Talk:Linear algebraic group

Thanks for the clarification wrt the Lie groups...I wasn't clear on that. Revolver

What I'm not quite clear on, myself, is the status of the unitary group. The 'unitarian trick' of Weyl is to say that it is Zariski-dense in GL(n,C) - I guess. But in what sense is the unitary group _not_ a real algebraic group?

Charles Matthews 20:32, 10 Dec 2003 (UTC)


 * The unitary group U(n) of matrices in GL(n,C) such that $$UU^*=I$$ is an algebraic group over the real numbers, as opposed to over the complex numbers. Its complex points are all of GL(n,C). Weyl's unitarian trick uses this fact in that the complexification of the real Lie algebra of U(n) is the complex Lie algebra gl(n), thus showing that GL(n,C) is semisimple (since U(n) is compact). It seems you asked this question a while ago... RobHar 02:06, 21 April 2007 (UTC)

Yes, there is a page on unitarian trick now. Charles Matthews 07:11, 24 May 2007 (UTC)

Problem with the Examples section
The Examples section contains this passage:

"For a positive integer n, the general linear group GL(n) over a field k, consisting of all invertible n-by-n matrices, is a linear algebraic group over k. It contains the subgroups
 * $$U \subset B \subset GL(n)$$

consisting of matrices of the form
 * $$\left ( \begin{array}{cccc} 1 & * & \dots & * \\ 0 & 1 & \ddots & \vdots \\ \vdots & \ddots & \ddots & * \\ 0 & \dots & 0 & 1\end{array} \right )$$ and $$\left ( \begin{array}{cccc} * & * & \dots & * \\ 0 & * & \ddots & \vdots \\ \vdots & \ddots & \ddots & * \\ 0 & \dots & 0 & *\end{array} \right )$$."

But thee section makes no statements whatsoever about the subgroups B and U of GL(n). So: What is their purpose here???216.161.117.162 (talk) 03:20, 10 September 2020 (UTC)


 * Maybe you should try reading the article before whining about it. --JBL (talk) 18:50, 27 September 2020 (UTC)