3
$\begingroup$

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?

$\endgroup$
4
  • $\begingroup$ What's the definition of $Lie(G)$? $\endgroup$ – Jason DeVito Jun 23 '20 at 15:47
  • $\begingroup$ $Lie(G)$ represents the left-invariant vector fields $\endgroup$ – roi_saumon Jun 23 '20 at 16:41
  • $\begingroup$ Do you know the result that the tangent bundle of a product splits as a sum of the tangent bundles to the factors? $\endgroup$ – Jason DeVito Jun 23 '20 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 '20 at 17:15
3
$\begingroup$

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

$\endgroup$
2
  • $\begingroup$ 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 '20 at 15:40
  • $\begingroup$ @roi_saumon: You should check that both give the same definition of $\widehat{X}_{(g,h)}$. $\endgroup$ – Jason DeVito Jun 26 '20 at 16:26
4
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.