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Let $\phi:\mathcal{G}\rightarrow \mathcal{K}$ and $\psi:\mathcal{H}\rightarrow \mathcal{K}$ be morphisms of Lie groupoids.

We define weak pullback/2-fibre product corresponding to $\phi:\mathcal{G}\rightarrow \mathcal{K}$ and $\psi:\mathcal{H}\rightarrow \mathcal{K}$ to be a groupoid $\mathcal{G}\times_{\mathcal{K}}\mathcal{H}$ whose object set is $$(\mathcal{G}\times_{\mathcal{K}}\mathcal{H})_0=\{(a,\alpha,b)| a\in \mathcal{G}_0, b\in \mathcal{H}_0,\alpha:\phi(a)\rightarrow \psi(b) \in \mathcal{K}_1\}.$$ Given $(a,\alpha,b ),(a',\alpha',b')\in (\mathcal{G}\times_{\mathcal{K}}\mathcal{H})_0$ we declare $$\text{Mor}((a,\alpha,b),(a',\alpha,b'))=\{u:a\rightarrow a'\in \mathcal{G}_1, v:b\rightarrow b'\in \mathcal{H}_1| \alpha'\circ \phi(u)=\psi(v)\circ \alpha\}.$$

We then see that $$(\mathcal{G}\times_{\mathcal{K}}\mathcal{H})_0= \mathcal{G}_0\times_{\phi,\mathcal{K}_0,s}\mathcal{K}_1\times_{t\circ pr_2,\mathcal{K}_0,\psi}\mathcal{H}_0$$

$$(\mathcal{G}\times_{\mathcal{K}}\mathcal{H})_1=\mathcal{G}_1\times_{t\circ \phi,\mathcal{K}_0,s}\mathcal{K}_1\times_{t,\mathcal{K}_0,s\circ \psi}\mathcal{H}_1.$$ Source, target maps are given by $$s(u,\gamma,v)=(s(u),\gamma\circ \phi(u),s(v))$$ $$t(u,\gamma,v)=(t(u),\psi(v)\circ \gamma, t(v))$$

Moerdijk (in page no $5$) says that assuming $t\circ pr_2:\mathcal{G}_0\times_{\mathcal{K}_0}\mathcal{K}_1\rightarrow \mathcal{K}_0$ is a submersion confirm that $\mathcal{G}\times_{\mathcal{K}}\mathcal{H}$ is a Lie groupoid. All I can see is $(\mathcal{G}\times_{\mathcal{K}}\mathcal{H})_0$ and $(\mathcal{G}\times_{\mathcal{K}}\mathcal{H})_1$ are manifolds (being pullbacks under submersions) if $t\circ pr_2:\mathcal{G}_0\times_{\mathcal{K}_0}\mathcal{K}_1\rightarrow \mathcal{K}_0$ is a submersion.

But, what I do not understand is, why does $s,t$ are smooth?

For source map, first and third coordinates are submersions. Second projection $\gamma\circ \phi(u)$ does not seem to be submersion, unless I assume $\phi:\mathcal{G}_1\rightarrow \mathcal{K}_1$ is a submersion. Similarly, to prove target map is submersion, first and third projections are submersions but second projection is $\psi(v)\circ \gamma$ does not seem to be submersion unless I assume $\psi:\mathcal{H}_1\rightarrow \mathcal{K}_1$ is a submersion.

I think we should also assume $\phi:\mathcal{G}_1\rightarrow \mathcal{K}_1$ and $\psi:\mathcal{G}_1\rightarrow \mathcal{K}_1$ are submersions.

Does it follow with out assuming $\phi,\psi$ are submersions?

