Let $k$ be an algebraically closed field (I am mostly interested in $k=\mathbb{C}$ if that matters).
Let $G$ be an affine algebraic subgroup of $GL_n(k)$ (ie $G$ is in fact linear). Let $H$ be a normal closed subgroup of $G$. It is classical that $G/H$ is itself an affine algebraic group and thus admits a faithful representation in $GL_m(k)$ for some $m$.
My question concerns the relationship between $m$ and $n$. Based on the proof of this fact, I have the feeling that $m$ cannot be bounded by $n$ unless we know more about $H$:
- is there is simple family of example for which $n$ is fixed but $m$ is unbounded?
- if we know that $H$ (as a variety) can be defined by polynomial equations of degree at most $d$, then I think we should have a bound of the form $$m\leqslant{p\choose d}^2$$ where $1\leqslant d\leqslant p$ and $p\leqslant{n^2+d\choose d}$. Is this indeed the case and does anyone have a reference for this fact? Probably this bound is very naive.
Maybe it is useful to explain my intuition for the bound of the second question: in the classical proof, we consider the coordinate ring of $G$ and, consider the action of $G$ on itself via multiplication, find a $G$-invariant subspace of $k[G]$ that contains the generators of $H$. To show that this subspace has finite dimension, we observe that it is contained in the subspace generated by all monomials of degree at most $d$ (the coordinate ring is generated by the $n^2$ entries of the matrix). Hence the bound for $p$. Then we use Grassmannian to convert this subspace into a subspace of dimension $1$, therefore increasing the overall dimension, and this gives the bound on $m$ (it is square because we encode matrices of size $k$ in $k^2$).