Lie group structure on some topological spaces

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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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.