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1. • CommentRowNumber1.
   • CommentAuthordomenico_fiorenza
   • CommentTimeDec 12th 2010
   • (edited Dec 12th 2010)
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   Author: domenico_fiorenza
   Format: MarkdownItexHere is where I'll blog my redaing of [[Topological Quantum Field Theories from Compact Lie Groups]] day 1. quickly read the intro and gone directly to section1. main statement there is proposition 1.2, identifying $H^2(B G,\mathbb{Z})$ with the group of abelian characters $Hom(G,U(1))$. in nLab we have a nice understanding of this identification as follows: the set of abelian characters is the set of cocycles $c:\mathbf{B}G\to \mathbf{B}U(1)$, and the above identification follows by the long fibration sequence of oo-Lie groupoids $$ \mathbf{B}\mathbb{Z}\to \mathbf{B}\mathbb{R}\to \mathbf{B}U(1)\simeq\mathbf{B}(\mathbb{Z}\hookrightarrow\mathbb{R})\to \mathbf{B}^2\mathbb{Z}\to\cdots $$ The description of the $0-1$ tqft associated with $c:\mathbf{B}G\to \mathbf{B}U(1)$ is given explicitely. Apparently it can be interpreted as follows: the vector space $F(pt)$ associated to a point is the space of sections of the line bundle on the groupoid $\mathbf{B}G$ induced by the cocycle $c$ (this space of sections will be $1$- or $0$-dimensional depending whether $c$ is the trivial cocycle or not). The complex number $F(S^1)$ associated to $S^1$ is the dimension of this vector space. This can be seen as the integral over $[S^1,\mathbf{B}G]$ of the morphism $[S^1,\mathbf{B}G]\to [S^1,\mathbf{B}U(1)]\to U(1)$ induced by $c$. That $F(S^1)$ is related this way to the space of maps from $S^1$ to $\mathbf{B}G$ and to the cocycle $c$ seems to be what is really crucial here. Yet in section 1 no deep reason for this being "the right answer" is given at this level and it is only remarked that the above described interpretation of $dim F(S^1)$ is possible.

   Here is where I’ll blog my redaing of Topological Quantum Field Theories from Compact Lie Groups

   day 1. quickly read the intro and gone directly to section1. main statement there is proposition 1.2, identifying with the group of abelian characters . in nLab we have a nice understanding of this identification as follows: the set of abelian characters is the set of cocycles , and the above identification follows by the long fibration sequence of oo-Lie groupoids

   The description of the tqft associated with is given explicitely. Apparently it can be interpreted as follows: the vector space associated to a point is the space of sections of the line bundle on the groupoid induced by the cocycle (this space of sections will be - or -dimensional depending whether is the trivial cocycle or not). The complex number associated to is the dimension of this vector space. This can be seen as the integral over of the morphism induced by .

   That is related this way to the space of maps from to and to the cocycle seems to be what is really crucial here. Yet in section 1 no deep reason for this being “the right answer” is given at this level and it is only remarked that the above described interpretation of is possible.

2. • CommentRowNumber2.
   • CommentAuthorUrs
   • CommentTimeDec 13th 2010
   • (edited Dec 13th 2010)
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   Author: Urs
   Format: MarkdownItexThanks, for starting this, Domenico. > The complex number $F(S^1)$ associated to $S^1$ is the dimension of this vector space. This can be seen as the integral over $[S^1,\mathbf{B}G]$ of the morphism $[S^1,\mathbf{B}G]\to [S^1,\mathbf{B}U(1)]\to U(1)$ induced by $c$. We said the following before elsewhere, but let's recall it: if we think of vector spaces and linear algebra along the lines of [[integral transforms on sheaves]], then the morphism $[S^1,\mathbf{B}G] \stackrel{\exp(S(-))}{\to} [S^1,\mathbf{B}U(1)]\to U(1)$ _is_ the path integral over the action, when regarded as an object in $$ \mathbf{H}/U(1) \,, $$ where $\mathbf{H} = \infty Grpd$ here: we have an $\infty$-groupoid over $U(1) \hookrightarrow \mathbb{C}$ and under groupoid-cardinality in $\mathbb{C}$ this is the Haar measure-weighted path integral over $\exp(S(-))$. Also just for the record, so that we get set up again for this discussion, this argument corresponds to theorem 6, <a href="http://arxiv.org/PS_cache/math/pdf/0503/0503266v1.pdf#page=16">page 16</a> of Simon Willerton's article.

   Thanks, for starting this, Domenico.

   The complex number associated to is the dimension of this vector space. This can be seen as the integral over of the morphism induced by .

   We said the following before elsewhere, but let’s recall it:

   if we think of vector spaces and linear algebra along the lines of integral transforms on sheaves, then the morphism

   is the path integral over the action, when regarded as an object in

   where here: we have an -groupoid over and under groupoid-cardinality in this is the Haar measure-weighted path integral over .

   Also just for the record, so that we get set up again for this discussion, this argument corresponds to theorem 6, page 16 of Simon Willerton’s article.

3. • CommentRowNumber3.
   • CommentAuthorUrs
   • CommentTimeDec 13th 2010
   • (edited Dec 13th 2010)
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   Author: Urs
   Format: MarkdownItexAn analogous statement should holds for the space assigned to the point: The identification $\mathbf{H}/\mathbf{B}G \simeq PSh(\mathbf{B}G)$ goes via the $(\infty,1)$-Grothendieck construction. So up to equivalence every element in the over $\infty$-category is given by a fibration $\psi : \Psi \to \mathbf{B}G$. This is like a flat bundle of $\infty$-groupoids over $\mathbf{B}G$: an $\infty$-groupoid fiber over the point and then autoequivalences for each $g \in G$. So the action functional again gives us a composite $\Psi \to [pt, \mathbf{B}G] \stackrel{\exp(S(-))}{\to} [pt, \mathbf{B}U(1)] \simeq \mathbf{B}U(1) \stackrel{\rho}{\to} Vect$ and we ought to be able to interpret this under a cardinality operation as a vector space (by passing to (co)invariants). So there should be this picture here: for $\Sigma_{in} \to \Sigma \leftarrow \Sigma_{out}$ we have $$ \array{ [\Sigma_{in}, \mathbf{B}G] &\leftarrow& [\Sigma, \mathbf{B}G] &\to& [\Sigma_{out}, \mathbf{B}G] \\ \downarrow && \downarrow && \downarrow \\ [\Sigma_{in}, \mathbf{B}U(1)] &\leftarrow& [\Sigma, \mathbf{B}U(1)] &\to& [\Sigma_{out}, \mathbf{B}U(1)] \\ \downarrow && \downarrow && \downarrow \\ [\Sigma_{in},Vect] &\leftarrow& [\Sigma, Vect] &\to& [\Sigma_{out}, Vect] } $$ and we should be pull-pushing through this int the corresponding over $\infty$-toposes.

   An analogous statement should holds for the space assigned to the point:

   The identification goes via the -Grothendieck construction. So up to equivalence every element in the over -category is given by a fibration . This is like a flat bundle of -groupoids over : an -groupoid fiber over the point and then autoequivalences for each .

   So the action functional again gives us a composite

   and we ought to be able to interpret this under a cardinality operation as a vector space (by passing to (co)invariants).

   So there should be this picture here: for we have

   and we should be pull-pushing through this int the corresponding over -toposes.

