Does there exist a complex Lie group G such that ... - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T19:13:23Z http://mathoverflow.net/feeds/question/77884 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/77884/does-there-exist-a-complex-lie-group-g-such-that Does there exist a complex Lie group G such that ... Alexander Moll 2011-10-12T01:50:50Z 2011-10-12T16:47:58Z <p>... every Riemann surface of genus $1$ appears as a complex one-parameter subgroup of $G$?</p> http://mathoverflow.net/questions/77884/does-there-exist-a-complex-lie-group-g-such-that/77886#77886 Answer by algori for Does there exist a complex Lie group G such that ... algori 2011-10-12T02:30:33Z 2011-10-12T03:06:04Z <p>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.</p> <p>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.</p> http://mathoverflow.net/questions/77884/does-there-exist-a-complex-lie-group-g-such-that/77890#77890 Answer by Tom Goodwillie for Does there exist a complex Lie group G such that ... Tom Goodwillie 2011-10-12T03:13:52Z 2011-10-12T16:47:58Z <p>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.</p> <p>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). </p> <p>Edit: Just to summarize what I have learned from this exercise: Let CCC="compact connected complex".</p> <p>(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.)</p> <p>(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). </p> <p>(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 $G$.</p> <p>(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.</p> <p>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.</p>