Understanding an Example of a Lie Groupoid

Question

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.

Answer 1

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.

Answer 2

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.