If $\phi$ is submersion, then, $$\mathcal{G}_1\times_{t\circ \phi,\mathcal{K}_0,s} \mathcal{K}_1\rightarrow \mathcal{K}_1\times_{t,\mathcal{K}_0,s}\mathcal{K}_1$$ given by $(u,\gamma)\mapsto (\phi(u),\gamma)$ is submersion. As multiplication map $\mathcal{K}_1\times_{t,\mathcal{K}_0,s}\mathcal{K}_1\rightarrow \mathcal{K}_1$ is submersion (thanks to David Roberts), the composition map given by $$(u,\gamma)\mapsto (\phi(u),\gamma)\mapsto \gamma\circ \phi(u)$$ is submersion.

So, for $(u,\gamma,v)\mapsto \gamma\circ \phi(u)$ to be submersion, we need $\phi$ to be submersion even after using multiplication map is submersion.

What am I missing here?

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    $\begingroup$ The composition map in a finite-dimensional Lie groupoid is a submersion. I've not seen this anywhere, I had to prove it myself. $\endgroup$ – David Roberts Feb 15 at 12:40
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    $\begingroup$ Oh.. Thanks. I will check it :) @DavidRoberts $\endgroup$ – Praphulla Koushik Feb 15 at 13:23
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    $\begingroup$ @DavidRoberts I see why it is submersion... Given a Lie group $G$ the multiplication map $m:G\times G\rightarrow G$ is a submersion... Same idea for Lie groupoid also.. the multiplication map (composition) is submersion... so source, target maps for groupoid mentioned above are submersions... $\endgroup$ – Praphulla Koushik Feb 15 at 15:48
  • $\begingroup$ Allow me to consider a special case of this question: is it necessary for $M \to N \leftarrow M'$ to be submersions for $M\times_N M'$ to be a manifold? In fact no, since transversality is a sufficient condition. And, if $M$ is a closed submanifold of $N$, and $M'=M$, then $M\times_N M = M\cap M = M$ is a manifold, and the derivative of both inclusions is nowhere surjective. $\endgroup$ – David Roberts Mar 27 at 21:37
  • $\begingroup$ @DavidRoberts that’s true. For pullback to be manifold, all we need is intersection is transversal.. Here I am trying to see if $t\circ pr_2: \mathcal{G}_0\times_{\mathcal{K}_0}\mathcal{K}_1\rightarrow \mathcal{K}_0$ being submersion is sufficient to prove $\mathcal{G}\times_{\mathcal{K}}\mathcal{H}$ is a Lie groupoid.... $\endgroup$ – Praphulla Koushik Mar 29 at 3:21
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This answer is too big to fit into a comment, and doesn't quite answer the question.

Lemma Given a diagram $$ \array{ U & \to & V & \leftarrow &W \\ \downarrow && \downarrow && \downarrow \\ X & \to & Y & \leftarrow & Z } $$ of finite dimensional manifolds and submersions, then the induced map $U\times_V W \to X\times_Y Z$ is a submersion.

This is proved by taking $(u,w) \in U\times_V W$ and looking at the diagram of tangent spaces and seeing that the induced linear map at $(u,w)$ is surjective.

You can iterate this lemma to deal with the case of maps between limits of longer zig-zags, like $A \to B \leftarrow C \to D \leftarrow E$.

The source and target maps for the groupoid you are looking for are almost examples of such maps between pullbacks. What you need is a generalisation of this above lemma that applies in your situation. Do the source map, without loss of generality, the target map then follows immediately. If you pick a point in the arrow space of the pullback, look at the diagram of tangent spaces at all the other points in the image of that point. You will get a diagram of vector spaces, and it should be possible to see why the map you are interested in is then surjective. As this was for a generic point, the source map has surjective derivative everywhere.

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  • $\begingroup$ In the diagram, it should be sufficient that the downward arrows are submersions, as well one pair of the horizontal arrows, that is, either both left-pointing or both right-pointing arrows, so that the pullbacks exist. $\endgroup$ – David Roberts Mar 29 at 6:25
  • $\begingroup$ I will write down the details and will reply soon. Thank you :) $\endgroup$ – Praphulla Koushik Apr 4 at 7:24

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