4. • CommentRowNumber4.
   • CommentAuthordomenico_fiorenza
   • CommentTimeDec 13th 2010
   • (edited Dec 14th 2010)
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   Author: domenico_fiorenza
   Format: MarkdownItexHi Urs, thanks for joining this! :) > We said the following before elsewhere, but let's recall it yes, recalling that here is a very good idea, thanks. and now... day 2 :) I've skipped to Section 3 to see how the 0-1 tqft case is handled by the general theory. adapting for $n=1$ the beginning of Section 3, the starting point is the existence of the $Bun_G$ functor $Bun_G:Bord_1^{SO}\to Fam_1$, which maps a manifold $M$ to the groupoid of principal $G$-bundles oner $M$ ($G$ is a finite group here). So this is nothing but the functor $M\mapsto \mathbf{H}(M,\mathbf{B}G)$ or, in the internal version $M\mapsto [M,\mathbf{B}G]$. Next, fixing $Vect$ to be the symmetric monoidal category $\mathcal{C}$ involved, the classical theory is given by a lift $I: Bord_1^{SO}\to Fam_1(Vect)$ of $Bun_G$.Here, objects in $Fam_1(Vect)$ are objects in $Fam_1$, i.e., finite groupoids, with a morphism to $Vect$; morphisms are _branes_ $$ \array{ && Z&&\\ &\swarrow && \searrow&\\ X && \Rightarrow && Y\\ &\searrow && \swarrow&\\ &&Vect&& } $$ In the case of a 0-1 tqft out of a cocycle $c:\mathbf{B}G\to \mathbf{B}U(1)$ the functor $I$ is defined by $I(pt_+)=\{\mathbf{B}G\to Vect\}$ where the morphism to $Vect$ is induced by $c$ and by the standard representation of $U(1)$. This definition of $I$ therefore relies on the [[cobordism hypotesis]] in the version given by Theorem 2.1 in the paper: since $SO(1)$ is the trivial group, giving $I:Bord_1^SO\to Fam_1(Vect)$ is the same thing that giving a fully dualizable object in $Fam_1(Vect)$. This leaves it completely implicit what is $I$ on 1-manifolds, which is not completely satisfying. I'll come back to this later.

   Hi Urs,

   thanks for joining this! :)

   We said the following before elsewhere, but let’s recall it

   yes, recalling that here is a very good idea, thanks.

   and now… day 2 :)

   I’ve skipped to Section 3 to see how the 0-1 tqft case is handled by the general theory. adapting for the beginning of Section 3, the starting point is the existence of the functor , which maps a manifold to the groupoid of principal -bundles oner ( is a finite group here). So this is nothing but the functor or, in the internal version .

   Next, fixing to be the symmetric monoidal category involved, the classical theory is given by a lift of .Here, objects in are objects in , i.e., finite groupoids, with a morphism to ; morphisms are branes

   In the case of a 0-1 tqft out of a cocycle the functor is defined by where the morphism to is induced by and by the standard representation of . This definition of therefore relies on the cobordism hypotesis in the version given by Theorem 2.1 in the paper: since is the trivial group, giving is the same thing that giving a fully dualizable object in .

   This leaves it completely implicit what is on 1-manifolds, which is not completely satisfying. I’ll come back to this later.

5. • CommentRowNumber5.
   • CommentAuthordomenico_fiorenza
   • CommentTimeDec 14th 2010
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   Author: domenico_fiorenza
   Format: MarkdownItexSill on the 0-1 tqft case. So far we have explicitly described only $I(pt_+)$. The definition of $I(pt_+)$ forces $I(pt_-)$ to be the representation of $G$ dual of the one associated with $pt_+$. So, explicitely, $g\cdot \varphi=c(g)^{-1}\varphi$, for $\varphi:\mathbb{C}\to \mathbb{C}$. Then $I([0,1])$ as a cobordism from $p_+$ to $p_+$ is the identity of $I(p_+)$ as a representation of $G$; similarly, $I([0,1])$ as a cobordism between $p_+\coprod p_-$ and $\emptyset$ is the canonical pairing $I(p_-)\otimes I(p_+)\to \mathbb{C}$ and $I([0,1])$ as a cobordism between $\emptyset$ and $p_+\coprod p_-$ is the copairing $\mathbb{C}\to I(p_+)\otimes I(p_-)$. Note that here we are using the fact that $I(\emptyset)$ is the trivial 1-dimensional representation of $G$, which is forced by the fact that $I(\emptyset)$ has to be the unit in the symmetric monoidal category of representations of finite groupoids. Note that $I$ maps a 1-manifold with boundary to a morphism between functors to $Vect$ rather than to a functor from a groupoid to $Vect$. This is obvious, since a 1-manifold is a 1-morphism in $Bord_1^{SO}$ and so it has to be mapped to a 1-morphism in $Fam_1(Vect)$. Next we come to $I(S^1)$. Since $I$ lifts $Bun_G$, $I(S^1)$, this has $I(S^1)$ has to be a functor $[S^1,\mathbf{B}G]\to 0Vect=\mathbb{C}$. that is $I(S^1)$ has to be a class function on $G$. This is is agreement with the brane interpretation of morphisms: from the cospan $\emptyset\to S^1\leftarrow \emptyset$ we get the span of groupoids $*\leftarrow [S^1,\mathbf{B}G]\to *$, and branes with this roof over $Vect$ are naturally identified with class functions on $G$. and with the given data there is no better candidate for this class function than the cocycle $c$ itself, seen as a function $c:G\to U(1)\subseteq \mathbb{C}$. With this choice, $$ F(S^1)=Sum_0(I(S^1))=\frac{1}{|G|}\sum_{g\in G}c(g) $$ in accordance with the formula on page 4 of the paper. Note that the fact that $I$ maps $S^1$ to a functor $G//G\to 0Vect$ has a simple and natural interpretation: since $S^1$ is a cobordism from the empty manifold to itself, $I(S^1)$ has to be a functor taking values in the space of morphisms of the unit of $1Vect$.

   Sill on the 0-1 tqft case. So far we have explicitly described only . The definition of forces to be the representation of dual of the one associated with . So, explicitely, , for . Then as a cobordism from to is the identity of as a representation of ; similarly, as a cobordism between and is the canonical pairing and as a cobordism between and is the copairing . Note that here we are using the fact that is the trivial 1-dimensional representation of , which is forced by the fact that has to be the unit in the symmetric monoidal category of representations of finite groupoids. Note that maps a 1-manifold with boundary to a morphism between functors to rather than to a functor from a groupoid to . This is obvious, since a 1-manifold is a 1-morphism in and so it has to be mapped to a 1-morphism in .

   Next we come to . Since lifts , , this has has to be a functor . that is has to be a class function on . This is is agreement with the brane interpretation of morphisms: from the cospan we get the span of groupoids , and branes with this roof over are naturally identified with class functions on . and with the given data there is no better candidate for this class function than the cocycle itself, seen as a function . With this choice,

   in accordance with the formula on page 4 of the paper.

   Note that the fact that maps to a functor has a simple and natural interpretation: since is a cobordism from the empty manifold to itself, has to be a functor taking values in the space of morphisms of the unit of .

6. • CommentRowNumber6.
   • CommentAuthordomenico_fiorenza
   • CommentTimeDec 14th 2010
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   Author: domenico_fiorenza
   Format: MarkdownItexI'm not satisfied with the ad hoc definition of $[S^1,\mathbf{B}G]\to 0Vect$. I'd like this to be $[S^1,\mathbf{B}G]\to [S^1,Vect]$, induced by $\mathbf{B}G\to Vect$, but I'm presently unable to give a rigorous meaning to $[S^1,Vect]$. Yet, I'm convinced precisely this is the key of the whole construction.

   I’m not satisfied with the ad hoc definition of . I’d like this to be , induced by , but I’m presently unable to give a rigorous meaning to . Yet, I’m convinced precisely this is the key of the whole construction.

