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    • CommentRowNumber1.
    • CommentAuthorUrs
    • CommentTimeMar 25th 2014
    • (edited Mar 27th 2014)

    Here is a question, probably for Charles (Rezk), if he sees it.

    The string orientation of tmf MStringtmf refines the un-twisted Witten genus on manifolds with String-structure. More generally, the Witten genus on a Spin-manifold is twisted by a complex vector bundle (“heterotic string”) and is a modular form for “String^c-structure”, characterized by 12p1c2=0.

    In the article

    • Qingtao Chen, Fei Han, Weiping Zhang, “Generalized Witten Genus and Vanishing Theorems” (arXiv:1003.2325)

    this twisted Witten genus on Stringc-structures is re-considered, and on p.2 an obvious question is mentioned: does the twisted Witten genus also have a “topological” lift to a map of spectra?

    Now, there is indeed an obious map of spectra MStringctmf, namely the hocolim over the left half of the diagram

    BStringcp*σcp*(12p1c2)BSpin×BSU12p1c2B3U(1)ρBGL1(tmf)tmfMod,

    where σc denotes the homotopy that exhibits BStringc as the homotopy fiber of 12p1c2, and where ρ is the twisting map exhibiting the plain string orientation of tmf as in Ando-Blumberg-Gepner 10.

    My exercise is to check if that map

    limσc:MStringctmf

    induces on homotopy groups the twisted Witten genus, correctly.

    While I am slowly chewing on this, I thought I’d ask if anyone has considered this before. Quite likely this is clear to experts such as Charles.

    Or rather, possibly the push should be rather along the left half of

    BStringcBString//SUpSpin(BSU*//SU)σcp*Spin(12p1)BSpinc212p1B3U(1)ρBGL1(tmf)tmfMod

    and hence land in SU-equivariant elliptic cohomology.

    ( Here I am using the pasting of homotopy pullbacks

    BStringBStringcBSpin*BSUB3U(1)

    in order to identify BStringcBString//SU. )

    I suppose that the twisted Witten genus should land in equivariant tmf this way is something that Matthew Ando has been suggesting, though I am not sure if I have seen the place where this is stated explicitly.

    • CommentRowNumber2.
    • CommentAuthorCharles Rezk
    • CommentTimeMar 28th 2014

    I am not an expert on this question. You should try Matt.