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    • CommentRowNumber1.
    • CommentAuthorRyan Thorngren
    • CommentTimeSep 29th 2013

    So I’ve been wondering in what capacity one can think of a spin structure as a trivialization of the second Stiefel-Whitney class. Let me begin with an analogy. There is a principal 2 bundle classified by w1 usually called the orientation double cover. An orientation is a trivialization of this bundle. Similarly, w2 defines a principal 2 2-bundle. If I’m not too confused, a trivialization of this should look like a connection with curvature w2. What sort of principal bundle is this a connection on? Can I see its relation to spin structures? Is it the spin connection in some sense?

    Thanks.

    • CommentRowNumber2.
    • CommentAuthorUrs
    • CommentTimeSep 30th 2013

    Yes, this is essentially right. I have just expanded a bit more at spin structure – Definition with more comments on this.

    Only notice that 2 is a discrete group, so all connections on 2-principal 2-bundles or their sections etc. are necessarily flat.

    If you want connection data to detect spin-like structures then you need to pass to spin^c structure. Is that maybe what you are after?

    • CommentRowNumber3.
    • CommentAuthorRyan Thorngren
    • CommentTimeOct 1st 2013

    Thanks, Urs. I’m not sure if what I actually want is a spin^c structure, since I don’t understand fermions that well yet. If I wanted to represent a spin^c structure, this would be the same as a line bundle whose Chern class is w2 when reduced mod 2?

    • CommentRowNumber4.
    • CommentAuthorUrs
    • CommentTimeOct 1st 2013

    Yes, that’s right.

    • CommentRowNumber5.
    • CommentAuthorRyan Thorngren
    • CommentTimeOct 18th 2013
    • (edited Oct 18th 2013)

    I’m still thinking about this and I realised that the right way to think about this sort of trivialization is the following. (Let me know if you agree). We can consider a string charged under w2 considered as a /2 2-connection. A trivialization of w2 defines a particle that this string can end on. This particle is of course a fermion. This explains the stringiness of fermions, why the w2 string is (essentially) topological, and I think gives a good way to think about what a trivialization of an n-connection gives you.

    Maybe spinning strings can be thought of as ends of 12p1 tubes…

    • CommentRowNumber6.
    • CommentAuthorUrs
    • CommentTimeOct 18th 2013
    • (edited Oct 18th 2013)

    Right, there is a general theorem about how trivializations of an n-bundle encode boundary conditions for n-dimensional field theories. This is discussed at Local prequantum field theory (schreiber).

    Generally, for G some -group and XBG regarded as a background field for some n-dimensional bulk field theory, then a boundary condition for this is a diagram of the form

    Q*XBG

    for some choice of Q, some choice of map QX and some choice of homotopy filling this. So this is a trivialization of the pullback of the bulk cocycle to Q and physically if X is the target space for a p-brane then Q is the target space of its boundary (p1)-brane.

    Specifically if QX is the identity, then it is just a trivialization of the bulk cocycle.

    And indeed, by this argument one finds that the Green-Schwarz anomaly cancellation trivializing 12p1 exhibits the heterotic string as the boundary field theory of the M2-brane ending on the Horava-Witten boundary of 11d spacetime.

    This last statement appears in Joost Nuiten’s thesis in the very last section 5.3.1. The bulk of the thesis discusses the general mechanism.

    • CommentRowNumber7.
    • CommentAuthorRyan Thorngren
    • CommentTimeNov 13th 2013

    Hi Urs, and anyone else who’s been following the discussion. I’ve been thinking about what we mean by a trivialization of w2 or other characteristic classes. It seems to me that this only makes sense if there is a preferred cocycle representing these classes. Is this true? What conditions do we want to hold for our preferred representative?

    • CommentRowNumber8.
    • CommentAuthorUrs
    • CommentTimeNov 13th 2013

    For two equivalent (cohomologous) cocycles, the spaces of their trivializations will be equivalent.

    But it is true that in some situations you want to remember the cocycle and not just the class, this is notably so in the context of twisted cohomology, which is just about the “relative trivialization” of one cocycle relative to another one.