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Since $T_p G$ is isomorphic to $T_e G$ for all $p\in G$, it makes sense that each vector in $T_p G$ can be identified with a vector in $T_e G$. Hence, to make the map from $TG$ one to one, we must identify each vector in $T_e G$ with a point in $G$, but I'm having trouble saying this precisely.

Thanks so much!

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    $\begingroup$ Careful: On any (connected) manifold, $T_p M$ and $T_q M$ are isomorphic for any choice of $p$ and $q$, but most manifolds don't have $TM\cong M\times T_p M$ for any $p$. Also, I don't understand your second sentence, why must we identify a vector in $T_e G$ with a point in $G$? $\endgroup$ Commented Feb 20, 2013 at 2:11
  • $\begingroup$ I guess I should be more precise: any $v\in TG$ belongs to $T_p G$ for some $p\in G$. But $T_p G \cong T_e G$, so $v=dLg_e(w)$ for some $w\in T_e G$ where $Lg$ is left multiplication by $g$. $\endgroup$ Commented Feb 20, 2013 at 2:22
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    $\begingroup$ Well, you have the key insight to solving the problem - where exactly are you struggling? $\endgroup$ Commented Feb 20, 2013 at 2:26

3 Answers 3

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I assume you mean to prove that $T_e G\times G$ and $TG$ are isomorphic as bundles over $G$. As Jason DeVito points out you have the right idea--you just use the translation maps to slide the local triviality at $e$ around the entire group so that it is globally trivial. So I'm guessing you are having difficulty just getting the details nailed out.

Define $\Phi:G\times T_e G\to TG$ by $\Phi(g,X)=(L_g)_\ast(X)$. Clearly $\Phi$ respects fibers and since $(L_g)_\ast$ is a linear isomorphism (since $L_g$ is a diffeomorphism) it follows that $\Phi$ is bijective.

Let us now prove that $\Phi$ is smooth. We shall see that everything shall be almost tautological once we choose charts correctly. Of course, the choosing of the charts is going to be slightly messy, but not too bad.

Choose any $(g,X)\in G\times T_e G$. Choose then a centered chart $(U,\varphi)$ of $G$ at $e$ and note that this naturally gives us a chart $(V,\psi)$ at $g$ with $V=L_g(U)$ and $\psi=\varphi\circ L_{g^{-1}}$. Now, the combination of these charts gives us a chart $(V\times T_e G,\phi)$ at $(g,X)$ defined by

$$\phi\left(h,\sum_{j=1}^{n}\alpha_i \left.\frac{\partial}{\partial x_i}\right|_e\right)=(\psi(g),\alpha_1,\cdots,\alpha_n)$$

Where, as usual, $\displaystyle \left.\frac{\partial}{\partial x_i}\right|_e$ is the pushforward of $\displaystyle \left.\frac{\partial}{\partial x_i}\right|_0\in T_0\mathbb{R}^n$ through $\varphi^{-1}$.

Recall that the choice of chart $(V,\psi)$ at $g$ gives us a $(\pi^{-1}(V),\tau)$ (where $\pi:TG\to G$ is the canonical projection) given on the fiber $T_h G$ by

$$\sum_{j=1}^{n}\alpha_i\left.\frac{\partial}{\partial x_i}\right|_h\ \longmapsto\ (\psi(h),\alpha_1,\cdots,\alpha_n)$$

where, as above, $\displaystyle \left.\frac{\partial}{\partial x_i}\right|_h$ is the pushforward of $\displaystyle \left.\frac{\partial}{\partial x_i}\right|_{\psi(h)}\in T_{\psi(h)}\mathbb{R}^n$.

So, now we want to show that $\tau\circ\Phi\circ\phi^{-1}$ is smooth. That said,

$$\begin{aligned}\tau\left(\Phi\left(\phi^{-1}\left(\psi(h),\alpha_1,\cdots,\alpha_n\right)\right)\right) &= \tau\left(\Phi\left(h,\sum_{j=1}^{n}\alpha_i\left.\frac{\partial}{\partial x_i}\right|_e\right)\right)\\ &=\tau\left((L_h)_\ast\left(\sum_{j=1}^{n}\alpha_i\left.\frac{\partial}{\partial x_i}\right|_e\right)\right)\\ &= \tau\left(\sum_{j=1}^{n}\alpha_i (L_h)_\ast\left(\left.\frac{\partial}{\partial x_i}\right|_e\right)\right)\\ &= \tau\left(\sum_{j=1}^{n}\alpha_i \left.\frac{\partial}{\partial x_i}\right|_h\right)\\ &= (\psi(h),\alpha_1,\cdots,\alpha_n)\end{aligned}$$

where we have used the fact that, as defined, $\displaystyle (L_h)_\ast\left(\left.\frac{\partial}{\partial x_i}\right|_e\right)=\left.\frac{\partial}{\partial x_i}\right|_h$ (check this!). Thus, it's pretty obvious that $\tau\circ\Phi\circ\phi^{-1}$ is smooth--it's the identity map!