7. • CommentRowNumber7.
   • CommentAuthorUrs
   • CommentTimeDec 14th 2010
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   Author: Urs
   Format: MarkdownItex> but I'm presently unable to give a rigorous meaning to $[S^1,Vect]$. The canonical interpretation would be as the [[functor category]] (or rather its [[core]] fr the purposes here) from the groupoid incarnation of $S^1$ to $Vect$. That is the groupoid whose objects are linear endomorphisms on a vector space, and whose morphisms are conjugations by invertible linear maps. What is important is, I think, that we regard $Vect$ as a pointed category, with point the ground field $k$ regarded as a vector space over itself. The component of $core(Func(S^1,Vect))$ on this component is $k//k^\times$ (since $k \simeq End_k(k)$). So its 0-truncation is indeed $k$.

   but I’m presently unable to give a rigorous meaning to .

   The canonical interpretation would be as the functor category (or rather its core fr the purposes here) from the groupoid incarnation of to . That is the groupoid whose objects are linear endomorphisms on a vector space, and whose morphisms are conjugations by invertible linear maps.

   What is important is, I think, that we regard as a pointed category, with point the ground field regarded as a vector space over itself. The component of on this component is (since ). So its 0-truncation is indeed .

8. • CommentRowNumber8.
   • CommentAuthordomenico_fiorenza
   • CommentTimeDec 15th 2010
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   Author: domenico_fiorenza
   Format: MarkdownItexHi Urs, the functor category interpretation works extremely well! and we do not need to regard $Vect$ as a pointed category nor to restrict to the core. Indeed, if $\mathbf{B}G\to Vect$ is the functor induced by $c$ composed with the standard representation of $U(1)$, then the induced fnctor $[S^1,\mathbf{B}G]\to [S^1,Vect]$ between functor categories maps the object $(*,g)$ to the object $(\mathbb{C},c(g))$ and the morphism $h: (*,g)\to (*,h^{-1}g h)$ to the morphism $c(h):(\mathbb{C},c(g))\to (\mathbb{C},c(h)^{-1} c(g) c(h))$ (where I left the conjugation explicit to have a form which is still valid for higher dimensional representations of $G$). So this part is ok, and to finally land in $\mathbb{C}$ we just need to take the trace, $Tr:[S^1,Vect] \to \mathbb{C}$.

   Hi Urs,

   the functor category interpretation works extremely well! and we do not need to regard as a pointed category nor to restrict to the core. Indeed, if is the functor induced by composed with the standard representation of , then the induced fnctor between functor categories maps the object to the object and the morphism to the morphism (where I left the conjugation explicit to have a form which is still valid for higher dimensional representations of ).

   So this part is ok, and to finally land in we just need to take the trace, .

9. • CommentRowNumber9.
   • CommentAuthorUrs
   • CommentTimeDec 15th 2010
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   Author: Urs
   Format: MarkdownItexAll right! :-) > we just need to take the trace Hopefully that will be said more intrinsically. At least for the 1-dimensional case, this step is 0-truncation of $[S^1, Vect]$.

   All right! :-)

   we just need to take the trace

   Hopefully that will be said more intrinsically. At least for the 1-dimensional case, this step is 0-truncation of .

10. • CommentRowNumber10.
    • CommentAuthordomenico_fiorenza
    • CommentTimeDec 15th 2010
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    Author: domenico_fiorenza
    Format: MarkdownItex> Hopefully that will be said more intrinsically. Yes, having a good intrinsic understanding of what is $Tr:[S^1,Vect] \to \mathbb{C}$ seems to be a crucial point here. I don't clearly see this as the truncation morphism $[S^1,Vect]\to \tau_{\leq 0}[S^1,Vect]$ unless "the 1-dimensional case" refers to the dimension of the vector space $V$ in the object $(V,\varphi)$ of $[S^1,Vect]$. Apart this, I was thinking that for $K$ a 1-dimensional manifold (with boundary), the functor category $[K,Vect]$ is a categorical verison of (a particular case of) the labelled diagrams appearing in Joyal-Street-Reshetikhin-Turaev graphical calculus for morphisms. So it seems that on the one hand we have a sort of "decorated cobordism" $Bord_1^{SO}(Vect)$ equipped with a natural morphism to $Vect$ given by the graphical calculus, and on the other we have the cocycle $c$ inducing $Bord_1^{SO}(\mathbf{B}G)\to Bord_1^{SO}(Vect)$. The composition of these is $Bord_1^{SO}(\mathbf{B}G)\to Vect$ and pushing forward to the point gives the tqft $Bord_1^{SO}\to Vect$. So the crucial step seems to understand $[K,n{}Vect]\to n{}Vect$ in arbitrary dimension.

    Hopefully that will be said more intrinsically.

    Yes, having a good intrinsic understanding of what is seems to be a crucial point here. I don’t clearly see this as the truncation morphism unless “the 1-dimensional case” refers to the dimension of the vector space in the object of .

    Apart this, I was thinking that for a 1-dimensional manifold (with boundary), the functor category is a categorical verison of (a particular case of) the labelled diagrams appearing in Joyal-Street-Reshetikhin-Turaev graphical calculus for morphisms. So it seems that on the one hand we have a sort of “decorated cobordism” equipped with a natural morphism to given by the graphical calculus, and on the other we have the cocycle inducing . The composition of these is and pushing forward to the point gives the tqft .

    So the crucial step seems to understand in arbitrary dimension.

11. • CommentRowNumber11.
    • CommentAuthordomenico_fiorenza
    • CommentTimeDec 16th 2010
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    Author: domenico_fiorenza
    Format: MarkdownItexThe morphism $Bord_1^{SO}(Vect)\to Vect$ should be completely natural and have the following interpretation: $Bord_1^{SO}(Vect)$ is the _free symmetric monoidal category with duals generated by $Vect$_; hence, for any symmetric monoidal category with duals $\mathcal{C}$ and any functor $F:Vect\to \mathcal{C}$, is induced a duality preserving symmetric monoidal functor $Z:Bord_1^{SO}(Vect)\to \mathcal{C}$ extending $F$. This appears to be a simple repharsing of Joyal-Street-Reshetikhin-Turaev diagrammatics, and suggests the following general statement: $Bord_n^{SO}(-)$ is the adjoint of the inclusion $\{$symmetric monoidal $(\infty,n)$-categories with duals$\}\hookrightarrow\{(\infty,n)$-categories$\}$ Given an arbitrary symmetric monoidal $(\infty,n)$-category $\mathcal{C}$, I expect that the full subcategory $fd(\mathcal{C})$ on fully dualizable objects is a symmetric monoidal $(\infty,n)$-category with duals, so the conjectured adjuction above would in paticular give an equivalence between duality preserving symmetric monoidal functors $Bord_n^{SO}(*)\to fd(\mathcal{C})\hookrightarrow \mathcal{C}$ and objects in $fd(\mathcal{C})$, i.e., the [[cobordism hypothesis]] in its original formulation. Coming back to the $\mathbf{B}G\to \mathbf{B}U(1)$ example, the above seems to say that a symmetric monoidal morphism $Bord_1^{SO}(\mathbf{B}G)\to Vect$ is completely determined (or better its homotopy class is) by the morphism $\mathbf{B}G\to Vect$ induced by the cocycle $c:\mathbf{B}G\to \mathbf{B}U(1)$. Next the problem of pushing this forward to a symmetric monoidal morphism $Bord_1^{SO}\to Vect$ seems to be reduced to the problem of pushing forward $\mathbf{B}G\to Vect$ along $\mathbf{B}G\to *$. This is a particular case of the problem of pushing forward a morphism $X\to Vect$ along $X\to *$, where $X$ is any garoupioid, and this can in turn be conveniently expressed in terms of left or right Kan extensions. This precisely transates the limit or colimit formulation of $Sum_1$ in the paper.

    The morphism should be completely natural and have the following interpretation: is the free symmetric monoidal category with duals generated by ; hence, for any symmetric monoidal category with duals and any functor , is induced a duality preserving symmetric monoidal functor extending .

