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1. • CommentRowNumber1.
   • CommentAuthordomenico_fiorenza
   • CommentTimeMay 20th 2010
   • (edited May 20th 2010)
   • PermaLink
   Author: domenico_fiorenza
   Format: MarkdownItexI've been thinking about the extended $n$-dimensional TQFT stuff.. what follows is an attemp towards an abstract nonsense path integral interpretation of vectors assigned to $n$-dimensional manifolds with boundary. recall that we have a complex number associated with a compact closed oriented $n$-dimensional manifold, and a complex vector space $V_M$ associated with a compact closed $(n-1)$-dimensional manifold $M$ (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 $n$-dimensional (compact oriented) manifold with boundary $(\Sigma,\partial\Sigma)$ a vector $v_{(\Sigma,\partial\Sigma)}$ in $V_{\partial \Sigma}$. the idea is to use relative cohomology, i.e., the homotopy fibre $\mathbf{H}(\Sigma,\partial\Sigma,\mathbf{B}^n U(1))$ of $\mathbf{H}(\Sigma,\mathbf{B}^n U(1))\to \mathbf{H}(\partial\Sigma,\mathbf{B}^n U(1))$. indeed, by definition of homotopy fibre, $\mathbf{H}(\Sigma,\partial\Sigma,\mathbf{B}^n U(1))$ is a brane over $\mathbf{H}(\partial\Sigma,\mathbf{B}^n U(1))$, and so a brane over $Vect$ by postcomposing with $$ \mathbf{H}(\partial\Sigma,\mathbf{B}^n U(1))\to \tau_1\mathbf{H}(\partial\Sigma,\mathbf{B}^n U(1))\simeq \mathbf{B}1 U(1) \to Vect $$ In other words, over $\mathbf{H}(\Sigma,\partial\Sigma,\mathbf{B}^n U(1))$ we have a flat line bundle with a section $\sigma$; moreover, this line bundle is the pull-back of a line bundle $E$ on $\mathbf{H}(\partial\Sigma,\mathbf{B}^n U(1))$, so we can push forward $\sigma$ to a section of $E$ by fiber integration along $\mathbf{H}(\Sigma,\partial\Sigma,\mathbf{B}^n U(1))\to \mathbf{H}(\partial\Sigma,\mathbf{B}^n U(1))$. more explicitly, fix any point $x$ in $\mathbf{H}(\partial\Sigma,\mathbf{B}^n U(1))$; then for any point $y$ in the fibre of $\mathbf{H}(\Sigma,\partial\Sigma,\mathbf{B}^n U(1))$ over $x$ we have a vector $\sigma_y$ in $E_x$. in other words, we have a morphism from the fibre over $x$ of $\mathbf{H}(\Sigma,\partial\Sigma,\mathbf{B}^n U(1))$ to the fibre over $x$ of $E$. taking fibrewise groupoid cardinality we obtain the desired section of $E$. now, recall that $V_{\partial \Sigma}$ is nothing but the space of sections of $E$, so what we have just constructed is the seeked element $v_{\Sigma,\partial\Sigma})$ 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 :)

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

   recall that we have a complex number associated with a compact closed oriented $n$-dimensional manifold, and a complex vector space $V_M$ associated with a compact closed $(n-1)$-dimensional manifold $M$ (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 $n$-dimensional (compact oriented) manifold with boundary $(\Sigma,\partial\Sigma)$ a vector $v_{(\Sigma,\partial\Sigma)}$ in $V_{\partial \Sigma}$.

