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I found this exercise in Tao’s book “Note on Hilbert fifth problem” . Note that if $G$ is connected then $H=G$ since $H$ is also closed and $H\neq\emptyset$ since the neutral element lies in $H$. But if $G$ is not connected I’m a little bit stuck. Tao suggest to take a coordinate chart $(U,\phi:U \to V)$ from the atlas on $H$ and translate it around to create an atlas on $G$ which makes $G$ a Lie group so i was trying to take the atlas $\{(gU,\psi_{g})\}_{g\in G}$ where $\psi_{g}:gU\to V , gu \mapsto \phi(u)$ for all $u$ in $U$. Unfortunately i don’t know how to prove the compatibility of the charts. Sorry for the bad English I’m from Italy :)

Marco Damele
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  • It depends on the exact formulation of the question (which you did not provide) and exact definition of a Lie group that Tao is using. Here is a counter-example when one uses the standard definition of Lie groups: Take $({\mathbb R}, +)$ with the discrete topology. It is not a Lie group but it contains an open Lie subgroup. – Moishe Kohan Feb 21 '24 at 12:16
  • Of course. G has to be a topological manifold, so T2,N2 and locally Euclidean. – Marco Damele Feb 21 '24 at 12:39

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If two charts have disjoint domains, then compatibility of these two charts is an empty condition. Suppose on the other hand $gU\cap hU\neq\emptyset$, ie the domains of two given charts intersect. Then $g^{-1}hU\cap U\neq\emptyset$ and $g^{-1}h\in H$.

Note by definition $\psi_g(u)=\phi(g^{-1}u)$, then $\psi_g\psi_h^{-1}(x) = (\phi g^{-1}h\phi^{-1})(x)$ and you get the coordinate expression of multiplication with $g^{-1}h$, which is an element of $H$ and so smooth.

s.harp
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  • Nice ! Thank you very much. How would you prove that the map μ: G x G —> G , (g,h) |—> g⁻¹h is smooth ? I was thinking μ is smooth as a map H x H —> H since H is a Lie group. This guarantee that μ:G x G—> G is smooth on (Id,Id) and so is smooth for all (g,h) in G x G. – Marco Damele Feb 21 '24 at 10:11
  • Yes. Close to $(g_0,h_0)$ write $(g,h)=(xg_0,yh_0)$ for $x,y$ close to $(1,1)$. Then under this diffeomorphism $\mu$ becomes $g_0^{-1} \mu(x,y) h_0 = (L_{g_0^{-1}}R_{h_0}\circ \mu ),(x,y)$. Since $\mu$ is smooth close to $(1,1)$ this checks smoothness close to $(g_0,h_0)$. Of course you must check smoothness of left and right multiplications $L_g, R_g$. – s.harp Feb 21 '24 at 11:35
  • Ok now is clear. Thank you Oskar :) – Marco Damele Feb 21 '24 at 13:10