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    • CommentRowNumber1.
    • CommentAuthordomenico_fiorenza
    • CommentTimeMay 20th 2010
    • (edited May 20th 2010)

    I’ve been thinking about the extended nn-dimensional TQFT stuff.. what follows is an attemp towards an abstract nonsense path integral interpretation of vectors assigned to nn-dimensional manifolds with boundary.

    recall that we have a complex number associated with a compact closed oriented nn-dimensional manifold, and a complex vector space V MV_M associated with a compact closed (n1)(n-1)-dimensional manifold MM (and we also have pretty clear abstract nonsense description of this constructions at least in Dijkgraaf-Witten theory). next, we want to associate to an nn-dimensional (compact oriented) manifold with boundary (Σ,Σ)(\Sigma,\partial\Sigma) a vector v (Σ,Σ)v_{(\Sigma,\partial\Sigma)} in V ΣV_{\partial \Sigma}.

    the idea is to use relative cohomology, i.e., the homotopy fibre H(Σ,Σ,B nU(1))\mathbf{H}(\Sigma,\partial\Sigma,\mathbf{B}^n U(1)) of H(Σ,B nU(1))H(Σ,B nU(1))\mathbf{H}(\Sigma,\mathbf{B}^n U(1))\to \mathbf{H}(\partial\Sigma,\mathbf{B}^n U(1)). indeed, by definition of homotopy fibre, H(Σ,Σ,B nU(1))\mathbf{H}(\Sigma,\partial\Sigma,\mathbf{B}^n U(1)) is a brane over H(Σ,B nU(1))\mathbf{H}(\partial\Sigma,\mathbf{B}^n U(1)), and so a brane over VectVect by postcomposing with

    H(Σ,B nU(1))τ 1H(Σ,B nU(1))B1U(1)Vect \mathbf{H}(\partial\Sigma,\mathbf{B}^n U(1))\to \tau_1\mathbf{H}(\partial\Sigma,\mathbf{B}^n U(1))\simeq \mathbf{B}1 U(1) \to Vect

    In other words, over H(Σ,Σ,B nU(1))\mathbf{H}(\Sigma,\partial\Sigma,\mathbf{B}^n U(1)) we have a flat line bundle with a section σ\sigma; moreover, this line bundle is the pull-back of a line bundle EE on H(Σ,B nU(1))\mathbf{H}(\partial\Sigma,\mathbf{B}^n U(1)), so we can push forward σ\sigma to a section of EE by fiber integration along H(Σ,Σ,B nU(1))H(Σ,B nU(1))\mathbf{H}(\Sigma,\partial\Sigma,\mathbf{B}^n U(1))\to \mathbf{H}(\partial\Sigma,\mathbf{B}^n U(1)). more explicitly, fix any point xx in H(Σ,B nU(1))\mathbf{H}(\partial\Sigma,\mathbf{B}^n U(1)); then for any point yy in the fibre of H(Σ,Σ,B nU(1))\mathbf{H}(\Sigma,\partial\Sigma,\mathbf{B}^n U(1)) over xx we have a vector σ y\sigma_y in E xE_x. in other words, we have a morphism from the fibre over xx of H(Σ,Σ,B nU(1))\mathbf{H}(\Sigma,\partial\Sigma,\mathbf{B}^n U(1)) to the fibre over xx of EE. taking fibrewise groupoid cardinality we obtain the desired section of EE. now, recall that V ΣV_{\partial \Sigma} is nothing but the space of sections of EE, so what we have just constructed is the seeked element v Σ,Σ)v_{\Sigma,\partial\Sigma})

    Urs should already have written this somewhere between the Lab and the Cafe’; my reconstructing it here is mostly to say that now I see it :)

    • CommentRowNumber2.
    • CommentAuthorUrs
    • CommentTimeMay 20th 2010
    • (edited May 20th 2010)

    Thanks.

    Let me see, is this here is the diagram that you are describing:

    H(Σ,Σ,B nU(1)) H(Σ,B nU(1)) σ * H(Σ,B nU(1)) τ 1 BU(1) ρ Vect \array{ && \mathbf{H}(\Sigma, \partial \Sigma, \mathbf{B}^n U(1)) \\ & \swarrow & & \searrow \\ \mathbf{H}(\Sigma, \mathbf{B}^n U(1)) &&\swArrow_{\sigma}&& * \\ & \searrow && \swarrow \\ && \mathbf{H}(\partial \Sigma, \mathbf{B}^n U(1)) \\ && \downarrow^{\tau_{\leq 1}} \\ && \mathbf{B} U(1) \\ && \downarrow^\rho \\ && Vect }

    ?