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

   $$\mathbf{H}(\partial\Sigma,\mathbf{B}^n U(1))\to \tau_1\mathbf{H}(\partial\Sigma,\mathbf{B}^n U(1))\simeq \mathbf{B}1 U(1) \to Vect$$

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

   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 :)

2. • CommentRowNumber2.
   • CommentAuthorUrs
   • CommentTimeMay 20th 2010
   • (edited May 20th 2010)
   • PermaLink
   Author: Urs
   Format: MarkdownItexThanks. Let me see, is this here is the diagram that you are describing: $$ \array{ && \mathbf{H}(\Sigma, \partial \Sigma, \mathbf{B}^n U(1)) \\ & \swarrow & & \searrow \\ \mathbf{H}(\Sigma, \mathbf{B}^n U(1)) &&\swArrow_{\sigma}&& * \\ & \searrow && \swarrow \\ && \mathbf{H}(\partial \Sigma, \mathbf{B}^n U(1)) \\ && \downarrow^{\tau_{\leq 1}} \\ && \mathbf{B} U(1) \\ && \downarrow^\rho \\ && Vect } $$ ? 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. $\mathbf{H}(\Sigma, X)$, where $X$ is target space, e.g. $X = \mathbf{B}G$ for DW theory. So probably what you have in mind is really this: $$ \array{ && \mathbf{H}(\Sigma, \partial \Sigma, X) \\ & \swarrow & & \searrow \\ \mathbf{H}(\Sigma, X) &&\swArrow_{\sigma}&& * \\ & \searrow && \swarrow_{\mathrlap{\phi|_{\partial\Sigma}}} \\ && \mathbf{H}(\partial \Sigma, X) \\ && \downarrow^{\mathrlap{\mathbf{H}(\partial \Sigma, \exp(i S))}} \\ && \mathbf{H}(\partial \Sigma, \mathbf{B}^n U(1)) \\ && \downarrow^{\mathrlap{\tau_{\leq 1}}} \\ && \mathbf{B} U(1) \\ && \downarrow^\rho \\ && Vect } $$ ? So this gives a canonical section on the pullback of the vector bundle on the space of boundary fields $\mathbf{H}(\partial \Sigma, X)$ to the space of bulk-and-boundary-fields $\mathbf{H}(\Sigma, \partial \Sigma, X)$. 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 $\sigma$ are homotopies= gauge transformations from the fixed boundary field configuration $\phi|_{\partial \Sigma}$ to the axctual boundary field configuration obtained by restricting a field configuration $\phi' \in \mathbf{H}(\Sigma,X)$ to the boundary. Hm, so $\sigma$ sees that homotopy= gauge transformation, but not the bulk field $\phi'$ itself. But the contribution of the action functional that we expect to see is that of the bulk field configuration $\exp(i S)(\phi' : \Sigma \to X)$. Hm, something is missing, maybe? Not sure.

   Thanks.

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

   $$\array{ && \mathbf{H}(\Sigma, \partial \Sigma, \mathbf{B}^n U(1)) \\ & \swarrow & & \searrow \\ \mathbf{H}(\Sigma, \mathbf{B}^n U(1)) &&\swArrow_{\sigma}&& * \\ & \searrow && \swarrow \\ && \mathbf{H}(\partial \Sigma, \mathbf{B}^n U(1)) \\ && \downarrow^{\tau_{\leq 1}} \\ && \mathbf{B} U(1) \\ && \downarrow^\rho \\ && Vect }$$

   ?

   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. $\mathbf{H}(\Sigma, X)$, where $X$ is target space, e.g. $X = \mathbf{B}G$ for DW theory. So probably what you have in mind is really this:

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

   ?

   So this gives a canonical section on the pullback of the vector bundle on the space of boundary fields $\mathbf{H}(\partial \Sigma, X)$ to the space of bulk-and-boundary-fields $\mathbf{H}(\Sigma, \partial \Sigma, X)$. 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 $\sigma$ are homotopies= gauge transformations from the fixed boundary field configuration $\phi|_{\partial \Sigma}$ to the axctual boundary field configuration obtained by restricting a field configuration $\phi' \in \mathbf{H}(\Sigma,X)$ to the boundary.

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

   Hm, something is missing, maybe? Not sure.

