G-modules and ideals of secant varieties

Question

Consider the action of $G = SL(n+1)$ on $\mathbb{P}^N$, and embed $\mathbb{P}^n$ in $\mathbb{P}^N$ via the degree two Veronese embedding. Let $V\subset\mathbb{P}^N$ be the corresponding Veronese variety. Then the ideal $I(Sec_k(V))$ of the $k$-secant variety of $V$ is a $G$-module.

Now, let $f_1,\dots, f_r$ be generators of $I(Sec_k(V))$ (therefore they are polynomials of degree $k+1$), take one of them say $f_1$. Then $g\cdot f_1\in I(Sec_k(V))_{k+1}$ (the degree $k+1$ part of $I(Sec_k(V))$) for any $g\in G$, and the linear span $H$ of $G\cdot f_1$ is a subspace of $I(Sec_k(V))_{k+1}$. Do we have that $I(Sec_k(V))_{k+1} = H$ ?

Answer 1

In the symmetric case the representation is not irreducible.

For instance, consider a $4\times 4$ symmetric matrix $Z^{+}$ with entries $z_{i,j}$. Then $\wedge^{2}Z^{+}$ is given by $$ \left(\begin{array}{cccccc} z_{0,0}z_{1,1}-z_{0,1}^2 & z_{0,0}z_{1,2}-z_{0,1}z_{0,2} & z_{0,0}z_{1,3}-z_{0,1}z_{0,3} & z_{0,1}z_{1,2}-z_{1,1}z_{0,2} & z_{0,1}z_{1,3}-z_{1,1}z_{0,2} & z_{0,2}z_{1,3}-z_{1,2}z_{0,3}\\ z_{0,0}z_{1,2}-z_{0,2}z_{0,1} & z_{0,0}z_{2,2}-z_{0,2}^2 & z_{0,0}z_{2,3}-z_{0,2}z_{0,3} & z_{0,1}z_{2,2}-z_{1,2}z_{0,2} & z_{0,1}z_{2,3}-z_{1,2}z_{0,3} & z_{0,2}z_{2,3}-z_{2,2}z_{0,3}\\ z_{0,0}z_{1,3}-z_{0,3}z_{0,1} & z_{0,0}z_{2,3}-z_{0,3}z_{0,2} & z_{0,0}z_{3,3}-z_{0,3}^2 & z_{0,1}z_{2,3}-z_{1,3}z_{0,2} & z_{0,1}z_{3,3}-z_{1,3}z_{0,3} & z_{0,2}z_{3,3}-z_{2,3}z_{0,3}\\ z_{0,1}z_{1,2}-z_{0,2}z_{1,1} & z_{0,1}z_{2,2}-z_{0,2}z_{1,2} & z_{0,1}z_{2,3}-z_{0,2}z_{1,3} & z_{1,1}z_{2,2}-z_{1,2}^2 & z_{1,1}z_{2,3}-z_{1,2}z_{1,3} & z_{1,2}z_{2,3}-z_{2,2}z_{1,3}\\ z_{0,1}z_{1,3}-z_{0,3}z_{1,1} & z_{0,1}z_{2,3}-z_{0,3}z_{1,2} & z_{0,1}z_{3,3}-z_{0,3}z_{1,3} & z_{1,1}z_{2,3}-z_{1,3}z_{1,2} & z_{1,1}z_{3,3}-z_{1,3}^2 & z_{1,2}z_{3,3}-z_{2,3}z_{1,3}\\ z_{0,2}z_{1,3}-z_{0,3}z_{1,2} & z_{0,2}z_{2,3}-z_{0,3}z_{2,2} & z_{0,2}z_{3,3}-z_{0,3}z_{2,3} & z_{1,2}z_{2,3}-z_{1,3}z_{2,2} & z_{1,2}z_{3,3}-z_{1,3}z_{2,3} & z_{2,2}z_{3,3}-z_{2,3}^2 \end{array}\right)$$ Let us interpret this matrix as a quadric on $\mathbb{P}(\bigwedge^2V)$. Fix homogeneous coordinates $[x_0:\dots:x_5]$ on $\mathbb{P}(\bigwedge^2V)$. Then we may write the quadric corresponding to $\wedge^{2}Z^{+}$ as $$Q_{\wedge^{2}Z^{+}}(x_0,\dots,x_5) = \sum_{0\leq i \leq 5}(\wedge^{2}Z^{+})_{(i,i)}x_i^2+2\sum_{0\leq i < j \leq 5}(\wedge^{2}Z^{+})_{(i,j)}x_ix_j$$ Note that the entries of $\wedge^{2}Z^{+}$ satisfy the relation $$(\wedge^{2}Z^{+})_{(0,5)}-(\wedge^{2}Z^{+})_{(1,4)}+(\wedge^{2}Z^{+})_{(2,3)} = z_{0,2}z_{1,3}-z_{1,2}z_{0,3}-z_{0,1}z_{2,3}+z_{1,2}z_{0,3}+z_{0,1}z_{2,3}-z_{1,3}z_{0,2}=0$$ and indeed $x_{0}x_5-x_{1}x_{4}+x_{2}x_{3}=0$ is the Pl\"ucker equation cutting out $\mathcal{G}(1,3)$ in $\mathbb{P}(\bigwedge^2V)$. The vector space $Sym^2(\bigwedge^2V)$ admits a decomposition $$Sym^2(\bigwedge^2V) = \left\langle x_{0}x_5-x_{1}x_{4}+x_{2}x_{3}\right\rangle\oplus H_{2}$$ into two irreducible representations of $SL(4)$, where $H_2\cong H^0(\mathcal{G}(1,3),\mathcal{O}_{\mathcal{G}(1,3)}(2))$ is the linear span of the $SL(4)$-orbit of $z_{2,2}z_{3,3}-z_{2,3}^2$ in $Sym^2(\bigwedge^2V)$.

More generally, there is always a splitting

$$Sym^2(\bigwedge^{k+1}V)\cong I(\mathcal{G}(k,n))_2\oplus H_{k+1}$$

where $I(\mathcal{G}(k,n))_2$ is the degree two part of the ideal of the Grassmannian of $k$-planes in $\mathbb{P}^n$ in its Plucker embedding.

You may find all the details for instance in Section 4 of the this paper:

https://arxiv.org/abs/1803.09161