1
$\begingroup$
  • Given a manifold $M$ we have a geometric stack associated to it namely $\underline{M}$ whose objects are smooth maps to $M$. For the sake of consistency I am writing $BM$ for $\underline{M}$.

  • Given a Lie group $G$ we have a geometric stack associated to it namely $BG$ whose objects are principal $G$ bundles.

  • Given a Lie groupoid $\mathcal{G}$ we have a geometric stack associated to it namely $B\mathcal{G}$ whose objects are principal $\mathcal{G}$ bundles.

These are called as Yoneda embedddings (I do not have precise reference where it is called so, except for manifolds corollary $4.16$).

Given a smooth map $f:M\rightarrow N$, if it is a submersion, then, $M\times_NM$ is a manifold. We have $2$-fibre product $\underline{M}\times_{\underline{N}}\underline{N}$ and the stack $\underline{M\times_NM}$.

I am able to see that $\underline{M\times_NM}\cong \underline{M}\times_{\underline{N}}\underline{M}$. We have $B(M\times_NM)\cong BM\times_{BN}BN$.

David Roberts say here that same holds in case of Lie groups and Dimitri Pavlov say here that same holds for Lie groupoids i.e., we have following.

  • Given a morphism of Lie groups $\theta:G\rightarrow H$ which is a surjective submersion (submersion is to ensure $G\times_H G$ is a Lie group), then $$B(G\times_HG)\cong BG\times_{BH}BG.$$

  • Given a morphism of Lie groupoids $f:\mathcal{G}\rightarrow \mathcal{H}$ such that the fibered product (in page no $5$, section $2.3$) $\mathcal{G}\times_{\mathcal{H}}\mathcal{G}$ is a Lie groupoid, then $$B(\mathcal{G}\times_{\mathcal{H}}\mathcal{G})\cong B\mathcal{G}\times_{B\mathcal{H}}B\mathcal{G}$$

Dmitri Pavlov said here that this has something to do with Preservation of limits by the Yoneda embedding and suggested this and this. But I am not familiar with $(\infty,1)$ categories. So, I am asking here (I am asking as a separate question).

How does one see that Yoneda embedding preserves limits in this setup? Please see my answer, I see the case in classical category theory.

Just an outline is also ok, just that it would be good if it is not mixed with $(\infty,1)$ categories.

$\endgroup$
  • $\begingroup$ They are not quite Yoneda embeddings: Yoneda is for presheaves. The 2-functor $LieGroupoids \to DiffStacks$ is not a full embedding, since not all maps between stacks come (even up to isomorphism) from functors between Lie groupoids. What you describe is the composite of $LieGroupoids \to Gpd^{Mfld^{op}}$, the 2-categorical Yoneda embedding, and stackification $Gpd^{Mfld^{op}} \to Stack(Mfld)$, and happens to factor through $DiffStacks \subset Stack(Mfld)$. $\endgroup$ – David Roberts Jan 22 at 23:25
  • $\begingroup$ Also, Yoneda preserves limits, and so does stackification, up to equivalence. Limits here should be suitably 2-categorical. And, in fact, what is usually referred to as the weak or homotopy pullback is in this setting the comma object (which coincides with the isocomma object as all 2-arrows in this setting are invertible). Because manifolds give representable presheaves, and the open cover topology is subcanonical, representable presheaves are sheaves and hence stacks, so the 2-functor $Mfld \to DiffStacks$ really is the Yoneda embedding: manifolds form a full subcategory of $DiffStacks$. $\endgroup$ – David Roberts Jan 22 at 23:30
  • $\begingroup$ Given a morphism of Lie groups $\theta\colon G\to H$ which is a submersion <-- no, when $\theta$ is a surjective submersion. When you say "it turns out", you should link to your source for this to aid other people who are learning, and so that people can check your assertion. :-) \\ Given a morphism of Lie groupoids $f\colon \mathcal{G} \to \mathcal{H}$ that intersects transversally <-- this is meaningless, what does it mean for a single morphism to "intersect transversally"? Section 2.3 of the notes of Moerdijk to which you link never uses the word "transversal[ly]". $\endgroup$ – David Roberts Jan 22 at 23:35
  • 1
    $\begingroup$ I should not have asked what is $\text{Gpd}^{\text{Mfld}^{op}}$.. I should have guessed it... they are functors $\text{Mfld}^{\text{op}}\rightarrow \text{Gpd}$... given Lie gorhpoid $\mathcal{G}$ we have a functor $B\mathcal{G}:\text{Mfld}^{\text{op}}\rightarrow \text{Gpd}$ given by $U\mapsto B\mathcal{G}(U)$ this is contravariant so you have written $\text{op}$.. @DavidRoberts this is for your functor category comment $\endgroup$ – Praphulla Koushik Jan 22 at 23:58
  • 1
    $\begingroup$ I'm done for now. $\endgroup$ – David Roberts Jan 23 at 1:28
1
$\begingroup$

