Chris L T521
Gold Member
MHB
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Here's this week's problem!
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Problem: Let $G$ and $H$ be two simply connected Lie groups with isomorphic Lie algebras. Show that $G$ and $H$ are isomorphic.
The following theorem can be used without proof in your solution:
Theorem: Suppose $G$ and $H$ are Lie groups with $G$ simply connected, and let $\mathfrak{g}$ and $\mathfrak{h}$ be their Lie algebras. For any Lie algebra homomorphism $\varphi:\mathfrak{g}\rightarrow\mathfrak{h}$, there is a unique Lie group homomorphism $\Phi:G\rightarrow H$ such that $\Phi_{\ast} = \varphi$ (where $\Phi_{\ast}$ denotes the pushforward of $\Phi$).
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Problem: Let $G$ and $H$ be two simply connected Lie groups with isomorphic Lie algebras. Show that $G$ and $H$ are isomorphic.
The following theorem can be used without proof in your solution:
Theorem: Suppose $G$ and $H$ are Lie groups with $G$ simply connected, and let $\mathfrak{g}$ and $\mathfrak{h}$ be their Lie algebras. For any Lie algebra homomorphism $\varphi:\mathfrak{g}\rightarrow\mathfrak{h}$, there is a unique Lie group homomorphism $\Phi:G\rightarrow H$ such that $\Phi_{\ast} = \varphi$ (where $\Phi_{\ast}$ denotes the pushforward of $\Phi$).
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\rightarrow G$ satisfying $\Phi_{\ast}=\varphi$ and $\Psi_{\ast}=\varphi^{-1}$. Both the identity map of $G$ and the composition $\Psi\circ\Phi$ are maps from $G$ to itself whose induced homomorphisms are equal to the identity, so the uniqueness part of the Theorem implies that $\Psi\circ\Phi= \mathrm{Id}_G$. Similarly, $\Phi\circ\Psi = \mathrm{Id}_H$ so $\Phi$ is a Lie group homomorphism.$\hspace{.25in}\blacksquare$[/sp]