# Product decomposition into semisimple and unipotent parts of an algebraic group (in Borel’s LAG)

Let $$G$$ be an algebraic group, i.e., an affine reduced, separated $$k$$-scheme of finite type with structure of a group. In Borel’s Linear Algebraic Groups Theorem III.10.6(4) says

Theorem 10.6 (3): Let $$G$$ be a connected, solvable algebraic group. Then $$G$$ is nilpotent if and only if $$G_s$$ is a subgroup of $$G$$. In this case, $$G_s$$ is a closed subgroup defined over $$k$$, and $$G$$ is the direct product $$G_s \times G_u$$.

The subgroups $$G_s, G_u \subset G$$ are the semisimple resp. the unipotent parts coming from the Jordan decomposition.

The proof’s strategy is to show firstly that $$G = G_s \cdot G_u$$ decomposes as direct product of abstract groups and make the observation that $$L(G_s) \cap L(G_u) = 0$$, where $$L(G_s)$$ and $$L(G_u)$$ are the Lie algebras of $$G_s$$ resp. $$G_u$$. Then the proof claims that this already implies that the product structure $$G = G_s \cdot G_u$$ holds as an algebraic group, because the canonical product map $$m:G_s \times G_u \to G$$ is bijective and separable.

(A similar strategy is also used in the proof of part (4) of the same theorem to show that $$G$$ decomposes as a semi-direct product $$G=T \cdot G_u$$ of algebraic groups, where $$T$$ is a maximal torus )

Question: Why does separability of the bijective product map $$G_s \times G_u \to G$$ imply that it’s an ismorphisms of algebraic groups in this proof?

The book uses two notations of separability: a morphism $$f: V \to W$$ of algebraic varieties is separable if the field extension $$K(W) \subset K(V)$$ is separable. There is also a notion for separability for homogeneous spaces: the canonical projection $$\pi: G \to G/H$$ is separable if the induced map $$(d \pi)_{1_G}: L(G) \to L(G/H)$$ on Lie algebras at the neutral element $$1_G$$ is surjective. (Cor. II.6.7)

The proof argues with Lie algebras, more concretely the proof implies that the induced map $$m_L: L(G_s \times G_u) \to L(G)$$ on Lie algebras is an isomorphism because $$\text{Ker}(m_L)= L(G_s) \cap L(G_u) = 0$$ applying a simple dimension count. So it suggest that the proof uses the second notion of separability, namely that one for homogeneous spaces.

But then, how to interpret $$G$$ somehow as an orbit space/homogeneous space of $$G_s \times G_u$$. At all, if we want to see $$G$$ as an orbit space, then we should have a transitive $$(G_s \times G_u)$$-action on $$G$$ which coincides with the product structure $$(u,v) \to u \cdot v:= m(u,v)$$. But with respect to which action?

The most natural "choices" to define a $$(G_s \times G_u)$$-action on $$G$$ might be given via $$((u,v), g) \mapsto u \cdot v \cdot g$$ or $$u \cdot g \cdot v$$.
Problem: an action $$\Phi \colon H \times S \to S$$ by a group $$H$$ on set $$S$$ must satisfy the “compatibility relation” with group multiplication $$\Phi(h_1 \cdot h_2, s)= \Phi(h_1, \Phi(h_2, s))$$ for all $$h_1,h_2 \in H, s \in S$$. But unfortunately the two natural guesses do not satisfy this rule. So either the argument in the book works differently or there is a less usual $$(G_s \times G_u)$$-action on $$G$$ applied. Which one? If the second case holds, I wonder why the book doesn’t write this action explicitly down.

And if there is some $$(G_s \times G_u)$$-action on $$G$$ established, how this would imply that $$G_s \times G_u \to G$$ is an isomorphism of algebraic groups? Does it boils down to something like that realizing the product map $$G_s \times G_u \to G$$ as homogeneous map the it is an isomorphism iff it is bijective (as set map) + separable?

Remark: This question is identical to this one I asked several weeks ago. Although there is an answer, I doubt that this is what Borel used in his argumentation, since it is expected that the information that $$L(G_s) \cap L(G_u) = 0$$ should be directly involved in the argument.

• The argument is straightforward: (1) $f:X \to Y$ is a map of irreducible smooth varieties of the same dimension and the map induced by $f$ on the tangent space at a point $x \in X(k)$ is an isomorphism then $f$ is separable. The statement about Lie algebras in Borel implies that this condition holds for $x$ the identity element. (2) Any separable map of smooth irreducible smooth varieties which is a bijection is an isomorphism.
– naf
Mar 6, 2023 at 4:06
• Proposition AG.18.2 in Borel implies: If $\alpha: V \rightarrow W$ is a dominant injective morphism of irreducible varieties, then $K(V)$ is purely inseparable over $K(W)$. Then note that if an extension is both purely inseparable and separable, it is trivial.
– spin
Mar 7, 2023 at 3:06

I don't always know how to make sense of Borel's somewhat old-fashioned approach, but I think the argument is simpler than it might seem: the subgroups $$G_s$$ and $$G_u$$ are mutually normalising and intersect trivially, hence commute, so they admit a direct product inside $$G$$; but then this direct product contains $$G(\overline k)$$, by the Jordan decomposition, so, since $$G$$ is smooth, must be all of $$G$$.
(I just noticed that you say only that $$G$$ is affine reduced over $$k$$, but I think you mean to assume either that $$G$$ is actually smooth, or that $$k$$ is algebraically closed (in which case reducedness implies smoothness). Borel only considers smooth groups.)