Since $\Phi$ is a bijective bundle map it follows from basic topology that $\Phi$ is a bundle isomorphism


Of course, if you are privy to a little more bundle theory than the definitions there is a simpler way to do all of the above. Namely, recall that a global frame for a vector bundle $\pi:B\to M^n$ is a set global sections $\sigma_i:M\to B$, where $i=1,\cdots,n$, such that for each $x\in M$ we have that $\{\sigma_i(x)\}$ is a basis for $B_x$. A common fact then is that a vector bundle is trivial if and only if it admits a global frame (see here if this is unknown to you).

So, the above ideal of sliding the local trivilizations around the entirety of $G$ becomes even more concrete since we can use local frames (local trivilizations) to construct global ones). Namely, let us a choose a chart $(U,\varphi)$ at $e\in G$ and note that we have a natural map $\sigma:U\to TG$ defined by taking a poing $h\in U$ to the basis $\displaystyle \left\{\frac{\partial}{\partial x_i}\mid_h\right\}$ (as defined above). Let us then extend $\sigma$ to a global section $\sigma:G\to TG$ by defining $\sigma$ on $L_g(U)$ for every $g\in G$ to send $x$ to $(L_g)_\ast(\sigma(g^{-1}x))$. It's not hard to see that this definition is well-defined, and moreover that $\sigma$ is then a global frame for $TG\to G$. Thus, $TG$ is trivial. You'll note that this is practically the same proof as above, just a lot of work hidden in the theorem "trivial bundles are exactly those with global frames".


One last thing I can't help to point out. This theorem (which really was not too hard to prove!) says that every Lie group is parallelizable (has a trivial tangent bundle). This precludes a lot of spaces from possessing a Lie group structure at all! For example, it is the statement of the famous Hairy Ball Theorem that every even dimensional sphere is not parallelizable and thus can't (for any attempted definition of an operation!) be given the structure of a Lie Group. That is pretty cool.

By the way, the only spheres which are Lie groups are $S^0$, $S^1$, and $S^3$.

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  • $\begingroup$ Wow! Thank you for such a detailed answer. I very much appreciate it. $\endgroup$ Commented Feb 20, 2013 at 17:46
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    $\begingroup$ I'm sorry to comment on this old answer, but I don't get the "(check this!)" part. I formulated a question on that in math.stackexchange.com/questions/663681/… $\endgroup$
    – Christoph
    Commented Feb 4, 2014 at 19:38
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    $\begingroup$ @Alex Youcis Your solution is not completely correct. In $\ \frac{\partial}{\partial x_{i}}|_{h}\ $, $\ x_{i} \ $ coordinate comes from the i-th coordinate of the chart map for the neighbourhood with base point h. So, when $\ \tau \ $ is applied on $\ \frac{\partial}{\partial x_{i}}|_{h}\ $ we'll not get $\ e_{i} \ $ because $\ \tau \ $ comes from the coordinate chart map based at g. $\endgroup$
    – Paladin
    Commented Aug 30, 2015 at 13:38
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    $\begingroup$ I agree with Neel. The invariance of the vector fields defined by the basis of the chart is definitely not true (it's true just for $g$ and not $h$!). However the solution of Alex Zorn is correct, namely, using that the inclusion by the zero section $G \hookrightarrow TG$ makes $G$ acts on $TG$ by left translation (via the differential of the multiplication). $\endgroup$
    – user40276
    Commented Oct 12, 2015 at 1:22
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    $\begingroup$ In fact, $(\tau \circ \Phi \circ \phi^{-1}) \big( \psi(h) , d \varphi |_e (u) \big) = \big( \psi(h) , d(\varphi \circ L_{g^{-1} h})|_e (u) \big) \ $ and is clearly not the identity. But is smooth. $\endgroup$
    – Gustavo
    Commented Jul 4, 2016 at 5:37
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Here's a more "category-y" way of seeing why the above map is smooth:

We have a smooth map: $\star\colon G \times G \rightarrow G$

This gives rise to a smooth map, the "derivative": $T{\star}\colon TG \times TG \rightarrow TG$

(recall that $f\colon M \rightarrow N$ gives rise to $Tf\colon TM \rightarrow TN$)

We also have a smooth map $F\colon G \times T_eG \rightarrow TG \times TG$, given by the two smooth maps $0\colon G \rightarrow TG$ (the $0$ vector field) and the inclusion $i\colon T_eG \rightarrow TG$.