    I see what you are getting at. I’d just have the same comment that I had recently: this is working is over the space of phases instead of that of fields. I think you want to pull back everything to the space of fields. H(Σ,X)\mathbf{H}(\Sigma, X), where XX is target space, e.g. X=BGX = \mathbf{B}G for DW theory. So probably what you have in mind is really this:

    H(Σ,Σ,X) H(Σ,X) σ * ϕ| Σ H(Σ,X) H(Σ,exp(iS)) H(Σ,B nU(1)) τ 1 BU(1) ρ Vect \array{ && \mathbf{H}(\Sigma, \partial \Sigma, X) \\ & \swarrow & & \searrow \\ \mathbf{H}(\Sigma, X) &&\swArrow_{\sigma}&& * \\ & \searrow && \swarrow_{\mathrlap{\phi|_{\partial\Sigma}}} \\ && \mathbf{H}(\partial \Sigma, X) \\ && \downarrow^{\mathrlap{\mathbf{H}(\partial \Sigma, \exp(i S))}} \\ && \mathbf{H}(\partial \Sigma, \mathbf{B}^n U(1)) \\ && \downarrow^{\mathrlap{\tau_{\leq 1}}} \\ && \mathbf{B} U(1) \\ && \downarrow^\rho \\ && Vect }

    ?

    So this gives a canonical section on the pullback of the vector bundle on the space of boundary fields H(Σ,X)\mathbf{H}(\partial \Sigma, X) to the space of bulk-and-boundary-fields H(Σ,Σ,X)\mathbf{H}(\Sigma, \partial \Sigma, X). Yes, good.

    Let’s see, is this the right section we get? Is its component the action on a configuration that we want?

    Hm, the componens of σ\sigma are homotopies= gauge transformations from the fixed boundary field configuration ϕ| Σ\phi|_{\partial \Sigma} to the axctual boundary field configuration obtained by restricting a field configuration ϕH(Σ,X)\phi' \in \mathbf{H}(\Sigma,X) to the boundary.

    Hm, so σ\sigma sees that homotopy= gauge transformation, but not the bulk field ϕ\phi' itself. But the contribution of the action functional that we expect to see is that of the bulk field configuration exp(iS)(ϕ:ΣX)\exp(i S)(\phi' : \Sigma \to X).

    Hm, something is missing, maybe? Not sure.

    • CommentRowNumber3.
    • CommentAuthorUrs
    • CommentTimeMay 20th 2010
    • (edited May 20th 2010)

    But here is maybe a way to fix it:

    write Bord n(X)Bord_n(X) for the category of bordisms in target space XX. Then consider

    Q * σ * ϕ 0:Σ n1X Bord n(X) exp(iS) Vect \array{ && Q \\ & \swarrow & & \searrow \\ * &&\swArrow_{\sigma}&& * \\ & {}_{\mathllap{\emptyset}}\searrow && \swarrow_{\mathrlap{\phi_0 : \Sigma_{n-1} \to X}} \\ && Bord_n(X) \\ && \downarrow^{\mathrlap{\exp(i S)}} \\ && Vect }

    where now the top square is a lax pullback. So now the components of σ\sigma are morphisms in Bord n(X)Bord_n(X) from the fixed boundary field configuration object ϕ 0:Σ n1X\phi_0 : \Sigma_{n-1} \to X to the object \emptyset, in other words these components are all possible field configurations ϕ:ΣX\phi : \Sigma \to X with Σ=Σ n1\partial \Sigma = \Sigma_{n-1} and ϕ| Σ=ϕ 0\phi|_{\partial \Sigma} = \phi_0.

    So by the mechanism of whiskering the whiskering with the action exp(iS)\exp(i S) will now indeed produce the section which on each ϕ:ΣX\phi : \Sigma \to X is the linear map

    exp(iS)(ϕ):V \exp(i S)(\phi) : \mathbb{C} \to V

    which picks the action of the given field configuration ϕ\phi as a vector in VV,the given vector space over the boundary.

    So the push-forward of this, as you describe, should be the right path integral.

    • CommentRowNumber4.
    • CommentAuthorUrs
    • CommentTimeMay 20th 2010
    • (edited May 20th 2010)

    Sorry, had to do some reformatting. Now the above comment should display roughly as intended.

  1. But here is maybe a way to fix it

    yes, that was precisely the picture I had in mind at the beginning. Then trying to formalize it I came to the first diagram you display in #2, being secretely interested (as you noticed) into the second diagram in the same post. What I miss in the description in #3 is that it seems to start from what I’d like to be an arrival point, namely, a linear rpresentation of bordism out of a background field.

