Let $G$ be a group，$k$ a field, and $V$ be a finite dimensional $k$-representation of $G$，consider $dim_L(V_L^G)$ for any field extension $L/k$. When will this become independent on $L$ i.e equal to $dim_k(V^G)$? If there is a left Haar measure on $G$ with $vol(G)=1$ and then this is true (at least when we can define integral of some functions on $G$ by smooth condition) because we can regard fixed vectors as those in the image of $v \mapsto \int_G g.v \ dg$ and rank of linear operator does not change, for example when $G$ is profinite and $k$ is characteristic zero and large.
I am interested in the general case, for example rational representation of algebraic group, continuous representation of topolpgical groups (non-compact case), the positive char case, small $k$ e.g number field case and so on.
This is always true. One way to see this is to view $V^G$ as the kernel of the linear map $T:V\to \prod_G V$ which on the $g$th coordinate is given by $g-1$. Now we tensor this with $L$ to get a map $T_L:V_L\to L\otimes\prod_G V$. In general, $L\otimes \prod_G V$ is not the same as $\prod_G V_L$. However, there is a canonical injective linear map $L\otimes\prod_G V\to \prod_G V_L$, and the composition of $T_L$ with this injection is just the map $V_L\to\prod_G V_L$ given by $g-1$ on the $g$th coordinate. Thus the kernel of $T_L$ is $V_L^G$.
But tensoring with $L$ is exact, and in particular preserves kernels. So $\ker(T_L)=L\otimes \ker(T)$, which means $V_L^G=L\otimes V^G$. Thus $V_L^G$ has the same dimension over $L$ as $V^G$ has over $k$.