In an arbitrary category $C$, if a functor $F: I\to C$ has a limit $L$ with projections $\pi_i, i\in \mathrm{Ob}(I)$, then we have a natural (in $X$) isomorphism in $\hom_C(X, L)\to \lim_i\hom_C(X,F(i))$.

In fact, this is better stated as : $(\pi_i\circ - : \hom_C(X,L)\to \hom_C(X, F(i)))_i$ is a limite cone.

Now if $I$ is small, $\mathbf{Set}^I$ makes sense, and limits in this category are computed pointwise, so that $(\pi_i\circ - : \hom_C(-,L)\to \hom_C(-, F(i)))_i$ is a limit cone in $\mathbf{Set}^I$ (a limit cone of functors).

What this says, with less precision on which maps we use, is $\hom_C(-,\lim F) \simeq \lim_i\hom_C(-,F(i))$.

But the Yoneda embedding is precisely $y : A\mapsto \hom_C(-,A)$, $f\mapsto f\circ -$ (for it to make sense as a functor we have to make sense of $\mathbf{Set}^C$, which can be problematic if $C$ isn't small, but let's not think about that), so what we said above can be restated : $y$ sends limit cones to limit cones, or again with less precision on the maps $y(\lim F) \simeq \lim y\circ F$, which is exactly the definition of preserving limits.

Now to prove the claim about the limit cone : let $(A,(p_i))$ be any cone over $i\mapsto \hom_C(X,F(i))$ in $\mathbf{Set}$. Fix $a\in A$. Then $(p_i(a))_i$ is a family of maps, and since $(p_i)$ is a cone, for each $f:i\to j$ it satisfies $\hom_C(X,F(f))(p_i(a)) = p_j(a)$, so $F(f)\circ p_i(a) = p_j(a)$.

enter image description here

Thus $(p_i(a))_i$ is actually a cone over $F$ with domain $X$; so it factors uniquely through some $p(a) : X\to L$, we then have $\pi_i\circ p(a) = p_i(a)$, in other words $(\pi_i\circ -)(p(a)) = p_i(a)$.

Define $p:A\to \hom_C(X,L)$ this way, and then the above equation tells us $(\pi_i\circ -)\circ p = p_i$, which tells us exactly that our initial cone factors through the intended one. It's not hard to check that this factorization is unique; so it is indeed a limit cone.

As stated, this has nothing to do with homotopy limits and $(\infty, 1)$-categories, though.

$\endgroup$
  • $\begingroup$ Please see my answer and let me know if I got it correctly... Thank you... This only says about $M\mapsto \underline{M}$ which is of the form $\text{Hom}(-,M)$.. Can you please say aomthing about $\mathcal{G}\mapsto B\mathcal{G}$.. How to (Do we) see them as $\text{Hom}(-,\mathcal{G})$? Can what you have written be extended for Lie groups? $\endgroup$ – Praphulla Koushik Jan 22 at 21:23
  • 1
    $\begingroup$ Yes you got it correctly; I don't know about $B\mathcal{G}$ though, I don't know what it is $\endgroup$ – Max Jan 22 at 21:36
  • $\begingroup$ Thank you Max. Are you familiar with what is called Homotopy limit? $\endgroup$ – Praphulla Koushik Jan 22 at 21:37
0
$\begingroup$

Let $F:\mathcal{I}\rightarrow \mathcal{C}$ is a functor. This is also called as diagram indexed by $\mathcal{I}$.

