# Understanding an Example of a Lie Groupoid

For the definition of Lie groupoid see https://en.wikipedia.org/wiki/Lie_groupoid .

In this question I want to understand Example 1.1.17 in "General theory of Lie groupoid and Lie algebroids" by Kirill C. H. Mackenzie (page 10).

Let $G$ be a Lie group with Lie algebra $\mathfrak{g}$. I want to see how $T^* G$ (the cotangent bundle) is a groupoid with base $\mathfrak{g}^*$ (the dual of Lie algebra? which is the dual of $\mathfrak{g}$ as a vector space endowed with Lie brackets?) Let $L_g : h \mapsto gh$ and $R_g : h \mapsto hg$ be left and right translations on $G$.

We consider $\theta \in T_g^\ast G$. This is a linear functional on $T_g G$ the tangent space of $G$ at the point $g$. $T_g G$ is isomorphic to $T_e G = \mathfrak{g}$.

Recall the for every arrow $f : x \to y$ we have source and target maps: $\alpha(f)=s(f)=x$ and $\beta(f)=t(f)=y$. In the example the book defines them as follows $$\alpha(\theta) = \theta \circ T(L_g) , \quad \beta(\theta) = \theta \circ T(R_g) \ .$$ I assume in $T(L_g)$ he means the differential map induced by $L_g$, that is $$T(L_g)(x) = g + x \ .$$ I don't understand the reasoning behind the $\alpha$ and $\beta$ he defined and how they return an object in $\mathfrak{g}^*$. What is happening there?

Finally, the book defines the multiplication of $\theta \in T_g^* G$ and $\varphi \in T_h^*G$ (there are no assumptions on $g$ and $h$) as $$\varphi \bullet \theta = \varphi \circ T(R_{g^{-1}}) = \theta \circ T(L_{h^{-1}}) \ .$$ I don't see how this product is a valid multiplication at all. The books then claim that with these definitions $T^* G \rightrightarrows \mathfrak{g}^*$ is a Lie groupoid. I don't understand how. First we need to show it is a groupoid, then that the relevant maps are smooth etc.

In this post, we'll worry about smoothness of everything. My original post already shows that what you have described is actually a groupoid, so this is the only step remaining in showing it's a Lie groupoid. I am much more comfortable with manifolds than categories, so this post will be much less computationally involved because I grok the big picture a lot better here.

To begin with, I'll use the standard fact:

Proposition: The following two maps $L,R:T^\ast G\rightarrow G\times \mathfrak{g}^\ast$ are vector bundle isomorphisms. With $\theta \in T_g^\ast G$, we have $$L(\theta) = (g, L_g^\ast \theta)$$ and $$R(\theta) = (g, R_g^\ast \theta).$$

We will let $\pi:G\times\mathfrak{g}^\ast\rightarrow \mathfrak{g}^\ast$ denote the other projection.

Using this, we can prove that the source and target maps, $s,t:T^\ast G\rightarrow \mathfrak{g}^\ast$ are smooth submersions. To begin with, we have

Proposition: We have $s = \pi \circ L$ and $t = \pi \circ R$.

Proof: We'll only prove the claim about the source map. For $\theta \in T^\ast_g G$, we have \begin{align} \pi(L(\theta)) &= \pi(g, L^\ast_g \theta) \\ &= L^\ast_g \theta\\ &= \theta \circ d_e L_g \\ &= s(\theta) \end{align}

Now, since $L$ is a bundle isomorphism, it is, in particular, a diffeomorphism. Hence, $\pi$ and $s$ have the same analyitical properties. But $\pi$ is obviously a smooth submersion, so this proves $s$ is. An analogous argument shows $t$ is.

We now argue the identity map $id:\mathfrak{g}^\ast \rightarrow T^\ast G$ is smooth. Recall that for $f\in \mathfrak{g}^\ast$, $\operatorname{id}(f) = f\in T_e^\ast G \subseteq T^\ast G$. Let $i$ denote the inclusion of $\mathfrak{g}^\ast= T^\ast_g G$ into $T^\ast G$. Then we have $\operatorname{id} = L^{-1}\circ i = R^{-1}\circ i$, so $\operatorname{id}$ is a composition of smooth functions so is smooth.

Finally, we argue that composition is smooth. Note that $\operatorname{comp}$ is only a partial function from $T^\ast G\times T^\ast G\rightarrow T^\ast G$, but we'll extend it total function in the following way: For $\theta \in T^\ast_g G$ and $\varphi\in T^\ast_h G$, we define $$\operatorname{comp}(\theta, \varphi) = \theta \circ d_{hg}L_{h^{-1}} = L_{h^{-1}}^\ast \theta.$$ (This may or may not equal $\varphi \circ d_{hg}R_{g^{-1}}$, depending on whether $t(\varphi)= s(\theta)$ or not). We'll show that this total function is smooth, hence, that the partial function is as well.

