mathfrak{g}}_{0}$-module giving rise to $gr(\xi)$ as an ${\mathfrak{f}}$ module and then compute the action of $Z_{i_{0}}$ to determine the coefficient on $\omega_{i_{0}}$ for each irreducible ${\mathfrak{f}}$-module appearing.

The ${\mathfrak{f}}$-module decomposition is straightforward with the aid of LiE , keeping in mind that $\text{dim}\,A=2$:

One then uses LiE to decompose these GL(U)-modules as ${\mathfrak{f}}$-modules. Next to determine the weight on the marked node (i.e., the coefficient of $\omega_{i_{0}}$), one uses the grading element $Z_{i_{0}}\in{\mathfrak{t}}$ which has the property that $Z_{i_{0}}(\alpha_{j})=\delta_{i_{0},j}$. Thus if $\lambda=\sum_{i\neq i_{0}}\lambda^{i}\omega_{i}$ is an irreducible ${\mathfrak{f}}$-module appearing in $W_{\mu}^{{\mathord{\,\otimes}\,}k}$, where the $\omega_{i}$ are fundamental weights of ${\mathfrak{g}}$, to find the coefficient of $\omega_{i_{0}}$ of the ${\mathfrak{g}}_{0}$-module, one calculates

where $(c{}^{-1})$ denotes the inverse of the Cartan matrix. In both our cases $Z_{i_{0}}(\mu)=-\frac{1}{3}$.

Now one calculates $H^{j}({\mathcal{B}},\Lambda^{p}gr(\xi))$. In practice we first calculated $H^{p-1}({\mathcal{B}},\Lambda^{p}gr(\xi))$, and only if this was nonzero did we calculate the other $H^{j}({\mathcal{B}},\Lambda^{p}gr(\xi))$.

The coordinate ring of σ​(X)𝜎𝑋\sigma(X)

The following proposition is due to F. Zak (, p. 51):

We work with the affine variety $\hat{\sigma}(X)\subset V$.

Notations as above. Let $H=Stab(v_{\lambda}+v_{\mu})$. Then

In particular, the irreducible $G$-module $V_{\lambda}$ occurs in $K[G]^{H}$ with multiplicity equal to the number of $H$-fixed points in $V_{\lambda}^{*}$.

Since $G/H\subseteq\hat{\sigma}(X)$ we obtain an inclusion $K[\hat{\sigma}(X)]\subseteq K[G/H]$ by restricting functions on $\hat{\sigma}(X)$. By , Theorem 3, Chapter II, section 3, the coordinate ring of $G$ has a left-right decomposition (as a $(G-G)$-bimodule)

$X=G(k,W)$, with $k>2$. Here without loss of generality we may take $\text{dim}\,W\geq 2k$. Indeed, if $dimW<2k$, we may pass to the dual Grassmannian $G(\text{dim}\,W-k,W^{*})$. If $dim(W)>2k$, $\hat{\sigma}(X)$ is contained in the subspace variety $R_{2k}(\Lambda^{k}W)$ of tensors that can be written using $\leq 2k$ basis vectors. Let $W^{\prime}\subset W$ be a $2k$ dimensional subspace. Consider the subgroup $H^{\prime}\subset H$,

The quotient $SL_{K}(W)/H^{\prime}$ can be identified with the variety $Hom_{K}^{inj}(W^{\prime},W)$ of injective linear maps from $W^{\prime}$ to $W$. Since the complement of $Hom_{K}^{inj}(W^{\prime},W)$ in $Hom_{K}(W^{\prime},W)$ has codimension $\geq 2$ every regular function on $Hom_{K}^{inj}(W^{\prime},W)$ extends to $Hom_{K}(W^{\prime},W)$. This means that we have the equalities

Here in the last equality we may view $\lambda$ as a partition. Note that the last equality states that the module $(S_{\lambda}W)^{*}$ appears with multiplicity $\text{dim}\,S_{\lambda}W^{\prime}$.

This reduces the calculation of $(S_{\lambda}W)^{H}$ to the case $W=W^{\prime}$. Assuming now that $dim(W)=2k$, with basis $e_{1},...,e_{2k}$, we may take $v_{\lambda}+v_{\mu}=e_{1}\wedge\cdots\wedge e_{k}+e_{k+1}\wedge\cdots\wedge e_{2k}$. Then

Considering $G(k,W)=SL(W)/P_{k}$, it is clear that no fundamental representation other than $W_{\omega_{k}}=\Lambda^{k}W$ has an $H$-fixed vector and in $W_{\omega_{k}}$ there is a $2$-dimensional subspace of such spanned by $e_{1}\wedge\cdots\wedge e_{k}$ and $e_{k+1}\wedge\cdots\wedge e_{2k}$. The corresponding two copies of $\Lambda^{k}W$ generate the ring of invariants in the following sense. We claim:

Two generating fundamental representations are in bidegrees $(1,0)$ and $(0,1)$. If we work instead with $GL(W)$, since $H$ acts trivially on the determinant, we get

To see this, write $E=\langle e_{1},...,e_{k}\rangle,F=\langle e_{k+1},...,e_{2k}\rangle$, we want to see how many instances of the trivial representation of $SL(E)\times SL(F)$ occurs in the irreducible $SL(W)$ module $S_{\lambda}W$. Now, since $W=E{\mathord{\,\oplus}\,}F$

This means this ring is the homomorphic image of the symmetric algebra on two copies of $\Lambda^{k}W$ corresponding to components in bidegrees $(1,0)$ and $(0,1)$ which is our claim.

so we need $i_{1}+b_{1}=i_{2}+b_{2}=\cdots=i_{k}+b_{k}$ and $a_{1}-b_{1}-i_{1}=a_{2}-b_{2}-i_{2}=\cdots=a_{k}-b_{k}-i_{k}$. This means the vectors $e_{j}^{i_{j}+b_{j}}f_{j}^{a_{j}-i_{j}}$ have to be all of the same weight, for $j=1,\ldots,k$.

the dimension of the subspace of $H$-invariant vectors is ${\rm min}\ a_{j}-{\rm max}\ b_{j}+1$.

References