... every Riemann surface of genus $1$ appears as a complex one-parameter subgroup of $G$?


No. In a connected complex Lie group all compact complex subgroups must be in the center, that is, in the kernel of the adjoint representation, because in a complex general linear group there is no nontrivial compact connected complex subgroup. And in a connected abelian complex Lie group there can be only countably many compact one-parameter complex subgroups.

Edit: So in fact in a complex Lie group there are only countably many compact connected one-parameter complex subgroups (not just up to isomorphism).

Edit: Just to summarize what I have learned from this exercise: Let CCC="compact connected complex".

(1) (I already knew this.) A CCC subgroup of $GL_n(\mathbb C)$ must be trivial. Compact implies contained in a conjugate of $U_n$; complex then implies $0$-dimensional; connected then implies trivial. (Alternatively, a holomorphic map from a CCC manifold to $\mathbb C$ (therefore to $GL_n(\mathbb C)$) must be constant.)

(2) (IAKT) A CCC group is always abelian. This follows from (1) using the adjoint representation. Therefore it must be a compact complex torus group, the quotient of a complex vector space by a full lattice (i.e. a subgroup generated by an $\mathbb R$-basis).

(3) (This had never occurred to me before, but it's obvious by the same argument.) In a connected complex Lie group a CCC subgroup is always in the center. Thus a complex Lie group $G$ has a maximal CCC subgroup in the strongest sense -- one which contains all others -- namely the (unique) maximal compact subgroup of the connected component of the center of the connected component of $G$.

(4) Therefore there cannot be a nonconstant family of CCC subgroups of a complex Lie group. (3) reduces this statement to the case where the ambient group is abelian, and that case is clear.

I don't know what you mean by infinite-dimensional Lie group, exactly, but it seems to me that the arguments above can be adapted to show that in the group of all complex automorphisms of a connected complex manifold there is a CCC containing all others (again contained in the center of the component) so that again there are no families.

  • $\begingroup$ I believe your answer agrees with algori's - thanks for your help! $\endgroup$ – Alexander Moll Oct 12 '11 at 3:25
  • $\begingroup$ Thanks for expanding your response - as of yet I don't have any systematic theory of "infinite dimensional complex Lie groups" in mind - I guess I'm looking for a way to relax the rigid behavior of these CCC Lie subgroups. $\endgroup$ – Alexander Moll Oct 12 '11 at 21:12

Let me show that such a group, if it exists, can't be algebraic. Any complex algebraic group $G$ is an extension of an abelian variety $A$ by an affine group $H$, i.e. we have an exact sequence $1\to H\to G\to A\to 0$ (Chevalley's theorem). Suppose $G$ had every elliptic curve as a subgroup. All of those would project isomorphically to $A$, since none of them cam intersect $H$. So we may just as well restrict ourselves to the case $G=A$. But $A$ has only countably many Lie subgroups.

upd: here is a proof in the general case: A subgroup of $G$ which is an elliptic curve is contained in a maximal (compact) torus; moreover, since all tori are conjugate, the isomorphism classes of the elliptic curves they contain are the same. Now we use the above argument: every torus contains countably many (closed) Lie subrgoups.

  • $\begingroup$ Thanks for your answer - it's very satisfying! I guess this means if there's any hope of finding such a $G$ it would have to be infinite-dimensional... $\endgroup$ – Alexander Moll Oct 12 '11 at 3:21
  • $\begingroup$ Dear Alexander -- welcome! I have to admit, I personally like Tom's answer better, since it doesn't use any heavy machinery. $\endgroup$ – algori Oct 12 '11 at 6:10
  • $\begingroup$ Alright - in that case I'll shift the pretty green check-mark downstairs to Tom's post. Thanks again! $\endgroup$ – Alexander Moll Oct 12 '11 at 21:02

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.