By considering $L\circ \operatorname{comp} \circ L^{-1}$, we get a map $$(G\times \mathfrak{g}^\ast) \times (G\times \mathfrak{g}^\ast) \rightarrow G\times \mathfrak{g}^\ast$$ which, chasing through all the definitions, maps $\big( (g, \theta), (h,\varphi)\big)$ to $(hg, \theta)$. Since group multiplication is smooth, $L\circ \operatorname{comp}\circ L^{-1}$ is smooth, and thus, so is $\operatorname{comp}$.

Since all the associated maps are smooth and the source and target maps are submersions, this (together with my previous post) shows that $T^\ast G\rightrightarrows \mathfrak{g}^\ast$ really is a Lie groupoid.

• Great answers. One more observation: if you identity $T^*G$ with $G \times \mathfrak g^*$ using your map $L$ and use this to transport the groupoid structure to one on $G \times \mathfrak g^* \rightrightarrows \mathfrak g^*$, you get nothing but the action groupoid for the coadjoint action of $G$ on $\mathfrak g^*$. – Santiago Canez Mar 7 '13 at 15:59
• @Santiago: I'm glad you like it. Honestly, I don't even know what an action groupoid is! Lie groups and their actions I know and love, but this is my first foray into anything involving groupoids. – Jason DeVito Mar 7 '13 at 20:08
• @5PM: Thanks for the bounty! – Jason DeVito Mar 8 '13 at 2:10

First, a disclaimer - I know almost nothing about Lie Groupoids. I'm just following my nose and verifying each of the axioms.

In this post, we just want to show that $T^\ast G \rightrightarrows \mathfrak{g}^\ast$ is a groupoid. Later on, in another post, I'll worry about showing it's actually a Lie groupoid.

First, the objects of our category are elements of $\mathfrak{g}^\ast$, that is, they are linear functionals from $\mathfrak{g}$ to $\mathbb{R}$. (These only depend on the linear structure of $\mathfrak{g}$, not on the algebra structure). Also, to be clear, I'll be identifying $\mathfrak{g}$ with $T_e G$ where $e\in G$ is the identity. The morphisms are, by definition, elements of the cotangent bundle of $G$. That is, each morphism is nothing but a linear functional from $T_g G$ to $\mathbb{R}$.

Now, let $\theta:T_g G \rightarrow \mathbb{R}$ be any such functional. What is $s(\theta)$, the source of $\theta$? By definition, this should be an element of $\mathfrak{g}^\ast$. That is, it should map an element of $\mathfrak{g}$ to $\mathbb{R}$. The author tells you exactly how to do it. If $v\in\mathfrak{g}$, then $$s(\theta)(v) = \theta(d_e L_g (v)).$$ (Be careful, in general, $d_e L_g(v) \neq g+ v$ - often the right hand side doesn't even make sense!)

Notice that this is well defined: First, since both $d_e L_g$ and $\theta$ are linear maps, the composition is linear, so $s(\theta)$ is linear. Second, the map $d_e L_g :\mathfrak{g}\cong T_e G \rightarrow T_g G$, so the image of $d_e L_g$ lies in the domain of $\theta$, so we can compose them. Lastly, the domain of $d_e L_g$ is $T_e G\cong \mathfrak{g}$ and the range of $\theta$ is $\mathbb{R}$, so overall, $\theta \circ d_e L_g$ is really an element of $\mathfrak{g}^\ast$.

What is $t(\theta)$, the target of $\theta$? Again, this should be something linear which accepts in elements of $\mathfrak{g}$ and spits out real numbers. Once again, the author has already given us the formula: $$t(\theta)(v) = \theta(d_e R_g(v)).$$ Just as in the previous paragraph, one can check that this really is well defined.

Now, suppose $\theta \in T^\ast_g G$ and $\varphi\in T^\ast_h G$ are two morphisms and assume further that $t(\theta) = s(\varphi)$. In other words, that $$\theta \circ d_e R_g = \varphi \circ d_e L_h.$$ Using the fact that, for any $k\in G$, $d_k R_g$ and $d_k L_h$ are isomorphisms with inverses $d_{kg} R_{g^{-1}}$ and $d_{hk} L_{h^{-1}}$ respectively, and using the fact that the left and right multiplication commute (and thus, so do their differentials), we get the following chain of equalities: \begin{align} \theta \circ d_e R_g &= \varphi\circ d_e L_h\\ \theta \circ d_e R_g \circ (d_e L_h)^{-1} &= \varphi \\ \theta \circ d_e R_g \circ d_h L_{h^{-1}} &= \varphi \\ \theta \circ d_h(R_g \circ L_{h^{-1}}) &= \varphi \\ \theta \circ d_h(L_h^{-1}\circ R_g) &= \varphi \\ \theta \circ d_{hg} L_{h^{-1}} \circ d_h R_g &= \varphi \\ \theta \circ d_{hg}L_{h^{-1}} &= \varphi \circ (d_h R_g)^{-1} \\ \theta \circ d_{hg}L_{h^{-1}} &= \varphi \circ d_{hg} R_{g^{-1}}\end{align}

The composition of elements of $T^\ast G$ should be another element of $T^\ast G$ and the author tells you that $\theta \bullet \varphi = \theta \circ d_{hg}L_{h^{-1}}$ (which just happens to equal $\varphi \circ d_{hg}R_{g^{-1}}$.) Notice that this is just another element of $T^\ast G$. More specifically, $\theta \bullet \varphi \in T^\ast_{hg} G$.

