Dimension of the GL-orbit of d-forms in one less variable Let $V:=k[x_0,\ldots,x_n]_d$ be the $k$-vector space of homogeneous polynomials of degree $d$. Let $G:=\mathrm{Gl}(n+1,k)$ act on $V$ induced by the canonical action on the linear forms: For $g=(g_{ij})\in G$, we have 
$$g.x_j=\sum_{i=0}^n g_{ij} x_i.$$
Now consider the subspace $W:=k[x_1,\ldots,x_j]_d\subseteq V$ of $d$-forms in one less variable. What is the dimension of $G.W$? In particular, is $G.W$ Zariski-dense in $V$? I think this should be well-known, so if someone ran across this before, I'd be glad just to get some pointers to papers or textbooks on this. 
 A: This was a really interesting question, thanks for asking! Two remarks and a result:
(i) For anyone wanting to check that the 'natural induced action' is indeed an action, one can do so 'for free' as follows: for $\sigma: V \rightarrow V$ in $GL(V)$, view $\sigma$ as $$V \rightarrow Sym^1 V \subset Sym V$$The universal property of $Sym$ feeds you $\sigma': Sym V \rightarrow Sym V$, by construction it sends $Sym^n V \rightarrow Sym^n V$ (in the desired way), the uniqueness of extension shows $G \rightarrow Aut(Sym V)$ hence $G \rightarrow Aut(Sym^n V)$ is a homomorphism as desired.    
(ii) The answer to the question will depend on $k$. For example, for $k[x,y]_2$, $W = k[x]_2 $  I claim that $G.W = k[x,y]_2$ iff char $k \neq 2$. By considering the basis $x^2, xy, y^2$, permuting $x$ and $y$, and the identity yields $x^2, y^2$, it suffices to show that $xy \in G.W$ iff char $k \neq 2$. Indeed, if char $k \neq 2$, consider linear maps sends $x \mapsto  \pm y$, the difference of their action on $x^2$ is $\pm 4xy$, and we may divide out by 4. On the other hand, if $char \; k = 2$, then if $\sigma$ sends $x \mapsto ax + by$, then $\sigma^*$ sends $x^2 \mapsto (ax+by)^2 = a^2x^2 + b^2y^2$ as desired.
(iii) I claim that if $k$ is algebraically closed and characteristic 0, then for just $W = k[x_0]_d$ already we have $G.W = k[x_0, \ldots, x_n]_d$! 
To see this, we first need a lemma, essentially borrowed from Dirichlet's theorem.
Consider a power series $$f(x) = \sum_{I \in \mathbb{Z}^n} a_I x^I$$and fix $M = (m_1, \ldots, m_n) \in \mathbb{Z}^n$. I claim we can write $\sum_{I \equiv J \mod M} a_I x^I$ as a linear combination of $f( \zeta_{m_i}^{k_i} x_i)$, for any $J \in \mathbb{Z}^n$. To see this, we use character theory.
Write $A = \oplus_i \mathbb{Z}/m_i \mathbb{Z}$, $A^* = Hom_{Ab}(A, k^*)$, using $$Hom(\oplus_i \mathbb{Z}/m_i, k^*) \simeq \prod_i Hom( \mathbb{Z}/m_i, k^*)$$and that each term in the product is exactly given by $1 \mapsto \zeta_{m_i}^k$ for $k = 0 \ldots m_i - 1$ and $\zeta_{m_i}$ a primitive $m_i^{th}$ root of unity in $k^*$, we see that every element of $A^*$ is given by $e_i \mapsto \zeta_{m_i}^{k_i}$, for $k_i = 0 \ldots m_i$. Recall the orthogonality relation $$\frac{1}{|A|} \sum_{\chi \in A^*} \chi(-a + b) = \frac{1}{|A|}\sum_{\chi \in A^*} \chi(-a) \chi(b) = \delta_{ab}$$
Finally, write $$f(\chi, x) = \sum_{I \in \mathbb{Z}^n} a_I \chi(I) x^I$$where $\chi(I)$ is simply the composition $\mathbb{Z}^n \rightarrow A \rightarrow k^*$. Then from the orthogonality relation we have $$ \frac{1}{|A|} \sum_{\chi \in A^*} \chi(-J) f(\chi, x) = \sum_{I \in \mathbb{Z}^n} a_I  \frac{1}{|A|} \sum_{\chi \in A^*} \chi(-J + I) x^I$$$$ = \sum_I a_I \delta_{IJ} x^I = \sum_{I \equiv J \mod M} a_I x^I$$
This proves the lemma (since $f(\chi, x) = f( x_i \zeta_{m_i}^{k_i})$). 
With this, let's see that $G.k[x_0]_d = k[x_0, \ldots , x_n]_d$. Indeed, the trick is to consider any $\sigma \in GL_n(k)$ which sends $x_0$ to $\sum_i x_i$, and hence $\sigma^*$ sends $x_0^d$ to $(\sum x_i)^d = \sum_I c_I x^I$, where $c_I$ are multinomial coefficients and $x^I$ runs over the monomial basis for $k[x_i]_d$. With that, using the previous lemma, applied to $f = (\sum x_i)^d$, $M = (d, d, d, \ldots, d)$, and any fixed $J$, and observing that $f(\chi, x)$ is simply $\sigma^* \circ \eta^*$, where $\eta: x_i \mapsto \zeta_{m_i}^{k_i} x_i$, we have that a linear combination of terms in the orbit of $x_0^d$ sums to $c_J x^J$, hence $x^J$ is in $G.k[x_0]_d$ for each $J$, as desired. 
