$\operatorname{Lie}(G \times H)\cong \operatorname{Lie}(G)\oplus \operatorname{Lie}(H)$

Question

I am trying to solve an exercise from Lee's Introduction to smooth manifolds book.

8-23. (a) Given Lie algebras $\mathfrak g$ and $\mathfrak h$, show that the direct sum $\mathfrak g\oplus \mathfrak h$ is a Lie algebra with the bracket defined by
$$[(X, Y),(X',Y')]=([X,X'],[Y,Y']).$$
(b) Suppose $G$ and $H$ are Lie groups. Prove that $\operatorname{Lie}(G \times H)$ is isomorphic to $\operatorname{Lie}(G)\oplus \operatorname{Lie}(H)$

The first question I could solve by showing linearity of the lie bracket and the Jacobi identity, using that the jacobi identity is true in $\mathfrak g$ and $\mathfrak h$. But how can I solve the second point?

Answer 1

I know this question already have an accepted answer but I want to post my answer here which maybe have a slightly different approach to the question.

We supposed to find an isomorphism $\phi : \text{Lie}(G) \oplus \text{Lie}(H) \to \text{Lie}(G \times H)$. Our fist guess would be the map $$ \widetilde{\phi} : \mathfrak{X}(G) \oplus \mathfrak{X}(H) \to \mathfrak{X}(G \times H) $$ defined by $\widetilde{\phi}(X,Y) = X\oplus Y$. This map is linear and preserve Lie bracket , with Lie bracket on $\mathfrak{X}(G) \oplus \mathfrak{X}(H)$ defined as in $(a)$ : for any $(X,Y) ,(X',Y') \in \mathfrak{X}(G) \oplus \mathfrak{X}(H)$ we have \begin{align*} \widetilde{\phi}\, \big[(X,Y),(X',Y') \big] &= \widetilde{\phi}\big( [X,X'],[Y,Y'] \big) \\ &= [X,X'] \oplus [Y,Y'] \\ &= [X \oplus Y, X' \oplus Y'] \\ &= [\widetilde{\phi}(X,Y), \widetilde{\phi}(X',Y')]. \end{align*}

So $\widetilde{\phi}$ is a Lie algebra homomorphism . Now we only need to show that the restriction map $ \phi : \text{Lie}(G) \oplus \text{Lie}(H) \to \text{Lie}(G \times H)$ is defined and invertible. If this map is defined (i.e., the image is indeed contained in $\text{Lie}(G \times H)$), then $\phi$ is a Lie algebra isomorphism since $\widetilde{\phi}$ one-to-one and the domain and codomain have the same dimension.

Before show that $\phi$ is defined, I will be a bit pedantic here and remind how vector field $X \oplus Y : G \times H \to T(G \times H)$ defined. For any $(g,h) \in G \times H$ the value $(X \oplus Y)_{(g,h)} \in T_{(g,h)}(G \times H)$ defined as $(X \oplus Y)_{(g,h)} = \alpha^{-1}(X_g,Y_h)$, where $$ \alpha : T_{(g,h)}(G \times H) \to T_gG \oplus T_hH $$ is the isomorphism $\alpha(v) := \Big(d(\pi_G)_g(v), d(\pi_H)_h(v)\Big)$.

So now we want to show that $\phi$ is defined, that is for any $X \in \text{Lie}(G)$ dan $Y \in \text{Lie}(H)$, $X \oplus Y$ is a left-invariant vector field. Denote $L_{(g,h)} : G \times H \to G \times H$ as the left translation on the product $$ L_{(g,h)} (g',h') = (gg',hh') = (L_g\times L_h) (g',h'). $$ Then we must show that for any $(g,h),(g',h')\in G \times H$ we have $$ d(L_{(g,h)})_{(g',h')}(X \oplus Y)_{(g',h')} =d(L_g \times L_h)_{(g',h')}(X \oplus Y)_{(g',h')} = (X \oplus Y)_{(gg',hh')}. $$ To show this, as usual denote $\alpha : T_{(g',h')}(G \times H) \to T_{g'}G \oplus T_{h'}H$ as isomorphism $\alpha(v) = \Big(d(\pi_G)_{g'}(v), d(\pi_H)_{h'}(v)\Big)$ and $\beta : T_{(gg',hh')}(G \times H) \to T_{gg'}G \oplus T_{hh'}H$ as isomorphism $\beta(v) = \Big(d(\pi_G)_{gg'}(v), d(\pi_H)_{hh'}(v)\Big)$. The whole point introducing $\alpha$ and $\beta$ is because the product vector field $X\oplus Y$ defined in terms of this and also for the reason to compute the differential of left-translation on product manifold $L_{(g,h)} = L_g \times L_h$ as we see in the computation below : \begin{align*} d(L_g\times L_h)_{(g',h')} (X \oplus Y)_{(g',h')} &= d(L_g\times L_h)_{(g',h')} \circ \alpha^{-1}(X_{g'},Y_{h'})\\ &=\beta^{-1} \circ \color{blue}{\Big( \beta \circ d(L_g\times L_h)_{(g',h')} \circ \alpha^{-1} \Big)} (X_{g'},Y_{h'}) \\ &=\beta^{-1} \circ \color{blue}{\Big(d(L_g)_{g'}, d(L_h)_{h'} \Big)} (X_{g'},Y_{h'}) \\ &=\beta^{-1}\Big(d(L_g)_{g'}(X_{g'}), d(L_h)_{h'}(Y_{h'}) \Big) \\ &= \beta^{-1}\big( X_{gg'}, Y_{hh'} \big) \\ &= (X \oplus Y)_{(gg',hh')}. \end{align*} Therefore $X \oplus Y$ is a left-invariant vector field on $G \times H$ and $\phi : \text{Lie}(G) \oplus \text{Lie}(H) \to \text{Lie}(G \times H)$ is defined. So $\phi (X,Y) = X \oplus Y$ is a Lie algebra isomorphism.

As you can see, without identification, this calculation is so pedantic (which is kind of a bad thing). But this is the only way i know.