3. • CommentRowNumber3.
   • CommentAuthorUrs
   • CommentTimeMay 20th 2010
   • (edited May 20th 2010)
   • PermaLink
   Author: Urs
   Format: MarkdownItexBut here is maybe a way to fix it: write $Bord_n(X)$ for the category of bordisms in target space $X$. Then consider $$ \array{ && Q \\ & \swarrow & & \searrow \\ * &&\swArrow_{\sigma}&& * \\ & {}_{\mathllap{\emptyset}}\searrow && \swarrow_{\mathrlap{\phi_0 : \Sigma_{n-1} \to X}} \\ && Bord_n(X) \\ && \downarrow^{\mathrlap{\exp(i S)}} \\ && Vect } $$ where now the top square is a lax pullback. So now the components of $\sigma$ are morphisms in $Bord_n(X)$ from the fixed boundary field configuration object $\phi_0 : \Sigma_{n-1} \to X$ to the object $\emptyset$, in other words these components are all possible field configurations $\phi : \Sigma \to X$ with $\partial \Sigma = \Sigma_{n-1}$ and $\phi|_{\partial \Sigma} = \phi_0$. So by the mechanism of whiskering the whiskering with the action $\exp(i S)$ will now indeed produce the section which on each $\phi : \Sigma \to X $ is the linear map $$ \exp(i S)(\phi) : \mathbb{C} \to V $$ which picks the action of the given field configuration $\phi$ as a vector in $V$,the given vector space over the boundary. So the push-forward of this, as you describe, should be the right path integral.

   But here is maybe a way to fix it:

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

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

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

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

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

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

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

4. • CommentRowNumber4.
   • CommentAuthorUrs
   • CommentTimeMay 20th 2010
   • (edited May 20th 2010)
   • PermaLink
   Author: Urs
   Format: MarkdownItexSorry, had to do some reformatting. Now the above comment should display roughly as intended.

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

5. • CommentRowNumber5.
   • CommentAuthordomenico_fiorenza
   • CommentTimeMay 20th 2010
   • PermaLink
   Author: domenico_fiorenza
   Format: MarkdownItex> 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.

   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.

6. • CommentRowNumber6.
   • CommentAuthordomenico_fiorenza
   • CommentTimeMay 22nd 2010
   • (edited May 22nd 2010)
   • PermaLink
   Author: domenico_fiorenza
   Format: MarkdownItexlet's give it a second try. starting with a background field $\alpha=e^{i S}:X\to \mathbf{B}^n U(1)$ we know how to transgress this to $\mathbf{H}(\Sigma_{n-k},X)\to\mathbf{H}(\Sigma_{n-k},\mathbf{B}^n U(1))\stackrel{\tau_k}{\to}\mathbf{B}^k U(1)$, for an $(n-k)$-dimensional compact closed oriented manifold $\Sigma_{n-k}$. If $\Sigma$ is an $n$-dimensional (compact oriented) with boundarty, and we split the boundaryo of $\Sigma$ into an incoming part and an outgoing part, so to have the cospan $\partial\Sigma_in\to \Sigma\leftarrow \partial\Sigma_{out}$, and consequently the span $\mathbf{H}(\partial\Sigma_{in},X)\leftarrow\mathbf{H}(\Sigma,X)\to\mathbf{H}(\partial\Sigma_{out},X)$. but, since both $\partial\Sigma_{in}$ and $\partial\Sigma_{out}$ are $(n-1)$-dimensional compact closed oriented manfolds, we also have $\mathbf{H}(\partial\Sigma_{in},X)\to \mathbf{B}^1 U(1)\leftarrow\mathbf{H}(\partial\Sigma_{out},X)$. So the basic claim is that $$ \array{ && \mathbf{H}(\Sigma,X) \\ & \swarrow & & \searrow \\ \mathbf{H}(\partial\Sigma_{in},X) &&\neArrow_{hol_\alpha}&& \mathbf{H}(\partial\Sigma_{out},X) \\ & \searrow && \swarrow \\ && \mathbf{B}^1 U(1) \\ && \downarrow \\ && Vect } $$ is a commutative diagram, where $hol_\alpha$ is $n$-dimensional parallel transport induced by $\alpha$. having this, the yoga of integration over $\mathbf{H}(\Sigma,\partial\Sigma_{in},\partial\Sigma_{out},X)$ to get an operator $\Phi_\Sigma:V_{\partial\Sigma_{in}}\to V_{\partial\Sigma_{out}}$ would follow by the brane-bibrane formalism.

