I am reading reading Lee's proof of the Closed Subgroup Theorem (Theorem $20.12$) and there is some point in the proof I really do not understand. As the proof is quite long, I'm not going to rewrite the whole proof, just the part I have a problem with. Let me recall the setting:
Let $G$ be a Lie group, $\mathfrak{g}$ its Lie algebra, $\mathfrak h \subseteq \mathfrak g$ a subspace and $\mathfrak b$ the complementary to $\mathfrak h$ in $\mathfrak g$ (so that $\mathfrak g = \mathfrak h \oplus \mathfrak b$). We know that there is a neighborhood $U_0$ of $0$ in $\mathfrak g$ such that $\exp|_{U_0}$ is a diffeomorphism and we can show that the map
$$\Phi: \mathfrak h \oplus \mathfrak b \to G : (X, Y) \mapsto \exp X\cdot \exp Y$$
is a diffeomorphism in some neighborhood $\widetilde{U}_0$ of $(0,0)$. Now the author says that we can choose a countable neighboorhood basis $(U_i)_i$ for $\mathfrak g$ at $0$ such that the $U_i$'s consist of coodinate balls whose radii approach zero (1. , here is my first question). After we set $V_i = \exp (U_i)$ and $\widetilde U_i = \Phi^{-1}(V_i)$ which are neighborhood bases for $G$ at $0$ and $\mathfrak h \oplus \mathfrak b$ at $(0,0)$ respectively (2., here is my second question). By discarding finitely many terms at the beginning of the sequence, we may assume that $U_i \subseteq U_0$ and $\widetilde U_i \subseteq \widetilde U_0$ for each (this is obvious if we assume that 1. is true). Finally, we consider $(X_i, Y_i) \in \widetilde{U}_i$ for each $i$ and as $\widetilde{U}_i$ is a neighborhood basis, $Y_i \to 0$ for $i \to \infty$ (3., here is my last question).
Now let met detail my questions:
- if I understand the sentence correctly, the sequence $(U_i)_i$ should generate every open of $\mathfrak g$ passing through $0$. How is it possible that the radius of $U_i$ goes to zero as the $U_i$'s should also generate the very big open sets of $\mathfrak g$ passing through $0$ ?
- I do not see why it should be bases as, at that point, we did not suppose $U_i \subseteq U_0$ so that $\exp$ and $\Phi$ are not diffeomorphisms.
- I really do not understand this point.. Can we deduce that the radii of $\widetilde U_i$ goes to zero if it is the case for $U_i$ ?
I hope this question is clear, any help is welcome!
Best Answer
All your questions actually concern general topology, they are not really related to manifolds or Lie groups.
The fact that $\{U_i\}$ is a neighborhood basis for $0$ by definition means that any neighborhood of $0$ contains one of the sets $U_i$. Otherwise put, any open subset that contains $0$ has to contain one of the $U_i$. This is different from the notion of a basis for the topology.
By construction, the neighborhood basis consist of nested sets, so you can assume that $U_j\subset U_i$ for $j>i$. This means that dropping the first $N$ sets $U_i$, one still has a neighborhood basis. Since one $U_{i_0}$ is contained in $U_0$, you can assume without loss of generality that they all are.
In a Hausdorff space, the intersection of all elements of any neighborhood basis of a point $x$ consists of the point $x$ only. By construction $Y_j\in U_i$ for any $j\geq i$. This readily implies convergence by definition.