It is straightforward to verify that $T{\star} \circ F$ is the same as what Alex Youcis specified above.

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This question is rather old. But since I think Alex Youcis solution is not quite right (or I'm interpreting it wrongly), I will post an answer with explicit computations that solves the case of the cotangent bundle too (by replacing $T$ by $T^{*}$), which cannot be solved by the categorial way presented by Alex Zorn (because multiplication turns into comultiplication!). Furthermore, often people consult this site, so this may be useful to anyone having difficults in this problem.

About my notation, I will use $T$ istead of $d$ for the differential and $T_p(f)$ instead of $d(f)_p$.

Let $m: G \times G \longrightarrow G$ be the multiplication, $\pi: TG \twoheadrightarrow G$ the tangent bundle, $i: G \longrightarrow G$ the inversion and $$\Phi: TG \longrightarrow TG$$ given by $$\Phi (x) = L_{i(\pi (x))}x$$, where $T l_g = L_g: TG \longrightarrow TG$ and $l_g = m\circ i_{g}: G \longrightarrow G$ and $i_g : G \longrightarrow \{ g \} \times G \hookrightarrow G \times G$ the inclusion.

Let $\phi_i : U_i \longrightarrow \mathbb{R}^n$ be charts of $G$ and $\varphi: T\mathbb{R}^n \longrightarrow \mathbb{R}^n \times \mathbb{R}^n$ a diffeomorphism. In this case, $\varphi \circ T\phi_i: TU_i \longrightarrow \mathbb{R}^n \times \mathbb{R}^n$ is a chart.

Locally, $\Phi$ is written as $$\varphi \circ T\phi_j \circ \Phi \circ T \phi_i^{-1} \circ \varphi^{-1} (x, v) = $$ $$= \varphi \circ T\phi_j(L_{i(\pi ( T \phi_i^{-1} \circ \varphi^{-1} (x, v)))} ( T \phi_i^{-1} \circ \varphi^{-1} (x, v))) =$$ $$= \varphi \circ T\phi_j(T(m \circ i_{ i(\pi (T \phi_i^{-1} \circ \varphi^{-1} (x, v) ))})) (T \phi_i^{-1} \circ \varphi^{-1} (x, v))) =$$ $$ =\varphi \circ T (\phi_j \circ \phi_i^{-1} \circ m \circ i_{ i(\pi (T \phi_i^{-1} \circ \varphi^{-1} (x, v) ))}\circ \phi_i^{-1}) \circ \varphi^{-1} (x, v)$$. The unique term that need to be checked if it's indeed smooth is $m \circ i_{ i(\pi (T \phi_i^{-1} \circ \varphi^{-1} (x, v) ))}$. However, for any smooth morphism $\psi: P \longrightarrow G$ (where $P$ is any smooth manifold), $m \circ i_{\psi (-)}: P \times G \longrightarrow G$ is smooth since $i_{\psi ( P)}$ is an inclusion for each $p$. Note that the derivative is not taken with respect to $m (\psi (-), -)$! In other words, for each $p$, $\psi (p) = g$ and just after this $m \circ i_{g} : G \longrightarrow G$ is differentiated. This is because we want $T(m\circ i_g) = L_g$ and not $T (m( \psi \times 1_G))_{p, g} = T_{(\psi( p ) ,g) } m (T_{p}\psi \times 1_{T_{g}G}) = L_{\psi (p )} (1_{T_{g}G}) + R_{g} T_p \psi$.

Now it remains to note that the map $\Phi: TG \longrightarrow TG$ have image in $T_e G$ since $L_{i (\pi (x))} : T_{\pi (x)} G \longrightarrow T_e G$. The surejctivity and injectivity follow easily by algebraic computations and the fact that $L_g$ is an isomorphism of vector bundles and the fact that this representation ($g \mapsto L_g$) of $G$ is a homomorphism.

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