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

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

   So the basic claim is that

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

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

7. • CommentRowNumber7.
   • CommentAuthordomenico_fiorenza
   • CommentTimeMay 24th 2010
   • (edited May 25th 2010)
   • PermaLink
   Author: domenico_fiorenza
   Format: MarkdownItexI was thinking that composition of bibranes is almost tautological: when we say that an $n$-bordism $\Sigma:\partial\Sigma_{in}\to \partial\Sigma_{out}$ is the composition of the $n$-bordism $\Sigma':\partial\Sigma_{in}\to M$ with the $n$-bordism $\Sigma'':M\to \parial\Sigma_{out}$, what we are saying is that we have a commutative diagram $$ \array{ &&&& \Sigma&& \\ &&& \nearrow & & \nwarrow&& \\ &&\Sigma' &&\nwArrow&& \Sigma''&& \\ &\nearrow&& \nwarrow && \nearrow&&\nwarrow \\ \partial\Sigma_{in}&&&& M&&&&\partial\Sigma_{out} } $$ so, applying $\mathbf{H}(-,X)$ and postcomposing with the morphisms $\mathbf{H}(\Sigma_{n-1},X)\to Vect$ obtained by transgression of the background field $X\to \mathbf{B}^n U(1)$, we obtain bibrane composition $$ \array{ &&&& \mathbf{H}(\Sigma,X)&& \\ &&& \swarrow & & \searrow&& \\ &&\mathbf{H}(\Sigma',X) &&\neArrow&& \mathbf{H}(\Sigma'',X)&& \\ &\swarrow&& \searrow && \swarrow&&\searrow \\ \mathbf{H}(\partial\Sigma_{in},X)&&\neArrow_{hol_\alpha}&& \mathbf{H}(M,X)&&\neArrow_{hol_\alpha}&&\mathbf{H}(\partial\Sigma_{out},X)\\ &\searrow&& \swarrow && \searrow&&\swarrow\\ &&Vect &&\neArrow_{Id}&& Vect&& \\ &&& \searrow & & \swarrow&&\\ &&&& Vect&& } $$ and one immediately sees associativity of bibrane product. Next step is considering the bibrane action on branes: we have the natural diagram $$ \array{ &&&& \mathbf{H}(\Sigma,\partial\Sigma_{in},X)&& \\ &&& \swarrow & & \searrow&& \\ &&\mathbf{H}(\partial\Sigma_{in},X) &&\neArrow&& \mathbf{H}(\Sigma,X)&& \\ &\swarrow&& \searrow_{Id} && \swarrow&&\searrow \\ *&&\neArrow_{\sigma}&& \mathbf{H}(\partial\Sigma_{in},X)&&\neArrow_{hol_\alpha}&&\mathbf{H}(\partial\Sigma_{out},X)\\ &\searrow&& \swarrow && \searrow&&\swarrow\\ &&Vect &&\neArrow_{Id}&& Vect&& \\ &&& \searrow & & \swarrow&&\\ &&&& Vect&& } $$ and we have to push it forward along $\mathbf{H}(\Sigma,\partial\Sigma_{in},X)\to \mathbf{H}(\partial\Sigma_{out},X)$. this should involve fibre-integration, and since the (homotopy) fibre of $\mathbf{H}(\Sigma,\partial\Sigma_{in},X)\to \mathbf{H}(\partial\Sigma_{out},X)$ is $\mathbf{H}(\Sigma,\partial\Sigma_{in},\partial\Sigma_{out},X)$ 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.. :)

