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Recently my professor told me that the usual definition of PSL(2,$\mathbb{R}$) = SL(2,$\mathbb{R}$)/{$\pm$I} does not give an algebraic group, but the following definition does:

PSL(2,$\mathbb{R}$) = {$A \in$ PSL(2,$\mathbb{C}$) = SL(2,$\mathbb{C}$)/{$\pm$I} : $\bar A = A$}.

Notice that $\left[ \begin{matrix} i & \\ & -i \end{matrix} \right]$ is in the group as per the second definition but not the first.

The same thing happens with other groups, for example PSp(4,$\mathbb{R}$) = Sp(4,$\mathbb{R}$)/{$\pm$I} (the usual definition) is not algebraic, but PSp(4,$\mathbb{R}$) = {$A \in$ PSp(4,$\mathbb{C}$) = Sp(4,$\mathbb{C}$)/{$\pm$I} : $\bar A = A$} is. Again, the matrix $\left[ \begin{matrix} i &&& \\ & i && \\ && -i & \\ &&& -i \end{matrix} \right]$ is in the second group but not the first.

The theme here is that if you take the real points of an algebraic group (which is still an algebraic group) then mod out by the center, you (may) no longer have an algebraic group. However, if you mod out by the center then look at the real points (elements satisfying $\bar A = A$) then you still get an algebraic group.

He didn't go into any detail about why that might be, since it was off topic from our discussion except to say that we were working with the wrong group (the first instead of the second one).

Could anyone please provide an explanation as to what makes the first group not algebraic and what makes the second algebraic? My not-very-developed intuition about algebraic groups tells me it has something to do with the first group not being defined over and algebraically closed field.

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  • $\begingroup$ do you mean PSL(4, $\mathbb{R}$)? $\endgroup$ – Loreno Heer Jul 12 '15 at 23:25
  • $\begingroup$ This MathOverflow question seems to be related. $\endgroup$ – Jim Belk Jul 13 '15 at 4:49
  • $\begingroup$ I didn't mean PSL(4,$\mathbb{R}$). I've edited the question to correct that and also (hopefully) make things more clear. $\endgroup$ – Émile Greer Jul 15 '15 at 15:35

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