    This appears to be a simple repharsing of Joyal-Street-Reshetikhin-Turaev diagrammatics, and suggests the following general statement: is the adjoint of the inclusion symmetric monoidal -categories with duals-categories

    Given an arbitrary symmetric monoidal -category , I expect that the full subcategory on fully dualizable objects is a symmetric monoidal -category with duals, so the conjectured adjuction above would in paticular give an equivalence between duality preserving symmetric monoidal functors and objects in , i.e., the cobordism hypothesis in its original formulation.

    Coming back to the example, the above seems to say that a symmetric monoidal morphism is completely determined (or better its homotopy class is) by the morphism induced by the cocycle . Next the problem of pushing this forward to a symmetric monoidal morphism seems to be reduced to the problem of pushing forward along . This is a particular case of the problem of pushing forward a morphism along , where is any garoupioid, and this can in turn be conveniently expressed in terms of left or right Kan extensions. This precisely transates the limit or colimit formulation of in the paper.

12. • CommentRowNumber12.
    • CommentAuthordomenico_fiorenza
    • CommentTimeDec 18th 2010
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    Author: domenico_fiorenza
    Format: MarkdownItexaccording to <a href="http://mathoverflow.net/questions/49732/free-symmetric-monoidal-infty-n-categories-with-duals">this MO answer</a>, the left adjoint point of view seems to be correct. I'll be presently not working on the details of this, but rather will try to read the constructions in Freed-Hopkins-Lurie-Teleman in the light of it. The $n=1$ case sketched above seems to say that this point of view is promising.

    according to this MO answer, the left adjoint point of view seems to be correct. I’ll be presently not working on the details of this, but rather will try to read the constructions in Freed-Hopkins-Lurie-Teleman in the light of it. The case sketched above seems to say that this point of view is promising.

13. • CommentRowNumber13.
    • CommentAuthorUrs
    • CommentTimeDec 18th 2010
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    Author: Urs
    Format: MarkdownItex> I expect that the full subcategory $fd(\mathcal{C})$ on fully dualizable objects is a symmetric monoidal $(\infty,n)$-category with duals Yes, that's part of claim 2.3.19 of _Classification of TFTs_ . I have finally started an entry [[symmetric monoidal (infinity,n)-category]] and recorded that statement there.

    I expect that the full subcategory on fully dualizable objects is a symmetric monoidal -category with duals

    Yes, that’s part of claim 2.3.19 of Classification of TFTs . I have finally started an entry symmetric monoidal (infinity,n)-category and recorded that statement there.

14. • CommentRowNumber14.
    • CommentAuthorUrs
    • CommentTimeDec 18th 2010
    • (edited Dec 18th 2010)
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    Author: Urs
    Format: MarkdownItexHi Domenico, I had been distracted by some other things and seem to have lost you a bit. Could you say again what you said before about $Bord_n^{SO}(Vect)$? I understand that you say you want to think of this as denoting the free "with dual"ization of $Vect$, but I am not sure if I see why and how. (I do remember that before we discussed elsewhere that $Bord_\infty(X)$ should be the free symmetric-monoidalization of $\Pi(X)$ in some sense.) Also you write the cocycle $c : \mathbf{B}G \to Vect$ induces $Bord_n^{SO}(\mathbf{B}G) \to Bord^{SO}_n(Vect)$. But by "$Bord_n^{SO}(\mathbf{B}G)$" we used to denote the $n$-category of cobordisms equipped with maps to $\mathbf{B}G$. Is that still what you ,mean? Sorry for not following, but if you could just restate in a paragraph what you are after now, that might help.

    Hi Domenico,

    I had been distracted by some other things and seem to have lost you a bit. Could you say again what you said before about ? I understand that you say you want to think of this as denoting the free “with dual”ization of , but I am not sure if I see why and how.

    (I do remember that before we discussed elsewhere that should be the free symmetric-monoidalization of in some sense.)

    Also you write the cocycle induces . But by “” we used to denote the -category of cobordisms equipped with maps to . Is that still what you ,mean?

    Sorry for not following, but if you could just restate in a paragraph what you are after now, that might help.

15. • CommentRowNumber15.
    • CommentAuthordomenico_fiorenza
    • CommentTimeDec 18th 2010
    • (edited Dec 18th 2010)
    • PermaLink
    Author: domenico_fiorenza
    Format: MarkdownItexHi Urs, let me use two distinct notations here not to make things foggier than they deserve. On the one side we should have a left adjoint to the inclusion of symmetric monoidal $(\infty,n)$-categories with duals into $(\infy,n)$-categories. Let us denote this by $Free^\otimes_n$. On the other side we have a notion of $\mathcal{C}$-decorated bordism, that is, precisely, $n$-cobordism with maps to $\mathcal{C}$, something we could denote by $Bord_n^{SO}(\mathcal{C})$ or more explicitly $[Bord_n^{SO},\mathcal{C}]$. The arguments in the final part of _Classification of TQFTs_ seem to suggest an equivalence $Free^\otimes_n(\mathcal{C})\simeq Bord^{SO}_n(\mathcal{C})$. Details of this are still unclear to me, except in the $n=1$ case where it is a familiar statement from Joyal-Street-Reshetikhin-Turaev. Assuming the statement is true, then by definition of adjointness we have, for any symmetric monoidal $(\infty,n)$-category with duals $\mathcal{C}$, a natural morphism $Bord^{SO}_n(\mathcal{C})\to \mathcal{C}$. What I'm saying is that (again at least in the $n=1$ case where I've gone through the details) this nicely gives the construction in _TQFTs from compact Lie groups_. Namely, $c:\mathbf{B}G\to Vect$ induces $Bord^{SO}_1(\mathbf{B}G)\to Bord^{SO}_1(Vect)$. Since $Vect$ is a symmetric monoidal category with duals (I'm assuming $Vect$ is denoting the category of finite dimensional vector spaces), then we have the natural symmetric monoidal duality preserving morphism $Bord^{SO}_1(Vect)\to Vect$ induced by the universal property of $Bord_1^{SO}$, and so we end up with a symmetric monoidal morphism $Bord^{SO}_1(\mathbf{B}G)\to Vect$. Now to end up witha tqft we are left with the task of pushing this forward along $\mathbf{B}G\to *$. Here the idea is: consider the Kan extension of $c:\mathbf{B}G\to Vect$ along $\mathbf{B}G\to *$, and apply to the homotopy commutative diagram you obtatin the hom-functor $[Bord_1^{SO},-]$. This, together with the natural morphism $Bord^{SO}_1(Vect)\to Vect$ gives the diagram $$ \array{ Bord_1^{SO}(\mathbf{B}G)&\to&Bord_1^{SO}(Vect)&\to&Vect\\ \downarrow&\nearrow&&&\\ Bord_1^{SO}(*) } $$ whose "lower profile" should be the seeked tqft. edit: is the lower profile the Kan extension $$ \array{ Bord_1^{SO}(\mathbf{B}G)&\to&Vect\\ \downarrow&\nearrow&&&\\ Bord_1^{SO}(*) } $$ ?

    Hi Urs,

    let me use two distinct notations here not to make things foggier than they deserve. On the one side we should have a left adjoint to the inclusion of symmetric monoidal -categories with duals into -categories. Let us denote this by . On the other side we have a notion of -decorated bordism, that is, precisely, -cobordism with maps to , something we could denote by or more explicitly .

    The arguments in the final part of Classification of TQFTs seem to suggest an equivalence . Details of this are still unclear to me, except in the case where it is a familiar statement from Joyal-Street-Reshetikhin-Turaev. Assuming the statement is true, then by definition of adjointness we have, for any symmetric monoidal -category with duals , a natural morphism . What I’m saying is that (again at least in the case where I’ve gone through the details) this nicely gives the construction in TQFTs from compact Lie groups.