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

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

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

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

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

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

   and we have to push it forward along $\mathbf{H}(\Sigma,\partial\Sigma_{in},X)\to \mathbf{H}(\partial\Sigma_{out},X)$. this should involve fibre-integration, and since the (homotopy) fibre of $\mathbf{H}(\Sigma,\partial\Sigma_{in},X)\to \mathbf{H}(\partial\Sigma_{out},X)$ is $\mathbf{H}(\Sigma,\partial\Sigma_{in},\partial\Sigma_{out},X)$ 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.. :)

8. • CommentRowNumber8.
   • CommentAuthordomenico_fiorenza
   • CommentTimeMay 25th 2010
   • PermaLink
   Author: domenico_fiorenza
   Format: MarkdownItexin the end, avoiding to display an unreadable diagram, what one asks to the groupoid integration is the factorization property $$ \int_{\mathbf{H}(\Sigma,\partial\Sigma_{in},\partial\Sigma_{out},X)}= \int_{\mathbf{H}(M,X)}\left(\int_{\mathbf{H}(\Sigma'',M,\partial\Sigma_{out},X)}\circ \int_{\mathbf{H}(\Sigma',\partial\Sigma_{in},M,X)}\right) $$ which is more nicely written as $$ \int_{{\mathbf{H}(\Sigma'',M,\partial\Sigma_{out},X)}\times_{\mathbf{H}(M,X)}{\mathbf{H}(\Sigma',\partial\Sigma_{in},M,X)}}= \int_{\mathbf{H}(M,X)}\left(\int_{\mathbf{H}(\Sigma'',M,\partial\Sigma_{out},X)}\circ \int_{\mathbf{H}(\Sigma',\partial\Sigma_{in},M,X)}\right) $$

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

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

   which is more nicely written as

   $$\int_{{\mathbf{H}(\Sigma'',M,\partial\Sigma_{out},X)}\times_{\mathbf{H}(M,X)}{\mathbf{H}(\Sigma',\partial\Sigma_{in},M,X)}}= \int_{\mathbf{H}(M,X)}\left(\int_{\mathbf{H}(\Sigma'',M,\partial\Sigma_{out},X)}\circ \int_{\mathbf{H}(\Sigma',\partial\Sigma_{in},M,X)}\right)$$
9. • CommentRowNumber9.
   • CommentAuthordomenico_fiorenza
   • CommentTimeMay 25th 2010
   • PermaLink
   Author: domenico_fiorenza
   Format: MarkdownItexmore in general, it seems one is asking for something like $$ \int_{\mathbf{A}\times_{\mathbf{B}}\mathbf{C}}hol_\alpha=\int_\mathbf{B}\left(\int_\mathbf{A} hol_\alpha\circ \int_\mathbf{C} hol_\alpha\right) $$ where the composition on the right hand side is the product in the [[category algebra]] of Vect.

   more in general, it seems one is asking for something like

   $$\int_{\mathbf{A}\times_{\mathbf{B}}\mathbf{C}}hol_\alpha=\int_\mathbf{B}\left(\int_\mathbf{A} hol_\alpha\circ \int_\mathbf{C} hol_\alpha\right)$$

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

10. • CommentRowNumber10.
    • CommentAuthorUrs
    • CommentTimeMay 25th 2010
    • PermaLink
    Author: Urs
    Format: MarkdownItexDomenico, 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.

    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.

11. • CommentRowNumber11.
    • CommentAuthordomenico_fiorenza
    • CommentTimeMay 25th 2010
    • PermaLink
    Author: domenico_fiorenza
    Format: MarkdownItexUrs, 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.. :)

    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.. :)

12. • CommentRowNumber12.
    • CommentAuthorUrs
    • CommentTimeMay 26th 2010
    • (edited May 26th 2010)
    • PermaLink
    Author: Urs
    Format: MarkdownItexre #6 > So the basic claim is that Yes, something like that. I usedd to draw that diagram for the case where $\Sigma$ 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.

    re #6

    So the basic claim is that Yes, something like that. I usedd to draw that diagram for the case where $\Sigma$ 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.

