3
$\begingroup$

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 $h_1 \in \operatorname{GL}_{n+1}(\mathbf C)$ such that $h_1 \cdot f=g$. Does there exist $h_2 \in \operatorname{GL}_n(\mathbf C)$ 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 $\rho_n: \operatorname{GL}_n \to \operatorname{End}(V_n)$ 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 $\operatorname{GL}_{n+1}$-orbits of elements of $V_n$?

$\endgroup$
1
  • $\begingroup$ If $h_1^k$ (for some $k$) leaves the last coordinate invariant, then yes. $\endgroup$
    – Igor Rivin
    Sep 21, 2011 at 12:07

1 Answer 1

4
$\begingroup$

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 $0 \le dim\le n$).

$\endgroup$
1
  • $\begingroup$ 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) $\endgroup$ Sep 22, 2011 at 10:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.