Let $\tau:GL(n,\mathbb{R}) \rightarrow GL(V)$ be a rational representation of the general linear group of degree $n$ on a finite-dimensional real vector space $V$. Let $C$ be an irreducible real algebraic set in $V$ such that the action of $GL(n,\mathbb{R})$ on $V$ induced by $\tau$ leaves the set $C$ invariant.

We endow $C$ with the subspace topology inherited from (Hausdorf) Euclidean topology of $V$. Supose there is a point $v\in C$ such that the orbit $GL(n,\mathbb{R})⋅v$ is an open subset of $C$.

My questions is: is it true that the orbit $GL(n,\mathbb{R})⋅v$ must be a Zariski-open subset of $C$?


1 Answer 1


No, not at all. Take for $V$ the space of quadratic forms on $\mathbb R^n$, let $v=\sum_ix_i^2$ be the standard form, and $C=V$. Then $GL(n,\mathbb R)v$ is the set of positive definite forms. So, it is Hausdorff open but not Zariski open. NB: This example even works for $n=1$.


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