    • CommentRowNumber6.
    • CommentAuthordomenico_fiorenza
    • CommentTimeMay 22nd 2010
    • (edited May 22nd 2010)

    let’s give it a second try. starting with a background field α=e iS:XB nU(1)\alpha=e^{i S}:X\to \mathbf{B}^n U(1) we know how to transgress this to H(Σ nk,X)H(Σ nk,B nU(1))τ kB kU(1)\mathbf{H}(\Sigma_{n-k},X)\to\mathbf{H}(\Sigma_{n-k},\mathbf{B}^n U(1))\stackrel{\tau_k}{\to}\mathbf{B}^k U(1), for an (nk)(n-k)-dimensional compact closed oriented manifold Σ nk\Sigma_{n-k}.

    If Σ\Sigma is an nn-dimensional (compact oriented) with boundarty, and we split the boundaryo of Σ\Sigma into an incoming part and an outgoing part, so to have the cospan Σ inΣΣ out\partial\Sigma_in\to \Sigma\leftarrow \partial\Sigma_{out}, and consequently the span H(Σ in,X)H(Σ,X)H(Σ out,X)\mathbf{H}(\partial\Sigma_{in},X)\leftarrow\mathbf{H}(\Sigma,X)\to\mathbf{H}(\partial\Sigma_{out},X). but, since both Σ in\partial\Sigma_{in} and Σ out\partial\Sigma_{out} are (n1)(n-1)-dimensional compact closed oriented manfolds, we also have H(Σ in,X)B 1U(1)H(Σ out,X)\mathbf{H}(\partial\Sigma_{in},X)\to \mathbf{B}^1 U(1)\leftarrow\mathbf{H}(\partial\Sigma_{out},X).

    So the basic claim is that

    H(Σ,X) H(Σ in,X) hol α H(Σ out,X) B 1U(1) Vect \array{ && \mathbf{H}(\Sigma,X) \\ & \swarrow & & \searrow \\ \mathbf{H}(\partial\Sigma_{in},X) &&\neArrow_{hol_\alpha}&& \mathbf{H}(\partial\Sigma_{out},X) \\ & \searrow && \swarrow \\ && \mathbf{B}^1 U(1) \\ && \downarrow \\ && Vect }

    is a commutative diagram, where hol αhol_\alpha is nn-dimensional parallel transport induced by α\alpha. having this, the yoga of integration over H(Σ,Σ in,Σ out,X)\mathbf{H}(\Sigma,\partial\Sigma_{in},\partial\Sigma_{out},X) to get an operator Φ Σ:V Σ inV Σ out\Phi_\Sigma:V_{\partial\Sigma_{in}}\to V_{\partial\Sigma_{out}} would follow by the brane-bibrane formalism.

    • CommentRowNumber7.
    • CommentAuthordomenico_fiorenza
    • CommentTimeMay 24th 2010
    • (edited May 25th 2010)

    I was thinking that composition of bibranes is almost tautological: when we say that an nn-bordism Σ:Σ inΣ out\Sigma:\partial\Sigma_{in}\to \partial\Sigma_{out} is the composition of the nn-bordism Σ:Σ inM\Sigma':\partial\Sigma_{in}\to M with the nn-bordism Σ:MparialΣ out\Sigma'':M\to \parial\Sigma_{out}, what we are saying is that we have a commutative diagram

    Σ Σ Σ Σ in M Σ out \array{ &&&& \Sigma&& \\ &&& \nearrow & & \nwarrow&& \\ &&\Sigma' &&\nwArrow&& \Sigma''&& \\ &\nearrow&& \nwarrow && \nearrow&&\nwarrow \\ \partial\Sigma_{in}&&&& M&&&&\partial\Sigma_{out} }

    so, applying H(,X)\mathbf{H}(-,X) and postcomposing with the morphisms H(Σ n1,X)Vect\mathbf{H}(\Sigma_{n-1},X)\to Vect obtained by transgression of the background field XB nU(1)X\to \mathbf{B}^n U(1), we obtain bibrane composition

    H(Σ,X) H(Σ,X) H(Σ,X) H(Σ in,X) hol α H(M,X) hol α H(Σ out,X) Vect Id Vect Vect \array{ &&&& \mathbf{H}(\Sigma,X)&& \\ &&& \swarrow & & \searrow&& \\ &&\mathbf{H}(\Sigma',X) &&\neArrow&& \mathbf{H}(\Sigma'',X)&& \\ &\swarrow&& \searrow && \swarrow&&\searrow \\ \mathbf{H}(\partial\Sigma_{in},X)&&\neArrow_{hol_\alpha}&& \mathbf{H}(M,X)&&\neArrow_{hol_\alpha}&&\mathbf{H}(\partial\Sigma_{out},X)\\ &\searrow&& \swarrow && \searrow&&\swarrow\\ &&Vect &&\neArrow_{Id}&& Vect&& \\ &&& \searrow & & \swarrow&&\\ &&&& Vect&& }

    and one immediately sees associativity of bibrane product. Next step is considering the bibrane action on branes: we have the natural diagram

