Exercise 8, § 6

Let $H$ be a complex Hilbert space of dimension $n_0$ and $G$ the unitary group of $H$ considered as a real Lie group. The group $G$ is simply connected.\footnote{Cf. N. H. Kuiper, The homotopy type of the unitary group of Hilbert space, Topology, 3 (1965), pp. 19–30. In this article it is even proved that $G$ is contractible.}

(a) Let $Z$ be the centre of $G$, which is isomorphic to $\mathbf{T}$. Let $a$ be an irrational number. Let $\mathfrak{s}$ be the Lie subalgebra of $L(G) \times L(G)$ consisting of the $(x, ax)$ where $x \in L(Z)$. Let S be the corresponding integral subgroup of $G \times G$. Then $\mathfrak{s}$ is an ideal of $L(G) \times L(G)$; however S is not closed and is dense in $Z \times Z$.

(b) Let $g = (L(G) \times L(G))/\mathfrak{s}$. There exists no Lie group with Lie algebra g. (Let H be such a group. As G is simply connected, there exists a morphism $\phi : G \times G \to H$ such that $L(\phi)$ is the canonical mapping of $L(G) \times L(G)$ onto g. Let N = Ker $\phi$. Then $L(N) \supset \mathfrak{s}$, hence N $\supset S$ and therefore N $\supset Z \times Z$ by (a). Then $L(N) \supset L(Z) \times L(Z)$, which is absurd.)