dimension of invariant subspaces Let $V$ be a complex vector space, $G\subset SL_n(V)$ a finite group. Define a function $f:V\to \mathbb{Z}$ by $f(v)=\text{dim} V^{G_v}$, where $G_v$ is the stabiliser subgroup of $v$, i.e. $g\in G$ such that $gv=v$. And $V^{G_v}$ is the invariant subspace of $V$ under this subgroup.
My question is: Do we know if this function is lower-semicontinuous (or maybe upper?)
Does it have any semi-continuity if we consider the Zariski topology, i.e. treating $V$ as spectrum of a complex polynomial ring?
 A: You have lower semi-continuity in the Zariski topology and this holds more generally for the restriction of $f$ to any $G$-subvariety $Z\subseteq V$ and we will need this for an inductive proof on $n:=\dim(Z)$ anyway. Note that the claim is trivial for $n=0$.
Let $K:=\bigcap_{v\in Z} G_v$ be the subgroup that stabilizes every point of the variety $Z$. This is a normal subgroup because for any $g\in G$, $h\in K$ and any $v\in Z$, we have $ghg^{-1}.v=gg^{-1}.v=v$ because $h$ stabilizes $g^{-1}.v$ and therefore $ghg^{-1}\in K$. We get an induced action of $H:=G/K$ on $Z$.
The set $U:=\{ v\in Z \mid H_v=\{1\} \}=\{v\in Z\mid G_v=K\}$ is Zariski-open and dense in $Z$. Indeed, the quotient morphism $Z\to Z/\!/H$ is a finite morphism, hence its fibers are generically of cardinality $|H|$ because $H$ acts faithfully. Note that $f(u)=\dim(V^K)$ for all $u\in U$.
The open set $U$ is $G$-invariant and therefore, the complement $Y:=Z\setminus U$ is also $G$-invariant and $\dim(Y)<\dim(Z)$. We can therefore consider the action of $G$ on $Y$ and we know by induction that $f$ is lower semi-continuous on $Y$. Take any point $y\in Y$, then we know $K\subsetneq G_y$ by construction. Hence, $V^{G_y}\subseteq V^{K}$. This inclusion is strict because $U\subseteq V^K$ but $U\nsubseteq V^{G_y}$. Therefore,
$$f(y)=\dim(V^{G_y})<\dim(V^K).$$
It follows that the set $\{ v\in Z \mid f(v) < \dim(V^K) \}=Y$ is closed and the other level sets (contained in $Y$) are also closed by induction. 

One might note that lower semi-continuity in the Zariski topology implies lower semi-continuity in the classical topology because the latter is finer. Since all the level sets are closed in the Zariski-topology, they are in particular closed in the classical topology.
