Quotient space of algebraic group

Question

Let $H \subset G$ closed subgroup of an algebraic group. We want to prove the existence of the quotient $G/H$ which is a quasi-projective variety and homogeneous G-space.

We can find a vector $0 \ne y_0 = [v] \in \mathbb{P}^n$ such that $\forall h \in H: h\cdot y_0 = y_0$ and thus $H = \mathrm{Stab}_G(y_0) =G_{y_0}$ (stabilizer) and $G/H =$ the orbit of $y_0$ under $G$.

We define a group action G-morphism $\varphi: G \to Y$ by $g \mapsto g\cdot y_0$. We want to prove that the differential $(d \varphi)_e$ is separable and surjective.

The tangent space of an alebraic group is a derivative module $\mathrm{Der}_k(k[G],k)$ where $D \in \mathrm{Der}_k(k[G],k)$ is a k-linear derivation (i.e. $\forall x \in k: D(x)=Dx = 0$). We also define a derivation $\delta \in \mathrm{Der}_k(k[G],k[G])$ which returns a derivative function. We call $$\mathcal{L}(G) = \left( \{ \delta \in \mathrm{Der}_k(k[G],k[G]) | \delta \mbox{ is left invariant}, \delta \lambda_g(f(x)) = \lambda_g \delta f(x) . \} \right) $$ There is isomorphism between these spaces, $D \mapsto \delta_D$.

According to [Springer, "Linear Algebraic Group", 4.4.7, p.72] we deine $J$ to be the ideal in $k[G]$ of functions that vanish on $H$. We want to prove $$T_eH = ( \{ D \in T_eG | \delta_D I \subseteq I \} ) $$ (this is a set and a tagent space).

After that we want to show that $T_eH = \ker (d \varphi)_e$ and by a theorem it is equivalent that $\varphi$ is separable.

I tried but didn't managed to prove the last two assertions (the two equalities about $T_eH$), so I need some help with this.

Answer 1

You want to show that $T_eH=\{D\in T_eG\mid\delta_DI\subset I\}$. This is something general about smooth subvarieties of a variety. The left hand side maps to the right hand side by functoriality of tangent spaces. Now $H$ is a smooth subvariety and the dimension of its tangent space is the dimension of $H$. So we must understand that the right hand side is no bigger. But for this one may look in local coordinates at $e$. Say $m$ is the ideal of functions vanishing at $e$. Then one may compute with $m/m^2$ to check that preserving $I$ imposes enough linear restrictions to bound the dimension. As an algebraist I would first pass to the $m$-adic completion which is a ring of power series. In that ring the ideal $I$ looks very simple.

The other equality cannot be proved without further data on the construction of $v$. But basically it is again a statement that things are OK in local coordinates, now around $v$. The condition on an element of $T_eG$ that it does not move $v$ must now impose enough linear restrictions to bound the dimension again.

Answer 2

@Wilberd van der Kallen : The comment option doesn't work for me right now (probably because I am logged from two different computers) so I write here.

I think I understand that by left invariance a function vanishing in $H$ can be checked for $e$ only, since $$ \delta \lambda_h f(x) = \lambda_h \delta f(x) $$ and we may factor $x = he$ for $x \in H$. Tell me if that is correct.

However, I don't understand the $m / m^2$ construction, and why it represents the module (I know that $m/m^2$ is a representive element with $\delta f(x) = f - f(e)$ but don't understand the implementation to a given derivative module, how to represent the general derivative by $\delta$).

(in the general case: let $B = A \oplus M$ direct sum of mudules where $A \subset B$ and $M$ is a maximal ideal in the ring. Then we defined $$\pi_A : B \to A \ ; \ \pi_A(a+m)=a$$ and $\delta (f) = (f - \pi_A(f))$ and then $M/M^2$ is a representative for a derivative module, since $\delta$ is a derivative.).

I also don't understand what do you mean by $m$-adic completion.

Thanks and sorry for the late answer.