MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $V_n$?

share|cite|improve this question
If $h_1^k$ (for some $k$) leaves the last coordinate invariant, then yes. – Igor Rivin Sep 21 '11 at 12:07
up vote 3 down vote accepted

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 $0 \le dim\le n$).

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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