    Namely, induces . Since is a symmetric monoidal category with duals (I’m assuming is denoting the category of finite dimensional vector spaces), then we have the natural symmetric monoidal duality preserving morphism induced by the universal property of , and so we end up with a symmetric monoidal morphism . Now to end up witha tqft we are left with the task of pushing this forward along . Here the idea is: consider the Kan extension of along , and apply to the homotopy commutative diagram you obtatin the hom-functor . This, together with the natural morphism gives the diagram

    whose “lower profile” should be the seeked tqft.

    edit: is the lower profile the Kan extension

    ?

16. • CommentRowNumber16.
    • CommentAuthordomenico_fiorenza
    • CommentTimeDec 18th 2010
    • PermaLink
    Author: domenico_fiorenza
    Format: MarkdownItexthe $n=2$ case seems to go straightforwardly along these lines: a cocylce $c:\mathbf{B}G\to \mathbf{B}^2 U(1)$ induces a morphism $\mathbf{B}G\to 2 Vect$, where $2 Vect$ is the 2-category whose objects are $\mathbb{C}$ algebras and whose morphisms are bimodules (this 2-category is denoted $Alg$ in FHLT). The defining representation $\mathbf{B}^2 U(1)\to 2 Vect$ maps the unique object of $\mathbf{B}^2 U(1)$ to the algebra $\mathbb{C}$, the unique 1-morphism of $\mathbf{B}^2 U(1)$ to vector space $\mathbb{C}$ (as a $(\mathbb{C},\mathbb{C})$-bimodule), and the element $x\in U(1)$ (as a 2-morphism in $\mathbf{B}^2 U(1)$) to the element $x\in U(1)\subseteq End(\mathbb{C})$. the Kan extension of $\mathbf{B}G\to 2 Vect$ along $\mathbf{B}G \to *$ is the twisted group algebra of $G$ (this is example 3.14 in FHLT).

    the case seems to go straightforwardly along these lines: a cocylce induces a morphism , where is the 2-category whose objects are algebras and whose morphisms are bimodules (this 2-category is denoted in FHLT). The defining representation maps the unique object of to the algebra , the unique 1-morphism of to vector space (as a -bimodule), and the element (as a 2-morphism in ) to the element . the Kan extension of along is the twisted group algebra of (this is example 3.14 in FHLT).

17. • CommentRowNumber17.
    • CommentAuthorUrs
    • CommentTimeDec 19th 2010
    • (edited Dec 19th 2010)
    • PermaLink
    Author: Urs
    Format: MarkdownItexHi Domenico, sorry to be slowing things down, but I have more questions about what precisely you are thinking of now. Of course I recognize plenty of ideas here, but to make progress I feel I need better grip on where you are thinking of precise statements and where you are sketching how the story ought to run. > let me use two distinct notations here not to make things foggier than they deserve. On the one side we should have a left adjoint to the inclusion of symmetric monoidal $(\infty,n)$-categories with duals into $(\infy,n)$-categories. Let us denote this by $Free^\otimes_n$. On the other side we have a notion of $\mathcal{C}$-decorated bordism, that is, precisely, $n$-cobordism with maps to $\mathcal{C}$, What kind of map from a cobordism $\Sigma$ to $\mathcal{C}$ are you thinking of here? Do you still map the groupoid $\Pi(\Sigma)$ to $\mathcal{C}$? > something we could denote by $Bord_n^{SO}(\mathcal{C})$ or more explicitly $[Bord_n^{SO},\mathcal{C}]$. The latter notation has an evident interpretation as the $n$-category of (symmetric monoidal) $n$-functors $Bord_n \to \mathcal{C}$ (i.e. an extended $n$d-TFT with values in $\mathcal{C}$). But I don't think that that's what you mean. I'd think it takes a bit more discussion to say what one might mean here. But maybe I am missing something. > except in the $n=1$ case where it is a familiar statement from Joyal-Street-Reshetikhin-Turaev There is a famous "Reshitikhin-Turaev construction" for $n = 3$ and $\mathcal{C}$ a modular tensor category. That's not what you mean, though? > the $n=2$ case seems to go straightforwardly along these lines: a cocylce $c:\mathbf{B}G\to \mathbf{B}^2 U(1)$ induces a morphism $\mathbf{B}G\to 2 Vect$, where $2 Vect$ is the 2-category whose objects are $\mathbb{C}$ algebras and whose morphisms are bimodules (this 2-category is denoted $Alg$ in FHLT). The defining representation $\mathbf{B}^2 U(1)\to 2 Vect$ maps the unique object of $\mathbf{B}^2 U(1)$ to the algebra $\mathbb{C}$, the unique 1-morphism of $\mathbf{B}^2 U(1)$ to vector space $\mathbb{C}$ (as a $(\mathbb{C},\mathbb{C})$-bimodule), and the element $x\in U(1)$ (as a 2-morphism in $\mathbf{B}^2 U(1)$) to the element $x\in U(1)\subseteq End(\mathbb{C})$. the Kan extension of $\mathbf{B}G\to 2 Vect$ along $\mathbf{B}G \to *$ is the twisted group algebra of $G$ (this is example 3.14 in FHLT). Yes, over a point, the quantization construction is a plain colimit. When extending this to all cobordisms, things tend to get trickier. Let me see...

    Hi Domenico,

    sorry to be slowing things down, but I have more questions about what precisely you are thinking of now. Of course I recognize plenty of ideas here, but to make progress I feel I need better grip on where you are thinking of precise statements and where you are sketching how the story ought to run.

    let me use two distinct notations here not to make things foggier than they deserve. On the one side we should have a left adjoint to the inclusion of symmetric monoidal -categories with duals into -categories. Let us denote this by . On the other side we have a notion of -decorated bordism, that is, precisely, -cobordism with maps to ,

    What kind of map from a cobordism to are you thinking of here? Do you still map the groupoid to ?

    something we could denote by or more explicitly .

    The latter notation has an evident interpretation as the -category of (symmetric monoidal) -functors (i.e. an extended d-TFT with values in ). But I don’t think that that’s what you mean. I’d think it takes a bit more discussion to say what one might mean here. But maybe I am missing something.

    except in the case where it is a familiar statement from Joyal-Street-Reshetikhin-Turaev

    There is a famous “Reshitikhin-Turaev construction” for and a modular tensor category. That’s not what you mean, though?

    the case seems to go straightforwardly along these lines: a cocylce induces a morphism , where is the 2-category whose objects are algebras and whose morphisms are bimodules (this 2-category is denoted in FHLT). The defining representation maps the unique object of to the algebra , the unique 1-morphism of to vector space (as a -bimodule), and the element (as a 2-morphism in ) to the element . the Kan extension of along is the twisted group algebra of (this is example 3.14 in FHLT).

    Yes, over a point, the quantization construction is a plain colimit. When extending this to all cobordisms, things tend to get trickier.

