Isomorphic Lie groups with isomorphic Lie algebras?

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SUMMARY

The discussion centers on proving that two simply connected Lie groups, $G$ and $H$, with isomorphic Lie algebras $\mathfrak{g}$ and $\mathfrak{h}$ are isomorphic. Utilizing the theorem that states for any Lie algebra homomorphism $\varphi:\mathfrak{g}\rightarrow\mathfrak{h}$, there exists a unique Lie group homomorphism $\Phi:G\rightarrow H$, the proof establishes that the compositions of the homomorphisms yield the identity maps on both groups. Therefore, $G$ and $H$ are indeed isomorphic as Lie groups.

PREREQUISITES
  • Understanding of Lie groups and Lie algebras
  • Familiarity with the concept of homomorphisms in algebra
  • Knowledge of simply connected spaces in topology
  • Comprehension of the uniqueness properties of homomorphisms
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  • Study the properties of simply connected Lie groups
  • Explore the implications of Lie algebra isomorphisms
  • Learn about the relationship between Lie groups and their representations
  • Investigate advanced topics in algebraic topology related to homotopy
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Mathematicians, particularly those specializing in algebra, topology, and differential geometry, as well as graduate students studying advanced topics in Lie theory.

Chris L T521
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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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No one answered this week's problem. You can find the solution below.

[sp]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.$\hspace{.25in}\blacksquare$[/sp]
 

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