When $n>3$ is even, how can I show that $PGL(n,\mathbb{R})$ has a faithful adjoint representation? Of course when n is even, $PGL(n,\mathbb{R})$ is not connected.
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$\begingroup$ Are you asking why $PGL_n(\mathbb{R})$ has trivial center? $\endgroup$– S. Carnahan ♦Commented Nov 8, 2010 at 10:02
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$\begingroup$ NO. I know that for connected lie groups, if it has a trivial center then it has a faithful adjoint representation. Are you saying that the same statement holds for nonconnected lie groups? $\endgroup$– user9552Commented Nov 8, 2010 at 13:23
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2$\begingroup$ Ah, but ${\rm{PGL}}_n$ is connected for Zariski topology over any field. For a connected adjoint semisimple gp $G$ over a field $k$ then its adjoint representation has trivial kernel (so it's even a closed immersion). For $k = \mathbf{R}$, the functor $G \rightsquigarrow G(\mathbf{R})$ from (perhaps disconnected) group varieties to Lie groups is clearly compatible with the formation of adjoint representations, so when $G$ is connected adjoint semisimple (connected for Zariski topology!) then $G(\mathbf{R})$ has faithful adjoint repn even if it is disconnected (for the classical topology). QED $\endgroup$– BCnrdCommented Nov 8, 2010 at 15:31
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Suppose $A \in PGL_n(\mathbb{R})$ lies in the kernel of the adjoint representation. Then for any lift $\tilde{A}$ of $A$ in $GL_n(\mathbb{R})$, and any traceless matrix $B$, we have $\tilde{A}B = B\tilde{A}$. This implies $\tilde{A}$ is scalar, and hence $A$ is the identity.
answered Nov 8, 2010 at 13:59
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$\begingroup$ How does $\tilde{A}B=B\tilde{A}$ imply $\tilde{A}$ is scalar? $\endgroup$– user9552Commented Nov 8, 2010 at 14:55
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1$\begingroup$ The equation holds for all matrices $B$. This implies $\tilde{A}$ is in the center of the matrix algebra. $\endgroup$– S. Carnahan ♦Commented Nov 9, 2010 at 11:22