    H(Σ,Σ in,X) H(Σ in,X) H(Σ,X) Id * σ H(Σ in,X) hol α H(Σ out,X) Vect Id Vect Vect \array{ &&&& \mathbf{H}(\Sigma,\partial\Sigma_{in},X)&& \\ &&& \swarrow & & \searrow&& \\ &&\mathbf{H}(\partial\Sigma_{in},X) &&\neArrow&& \mathbf{H}(\Sigma,X)&& \\ &\swarrow&& \searrow_{Id} && \swarrow&&\searrow \\ *&&\neArrow_{\sigma}&& \mathbf{H}(\partial\Sigma_{in},X)&&\neArrow_{hol_\alpha}&&\mathbf{H}(\partial\Sigma_{out},X)\\ &\searrow&& \swarrow && \searrow&&\swarrow\\ &&Vect &&\neArrow_{Id}&& Vect&& \\ &&& \searrow & & \swarrow&&\\ &&&& Vect&& }

    and we have to push it forward along H(Σ,Σ in,X)H(Σ out,X)\mathbf{H}(\Sigma,\partial\Sigma_{in},X)\to \mathbf{H}(\partial\Sigma_{out},X). this should involve fibre-integration, and since the (homotopy) fibre of H(Σ,Σ in,X)H(Σ out,X)\mathbf{H}(\Sigma,\partial\Sigma_{in},X)\to \mathbf{H}(\partial\Sigma_{out},X) is H(Σ,Σ in,Σ out,X)\mathbf{H}(\Sigma,\partial\Sigma_{in},\partial\Sigma_{out},X) one sees that integration over relative cohomology enters the picture (as it should be, e.g., from the combinatorial description of Djikgraaf-Witten theory). Final check to be done is to show that this actually defines a bibrane action on branes, i.e. that (brane * bibrane) * bibrane=brane * (bibrane * bibrane). this comes down to drawing some gigantic diagram, I’ll try to display tomorrow.. :)

  2. in the end, avoiding to display an unreadable diagram, what one asks to the groupoid integration is the factorization property

    H(Σ,Σ in,Σ out,X)= H(M,X)( H(Σ,M,Σ out,X) H(Σ,Σ in,M,X)) \int_{\mathbf{H}(\Sigma,\partial\Sigma_{in},\partial\Sigma_{out},X)}= \int_{\mathbf{H}(M,X)}\left(\int_{\mathbf{H}(\Sigma'',M,\partial\Sigma_{out},X)}\circ \int_{\mathbf{H}(\Sigma',\partial\Sigma_{in},M,X)}\right)

    which is more nicely written as

    H(Σ,M,Σ out,X)× H(M,X)H(Σ,Σ in,M,X)= H(M,X)( H(Σ,M,Σ out,X) H(Σ,Σ in,M,X)) \int_{{\mathbf{H}(\Sigma'',M,\partial\Sigma_{out},X)}\times_{\mathbf{H}(M,X)}{\mathbf{H}(\Sigma',\partial\Sigma_{in},M,X)}}= \int_{\mathbf{H}(M,X)}\left(\int_{\mathbf{H}(\Sigma'',M,\partial\Sigma_{out},X)}\circ \int_{\mathbf{H}(\Sigma',\partial\Sigma_{in},M,X)}\right)
  3. more in general, it seems one is asking for something like

    A× BChol α= B( Ahol α Chol α) \int_{\mathbf{A}\times_{\mathbf{B}}\mathbf{C}}hol_\alpha=\int_\mathbf{B}\left(\int_\mathbf{A} hol_\alpha\circ \int_\mathbf{C} hol_\alpha\right)

    where the composition on the right hand side is the product in the category algebra of Vect.

    • CommentRowNumber10.
    • CommentAuthorUrs
    • CommentTimeMay 25th 2010

    Domenico,

    this is great stuff you are posting here. I want to look at it in detail, but at the moment feel a bit swamped with some tasks that I do need to look after. So this here is for the moment just to say that I am still very much interested, but might be busy working on some other parts of the Lab for some other things I need to do for a while. For almost three weeks maybe. But I’ll see what I can do.