Now that composition is defined, we need to check several things - that the operation is associative, that every point has a correspond identity morphism, and that every morphism is invertible.

Associativity. Given $\theta,\varphi,$ and $\eta$ with basepoints $g,h,$ and $k$ respectively, we have \begin{align} \theta \bullet (\varphi \bullet \eta) &= \theta \bullet (\varphi \circ d_{kh} L_{k^{-1}}) \\ &= \theta \circ d_{khg}L_{(kh)^{-1}} \\ &= \theta \circ d_{khg}(L_{h^{-1}} L_{k^{-1}}) \\ &= \theta \circ d_{hg}L_{h^{-1}}\circ d_{khg} L_{k^{-1}}\\ &= (\theta \circ d_{hg}L_{h^{-1}})\bullet \eta \\ &= (\theta \bullet \varphi) \bullet \eta \end{align}

Identity. Given $f\in \mathfrak{g}^\ast$, we seek an element $\theta \in T^\ast G$ to act as the identity morphism at $f$, meaning $s(\theta) = t(\theta) = f$. Well, we can think of $\mathfrak{g}^\ast$ as $T_e^\ast G$, so this gives a natural guess for $\theta$: pick $\theta = f\in T_e^\ast G$. Then $s(\theta) = \theta \circ d_e L_e = \theta = f$ and likewise for $t(\theta)$. We also have $\theta \bullet \varphi = \varphi \circ d_h R_{e^{-1}} = \varphi$ and likewise for compositions on the other side.

Inverses. Given a morphism $\theta\in T_g^\ast G$ we want to find an inverse. I claim that $\varphi\in T_{g^{-1}}^\ast G$ with $\varphi = \theta \circ d_{g^{-1}} L_{g^2}$ does the trick. First, note that $\theta \bullet \varphi = \theta \circ d_e L_g$ is a linear map from $\mathfrak{g}^\ast$ to $\mathbb{R}$, that is, it's the identity at the point $\theta\circ d_e L_g \in \mathfrak{g}^\ast = s(\varphi)$. Likewise, composition in the other order works as well.

All this just verifies we actually have a groupoid.

• Thanks for the answer, I just saw it and went over it in a brief. I will go over it more carefully later. – LinAlgMan Mar 11 '13 at 13:27
• I understand from your answer that $\theta \in T_g^*G$ is a morphism $$\theta : \theta \circ d_eL_g \to \theta \circ d_e R_g$$ but the choice of source and target seems somewhat arbitrary. Why couldn't be the other direction: $$\theta : \theta \circ d_e R_g \to \theta \circ d_e L_g ?$$ – LinAlgMan Mar 12 '13 at 12:25
• @LinAlgMan: I actually have no intuition for what I did. I suggest you try it the other direction and see what happens! (If it turns out not to work the other way, one possible source of the asymmetry is that often, the identification between $T_e G$ and $\mathfrak{g}$ is done by using left-invariant vector fields. I used that identification many times.) – Jason DeVito Mar 12 '13 at 12:50
• $$s( \theta \bullet \varphi ) = \theta \circ d_{hg} L_{h^{-1}} \circ d_e L_{hg} = \theta \circ d_e L_g = s(\theta)$$ since the composition of differentials satisfy $$e \to hg \to h^{-1}(hg) = h^{-1} h g = g$$ and thus $$d_{hg} L_{h^{-1}} \circ d_e L_{hg} = d_e L_g$$ and similarly $t( \theta \bullet \varphi) = t(\varphi)$. So $$T_{hg}^* G \ni \theta \bullet \varphi : s(\theta) \to t(\theta) = s(\varphi) \to t(\varphi)$$ where in the bullet multiplication the $\theta$ acts first. – LinAlgMan Mar 12 '13 at 13:02
• Inverses: If $\varphi \in T^*_{g^{-1}}G$ as you wrote then $\theta \bullet \varphi = \theta \circ d_e L_g$ is the identity on $\theta \circ d_e L_g = s(\theta)$ and in the other direction $\varphi \bullet \theta = \theta \circ d_e R_g$ is the identity on $t(\theta)$. Thank you. – LinAlgMan Mar 12 '13 at 13:36