    Let me see…

18. • CommentRowNumber18.
    • CommentAuthordomenico_fiorenza
    • CommentTimeDec 19th 2010
    • (edited Dec 19th 2010)
    • PermaLink
    Author: domenico_fiorenza
    Format: MarkdownItexHi Urs, no need to apologize: I see what I'm writing here is really foggy. What I'm trying to fix is an essential version of the constructions in FHLT: my impression there is that in many occasions they describe some object coming out of nowhere as the "right" object, and only after they give a simple general recipe which would have produced that specific object. This is particularly evident in the case of the tensor category $Vect^\tau [G}$ defined in section 4 of FHLT: the general argument of the paper suggest that $Vect^\tau [G}$ should arise very simply as the colimit of the functor $\mathbf{B}G\to 3 Vect$ induced by the cocycle $c:\mathbf{B}G\to \mathbf{B}^3 U(1)$, and by the defining representation $\mathbf{B}^3 U(1)\to 3 Vect$ mapping the unique object of $\mathbf{B}^3 U(1)$ to the tensor category of finite dimensional vector spaces over $\mathbb{C}$. Yet, this seems not to be stated explicitely in the paper. So, for how trivial it is, I'm now going to spend some time to convince myself that what I've just written is correct :) >What kind of map from a cobordism $\Sigma$ to $\mathcal{C}$ are you thinking of here? Do you still map the groupoid $\Pi(\Sigma)$ to $\mathcal{C}$? Yes, probably that's the correct way of looking at what I'm at the moment thinking in more naive terms: namely, I'm presently thinking in terms of functors from a triangulation of $\Sigma$ to $\mathcal{C}$. Extremely rough, as you see. I guess you are right and $\Pi(\Sigma)$ is the right object to be considered. > something we could denote by $Bord_n^{SO}(\mathcal{C})$ or more explicitly $[Bord_n^{SO},\mathcal{C}]$. You are right, teh second notation is absolutely confusing. What I had in mind is the following: $Bord_n^{SO}(\mathcal{C})$ maps a triangulated manifold $\Sigma$ to the $[\Sigma,\mathcal{C}]$. So, adoping the $\Pi(\Sigma)$ point of view (which I'm conviced should be the correct one), this should be $$ Bord_n^{SO}(\mathcal{C}): \Sigma \to [\Pi(\Sigma),\mathcal{C}] $$ at least as a first approximation. >There is a famous "Reshitikhin-Turaev construction" for $n = 3$ and $\mathcal{C}$ a modular tensor category. That's not what you mean, though? I mean the following: given an arbitrary category $\mathcal{C}$, one can consider the category whose objects are finite sequences of objects of $\mathcal{C}$ (each one equipped with a $+/-$ sign), and whose morphisms are oriented links between these sequences, with strands decorated by morphisms in $\mathcal{C}$. Juxtaposition of links and the "obvious" braiding makes this a symmetric monoidal category, and if i'm not wrong this should be the free symmetric monoidal category generated by $\mathcal{C}$. I'm not sure of the source for this construction, but I think I've learned it from works of Joyal-Street and Reshetikhin-Turaev. > Yes, over a point, the quantization construction is a plain colimit. When extending this to all cobordisms, things tend to get trickier. Absolutely. But I'm confident we can get to the point where tricks disappear and things become completely natural. Having a good understandng of what happens over the point is a good beginning..

    Hi Urs,

    no need to apologize: I see what I’m writing here is really foggy. What I’m trying to fix is an essential version of the constructions in FHLT: my impression there is that in many occasions they describe some object coming out of nowhere as the "right" object, and only after they give a simple general recipe which would have produced that specific object. This is particularly evident in the case of the tensor category defined in section 4 of FHLT: the general argument of the paper suggest that should arise very simply as the colimit of the functor induced by the cocycle , and by the defining representation mapping the unique object of to the tensor category of finite dimensional vector spaces over . Yet, this seems not to be stated explicitely in the paper.

    So, for how trivial it is, I’m now going to spend some time to convince myself that what I’ve just written is correct :)

    What kind of map from a cobordism to are you thinking of here? Do you still map the groupoid to ?

    Yes, probably that’s the correct way of looking at what I’m at the moment thinking in more naive terms: namely, I’m presently thinking in terms of functors from a triangulation of to . Extremely rough, as you see. I guess you are right and is the right object to be considered.

    something we could denote by or more explicitly .

    You are right, teh second notation is absolutely confusing. What I had in mind is the following: maps a triangulated manifold to the . So, adoping the point of view (which I’m conviced should be the correct one), this should be

    at least as a first approximation.

    There is a famous "Reshitikhin-Turaev construction" for and a modular tensor category. That’s not what you mean, though?

    I mean the following: given an arbitrary category , one can consider the category whose objects are finite sequences of objects of (each one equipped with a sign), and whose morphisms are oriented links between these sequences, with strands decorated by morphisms in . Juxtaposition of links and the "obvious" braiding makes this a symmetric monoidal category, and if i’m not wrong this should be the free symmetric monoidal category generated by . I’m not sure of the source for this construction, but I think I’ve learned it from works of Joyal-Street and Reshetikhin-Turaev.

    Yes, over a point, the quantization construction is a plain colimit. When extending this to all cobordisms, things tend to get trickier.

    Absolutely. But I’m confident we can get to the point where tricks disappear and things become completely natural. Having a good understandng of what happens over the point is a good beginning..

19. • CommentRowNumber19.
    • CommentAuthorUrs
    • CommentTimeDec 20th 2010
    • (edited Dec 20th 2010)
    • PermaLink
    Author: Urs
    Format: MarkdownItex> this should be [...] Yes. This is along the lines indicated in the entry on [[Topological Quantum Field Theories from Compact Lie Groups]]: we imagine that we have an $\infty$-topos $\mathbf{H}$ into which manifolds faithfully embed $Manifolds \hookrightarrow \mathbf{H}$. Then for every object $X \in \mathbf{H}$ there ought to be an $(\infty,n)$-functor $$ Bord_n(*) \to Fam_n(*) $$ given by sending a manifold $\Sigma$ with boundaries $\Sigma_{in} \to \Sigma \leftarrow \Sigma_{out}$ to the span of $\infty$-groupoids $$ \mathbf{H}(\Sigma_{in}, X) \leftarrow \mathbf{H}(\Sigma, X) \to \mathbf{H}(\Sigma_{out}, X) \,. $$ If $\mathbf{H}$ is cohesive and $X$ is of the form $LConst A$ then we have $\mathbf{H}(\Sigma, X) \simeq \infty Grpd(\Pi(\Sigma), A)$. Now, given a morphism $X \to A$ (an "action functional") there ought to be an $n$-stack object $QC_n$ which assigns an $(\infty,n)$-category of quasicoherent sheaves of $n$-modules, and a canonical representation $A \to QC_n$. Then the composite $X \to A \to QC_n$ ought to be the full action functional. And using this morphism there ought to be a lift of the above $$ Z_X : Bord_n \to Fam_n(*) $$ to $Bord_n \to Fam_n(n Vect)$ or the like. And then quantization ought to be postcomposition with a canonical morphism $$ \int \;\; : \;\; Fam_n(n Vect) \to n Vect \,. $$

    this should be […]

    Yes. This is along the lines indicated in the entry on Topological Quantum Field Theories from Compact Lie Groups:

    we imagine that we have an -topos into which manifolds faithfully embed . Then for every object there ought to be an -functor

    given by sending a manifold with boundaries to the span of -groupoids

    If is cohesive and is of the form then we have .

    Now, given a morphism (an “action functional”) there ought to be an -stack object which assigns an -category of quasicoherent sheaves of -modules, and a canonical representation . Then the composite ought to be the full action functional. And using this morphism there ought to be a lift of the above

    to or the like.

    And then quantization ought to be postcomposition with a canonical morphism

20. • CommentRowNumber20.
    • CommentAuthordomenico_fiorenza
    • CommentTimeDec 20th 2010
    • PermaLink
    Author: domenico_fiorenza
    Format: MarkdownItexHi Urs, ok, so the $(\infty,n)$ functor $Bord_n(*)\to Fam_n(*)$ associated with an object $X$ is what I'm calling $Bord_n(X)$, while the $n$-stack object $QC_n$ is what I've implicitly been denoting $n Vect$. So the way I figure things is that a morphism $X\to QC_n$ induces a morphism $Bord_n(X)\to Bord_n(QC_n)$; next $Bord_n(QC_n)$ should have a natural lift $\widetilde{Bord}_n(QC_n)$ to a functor taking values in $Fam_n(n Vect)$, and this should be the crucial point of the whole construction. Then the composition of these two maps should give $$ Bord_n(X)\to Fam_n(n Vect) $$ This is simply restating what you have just written above, stressing attention on $Bord_n(QC_n)$. I mean: it doesn't really matter how do we get to $Bord_n(QC_n)$ (a way could be, e.g., via a cocycle $\mathbf{B}G\to \mathbf{B}^n U(1)\to QC_n$ as we did several times); what really matters is wahat we are able to do once we are on $Bord_n(QC_n)$, i.e., how do we canonically endow the oo-groupoids $\mathbf{H}(\Sigma,QC_n)$ with morphisms to $n Vect$. Push-forward along $\Sigma\to *$?

