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On page 88 of Hall's Lie Groups, Lie Algebras, and Representations the following is stated:

It is easy to check that the Lie algebra of $G\times H$ is isomorphic to the direct sum of the Lie algebra of $G$ and the Lie algebra of $H$.

$G$ is a closed subgroup of $GL(n;\mathbb{C})$ and $H$ is a closed subgroup of $GL(m;\mathbb{C})$.

How can I check this statement without resorting to differential geometry arguments? (Hall uses a minimal amount of differential geometry in this book.)

I'm imagining $G\times H$ as a closed subgroup of $GL(n+m,\mathbb{C})$ with block diagonal components of size $n$ and $m$. I've tried to see how I could use the exponential map to relate the Lie algebra to the Lie group but I'm not sure how to proceed.

(Note that there is a related answer here but this uses differential geometry language that Hall does not rely on and I do not want to use.)

Thanks for the help!

fred
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  • The answer you have linked is perfectly fine. You cannot assume that $G$ and $H$ are already matrix groups. – Dietrich Burde May 24 '22 at 16:33
  • @DietrichBurde, thanks for the comment. I'd edited my question to reflect that I'm looking for an explanation that doesn't leverage notions from differential geometry. The answers that are linked discuss vector fields, tangent spaces, left-translations, push forwards, and so on but I have the impression that Hall has in mind an explanation that doesn't require this machinery. (Of course, hopefully I'll feel more comfortable with those diff'l geom ideas before too long!) – fred May 24 '22 at 17:36
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    Hall avoids differential geometry machinery by only considering matrix Lie groups. In that setting the proof is easy: $e^{X+Y} = (e^{X},e^{Y}) \in G \times H$ for $X \in \mathfrak{g}$, $Y \in \mathfrak{h}$. It is a good idea to understand some of the differential geometry machinery though (at least as a rough sketch). – Callum May 24 '22 at 18:05
  • @Callum, that's helpful. Just to make sure I'm 100% clear, I want to be careful. When you write $X+Y$, this means the direct sum $(X,Y)$. Applying the rules for direct sums to the series yields the RHS of the equation. It has been shown that exponentiating a direct sum of elements of the two Lie algebras gives a direct sum of exponentiated Lie algebra elements. Correct? I'm confused by the transition from direct sum for the algebra and direct product for the groups. (The categorical notions of product and coproduct are familiar to me but I haven't been able to use this to help.) – fred May 24 '22 at 18:44
  • @Callum Also, in order to get the isomorphism in the quoted statement of the question, I would need to show that ${X+Y}$ with $X\in \mathfrak{G}$ and $Y\in \mathfrak{H}$ captures all the elements in the Lie algebra of $G\times H$. So I think I must show something more than $(e^X, e^Y)\in G\times H$. My idea is that the dimension of the Lie alg for $G\times H$ is equal to the dim of the Lie alg of $G$ times the dim of the Lie alg of $H$. But this is just the same as the dimension of the direct sum of those to Lie algebras. Does this seem correct? – fred May 24 '22 at 18:55
  • Try exponentiating a block matrix and see what you get. You should find that to give you a block diagonal matrix you need to also start with one. Also the dimension is not the product but the sum of the two dimensions. – Callum May 24 '22 at 20:37
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    @fred It is also worth noting that the direct sum of Lie algebras and the direct product of Lie groups are both just Cartesian products so there isn't really anything to transition between. Alternatively you can view both via block diagonal matrices – Callum May 25 '22 at 07:38
  • Ah, because block diagonal matrices have invariant subspaces, there is an isomorphism that allows us to study Cartesian products of matrices as block diagonal matrices. I can then form the Lie algebra for $G\times H$ as a block diagonal matrix which is the direct sum of the respective Lie algebras of $G$ and $H$. Thanks @Callum, this has been really useful for me. – fred May 25 '22 at 13:10

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