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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.

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1 Answer 1

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.

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You have shown that all monomials $x^J$ are in $G.W$, and this is interesting - but unfortunately, it does not say much about the dimension of $G.W$ as a subvariety of $V$: Note that $G$ does not act linearly on $V$, and $G.W$ is not a linear subspace (in general). In other words, showing that a basis is contained in it does not imply that it is equal to $V$. –  Jesko Hüttenhain Nov 25 '12 at 18:55
Thanks for the reply! Okay, (i) I could have sworn $G$ would act $k$ linearly on $V$, and based on this indeed (ii) I assumed (wrongly :p) that $G.W$ denoted the span of the orbit, not the orbit itself - I agree this in general isn't a subspace. In fact I showed that $x^J$ are in the span of the orbit, not the orbit itself $(\frac{1}{|A|} \sum_\chi \chi(-J) f(\chi, x)$ is the crucial linear combo) :(. Would you mind explaining (i)? Cheers –  uncookedfalcon Nov 25 '12 at 22:59
I just came back to correct my mistake, of course $G$ does act linearly, so that's not the problem here. I jumped the gun on that one because something cannot be quite right about your proof: I know there are certain polynomials that can not be made to require less variables under any linear transformation, namely the determinant. It's an old result by Frobenius. I will need to reed your answer more carefully. –  Jesko Hüttenhain Nov 25 '12 at 23:27
Alright. $G$ acts linearly, but that only means that $g\in G$ acts as a linear map. However, the action $x_0\mapsto g.x_0$ is not "linear" in $g$, it isn't even clear what that should mean. In other words, when $x^{J_1}=g_1.x_0$ and $x^{J_2}=g_2.x_0$, then what is $x^{J_1}+x^{J_2}$? –  Jesko Hüttenhain Nov 26 '12 at 16:04
Hey man - all I use is that the "action" is a group homomorphism $G \rightarrow Aut_k(k[x_i]_d)$; that is each $g$ acts linearly and composition goes to composition. I agree "linear" in $g$ doesn't make sense, I certainly do not use it! Let me stress again what I think the issue might be: there's two possible interpretations of $G.W$; (i) $$G.W = \{ g \cdot w : \forall g \in G, w \in W \}$$and (ii) is the vector space spanned by (i). I proved that (ii) is all of $V$, not (i). Does this address your concern? –  uncookedfalcon Nov 27 '12 at 0:05

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