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There is a theorem of Rosenlicht ("Some basic theorems on algebraic groups", 1956, Theorem 13) asserting that a quotient of a connected algebraic group by its center is linear. So a connected algebraic group with trivial center is linear.

Is it true of connected complex Lie groups? I.e. is a connected complex Lie group with a trivial center a subgroup of $GL(n,\mathbb{C})$? Is it algebraic?

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Rosenlicht's term "algebraic groups" includes non-affine examples such as abelian varieties. But a "complex Lie group" is more narrowly defined. So your last formulation of the question is unclear. Early in Chevalley's treatment of affine algebraic groups he shows that such a group is linear, whereas some familiar real Lie groups are not. The detailed structure/classification shows that complex semisimple Lie groups are indeed linear, but for solvable Lie groups you'd have to look further into Hochschild's work including his old book Structure of Lie Groups, etc. – Jim Humphreys Sep 21 '11 at 13:53
I have just noticed that I have written "algebraic" where I should have written "complex Lie". I am sorry, the question should read: are connected complex Lie groups with trivial center linear algebraic? – Dima Sustretov Sep 21 '11 at 18:03
@Jim Humphreys: I must be missing something. In what respect is "complex Lie group" more narrowly defined? Doesn't "complex Lie group" meean a group which is a complex manifold with the group operation a holomorphic map? This definition seems to cover groups that are not subgroups of $GL(n,\mathbb{C})$, like complex tori. I wondered if the Rosenlicht's result generalised from algebraic varieties to complex manifolds. I perhaps mixed two notions together. First, I wonder if a connected complex Lie group with a trivial center is a subgroup of $GL(n,\mathbb{C})$ [continued] – Dima Sustretov Sep 22 '11 at 19:41
@ Dmitry: The answer to your first question ("is a connected complex Lie group with a trivial center, a subgroup of $GL_n(\mathbb{C})$?") is clearly yes: look at the adjoint representation of a Lie group $G$, it is classical that its kernel is $Z(G)$. So if $Z(G)$ is trivial, $G$ is linear. – Alain Valette Sep 22 '11 at 21:00
@Alain: this only gives an injective homomrphism into $GL_n$. Is it clearly an embedding, i.e. an isomorphism onto a closed subgroup? – Laurent Moret-Bailly Sep 23 '11 at 5:52
up vote 4 down vote accepted

As Alain Valette says, a centreless connected complex Lie group $G$ has an injective homomorphism into $GL_n({\mathbb C})$. However, it need not be algebraic. To see this, consider the semi-direct product $G={\mathbb C}^2 \rtimes {\mathbb C}$. Here $z\in {\mathbb C}$ acts on the standard basis $e_1,e_2$ by the characters $e^{2\pi i z}$ and $e^{2\pi i z/\sqrt{2}}$ respectively. If $G$ can be given the structure of an algebraic group, then these two characters on ${\mathbb C}$ would become algebraically dependent, which cannot be.

Incidentally, this $G$ is not closed in its adjoint "embedding", since the closure contains $S^1\times S^1$ in the diagonal part. That is $G\subset {\mathbb C}^2\rtimes D_2$ where $D_2$ is the group of diagonals in $GL({\mathbb C}^2)$.

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Dear Aakumadula, could provide a bit more detail? I do not understand what you mean by having a structure of an algegbraic group. Does it mean being a closed subgroup of $GL_n(\mathbb C)$? How does that imply that the two characters should be algebraically independent? – Dima Sustretov Dec 8 '12 at 22:13
Dear Dima, the question is "is a connected complex Lie group algebraic"? I understand it to mean that given a complex Lie group $G$, can you put the structure of a complex algebraic group $G'|$ such that as complex LIE groups, $G$ and $G'$ are isomorphic? Now suppose that the $G$ in the example is a complex ALGEBRAIC group (in the foregoing sense). $e_1,e_2$ may be thought of as elements of the Lie algebra, which is necessarily an algebraic representation of $G$, and are eigenvectors for $G$. But, $${\mathbb C}^2$ acts trivially, and hence $G$ acts via a one dimensional group on the span – Venkataramana Dec 8 '12 at 22:38
of $e_1,e_2$. Hence the two algebraic characters $\chi _1, \chi _2$ of the image of $G$ (which is one dimensional) must be algebraically dependent i.e. there exist integers $m,m'$ such that $\chi ^m=\chi ' ^{m'}$. – Venkataramana Dec 8 '12 at 22:41
Perhaps I should add that the image o an algebraic group under an algebraic homomorphism is also an algebraic group – Venkataramana Dec 8 '12 at 22:49

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