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 $I(Sec\_k(V))\_{k+1} = H$ ? 



G-modules and ideals of secant varieties
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>


