MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
Are you asking why $PGL_n(\mathbb{R})$ has trivial center? – S. Carnahan Nov 8 '10 at 10:02
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? – user9552 Nov 8 '10 at 13:23
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 – BCnrd Nov 8 '10 at 15:31

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.

share|cite|improve this answer
How does $\tilde{A}B=B\tilde{A}$ imply $\tilde{A}$ is scalar? – user9552 Nov 8 '10 at 14:55
The equation holds for all matrices $B$. This implies $\tilde{A}$ is in the center of the matrix algebra. – S. Carnahan Nov 9 '10 at 11:22

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.