Exercise 7, § 3

Let $G$ be a real Lie group, denote by $G_0$ its identity component, by $Z_0$ the center of $G_0$ and by $\pi_0(G) = G/G_0$. Let $(\widetilde{G}_0, p_0)$ be a universal covering of $G_0$, let $\widetilde{Z}_0$ be the center of $\widetilde{G}_0$ and let $\widetilde{e}$ be its identity element.

a) Let $\theta :\pi_0(G)\rightarrow$ Out(G$_0$) be the homomorphism associated with the extension $\mathscr{E}: G_0\rightarrow G\rightarrow \pi_0(G)$ (A, X, §7, Exercise 5). Prove that the cohomology class $\omega (\pi_0(G),Z_0, \theta )$ of $H^3(\pi_0(G),Z_0)$ defined in loc. cit. is zero. b) Suppose that there exists a Lie group $\widetilde{G}$, a surjective mapping $p:\widetilde{G}\rightarrow G$ and an isomorphism $j$ of $\widetilde{G}_0$ onto the identity component of $\widetilde{G}$ such that $p_0=p\circ j$. Prove that one has $\omega (\pi_0(G),\widetilde{Z}_0, \psi \circ \theta ) = 0$ in $H^3(\pi_0(G),\widetilde{Z}_0)$, where $\psi :$ Out(G$_0$)$\rightarrow$ Out($\widetilde{G}_0$) is the homomorphism defined in Exercise 5.

c) Suppose that $\omega (\pi_0(G),\widetilde{Z}_0, \psi \circ \theta ) = 0$. Let $\widetilde{\mathscr{E}}:\widetilde{G}_0\rightarrow E\rightarrow \pi_0(G)$ be an extension of $\pi_0(G)$ by $\widetilde{G}_0$ whose associated homomorphism is equal to $\psi \circ \theta$. Prove that $E/\pi_1(G_0)$ is an extension of $\pi_0(G)$ by $G_0$ whose associated homomorphism is equal to $\theta$.

Let then $\gamma$ be the unique element of $H^2(\pi_0(G),G_0)$ such that $\gamma \cdot [E/\pi_1(G_0)] =$ [G] (loc. cit., e)). Consider the homomorphism $\delta : H^2(\pi_0(G),Z_0)\rightarrow$ $H^3(\pi_0(G), \pi_1(G))$ associated with the exact sequence of abelian groups $1\rightarrow \pi_1(G_0)\rightarrow$ $\widetilde{Z}_0\rightarrow Z_0\rightarrow 1$ (Exercise 6). Prove that the element $\delta (\gamma )$ of $H^3(\pi_0(G), \pi_1(G))$ does not depend on the choice of the extension $\widetilde{\mathscr{E}}$; denote it by $\delta (G)$.

d) Suppose that $\omega (\pi_0(G),\widetilde{Z}_0, \psi \circ \theta ) = 0$. In order that there exist a Lie group $\widetilde{G}$, a surjective mapping $p:\widetilde{G}\rightarrow G$ and an isomorphism $j$ of $\widetilde{G}_0$ onto the identity component of $\widetilde{G}$ such that $p_0=p\circ j$, it is necessary and sufficient that one have $\delta (G) = 0$. [^1]

[^1]: For examples, see the article of R. L. Taylor, “Covering groups of nonconnected topological groups”, Proc. Amer. Math. Soc 5 (1954), p. 753–768.