  4. Urs,

    thanks. Please, do not feel you should read this before you’re over with what you’re after at the moment: since things are posted here, they will calmly wait for you or anyone else to read them. For instance, this morning I was reading stuff from you in the Cafe’ posted there in 2007.. :)

    • CommentRowNumber12.
    • CommentAuthorUrs
    • CommentTimeMay 26th 2010
    • (edited May 26th 2010)

    re #6

    So the basic claim is that Yes, something like that. I usedd to draw that diagram for the case where Σ\Sigma is the cylinder over one of its boundary components. With sufficient care it should work for arbitrary topologies, following the cobordism yoga further above. yes.

    re #7

    what we are saying is that we have a commutative diagram

    Yes, in fact this should be a pushout diagram if things are set up suitably nicely, exhibiting the composition of cospans in the 2-category of cospans.

    • CommentRowNumber13.
    • CommentAuthordomenico_fiorenza
    • CommentTimeMay 28th 2010
    • (edited May 28th 2010)

    here is another instance of how the compatibility of groupoid measure with fibrations is secretely used in the path-integral stuff.

    consider a complex vector bundle EE endowed with a connection \nabla on a manifold XX. then we have a 1-dimensional TQFT mapping a point to the vector space VV of global sections of EXE\to X and a closed interval to the identity operator for VV. by the abstract nonsense brane-bibrane construction, this means that we have the following path integral representation for the operator id Vid_V:

    σ(x)= γ:yxdγhol (γ)σ(y), \sigma(x)= \int_{\gamma:y\to x} d \gamma \, hol_\nabla(\gamma) \sigma(y),

    where σ\sigma is a section of EE. The dimension of VV is therefore

    dim(V)=tr( γ:yxdγhol (γ))= Xdx Ω xXdγtrhol (γ) dim(V)=tr \left( \int _{\gamma:y\to x} d \gamma \, hol_\nabla(\gamma)\right)=\int_X d x \int _{\Omega_x X} d \gamma \,tr hol_\nabla(\gamma)

    And here a fibration sequence comes in:

    Ω xX X * x X \array{ \Omega_x X &\to & \mathcal{L}X\\ \downarrow && \downarrow\\ *&\stackrel{x}{\to}& X }

    so we find

    dim(V)= Xdγtrhol (γ), dim(V)=\int_{\mathcal{L}X} d \gamma tr hol_\nabla (\gamma),

    and this integral over a loop space is precisely the complex number (a 0-vector space over \mathbb{C}) associated by our TQFT to the closed 1-manifold S 1S^1.

    • CommentRowNumber14.
    • CommentAuthorUrs
    • CommentTimeMay 28th 2010

    Could you amplify which of these steps you are now thinking of as formally well-defined steps and which not? I do understand what you are saying on general grounds, but I am not sure I see if you are pointing out a formalization now or a general strategy.

  5. I think we have well defined steps only for finite models, by now. what I’m trying to amplify here are the properties of the groupoid measure one uses on general ground, to focus attention on them. an emerging feature groupoid cardinality seems to have (or at least should have) is compatibility with fibration sequences (the groupoid cardinality version of Fubini’s theorem). I’m remarking this since compatibility with fibration sequences is not mentioned at groupoid cardinality, and could also be relevant to fiber integration

    • CommentRowNumber16.
    • CommentAuthorUrs
    • CommentTimeMay 28th 2010
    • (edited May 28th 2010)

    By the way, is that fiber sequence clear, that you gave?

    There is

    Ω xX * * X, \array{ \Omega_x X &\to& * \\ \downarrow && \downarrow \\ * &\to& X } \,,

    which is equivalent to

    Ω xX P xX * X, \array{ \Omega_x X &\to& P_x X \\ \downarrow && \downarrow \\ * &\to& X } \,,

    I am slightly worried about the X\mathcal{L} X that you have there. But I also haven’t thought about it hard enough.

    an emerging feature groupoid cardinality seems to have (or at least should have) is compatibility with fibration sequences (the groupoid cardinality version of Fubini’s theorem). I’m remarking this since compatibility with fibration sequences is not mentioned at groupoid cardinality, and could also be relevant to fiber integration

    Yes, okay. I need to understand it better, too. That’s why I am asking: I am trying to find out what exactly you have understood! :-)

  6. By the way, is that fiber sequence clear, that you gave?

    it is the one at free loop space object. of course, this does not mean that it is correct, only that we have to check and eventually correct there, too :-)

    • CommentRowNumber18.
    • CommentAuthorUrs
    • CommentTimeMay 28th 2010

    Ah, right, stupid me. I think I even typed that computation back then!? :-)

    Good that we have the Lab, it has a better memory than I have.