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I was reading Fulton Harris' Representation theory, A first course, where I came across the following:

Let $H$ be a Lie group and $T$ be a discrete subgroup of its center $Z(H)$. Then there exists a unique Lie group structure on $H/T$ such that the quotient map $\pi:H \rightarrow H/T$ is a Lie group map.

By unique, does it means that there is no other Lie group structure in $H/T$ such that the map becomes a Lie group map ?

The book says that it is straightforward but I am not getting it all. How to prove this?

PS: Is this true that that quotient topology makes the projection map continuous and being a homomorphism too, it becomes smooth and hence a Lie group map?

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Just to make things precise: The group structure on $H/T$ is the usual quotient group structure. The map $\pi: H \to H/T$ is the usual projection map from a group onto its quotient. The only missing part is the smooth structure on $H/T$. The claim is that there is a unique smooth structure on $H/T$ such that $H/T$ is a Lie group and $\pi$ is a map of Lie groups.

Taking the quotient topology on $H/T$ is a good start, but it is not enough, since you need a smooth structure to have a Lie group.

Here is a sketch of the argument. If $x \in H$, we write $\overline{x}$ for $\pi(x)$.

Since $T$ is discrete, $\pi$ is a covering map (we are using the quotient topology on $H/T$). So if $x \in H$, then we may choose a neighborhood $U_\overline{x}$ of $\overline{x} \in H/T$ such that $\pi^{-1}(U_\overline{x})$ is the disjoint union of neighborhoods $\bigcup_{y \in \pi^{-1}(\overline{x})} U_y$, and $\pi$ restricted to each $U_y$ is a homeomorphism. Since $H$ is a Lie group, all the neighborhoods $U_y$ have the same smooth structure. We pick any $U_y$ and transport the smooth structure from $U_y$ to $U_\overline{x}$. Now we only have to check that the smooth structure we have defined on $H/T$ is compatible on overlaps, but this is so because the original smooth structure on $H$ was compatible on overlaps and $T$ is discrete so the quotient cannot mess things up.

For uniqueness, note that if we have any smooth structure on $H/T$ making $\pi$ into a smooth map, then any smooth function $U_\overline{x} \to \mathbb{R}$ must pull back to a smooth function $\bigcup_{y \in \pi^{-1}(\overline{x})}U_y \to \mathbb{R}$, but the latter is just many copies of a single function $U_y \to \mathbb{R}$ (pick an arbitrary $y$) where $U_y$ is homeomorphic to $U_\overline{x}$, so we have no choice but to copy the smooth structure of $U_y$ onto $U_\overline{x}$.

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    $\begingroup$ Thanks. I am still reading on this so it took me a while to understand. Your comment on "T is discrete so the quotient can't mess things up." is exactly what I am trying to understand. Can you elucidate it a bit more? $\endgroup$ – NeerajKumar Apr 20 '16 at 8:39
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    $\begingroup$ If you choose the neighborhoods for the overlaps small enough, then no two points in the neighborhoods will be equivalent under $T$, so you are back to the case of $H$. $\endgroup$ – Ted Apr 20 '16 at 15:19

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