By the Limit of this diagram, we mean an object (universal ) $L$ of $\mathcal{C}$ and a collection of arrows (universal again) $\pi_i:L\rightarrow F(i)$ such that, for each arrow $m:i\rightarrow j$ in $\mathcal{I} $ the following diagram is commutative. This is usually denoted by $\varprojlim_{\mathcal{I}}F(i)$ or simply by $\varprojlim_{\mathcal{I}}F$.

enter image description here

Fixing an object $X$ in $\mathcal{C}$, I want to prove that $$\varprojlim_{\mathcal{I}}(\text{Hom}_{\mathcal{C}}(X,F(i))) =\text{Hom}_{\mathcal{C}}(X,\varprojlim_{\mathcal{I}}F(i))$$ i.e., an isomorphism $$\varprojlim_{\mathcal{I}}(\text{Hom}_{\mathcal{C}}(X,F(i))) =\text{Hom}_{\mathcal{C}}(X,L)$$ Let $(A,(p_i))$ be cone for the functor $\mathcal{I}\rightarrow \text{Set}$ given by $i\mapsto \text{Hom}_{\mathcal{C}}(X,F(i))$ i.e., we have following commutative diagrams

enter image description here

To prove that $\text{Hom}_{\mathcal{C}}(X,L)$ is equal to $\varprojlim_{\mathcal{I}}(\text{Hom}_{\mathcal{C}}(X,F(i)))$ it suffices to prove that there exists unique arrow $p:A\rightarrow \text{Hom}_{\mathcal{C}}(X,L)$ such that the following diagram is commutative.

enter image description here

So, we define an arrow $p:A\rightarrow \text{Hom}_{\mathcal{C}}(X,L)$ i.e., given $a\in A$ we define arrow $p(a):X\rightarrow L$ in $\mathcal{C}$. How to one get such arrow? See above diagram.

For each $a\in A$, we have $p_i(a):X\rightarrow F(i)$ such that $F(m)\circ p_i(a)=p_j(a)$ giving following diagram which gives an arrow $X\rightarrow L$ by universal property

enter image description here

This is the $p(a):X\rightarrow L$ that we associate for each $a\in L$.

This gives the map $p:A\rightarrow \text{Hom}_{\mathcal{C}}(X,L)$ satisfying conditions mentioned above. Thus, we have $$\varprojlim_{\mathcal{I}}(\text{Hom}_{\mathcal{C}}(X,F(i))) =\text{Hom}_{\mathcal{C}}(X,L)=\text{Hom}_{\mathcal{C}}(X,\varprojlim_{\mathcal{I}}F(i))$$

Consider the functor $\text{Man}\rightarrow \text{Stacks}$ that sends $M$ to $\underline{M}$ which is precisely $\text{Hom}(-,M)$.

Let $M_1,M_2,M_3$ be manifolds and $M_1\rightarrow M_2,M_3\rightarrow M_2$ be arrows.

We have $\varprojlim M_i = M_1\times_{M_2}M_3$. We have $$\text{Hom}(-,\varprojlim M_i)=\varprojlim \text{Hom}(-,M_i)$$ $$\underline{M_1\times_{M_2}M_3}=\text{Hom}(-,M_1\times_{M_2}M_3)=\varprojlim \underline{M_i}=\underline{M_1}\times_{\underline{M_2}}\underline{M_3}$$

$\endgroup$
  • 1
    $\begingroup$ This still does not answer the question of Lie groupoids.. I am trying to make sense of how $\mathcal{G}\rightarrow B\mathcal{G}$ is said as Yoneda embeddding... This does not seem to be of the form $Hom(-,\mathcal{G})$ wherever this makes sense.. $\endgroup$ – Praphulla Koushik Jan 22 at 21:14

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.