I am reading Brian C Hall's Lie Groups, Lie Algebras, and Representations and trying to do one of the exercises about showing that isomorphic matrix Lie groups give rise to isomorphic Lie Algebras. The book notes that point 2 of Theorem 3.28 gives that $\phi$ is a Lie Algebra homomorphism and I've worked out the proof of isomorphism from there, but I became confused about one point. Def 3.6 requires a Lie algebra isomorphism to be linear, but we only show real-linearity of $\phi$ in Thm 3.28. That is $\phi(tX+Y) = t\phi(X) + \phi(Y)$ for $t \in \mathbb{R}$, but we don't have $\phi(iX) = i\phi(X)$. So how is $\phi$ a Lie algebra homomorphism?

Also is the bilinearity requirement for $[\cdot, \cdot]$ in Def 3.1 saying that $[\cdot, \cdot]$ is complex-bilinear (we can pull out factors of $i$ and factors of $t$ for $t\in \mathbb{R}$) even if we are working with a real Lie algebra or is it saying that $[\cdot, \cdot]$ is real-bilinear (we can pull out $t \in \mathbb{R}$ but not $i$) for real Lie algebras and complex-bilinear for complex Lie algebras?

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    $\begingroup$ But $i\not\in \Bbb R$. A Lie algebra homomorphism is a $K$-linear map, and here $K=\Bbb R$. But I agree, it should say "Let $G$ and $H$ be real linear groups" in Theorem $3.28$. Perhaps this was said before. $\endgroup$ Commented May 8, 2019 at 18:47
  • $\begingroup$ So the linearity requirement on Lie algebra homomorphisms is that the map must be real (complex) linear for real (complex) Lie algebras? $\endgroup$ Commented May 8, 2019 at 18:51
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    $\begingroup$ It must be $K$-linear for a Lie algebra over $K$ (which is a $K$-vector space), for an arbitrary field $K$. $\endgroup$ Commented May 8, 2019 at 18:52

1 Answer 1


In general, the Lie algebra of a matrix Lie group is only a real Lie algebra--it may not be closed under multiplication by i. Thus, it really only makes sense for the homomorphism to be real linear.


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