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.

  • 3
    $\begingroup$ 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. $\endgroup$ – YCor Mar 12 at 8:29
  • $\begingroup$ Thanks for the comment. Could you give a bit more detail? $\endgroup$ – unknownymous Mar 12 at 16:24
  • 1
    $\begingroup$ See Proposition D.2.1 in these notes of Brian Conrad for a proof: math.stanford.edu/~conrad/papers/luminysga3.pdf $\endgroup$ – Sam Gunningham Mar 14 at 22:06

I upvoted the two answers in the comments but the fact that you are asking also follows easily from algebraicity of holomorphic representations. Four possible proofs are outlined in the link above. The difference between reductive and semisimple is immaterial: just use the fact that a connected reductive group is a product of a semisimple group and a torus to conclude that it acts algebraically.

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.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.