# $Lie(G \times H)\cong Lie(G)\oplus Lie(H)$

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 $$Lie(G \times H)$$ is isomorphic to $$Lie(G)\oplus Lie(H)$$

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?

• What's the definition of $Lie(G)$? – Jason DeVito Jun 23 '20 at 15:47
• $Lie(G)$ represents the left-invariant vector fields – roi_saumon Jun 23 '20 at 16:41
• Do you know the result that the tangent bundle of a product splits as a sum of the tangent bundles to the factors? – Jason DeVito Jun 23 '20 at 16:58
• No, I didn't. How do you write this result and how do you use it in this case? – roi_saumon Jun 23 '20 at 17:15

Given a left invariant vector field (livf) $$X$$ on $$G$$, we can create a livf on $$G\times H$$ as follows. We define $$\widehat{X}_{(g,h)} = (L_{(e,h)})_\ast i_\ast X_g$$ where $$i$$ is the inclusion $$i:G\rightarrow G\times \{e\}\subseteq G\times H$$ and $$L$$ is left multiplication. I'll leave it to you to show $$\hat{X}$$ really is left invariant.

Similarly, we can push livfs on $$H$$ forward to $$G\times H$$. I'll write this as $$Y\mapsto \widetilde{Y}$$.

This gives a map $$\phi:Lie(G)\times Lie(H)\rightarrow Lie(G\times H)$$ given by $$\phi(X,Y) = \widehat{X} + \widetilde{Y}$$. Since $$\phi$$ is given by pushforwards, it is obviously linear. We claim that, in fact, $$\phi$$ is a Lie algebra isomorphism.

To see $$\phi$$ is bijective, note that since the dimension of the source and target match, we need only show that $$\phi$$ is injective. So, assume $$(X,Y)\in Lie(G)\times Lie(H)$$ and $$\phi(X,Y) = 0$$. If we specialize to the point $$(g,h) = (e,e)$$, we see that $$\widehat{X}_{(e,e)} = i_\ast X_e\subseteq T_e G\times \{0\}\subseteq T_{(e,e)}(G\times H)$$. (Here, I am using the fact that on any product manifold $$M\times N$$, we have a natural splitting $$T_{(m,n)} (M\times N) \cong T_m M\oplus T_n N$$, which I alluded to in a comment above.)

In the same fashion, we see that $$\widetilde{Y}_{(e,e)} \in \{0\}\times T_e H\subseteq T_{(e,e)} (G\times H)$$. Since $$\phi(X,Y) = 0$$, $$\widetilde{Y} = -\widehat{X}\in T_e G\times \{0\}$$. Thus, $$\widetilde{Y}\in \left( T_e G\times \{0\}\right) \cap \left( \{0\}\times T_e H\right)$$, so $$\widetilde{Y} = 0$$. Since $$\widehat{X} = -\widetilde{Y}$$, $$\widehat{X} = 0$$ as well. This shows that $$\phi$$ is injective, hence bijective.

Finally, we need to check that $$\phi$$ preserves the bracket. Because $$\phi$$ is given by pushforwards on each factor, it preserves the Lie bracket on pairs of the form $$(X_1,0)$$ and $$(X_2,0)$$, and it also preserves the bracket on pairs of the form $$(0,Y_1)$$ and $$(0,Y_2)$$.

By linearity, it sufficies to check that that $$\phi$$ preserves the Lie bracket on pairs of the form $$(X,0), (0,Y)$$. Of course, in the domain of $$\phi$$, $$[(X,0), (0,Y)]=0$$, so must shows that $$[\widehat{X},\widetilde{Y}] = 0$$. To that end, if suffices to show that the flows of $$\widehat{X}$$ and $$\widetilde{Y}$$ commute.

Fix a point $$(g,h)\in G\times H$$. The $$\widehat{X}$$ flow through $$(g,h)$$ is simply $$\alpha(t) = (g,h)(\exp(tX), e)$$ (because at time $$0$$, we get $$(g,h)$$, and the derivative at time $$0$$ is $$(L_{(g,h)})_\ast (i_\ast X) = \widehat{X}.$$)

Similarly, the $$\widetilde{Y}$$ flow through $$(g,h)$$ is $$\beta(t) = (g,h)(e,\exp(tY))$$. Since $$(e,\exp(tY)$$ and $$(\exp(tX),e)$$ commute (because the identity $$e$$ commutes with everything), the flows commute, so the Lie bracket $$[\widehat{X},\widetilde{Y}] = 0$$.

• Thank you, this helped me a lot. Isn't there a typo in the beginning "$\widehat{X}_{(g,h)} = (L_{(e,h)})_\ast i_\ast X_g$", shouldn't it be "$\widehat{X}_{(g,h)} = (L_{(g,h)})_\ast i_\ast X_e$" – roi_saumon Jun 26 '20 at 15:40
• @roi_saumon: You should check that both give the same definition of $\widehat{X}_{(g,h)}$. – Jason DeVito Jun 26 '20 at 16:26

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.