I am working through Brocker & tom Dieck's Representations of Compact Lie Groups, and this is one of the early exercises. The unitary group's Lie algebra $\mathfrak{u}(n)$ is formed from skew-Hermitian matrices; the Special Unitary group's lie algebra $\mathfrak{su}(n)$ is formed from traceless antihermitian matrices. These properties seem to fall out of some basic calculations with the exponential map.

However, I am having a hard time working out a similar description for the lie algebra of $PGL(n)$ -- maybe because I do not have a good picture of the properties of its matrices to begin with. Can someone help me get a concrete idea of what $\mathfrak{pgl}(n)$ looks like?


The Lie algebra of $PGL(n,\mathbb{C})=GL(n,\mathbb{C})/Z$ is the simple Lie algebra $\mathfrak{sl}(n,\mathbb{C})$ of traceless matrices with the Lie bracket $[A,B]=AB-BA$. For references containing a proof see here.

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    $\begingroup$ Thanks. Knowing that, I'll try to construct a proof. I can't quite find a proof that Lie$(PGL(n, \mathbb{C}))$ is $\mathfrak{sl}(n,\mathbb{C})$ in Wikipedia or its references -- just that Lie$(PSL(n,\mathbb{C}))$ = $\mathfrak{sl}(n,\mathbb{C})$, and that every group with the lie group with the same lie algebra covers PSL. $\endgroup$ – Andrew Tindall Oct 31 '17 at 19:37
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    $\begingroup$ And there is also written: "Note that $PGL(n, F)$ and $PSL(n, F)$ are equal if and only if every element of $F$ has an $n$-th root in $F$. " So you are done, since this is true for $F=\mathbb{C}$. $\endgroup$ – Dietrich Burde Oct 31 '17 at 19:46
  • $\begingroup$ OK, got it. Thanks! $\endgroup$ – Andrew Tindall Oct 31 '17 at 19:47

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