# Lectures on Lie Groups by Dragan Milicic

By Dragan Milicic

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Extra resources for Lectures on Lie Groups

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It follows that Un , n ∈ N, are open in M . Moreover, we have ∞ ∞ Un = n=1 ∞ (M − (gn V ) · m) = M − n=1 (gn V ) · m = ∅. 4, at least one Un cannot be dense in M . Hence M − V · m = τ (gn−1 )(M − (gn V ) · m) is not dense in M . It follows that V · m has a nonempty interior. Assume that g · m, g ∈ V , is an interior point of V · m. Then (g −1 V ) · m is a neighborhood of m. Therefore, (g −1 V ) · m ⊂ V 2 · m ⊂ U · m = ρ(m)(U ) is a neighborhood of m ∈ M . This establishes our claim. Assume now that U is an arbitrary open set in G.

Proof. By our assumption, H/H0 = {1}. 4 by induction in n ∈ N. If n = 1, SO(1) = {1} and the statement is obvious. Moreover, S n−1 is connected for n ≥ 2. By inducition, we can assume that SO(n − 1) is connected. As we remarked above, SO(n)/ SO(n − 1) is diffeomorphic to S n−1 . 6. Consider now V = Cn . The algebra of complex linear transformations on V can be identified with the algebra Mn (C) of n × n complex matrices. It can be viewed as a real algebra with identity. The corresponding group of regular elements is the group GL(n, C) of regular matrices in Mn (C).

Therefore, H is closed under multiplication. , H is a subgroup of G. Since V ⊂ H, H is a neighborhood of the identity in G. , H is open in G. This implies that the complement of H in G is a union of H-cosets, which are also open in G. Therefore, H is also closed in G. Since G is connected, H = G. This result has the following consequence. 2. Corollary. Let G be a connected Lie group. Then G is separable. Proof. Let U be a neighborhood of 1 which is domain of a chart. Then, U contains a countable dense set C.