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    • CommentRowNumber1.
    • CommentAuthorMike Shulman
    • CommentTimeOct 3rd 2011

    Hi everyone! I’m back from a 2-week vacation without Internet access. It’ll take me a little while to catch up on everything, but just so you know, I’m here again now….

    • CommentRowNumber2.
    • CommentAuthorUrs
    • CommentTimeOct 3rd 2011
    • (edited Oct 3rd 2011)

    Good to hear that you are back!

    I was in Pittsburgh last week and spoke with some of the HoTT people there. Only through this conversation has it become clear to me that the HoTT derivation of

    π 1(S 1)= \pi_1 (S^1) = \mathbb{Z}

    that you had posted to the blog a while back is regarded as a big achievement in the field! I hear that Joyal is currently following a project for next computing π 3(S 2)\pi_3(S^2). Sounds like this is something that will keep HoTTists busy for a while. :-)

    • CommentRowNumber3.
    • CommentAuthorTodd_Trimble
    • CommentTimeOct 3rd 2011

    Welcome back! I figured that you were just hunkering down to research and avoiding getting distracted by internet stuff, forgetting that I too take internet-free vacations on occasion. :-)

    • CommentRowNumber4.
    • CommentAuthorMike Shulman
    • CommentTimeOct 3rd 2011

    the HoTT derivation of π 1(S 1)=\pi_1 (S^1) = \mathbb{Z} that you had posted to the blog a while back is regarded as a big achievement in the field!

    That’s nice to hear; I haven’t actually talked to any of them in person since the HoTT Oberwolfach meeting. (-: I wonder what approach Joyal is taking to π 3(S 2)\pi_3(S^2). The “obvious” approach to me is to construct the Hopf fibration and prove the homotopy long exact sequence of a fibration, but it’s not obvious to me how to identify the total space of the Hopf fibration as being S 3S^3 (or, at least, 2-simply-connected). I guess that that approach would also first involve proving π n(S n)=\pi_n(S^n)=\mathbb{Z} at least for small nn, which one could maybe do with the Freudenthal suspension theorem.

    • CommentRowNumber5.
    • CommentAuthorUrs
    • CommentTimeOct 3rd 2011
    • (edited Oct 3rd 2011)

    Yes, apparently Joyal is going via the Hopf fibration. But that’s already all I have heard about it. (Apart from giving a talk to the whole group, I mostly chatted with Krzysztof Kapulkin, who told me these things.)