1
$\begingroup$

Today I was explaining to some one the notion of $\mathcal{G}$ spaces, covering spaces over orbifolds from Orbifolds as Groupoids: an Introduction.

Definition : Let $X$ be a locally compact Hausdorff space. An orbifold structure on $X$ is given by an orbifold groupoid $\mathcal{G}$ and a homeomorphism $f:|\mathcal{G}|\rightarrow X$. If $\mathcal{H}\rightarrow\mathcal{G}$ is a equivalnece, then $|\phi|:|\mathcal{H}|\rightarrow |\mathcal{G}|$ is a homeomorphism and the composition $f\circ|\phi|:|\mathcal{H}|\rightarrow |\mathcal{G}|\rightarrow X$ is viewed as defining an equivalent orbifold structure on $X$. An orbifold $\underline{X}$ is a space $X$ equipped with an equivalnece class of orbifold structures.

I have given definition of a $\mathcal{G}$-space as a smooth manifold $E$ with smooth maps $\pi:E\rightarrow \mathcal{G}_0$ and $\mu:E\times_{\mathcal{G}_0}\mathcal{G}_1\rightarrow E$ behaving like a group action map.

Then, I said, given a morphism of Lie groupoids $\phi:\mathcal{H}\rightarrow \mathcal{G}$ there is a functor from category of $\mathcal{G}$ spaces to the category of $\mathcal{H}$ spaces $\phi^*:(\mathcal{G}-\text{spaces})\rightarrow (\mathcal{H}-\text{spaces})$.

Definition : A covering space over a groupoid $\mathcal{G}$ is a $\mathcal{G}$ such that the map $\pi:E\rightarrow \mathcal{G}_0$ is a covering projection.

Then, I said the equivalnece of categories $\phi^*:(\mathcal{G}-\text{spaces})\rightarrow (\mathcal{H}-\text{spaces})$ restircted to covering spaces is still an equivalence. So, I can talk about covering space over an orbifold.

Definition : A covering space over an orbifold $\underline{X}$ is a covering space over Lie groupoid $\mathcal{G}$ representing $\underline{X}$ (in the sense defined above).

This notion is well defined : Suppose there is another Lie groupoid representing $\underline{X}$ in the equivalnece class, say $\mathcal{H}$ then, this $\mathcal{H}$ has to be morita equivalent with $\mathcal{G}$. So, category of covering spaces over $\mathcal{G}$ would be same(equivalent) as that of category of covering spaces over $\mathcal{H}$. So, the notion makes sense.

Then, he asked

"why do you do this much just to define the notion of covering space over a space $X$?"...

I did not think about this before and said at that moment that, $X$ is not merely a topological space in which case you can define covering map to be just local homeomorphism plus something, the usual definition.

One standard example in the classical notion of orbifold is quotient space of a manifold by a Lie group $M/G$.

Suppose you want to define the notion of covering space over $M/G$, you can not treat $M/G$ as simply a topological space, in which case you can define covering space to be a map $\pi:E\rightarrow X=M/G$ such that given $x\in X$ there is an open set $U$ containg $x$ and $\pi^{-1}(U)$ is disjoint union $\bigsqcup V_\alpha$ where $V_\alpha$ is mapped homeomorphically onto $U$ under $\pi$. You can not even treat $M/G$ as a smooth manifold where you would define smooth covering map just by replacing homeomorphically in above definition by diffeomorphically. In general $M/G$ is not a smooth manifold. It is more than a topological space and less than a smooth manifold. It is not easy/obvious to define what is a covering space over $M/G$ would be.. So, to make sense of covering spaces, we have to go to orbifold groupoid setup where notion of covering space would be just simple as above.

To summarize I said the following :

Orbifolds are neither just topological space where you can define covering maps to be local homeomorphisms plus something nor smooth manifolds where you can define covering maps to be local diffeomorphisms plus something. Orbifolds are something more than just a topological space and less than a smooth manifold. So, you need a separate approach to define covering spaces and groupoid approach is useful/straight-forward.

I just want to know if what I have said is actually a reason or did I misunderstood the idea.

Any comments are welcome.

Edit: This is just to bump this question up so that it gets some attention and in turn some comments.

$\endgroup$

Your Answer

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

Browse other questions tagged or ask your own question.