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Problem of the Week #87 - January 27th, 2014

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Chris L T521

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Jan 26, 2012
995
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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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Chris L T521

Well-known member
Staff member
Jan 26, 2012
995
No one answered this week's problem. You can find the solution below.

Proof: Let $\mathfrak{g}$ and $\mathfrak{h}$ be the Lie algebras, and let $\varphi:\mathfrak{g}\rightarrow\mathfrak{h}$ be a Lie algebra isomorphism between them. Then by the theorem mentioned, there are Lie Algebra $\Phi:G\rightarrow H$ and $\Psi:H\rightarrow G$ 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.$\hspace{.25in}\blacksquare$
 
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