most recent 30 from http://mathoverflow.net 2013-05-25T16:58:53Z http://mathoverflow.net/feeds/question/76050 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/76050/is-connected-complex-lie-group-with-a-trivial-center-linear Dima Sustretov 2011-09-21T11:30:34Z 2012-12-07T23:09:15Z

<p>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.</p> <p>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?</p>

http://mathoverflow.net/questions/76050/is-connected-complex-lie-group-with-a-trivial-center-linear/115754#115754 Aakumadula 2012-12-07T22:51:44Z 2012-12-07T23:09:15Z

<p>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. </p> <p>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 $GL({\mathbb C}^2)$. </p>