    Hi Urs,

    ok, so the functor associated with an object is what I’m calling , while the -stack object is what I’ve implicitly been denoting .

    So the way I figure things is that a morphism induces a morphism ; next should have a natural lift to a functor taking values in , and this should be the crucial point of the whole construction. Then the composition of these two maps should give

    This is simply restating what you have just written above, stressing attention on . I mean: it doesn’t really matter how do we get to (a way could be, e.g., via a cocycle as we did several times); what really matters is wahat we are able to do once we are on , i.e., how do we canonically endow the oo-groupoids with morphisms to .

    Push-forward along ?

21. • CommentRowNumber21.
    • CommentAuthorUrs
    • CommentTimeDec 20th 2010
    • (edited Dec 20th 2010)
    • PermaLink
    Author: Urs
    Format: MarkdownItex> ok, so the $(\infty,n)$ functor $Bord_n(*)\to Fam_n(*)$ associated with an object $X$ is what I'm calling $Bord_n(X)$, Okay, good. > while the $n$-stack object $QC_n$ is what I've implicitly been denoting $n Vect$. All right, this is less important, as this is just notation for an object that is kind of obvious. The $QC$-notation is a bit more suggestive for an object in a $(\infty,n)$-sheaf topos, but maybe that need not be our primary concern at the moment. > So the way I figure things is that a morphism $X\to QC_n$ induces a morphism $Bord_n(X)\to Bord_n(QC_n)$; Hm, okay, I see. That's a slight variant of saying that the description in terms of that lift. Let me think about it. > how do we canonically endow the oo-groupoids $\mathbf{H}(\Sigma,QC_n)$ with morphisms to $n Vect$. I think we should come back to that nice fact that $\tau_{n-d}$\mathbf{H}(\Pi(\Sigma_d),\mathbf{B}^n U(1)) \simeq \mathbf{B}^{n-d}U(1)$. This means that over $\Sigma_d$ we get the data suitable for a transformation between the data over $\Sigma_{d-1}$ $$ \array{ && \mathbf{H}(\Pi\Sigma_d, \mathbf{B}^n U(1)) \\ & \swarrow && \searrow \\ \mathbf{H}(\Pi\Sigma_{d-1}, \mathbf{B}^n U(1)) &&\swArrow&& \mathbf{H}(\Pi\Sigma'_{d-1}, \mathbf{B}^n U(1)) \\ & \searrow && \swarrow \\ && \mathbf{B}^{n-(d-1)}U(1) \\ && \downarrow \\ && (n-(d-1)Vect) } \,. $$ However, I am not yet sure how to organize this properly.

    ok, so the functor associated with an object is what I’m calling ,

    Okay, good.

    while the -stack object is what I’ve implicitly been denoting .

    All right, this is less important, as this is just notation for an object that is kind of obvious. The -notation is a bit more suggestive for an object in a -sheaf topos, but maybe that need not be our primary concern at the moment.

    So the way I figure things is that a morphism induces a morphism ;

    Hm, okay, I see. That’s a slight variant of saying that the description in terms of that lift. Let me think about it.

    how do we canonically endow the oo-groupoids with morphisms to .

    I think we should come back to that nice fact that \mathbf{H}(\Pi(\Sigma_d),\mathbf{B}^n U(1)) \simeq \mathbf{B}^{n-d}U(1)$.

    This means that over we get the data suitable for a transformation between the data over

    However, I am not yet sure how to organize this properly.

22. • CommentRowNumber22.
    • CommentAuthorUrs
    • CommentTimeDec 21st 2010
    • (edited Dec 21st 2010)
    • PermaLink
    Author: Urs
    Format: MarkdownItexDomenico, please remind me: had you worked out the relative version of the truncation formula that is needed for what I said in the above comment? For $\Sigma$ an $d$-dimensional monifold with boundary $\Sigma_{in} \coprod \Sigma_{out} \hookrightarrow \Sigma$, do we know how to get the natural transformation as above?

    Domenico,

    please remind me: had you worked out the relative version of the truncation formula that is needed for what I said in the above comment?

    For an -dimensional monifold with boundary , do we know how to get the natural transformation as above?

23. • CommentRowNumber23.
    • CommentAuthordomenico_fiorenza
    • CommentTimeDec 21st 2010
    • (edited Dec 21st 2010)
    • PermaLink
    Author: domenico_fiorenza
    Format: MarkdownItexHi Urs, yes and no: I had tried to think to that in terms of relative cohomology but was no able to get to a clean result. What confuses me about the truncation formula is that it seems to be much more restrictive than one would expect the theory to be: e.g., for $n=1$ one is delaing with line bundles, whereas things seem to go through for arbitrary vector bundles. That's why I'd try to think directly in terms of $\mathbf{H}(-,QC_n)$ rather than in terms of $\mathbf{H}(-,\mathbf{B}^n U(1))$. The basic construction we need to understand, I think, is the following: given a flat vector bundle (or more in general a vector bundle equipped with a connection) on a space $X$, taking traces of holonomy gives a section of the $End(\mathbf{1}_\mathbb{C})$-bundle over the loop space $\mathcal{L}X$, where $\mathbf{1}_\mathbb{C}$ is the trivial $\mathbb{C}$-bundle. Here we are hiddenly using the fact that the trace gives a natural brane $$ \array{ && \mathbf{H}(\Pi S^1, QC_1) \\ & \swarrow && \searrow \\ * &&\Rightarrow&& * \\ & \searrow && \swarrow \\ && Vect } \,. $$ where the bottom arrows pick out the algebra object $\mathbb{C}$ in $Vect$. we have been around this several times, already, but I still think we have not got into the real mening of this; and indeed we are not able to see clearly which is the natural higher dimensional generalization. or at least I am not :)

    Hi Urs,

    yes and no: I had tried to think to that in terms of relative cohomology but was no able to get to a clean result.

    What confuses me about the truncation formula is that it seems to be much more restrictive than one would expect the theory to be: e.g., for one is delaing with line bundles, whereas things seem to go through for arbitrary vector bundles. That’s why I’d try to think directly in terms of rather than in terms of .

    The basic construction we need to understand, I think, is the following: given a flat vector bundle (or more in general a vector bundle equipped with a connection) on a space , taking traces of holonomy gives a section of the -bundle over the loop space , where is the trivial -bundle. Here we are hiddenly using the fact that the trace gives a natural brane

    where the bottom arrows pick out the algebra object in .

    we have been around this several times, already, but I still think we have not got into the real mening of this; and indeed we are not able to see clearly which is the natural higher dimensional generalization. or at least I am not :)

24. • CommentRowNumber24.
    • CommentAuthordomenico_fiorenza
    • CommentTimeDec 21st 2010
    • PermaLink
    Author: domenico_fiorenza
    Format: MarkdownItexHi Urs, I'm thinking to what you wrote in #19 again. That's extremely neat, let me rewrite it in my words to see if I got the point: we have _classical theories_ $Bord_n(*)\to Fam_n( n Vect)$ and _quantum theories_ $Bord(*)\to n Vect$. The quantization of a classical theory is dealt with in FHLT, and is the natural functor $$ \int: Fam_n( n Vect)\to n vect $$ The crucial question, then is: given any object $X$ in a suitable $(\infty,1)$-topos $\mathbf{H}$, this gives an obvious funtor $Z_X: Bord_n(*)\to Fam_n(*)$; can this functor be lifted to a classical theory? how? these lift I would call _classical theories over $X$_. Each of these gives, in particular a functor $X\to n Vect$, corresponding to $\mathbf{H})*,X)\simeq X$. The probelm is how much does this functor tell us of the eventual classical theory over $X$?

