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I have some basic background in Lie theory and I have some difficulties to show that some topological spaces admits a Lie group structure. More precisely, for a given Lie group $G$:

1) Why its tangent (and cotangent) bundle admits a Lie group structure? I believe it's not too difficult !

2) Why its universal covering is a Lie group?

Thanks for any help!

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Short answer: 1) Lie groups are paralellizable, hence $TG = \mathfrak{g} \times G$ as manifolds 2) the universal covering space of a manifold is a manifold and the universal covering space of a topological group has a canonical group structure (if the universal covering space exists). This group structure is smooth/analytic because these are local properties and are thus smooth upstairs because the covering projection is a local diffeomorphism. Could you maybe say for which of these points you want/need more detail? –  t.b. Sep 2 '11 at 10:03
    
@Theo: Is there an easy example of a topological group whose universal covering space doesn't exist? –  Jason DeVito Sep 2 '11 at 18:20
    
@Jason: I don't know of any non-trivial example, i.e., one in which something more subtle than mere connectedness is the issue. It was just poor phrasing with the goal to make it clear that some minimal requirements must be fulfilled. I didn't have any specific example in mind. –  t.b. Sep 2 '11 at 19:02
    
Many thanks Theo for tour short but useful answer! $TG$ is isomorphic to $G\times\mathcal{G}$ so its a Lie group for the product $(g,\xi)\cdot(h,\eta)=(gh,\xi+\eta)$? For the universal covering I have to check its canonical group structure and to show the smoothness of the multiplication. I hope I can look up! –  amine Sep 3 '11 at 9:57
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amine: sorry, I didn't see your question in the comments up to now. If you want me to see a comment, write @Theo (as Jason did in his comment above). Your question should be addressed in my answer below. –  t.b. Sep 5 '11 at 15:24
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1 Answer

up vote 6 down vote accepted

I thought I should elaborate my comment into an answer.

First of all, let me make some remarks on the Lie group structure on the tangent bundle:

The multiplication map $m: G \times G \to G$ yields (after identification of $T(G \times G)$ with $TG \times TG$) a map $Tm: TG \times TG \to TG$.

It is then straightforward to check that this yields a Lie group structure on $TG$.

For instance, associativity follows from the one of $G$ as follows: We have $m \circ (m \times \operatorname{id}_G) = m \circ (\operatorname{id}_G \times m)$, so $$\begin{align*} T(m \circ (m \times \operatorname{id}_G)) & = Tm \circ T(m \times \operatorname{id}_G) & T(m \circ (\operatorname{id}_G \times m)) & = Tm \circ T(\operatorname{id}_G \times m) \\ & = Tm \circ (Tm \times \operatorname{id}_{TG}) & & = Tm \circ (\operatorname{id}_{TG} \times Tm) \end{align*} $$ and thus $Tm \circ (Tm \times \operatorname{id}_{TG}) = Tm \circ (\operatorname{id}_{TG} \times Tm)$ which is associativity of $Tm$.

Similarly, denoting $\varepsilon: G \to G$ the map $\varepsilon(g) = 1_{G}$ we have the unit axiom for $G$ telling us that $m \circ (\varepsilon \times \operatorname{id}_{G}) = \operatorname{id}_G = m \circ (\operatorname{id}_G \times \varepsilon)$. Applying the functor $T$ yields the unit axiom for $TG$ with unit $T\varepsilon = 0 \in T_{1}G$, and, finally, the inversion map $i: G \to G$, $i(g) = g^{-1}$ yields that $Ti$ is the inversion of $TG$.

The abstract nonsense going on here is that a functor preserving finite products (hence terminal objects) carries group objects to group objects.

Now you should work out the group structure on $TG$ explicitly. The bundle projection $\pi: TG \to G$ will turn out to be a homomorphism of Lie groups with kernel $\mathfrak{g} = T_1G$. This gives rise to a short exact sequence $$ 0 \to \mathfrak{g} \to TG \to G \to 1$$ The zero section $s(g) = 0 \in T_gG$ yields a semi-direct product decomposition $TG \cong \mathfrak{g} \rtimes G$ where $G$ acts on $\mathfrak{g}$ via the adjoint action.


As for the universal covering, we can do it essentially as I outlined in the comment above:

  1. The universal covering $\widetilde{M}$ of a manifold $M$ has a unique manifold structure making the covering projection $\pi:\widetilde{M} \to M$ a local diffeomorphism.
  2. Let $G$ be a connected Lie group. Choose a base point $1_{\widetilde{G}} \in \widetilde{G}$ in the fiber $\pi^{-1}(1_G)$ above $1_G$. Identifying $\widetilde{G \times G}$ with $\widetilde{G} \times \widetilde{G}$ the map $m \circ (\pi \times \pi): \widetilde{G} \times \widetilde{G} \to G \times G \to G$ lifts uniquely to a map $\widetilde{m}: \widetilde{G} \times \widetilde{G} \to \widetilde{G}$ such that $\widetilde{m}(1_{\widetilde{G}},1_{\widetilde{G}}) = 1_{\widetilde{G}}$. Associativity of $\widetilde{m}$ then follows from a short verification that $\widetilde{m} \circ (\operatorname{id}_{\widetilde{G}} \times \widetilde{m})$ and $\widetilde{m} \circ (\widetilde{m} \times \operatorname{id}_{\widetilde{G}})$ are both lifts of the same map $\widetilde{G} \times \widetilde{G} \times \widetilde{G} \to G$ sending $(1_{\widetilde{G}},1_{\widetilde{G}},1_{\widetilde{G}})$ to $1_{\widetilde{G}}$, so they must be equal. Similarly for inversion and unit. This implies that $\widetilde{G}$ (after the choice of a base point) has a unique structure of a topological group.
  3. It remains to argue that the group structure on $\widetilde{G}$ is smooth. This is relatively easy by using the fact that the covering projections $\pi: \widetilde{G} \to G$ and $\pi \times \pi: \widetilde{G} \times \widetilde{G} \to G \times G$ are local diffeomorphisms. You should also check that $\pi_1(G) \cong \ker{\pi}$ and it follows easily that $\pi_1(G)$ is abelian and central.

Finally, let me point out that there is also the following result:

Let $\varrho: G \to H$ be a continuous homomorphism of topological groups and assume that it is a covering map. If either one among $G$ or $H$ is a Lie group. Then there is a unique smooth structure on the other one such that $\varrho$ is a smooth homomorphism of Lie groups and a local diffeomoriphism.

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Very nice answer! Thank you! –  amine Sep 6 '11 at 12:19
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