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?

  • $\begingroup$ What's the definition of $Lie(G)$? $\endgroup$ Jun 23, 2020 at 15:47
  • 1
    $\begingroup$ $Lie(G)$ represents the left-invariant vector fields $\endgroup$
    – roi_saumon
    Jun 23, 2020 at 16:41
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    $\begingroup$ Do you know the result that the tangent bundle of a product splits as a sum of the tangent bundles to the factors? $\endgroup$ Jun 23, 2020 at 16:58
  • $\begingroup$ No, I didn't. How do you write this result and how do you use it in this case? $\endgroup$
    – roi_saumon
    Jun 23, 2020 at 17:15

2 Answers 2


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.


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 $[\widehat{X},\widetilde{Y}] = 0$.

  • $\begingroup$ 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 "$\widehat{X}_{(g,h)} = (L_{(g,h)})_\ast i_\ast X_e$" $\endgroup$
    – roi_saumon
    Jun 26, 2020 at 15:40
  • $\begingroup$ @roi_saumon: You should check that both give the same definition of $\widehat{X}_{(g,h)}$. $\endgroup$ Jun 26, 2020 at 16:26

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