13. • CommentRowNumber13.
    • CommentAuthordomenico_fiorenza
    • CommentTimeMay 28th 2010
    • (edited May 28th 2010)
    • PermaLink
    Author: domenico_fiorenza
    Format: MarkdownItexhere 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 $E$ endowed with a connection $\nabla$ on a manifold $X$. then we have a 1-dimensional TQFT mapping a point to the vector space $V$ of global sections of $E\to X$ and a closed interval to the identity operator for $V$. by the abstract nonsense brane-bibrane construction, this means that we have the following path integral representation for the operator $id_V$: $$ \sigma(x)= \int_{\gamma:y\to x} d \gamma \, hol_\nabla(\gamma) \sigma(y), $$ where $\sigma$ is a section of $E$. The dimension of $V$ is therefore $$ dim(V)=tr \left( \int _{\gamma:y\to x} d \gamma \, hol_\nabla(\gamma)\right)=\int_X d x \int _{\Omega_x X} d \gamma \,tr hol_\nabla(\gamma) $$ And here a fibration sequence comes in: $$ \array{ \Omega_x X &\to & \mathcal{L}X\\ \downarrow && \downarrow\\ *&\stackrel{x}{\to}& X } $$ so we find $$ dim(V)=\int_{\mathcal{L}X} d \gamma tr hol_\nabla (\gamma), $$ 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 $S^1$.

    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 $E$ endowed with a connection $\nabla$ on a manifold $X$. then we have a 1-dimensional TQFT mapping a point to the vector space $V$ of global sections of $E\to X$ and a closed interval to the identity operator for $V$. by the abstract nonsense brane-bibrane construction, this means that we have the following path integral representation for the operator $id_V$:

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

    where $\sigma$ is a section of $E$. The dimension of $V$ is therefore

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

    And here a fibration sequence comes in:

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

    so we find

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

    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 $S^1$.

14. • CommentRowNumber14.
    • CommentAuthorUrs
    • CommentTimeMay 28th 2010
    • PermaLink
    Author: Urs
    Format: MarkdownItexCould 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.

    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.

15. • CommentRowNumber15.
    • CommentAuthordomenico_fiorenza
    • CommentTimeMay 28th 2010
    • PermaLink
    Author: domenico_fiorenza
    Format: MarkdownItexI 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]]

    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

16. • CommentRowNumber16.
    • CommentAuthorUrs
    • CommentTimeMay 28th 2010
    • (edited May 28th 2010)
    • PermaLink
    Author: Urs
    Format: MarkdownItexBy the way, is that fiber sequence clear, that you gave? There is $$ \array{ \Omega_x X &\to& * \\ \downarrow && \downarrow \\ * &\to& X } \,, $$ which is equivalent to $$ \array{ \Omega_x X &\to& P_x X \\ \downarrow && \downarrow \\ * &\to& X } \,, $$ I am slightly worried about the $\mathcal{L} X$ 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! :-)

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

    There is

    $$\array{ \Omega_x X &\to& * \\ \downarrow && \downarrow \\ * &\to& X } \,,$$

    which is equivalent to

    $$\array{ \Omega_x X &\to& P_x X \\ \downarrow && \downarrow \\ * &\to& X } \,,$$

    I am slightly worried about the $\mathcal{L} X$ 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! :-)

17. • CommentRowNumber17.
    • CommentAuthordomenico_fiorenza
    • CommentTimeMay 28th 2010
    • PermaLink
    Author: domenico_fiorenza
    Format: MarkdownItex> 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 :-)

    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 :-)

18. • CommentRowNumber18.
    • CommentAuthorUrs
    • CommentTimeMay 28th 2010
    • PermaLink
    Author: Urs
    Format: MarkdownItexAh, 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.

    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.