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Is the projective special unitary group a lie group? If so then what is its Lie algebra and where can I find out more about these objects?

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    $\begingroup$ Given the complicated history of the concept of "Lie group", this is a reasonable question. But it's also quite elementary and might fit better on Stack Exchange. $\endgroup$
    – Jim Humphreys
    Oct 21, 2013 at 17:29

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Yes, $PSU(n)$ is a Lie group. Indeed, the quotient of a Lie group by a closed normal subgroup is itself a Lie group in a natural way. Furthermore, $PSU(n)$ is the quotient of $SU(n)$ by the center $\mathbb{Z}/n\mathbb{Z}$. Since $\mathbb{Z}/n\mathbb{Z}$ is a finite group, its Lie algebra is $\{0\}$. Therefore, the Lie algebra of $PSU(n)$ is the quotient of $\frak{su}(n)$ by $\{0\}$, and so is just $\frak{su}(n)$.

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    $\begingroup$ Thanks for the quick and concise answer. Do you also know if $PSU(N)$ is compact and connected? Should this be obvious is some way? I would be tempted to guess that it as N connected components but I can't really justify this. Also, in what sense can any finite group have a Lie algebra? You say: Since $Z/nZ$ is a finite group, its Lie algebra is ${0}$. $\endgroup$
    – Benjamin
    Oct 21, 2013 at 17:48
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    $\begingroup$ That $PSU(n)$ is compact and connected follows from the fact that it is the continuous image of the quotient map $SU(n)\rightarrow PSU(n)$ (since $SU(n)$ is compact and connected). You can endow any finite group with the discrete topology, as well as a Lie group structure. The exponential map is a local diffeomorphism from a neighbourhood of $0$ to a neighbourhood of the group identity. Since the identity element is itself open in the discrete topology, you can show that the Lie algebra must have been $\{0\}$. $\endgroup$ Oct 21, 2013 at 17:57

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