GL(n+1)- vs. GL(n)-orbits

Let $f,g \in \mathbf {C}[x_1, \ldots, x_n]\subseteq \mathbf {C}[x_1, \ldots, x_{n+1}]$ be two polynomials with complex coefficients and suppose that there exists $h_1 \in \mathop{GL}_{n+1}(\mathbf C)$ such that $h_1 \cdot f=g$. Does there exist $h_2 \in \mathop{GL}_n(\mathbf C)$ such that $h_2\cdot f=g$?

This is a toy example of a more general question I'm interested in: given a family of representations $\rho_n: \mathop{GL}_n \to \mathop{End}(V_n)$ such that $V_n$ embeds naturally in $V_{n+1}$ (e.g., $V_n=\Lambda^r(\mathbf C^n)$), what can be said about the $\mathop{GL}_{n+1}$-orbits of elements of $V_n$?

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If $h_1^k$ (for some $k$) leaves the last coordinate invariant, then yes. – Igor Rivin Sep 21 '11 at 12:07

Yes, this is true. Let me consider the case when there is no linear change of coordinates $(x_1,...,x_n)$ such that $g=g(x_1,...,x_{n-1})$. In this case it is clear that the vector $v=h_1(0,...,1)$ does not belong to the plane $(x_{n+1}=0)\subset \mathbb C^{n+1}$. Hence we can find a linear transformation $h_3$ of $\mathbb C^{n+1}$ that lives invariant all lines in $\mathbb C^{n+1}$ parallel to $v$ and sends the plane $h_1((x_{n+1}=0))$ to $(x_{n+1}=0)$. This transformation leaves $h_1(f)$ invariant. Finally chose $h_2$ to be the restriction of $h_3\circ h_1$ to the plane $x_{n+1}=0$.

The general case is similar. We just need to play with the largest subgroups in the group of translations of $\mathbb C^n$ that leave $f$ and $g$ invariant (respectively) - such groups are unique (and have the same dimension $0 \le dim\le n$).

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Thank you for your answer. – Guntram Sep 22 '11 at 10:04
Guntram, you are welcome! (I think the main (simple) idea in the answer is that for each polynomial one can associate a canonical subgroup in $\mathbb C^n$ of parallel translations that leaves the polynomial invariant) – Dmitri Sep 22 '11 at 10:13