## Quotient space of algebraic group

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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.

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I would change the title to "Quotient space of algebraic group," since you are asking about taking the quotient by an arbitrary (not necessarily normal) closed subgroup. And why do you say $G/H$ is projective? This is often false: e.g. take $G$ to be the Heisenberg group and $H$ its center. – Justin Campbell Jun 14 at 19:32
We build $Y=G/H$ as a projective space in order to have a H-invariant vector $0\ne v \in k^n$ such that $\forall h \in H : h \cdot v = kv \in \mathrm{Span}_k(v)$. However, this is not the important thing in my question. – Zachi Evenor Jun 14 at 19:43

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There is the problem that $H$ had better be the scheme theoretic stabilizer of the vector $v$, not just the algebraic group stabilizer. In other words, things may go wrong if you do not require them to go right. It is a condition on the construction of $v$ that the Lie algebra of $H$ is the full stabilizer of the point $v$ in the Lie algebra of $G$. It can fail. – Wilberd van der Kallen Jun 15 at 10:15
@Wilberd van der Kallen We proved a Lemma that said that the ring of regular functions on the relevant open sets is stabilized by $H$. Let $U \subset Y$ be an open set, then one can show (it is a long proof, though), that $$\mathcal{O}_Y(U) \circ \varphi = \left( \mathcal{O}_G(\varphi^{-1}(U)) \right)^H$$ so I guess this satisfies the demands regarding the scheme. Note that in class we havn't used the "schemes" and "sheaves" concepts and use more basic tools (Alg. group, Lie algebras, ringed spaces, differential etc). – Zachi Evenor Jun 15 at 11:18

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@ Zachi Evenor. I am not worried if $H$ stabilizes, I am worried if there may be more elements in the Lie algebra of $G$ that also stabilize $v$. In that case we say that the scheme theoretic stabilizer is bigger. So my problem is that what you try to prove is not always true. For example, suppose $G$ is the general linear group and we are in positive characteristic $p$. If you compose the action of $G$ with the Frobenius map from $G$ to $G$ (that raises all entries to the $p$-th power), then suddenly the full Lie algebra of $G$ acts trivially and there is no way to prove the desired formula. – Wilberd van der Kallen Jun 15 at 12:00

## 2 Answers

@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.

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@Zachi Evenor  The definition of a tangent space in terms of $m/m^2$ is standard and the notion of $m$-adic completion is also standard. The latter just means that one computes with power series expansions in terms of local coordinates. Ask anyone knowing enough algebraic geometry. I am not using the description of the tangent space in terms of left invariant derivations because for me it obscures what is happening. – Wilberd van der Kallen Jun 30 at 12:36

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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.

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I have to think about it. – Zachi Evenor Jun 17 at 19:57
@Wilberd van der Kallen I replied your answer in a different answer (see below) due to technical reasons. I still have questions as written there. Thanks, – Zachi Evenor Jun 24 at 12:53
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