    Hi Urs,

    I’m thinking to what you wrote in #19 again. That’s extremely neat, let me rewrite it in my words to see if I got the point: we have classical theories and quantum theories . The quantization of a classical theory is dealt with in FHLT, and is the natural functor

    The crucial question, then is: given any object in a suitable -topos , this gives an obvious funtor ; can this functor be lifted to a classical theory? how?

    these lift I would call classical theories over . Each of these gives, in particular a functor , corresponding to . The probelm is how much does this functor tell us of the eventual classical theory over ?

25. • CommentRowNumber25.
    • CommentAuthordomenico_fiorenza
    • CommentTimeDec 21st 2010
    • (edited Dec 21st 2010)
    • PermaLink
    Author: domenico_fiorenza
    Format: MarkdownItexHey, wait, the answer is: _fully dualizable object_! Namely, $Fam_n(n Vect)$ is a symmetric monoidal $(\infty,n)$-category, so by the cobordism hypotesis giving a classical field theory, i.e., a symmetric monoidal functor $I:Bord_n(*)\to Fam_n(n Vect)$ is the same thing as giving a fully dualizable object in $Fam_n(n Vect)$, i.e. a suitable functor $X\to n Vect$ with $X$ an oo-groupoid. More precisely, the datum of $I$ is equivalent to the datum of $I(*)=\{X\to n Vect\}$ Now, composing $I$ with the forgetful morphism $Fam_n(n Vect)\to Fam_n(*)$ gives a functor $F:Bord_n(*)\to Fam_n(*)$ which (trivially) is lifted by $I$. If the functor $F$ is representable, then it is represented by $X$, since $F(*)=X$. This would mean that $I$ lifts a functor of the form $\mathbf{H}(-,X)$ as in the posts above. On the other hand, making the fully dualizable object $I(*)$ be the cornerstone of the construction makes the issue of the representability of $F$ an interesting but probably not crucial question. So, coming back to the FHLT paper, I think we should ask ourselves the following questions: a) does $Sum_n: Fam_n(\mathcal{C})\to \mathcal{C}$ preserve full dualizable objects? b) are the "obvious" functors $\mathbf{B}G\to n Vect$ induced by cocycles $\mathbf{B}G\to \mathbf{B}^n U(1)$ fully dualizable objects in $Fam_n(n Vect)$? c) is the defining representation $\mathbf{B}^n U(1)\to n Vect$ a fully dualizable objects in $Fam_n(n Vect)$? d) if $X\to \mathcal{C}$ is a fully dualizable object in $Fam_n(\mathcal{C})$ and $Y\to X$ is an $n$-functor, is the composition $Y\to \mathcal{C}$ a fully dualizable object in $Fam_n(\mathcal{C})$ (i.e., does c) implies b)? )

    Hey, wait, the answer is: fully dualizable object! Namely, is a symmetric monoidal -category, so by the cobordism hypotesis giving a classical field theory, i.e., a symmetric monoidal functor is the same thing as giving a fully dualizable object in , i.e. a suitable functor with an oo-groupoid. More precisely, the datum of is equivalent to the datum of

    Now, composing with the forgetful morphism gives a functor which (trivially) is lifted by . If the functor is representable, then it is represented by , since . This would mean that lifts a functor of the form as in the posts above. On the other hand, making the fully dualizable object be the cornerstone of the construction makes the issue of the representability of an interesting but probably not crucial question.

    So, coming back to the FHLT paper, I think we should ask ourselves the following questions:

    a) does preserve full dualizable objects?

    b) are the “obvious” functors induced by cocycles fully dualizable objects in ?

    c) is the defining representation a fully dualizable objects in ?

    d) if is a fully dualizable object in and is an -functor, is the composition a fully dualizable object in (i.e., does c) implies b)? )

26. • CommentRowNumber26.
    • CommentAuthorUrs
    • CommentTimeDec 21st 2010
    • PermaLink
    Author: Urs
    Format: MarkdownItexre #23, you write: > What confuses me about the truncation formula is that it seems to be much more restrictive than one would expect the theory to be: e.g., for n=1 one is delaing with line bundles, whereas things seem to go through for arbitrary vector bundles. Yes, true, but I would be willing to believe that the case $\mathbf{B}^n U(1)$ plays a special role that justifies it having its tailor-made theorems. After all, the Lagrangians for $\infty$-Chern-Simons theory that we are trying to figure out the extended quantization of always take values in $\mathbf{B}^n U(1)_{diff}$. So I wouldn't be shocked if that case has a good general abstract quantization story, while a more general case does not. I think the truncation formula we have works too well to easily discard it. After all, it is a general abstract way of saying "the action is the integral over the Lagrangian". That should not be light-heartedly be discrded! :-)

    re #23, you write:

    What confuses me about the truncation formula is that it seems to be much more restrictive than one would expect the theory to be: e.g., for n=1 one is delaing with line bundles, whereas things seem to go through for arbitrary vector bundles.

    Yes, true, but I would be willing to believe that the case plays a special role that justifies it having its tailor-made theorems. After all, the Lagrangians for -Chern-Simons theory that we are trying to figure out the extended quantization of always take values in . So I wouldn’t be shocked if that case has a good general abstract quantization story, while a more general case does not.

    I think the truncation formula we have works too well to easily discard it. After all, it is a general abstract way of saying “the action is the integral over the Lagrangian”. That should not be light-heartedly be discrded! :-)

27. • CommentRowNumber27.
    • CommentAuthorUrs
    • CommentTimeDec 21st 2010
    • PermaLink
    Author: Urs
    Format: MarkdownItexConcerning your other comments: 'll think about it and come back to it a little later. Need to get something else done first.

    Concerning your other comments: ’ll think about it and come back to it a little later. Need to get something else done first.

28. • CommentRowNumber28.
    • CommentAuthordomenico_fiorenza
    • CommentTimeDec 21st 2010
    • PermaLink
    Author: domenico_fiorenza
    Format: MarkdownItex> the truncation formula we have works too well to easily discard it. After all, it is a general abstract way of saying "the action is the integral over the Lagrangian". That should not be light-heartedly be discrded! :-) a very good point! absolutely! :) concerning #25, the answer to a) is: _trivially true_. indeed that is an immediate consequence of the [[cobordism hypothesis]] and of the fact $Sum_n$ is a symmetric monoidal functor.

    the truncation formula we have works too well to easily discard it. After all, it is a general abstract way of saying “the action is the integral over the Lagrangian”. That should not be light-heartedly be discrded! :-)

    a very good point! absolutely! :)

    concerning #25, the answer to a) is: trivially true. indeed that is an immediate consequence of the cobordism hypothesis and of the fact is a symmetric monoidal functor.

29. • CommentRowNumber29.
    • CommentAuthordomenico_fiorenza
    • CommentTimeDec 27th 2010
    • PermaLink
    Author: domenico_fiorenza
    Format: MarkdownItexconcerning question b) in #25, the answer is _yes_ at least for $n=1$ and $n=2$. Indeed, by the cobordism hypothesis, the value assigned to the point is a fully dualizable object, and in FHLT, what the functor $I$ does on the point in the 1- and 2- dimensional case is said explicitely: it maps the point to the representation of $\mathbf{B}G$ induced by the cocycle $c:\mathbf{B}G\to \mathbf{B}^n U(1)$. I think one should expect the answer to be yes regardless of the dimension.

    concerning question b) in #25, the answer is yes at least for and . Indeed, by the cobordism hypothesis, the value assigned to the point is a fully dualizable object, and in FHLT, what the functor does on the point in the 1- and 2- dimensional case is said explicitely: it maps the point to the representation of induced by the cocycle . I think one should expect the answer to be yes regardless of the dimension.