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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 $I(Sec_k(V))_{k+1} = H$ ?

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  • $\begingroup$ Hmm, I don't think so. Here $\mathbb{P}^N$ is the space of $(n+1) \times (n+1)$ symmetric matrices ($N = \binom{n+2}{2}$). The Veronese is the variety of rank one symmetric matrices. Its $k$th secant variety is the variety of rank $k$ symmetric matrices. It is defined by the vanishing of the $(k+1) \times (k+1)$ minors. If I understand correctly, you are asking whether the $G$-representation on the vector space of size $k+1$ minors of a symmetric matrix is an irreducible representation. But there is a subrepresentation spanned by the principal minors (same row and column indices). $\endgroup$
    – Zach Teitler
    Commented Nov 24, 2017 at 17:53
  • $\begingroup$ Thanks. Do you have a reference? If instead of considering symmetric matrices you consider general matrices it seems to me that the representation of $G\times G$ should be irreducible. $\endgroup$
    – user117617
    Commented Nov 24, 2017 at 18:38
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    $\begingroup$ Why do you write, "But there is a subrepresentation spanned by the principal minors (same row and column indices)"? When $k$ equals $1$, this span is not a subrepresentation. Note, in this case, the question is simply whether the degree $2$ graded piece of the homogeneous ideal of the Veronese variety is an irreducible representation, which it is. $\endgroup$
    – Jason Starr
    Commented Nov 24, 2017 at 18:41
  • $\begingroup$ @J_Cole I agree that with $G \times G$ acting on general matrices, it would be irreducible. $\endgroup$
    – Zach Teitler
    Commented Nov 24, 2017 at 19:18
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    $\begingroup$ @AbdelmalekAbdesselam. Presumably the OP wants the generators $(f_1,\dots,f_r)$ to be a minimal set of generators, which forces the elements to be homogeneous. As Zach Teitler states, the issue is whether or not the graded ideal in degree $k+1$ is an irreducible representation. A related question is Exercise 15.45, p. 230 of Fulton-Harris, "Representation Theory, A First Course." $\endgroup$
    – Jason Starr
    Commented Nov 24, 2017 at 20:02

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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

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