# Is the Jordan decomposition for reductive groups algebraic?

Let $$G$$ be a connected affine algebraic group over $$\mathbb{C}$$. It's a known fact that elements of $$G$$ admit a decomposition into semisimple and unipotent elements. Namely, choose a faithful representation $$G \subseteq GL(n,\mathbb{C})$$. Then given $$g \in G$$, the Jordan decomposition states that $$g = su$$, where $$s$$ is semisimple, $$u$$ is unipotent, and $$su = us$$. The crucial fact is then that $$G$$ contains both $$s$$ and $$u$$, and this decomposition is independent of the representation. Therefore, the decomposition $$g = su$$, as well as the notions of being semisimple and unipotent, are intrinsic to $$G$$. Furthermore, the decomposition is preserved by morphisms of algebraic groups.

However, if we allow holomorphic maps of Lie groups, the decomposition is not respected. For example, the exponential map $$exp : \mathbb{C} \to \mathbb{C}^{*}$$ sends unipotent elements to semisimple elements!

So this makes me wonder whether the decomposition depends on the algebraic structure of $$G$$.

For definiteness, here's my question: Let $$G_{1}$$ and $$G_{2}$$ be connected complex reductive groups over $$\mathbb{C}$$, and let $$\phi: G_{1} \to G_{2}$$ be a holomorphic group isomorphism. Does $$\phi$$ preserve the Jordan decomposition?

My feeling is that the answer is yes, and that it will come down to something like $$G_{i}$$ admitting a unique algebraic structure compatible with the group structure, and any holomorphic isomorphism being automatically algebraic.

• Yes, it's easy to check that any biholomorphic isomorphism $G_1\to G_2$ between reductive complex groups is algebraic. One can suppose $G_1,G_2$ connected. Then one reduces to the case of semisimple groups on the one hand (which is essentially immediate) and the case of tori. Since all $n$-dim tori are isomorphic, we can consider biholomorphic automorphisms of the $n$-tori. At the Lie algebra level we have $GL_n(\mathbf{C})$. To pass to the group it has to preserve the lattice kernel, so we get $GL_n(\mathbf{Z})$ as biholomorphism group and this is algebraic. – YCor Mar 12 at 8:29
• Thanks for the comment. Could you give a bit more detail? – unknownymous Mar 12 at 16:24
• See Proposition D.2.1 in these notes of Brian Conrad for a proof: math.stanford.edu/~conrad/papers/luminysga3.pdf – Sam Gunningham Mar 14 at 22:06

Now realize $$G_2$$ as a Zariski-closed subgroup of $$GL_n({\mathbb C})$$. This gives you a holomorphic representation $$\phi: G_1 \rightarrow GL_n({\mathbb C})$$. Its algebraicity explained above seals the deal.