This question is related to one I asked previously. Sorry if some of the notation or discussion below is unwieldy or nonstandard. I am still trying to learn the relevant terminology, so it's likely that what follows could be expressed more clearly.
Consider the Veronese embedding of degree 2: $\nu_2:\mathbb{RP}^n\rightarrow\mathbb{RP}^{N-1}$, where $N=(n+1)(n+2)/2$ and $n\geq 2$. The tangent space to the Veronese embedding $V_n=\nu_2(\mathbb{RP}^n)$ naturally lies in the Grassmannian $\mathbb{G}(n,N-1)$ (isomorphic to the Grassmannian of $n+1$-dimensional linear subspaces of $\mathbb{R}^{N}$, which I would write $Gr(n+1,N$). This defines the Gauss map of $V_n$: $\gamma_n:V_n\rightarrow\mathbb{G}(n,N-1)$.
Let us say that a point of $\mathbb{G}(N-n-2,N-1)$ is "V-degenerate" if the corresponding $N-n-2$-plane in $\mathbb{RP}^{N-1}$ intersects every $n$-plane $\gamma_n(x)$ for $x\in V_n$. Equivalently, the corresponding $N-n-1$-dimensional linear subspace of $\mathbb{R}^{N}$ is linearly dependent with every $n+1$-dimensional linear subspace arising from $\gamma_n(x)$ for all $x\in V_n$. I am interested in the set of all V-degenerate points, particularly in the case $n=2$, where $N=6$ and $N-n-2=2$ again. By a computer algebra calculation, I believe the following is true:
Let $\gamma_2(V_2)$ be the image of the Gauss map of $V_2$. Then the set of all V-degenerate points is precisely $\gamma_2(V_2)\subset\mathbb{G}(2,5)$.
To see that this is true in one direction (that the desired set contains $\gamma_2(V_2)$) one can check that the following determinant is identically zero:
$$\det\begin{pmatrix} 2x & 0 & 0 & 2l & 0 & 0\\ 0 & 2y & 0 & 0 & 2m & 0\\ 0 & 0 & 2z & 0 & 0 & 2n\\ y & x & 0 & m & l & 0\\ 0 & z & y & 0 & n & m\\ z & 0 & x & n & 0 & l \end{pmatrix}.$$
The first three columns and the last three columns are the Jacobian matrix of the Veronese embedding which give us the points $\gamma_2\circ\nu_2([x:y:z])$ and $\gamma_2\circ\nu_2([l:m:n])$ (where $[x:y:z]$ and $[l:m:n]$ are homogeneous coordinates in $\mathbb{RP}^2$). The fact that this determinant vanishes identically means that if we fix $[l:m:n]$ and let $[x:y:z]$ vary, the projective 2-plane defined by the Gauss map at that point on $V_2$ is linearly dependent with the planes arising from the Gauss map for all other points $\nu_2([x:y:z])$.
How can I see that there can be no other V-degenerate points in $\mathbb{G}(2,5)$ without resorting to a Groebner basis computation with Grassmann-Plücker coordinates?
I have been looking at the book "Differential Geometry of Varieties with Degenerate Gauss Maps" by Akivis and Goldberg and it seems that this question might be related to the fact that the singular points on the cubic symmetroid (the dual variety to the Veronese surface) are precisely the points on a Veronese surface (sections 2.5.3 and 2.5.4). However, I can't seem to connect these facts to my question on my own yet. Any hints would be appreciated.
Finally, I am curious about the situation for $n\geq3$ as well.
Is there a similarly nice / explicit description of the V-degenerate points for higher $n$? Of course we still have an analogous determinant formula but is there a simple geometric picture?
