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    • CommentRowNumber1.
    • CommentAuthorhilbertthm90
    • CommentTimeFeb 6th 2013

    I just learned about rigidification and decided to record it somewhere.

    1. I’m not sure if the title is good, because there is the notion of the rigidification of quasi-categories.

    2. Surely this notion has a higher analogue that maybe someone knows more about. Surely you could take an nn-stack and consider the nn-categorical fiber product to make a notion of inertia, and then rigidify with respect to some subgroup object inside…

    • CommentRowNumber2.
    • CommentAuthorMike Shulman
    • CommentTimeFeb 7th 2013

    I don’t quite understand…. it seems to me that if HH were a substack of the inertia, then each object of H(T)H(T) would be assigned to a specified loop at a specified object ξ\xi in 𝒮(T)\mathcal{S}(T), whereas at rigidification you wrote that for all ξ\xi there is an embedding of H(T)H(T) into Aut T(ξ)Aut_T(\xi).

    • CommentRowNumber3.
    • CommentAuthorhilbertthm90
    • CommentTimeFeb 7th 2013

    Ah yes, it shouldn’t say substack. I was asking someone about this and they were referring to it as a “subgroup,” but I realize now they probably were just using that term colloquially to mean that condition listed and didn’t mean that it is literally a sub-group-stack (or whatever order that should be).

    • CommentRowNumber4.
    • CommentAuthorUrs
    • CommentTimeFeb 7th 2013
    • (edited Feb 7th 2013)

    Thanks for starting this entry!

    I think the simple way to say it is that given a pointed group stack homomorphsim

    ι:BHAut(X) \iota \;\colon\; \mathbf{B}H \to \mathbf{Aut}(X)

    the rigidification is the induced homotopy quotient X//BHX//\mathbf{B}H.

    Agreed?

    I have expanded the

    in the entry to reflect this.

    Also, I have added to the

    discussion of the case for the trivial site, such as to extract the basic mechanism without distraction by the site dependence.

    Matt, please check if you can live with these additions. Let me know if there is anything you’d rather want to have changed.

    Finally I have a question to you: how strongly do you feel about the Idea-sentence that you wrote into the entry? I am not sure if I see what you have in mind with it. If you feel that this has to be the idea of rigidification, then maybe I’d ask you to expand on your sentence. But I have taken the liberty of adding an alternative (or maybe just additional?) description of the idea of rigidification to the

    Please have a look and let me know what you think.

    • CommentRowNumber5.
    • CommentAuthorMike Shulman
    • CommentTimeFeb 7th 2013

    Urs, that’s the sort of description I was hoping for, but I don’t see how to get it out of the definition that he wrote. Do you mean by Aut(X)Aut(X) the stack of automorphisms of XX? The original definition assigned to every element of H an automorphism in XX at each of its objects.

    • CommentRowNumber6.
    • CommentAuthorMike Shulman
    • CommentTimeFeb 7th 2013

    Matt, is the inertia even a group stack? I don’t see how.

    • CommentRowNumber7.
    • CommentAuthorUrs
    • CommentTimeFeb 7th 2013
    • (edited Feb 7th 2013)

    Urs, that’s the sort of description I was hoping for, but I don’t see how to get it out of the definition that he wrote.

    Mike, yes, look at it: a pointed homomorphism

    BHAut(X) \mathbf{B}H \to \mathbf{Aut}(X)

    sends the unique point of BH\mathbf{B}H to the identity functor on XX, and then each element of HH to an autoequivalence of the identity functors. That’s a natural transformation which to each object in XX assigns an automorphism in XX of that object.

    I have spelled that out in some detail in the Examples-section.

    • CommentRowNumber8.
    • CommentAuthorUrs
    • CommentTimeFeb 7th 2013
    • (edited Feb 7th 2013)

    Matt, is the inertia even a group stack? I don’t see how.

    It’s a group object over XX, I suppose that’s what he means.

    But this is related to my above request: the original Idea-sentence maybe deserves some fine-tuning.

    • CommentRowNumber9.
    • CommentAuthorUrs
    • CommentTimeFeb 7th 2013
    • (edited Feb 7th 2013)

    I see that the cache bug was in effect and hiding my edits. Sorry, I only realize that now: I had renamed the entry to

    (cache now cleared, so now you can actually see this even when looking at rigidification).

    By the way, I see that this statement about ι:BHAut(X)\iota \colon \mathbf{B}H \to \mathbf{Aut}(X) is essentially also in

    On page 16 they say that the rigidification of XX is X//BHX//\mathbf{B}H. For that to make sense we must have had an action of BH\mathbf{B}H, hence a 2-group homomorphism BHAut(X)\mathbf{B}H \to \mathbf{Aut}(X). Indeed, it’s straightwarward to check that this is equivalently the data that they consider.

    • CommentRowNumber10.
    • CommentAuthorhilbertthm90
    • CommentTimeFeb 7th 2013

    So much has happened while I was asleep! I like the edits. I don’t know very much about this, because I just learned of its existence a few days ago.

    • CommentRowNumber11.
    • CommentAuthorMike Shulman
    • CommentTimeFeb 8th 2013

    Okay… so when you say a “group homomorphism” BHAut(X)\mathbf{B}H \to Aut(X), are you using an implicit assumption that HH is at least E 2E_2 in order that BH\mathbf{B}H is a group? Or do you just mean an arbitrary morphism of stacks? I think I’m confused partly because I don’t know what it means for a group homomorphism to be “pointed” — or, more precisely, I don’t know what it would mean for it not to be pointed.

    • CommentRowNumber12.
    • CommentAuthorUrs
    • CommentTimeFeb 8th 2013
    • (edited Feb 8th 2013)

    True, I don’t have to say “pointed”.

    And yes, for the general case I assumed HH to be a “braided infinity-group”.

    • CommentRowNumber13.
    • CommentAuthorMike Shulman
    • CommentTimeFeb 8th 2013

    Okay, I get it, thanks!

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