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1. • CommentRowNumber1.
   • CommentAuthordomenico_fiorenza
   • CommentTimeOct 4th 2010
   • PermaLink
   Author: domenico_fiorenza
   Format: MarkdownItexI'm confused by the definition of $\mathbf{B}^n U(1)_{diff,simp}$ at <a href="http://ncatlab.org/nlab/show/circle+n-bundle+with+connection#DeligneCohomologyTheorem">circle n-bundle with connection</a>. Is there a "modulo $\mathbf{B}^n\mathbb{Z}$" missing? and, if so, which sense we quotient by $\mathbf{B}^n\mathbb{Z}$ there?

   I’m confused by the definition of $\mathbf{B}^n U(1)_{diff,simp}$ at circle n-bundle with connection. Is there a “modulo $\mathbf{B}^n\mathbb{Z}$” missing? and, if so, which sense we quotient by $\mathbf{B}^n\mathbb{Z}$ there?

2. • CommentRowNumber2.
   • CommentAuthorUrs
   • CommentTimeOct 4th 2010
   • (edited Oct 4th 2010)
   • PermaLink
   Author: Urs
   Format: MarkdownItexThere must be a quotient by $\mathbf{B}^n \mathbb{Z}$, yes. If it is not there, that's a typo. I'll look into it as soon as I got to grab some dinner. We take the quotient in the sense of simplicial abelian groups (of forms on simplices, under addition of forms). For that we embed $\mathbf{B}^n \mathbb{Z}_{simp} \hookrightarrow \mathbf{B}^{n}\mathbb{R}_{simp,diff}$ by identifying $k \in \mathbb{Z}$ with the unique $n$-form on $U \times \Delta^n$ that is entirely constant and whose integral over the $n$-simplex equals $k \in \mathbb{Z} \hookrightarrow \mathbb{R}$.

   There must be a quotient by $\mathbf{B}^n \mathbb{Z}$, yes. If it is not there, that’s a typo. I’ll look into it as soon as I got to grab some dinner.

   We take the quotient in the sense of simplicial abelian groups (of forms on simplices, under addition of forms).

   For that we embed $\mathbf{B}^n \mathbb{Z}_{simp} \hookrightarrow \mathbf{B}^{n}\mathbb{R}_{simp,diff}$ by identifying $k \in \mathbb{Z}$ with the unique $n$-form on $U \times \Delta^n$ that is entirely constant and whose integral over the $n$-simplex equals $k \in \mathbb{Z} \hookrightarrow \mathbb{R}$.

3. • CommentRowNumber3.
   • CommentAuthordomenico_fiorenza
   • CommentTimeOct 4th 2010
   • PermaLink
   Author: domenico_fiorenza
   Format: MarkdownItexok, fine. At the same page there is: Observation. Under the Dold-Kan correspondence the normalized chain complex of $\mathbf{B}^n U(1)_{sim}$ is... in which there seems to be a homologically trivial part in the complex on the left. this gives me some problem, since it seems to me differential forms with sitting instants on simplices are not an homologically trivial complex, so I'm missing something here.

   ok, fine. At the same page there is:

   Observation. Under the Dold-Kan correspondence the normalized chain complex of $\mathbf{B}^n U(1)_{sim}$ is…

   in which there seems to be a homologically trivial part in the complex on the left. this gives me some problem, since it seems to me differential forms with sitting instants on simplices are not an homologically trivial complex, so I’m missing something here.

4. • CommentRowNumber4.
   • CommentAuthorUrs
   • CommentTimeOct 4th 2010
   • (edited Oct 4th 2010)
   • PermaLink
   Author: Urs
   Format: MarkdownItexHi Domenico, I am now looking at the entry. To my relief I see that where I define $\mathbf{B}^n U(1)_{simp}$ I _do_ write the quotient by $\mathbf{B}^n \mathbb{Z}$. But you are right that the same was omitted in the definition of the refinement $\mathbf{B}^n U(1)_{diff}$. I fixed that now. Concerning the second question: if I am looking at the same line that you are, then there are the symbols $$ (...) / \sim $$ behind the terms in the chain complex. Could you check that for the line that you are looking at? That's supposed to be the $\mathbb{Z}$-quotient, again, that makes this homologically non-trivial. But I see that it is easy to overlook these symbols and more amplification of what is meant is in order anyway. As soon as you confirm that this is the line you are looking at, I'll edit that bit to make it read better.

   Hi Domenico,

   I am now looking at the entry. To my relief I see that where I define $\mathbf{B}^n U(1)_{simp}$ I do write the quotient by $\mathbf{B}^n \mathbb{Z}$. But you are right that the same was omitted in the definition of the refinement $\mathbf{B}^n U(1)_{diff}$. I fixed that now.

   Concerning the second question: if I am looking at the same line that you are, then there are the symbols

   $$(...) / \sim$$

   behind the terms in the chain complex. Could you check that for the line that you are looking at? That’s supposed to be the $\mathbb{Z}$-quotient, again, that makes this homologically non-trivial.

   But I see that it is easy to overlook these symbols and more amplification of what is meant is in order anyway. As soon as you confirm that this is the line you are looking at, I’ll edit that bit to make it read better.

5. • CommentRowNumber5.
   • CommentAuthordomenico_fiorenza
   • CommentTimeOct 4th 2010
   • PermaLink
   Author: domenico_fiorenza
   Format: MarkdownItexyes, it's that's line. what confuses me (apart the $\mathbb{Z}$ quotient which is now ok) is the part in the first complex which is denoted by the ellipsis: what is the complex there, explicitly? is that part meant to be a homologically trivial complex?

   yes, it’s that’s line. what confuses me (apart the $\mathbb{Z}$ quotient which is now ok) is the part in the first complex which is denoted by the ellipsis: what is the complex there, explicitly? is that part meant to be a homologically trivial complex?

6. • CommentRowNumber6.
   • CommentAuthorUrs
   • CommentTimeOct 4th 2010
   • (edited Oct 4th 2010)
   • PermaLink
   Author: Urs
   Format: MarkdownItexSo over $U$ in degree $k$ the complex consists of closed vertical $n$-forms (with sitting instants) on $U \times \Delta^k$ modulo forms in the image of the degenracy maps (which are forms coming from pullback of forms on $U \times \Delta^{k-1}$ via the degeneracy maps). We see that this has no homology groups in higher degree by observing (under Dold-Kan) that the original simplicial set has no homotopy groups there: if you have a closed $n$-form on the $k \gt n$-sphere, it always extends to a closed $n$-form on the $k+1$-ball. An explicit proof of this last statement is currently not spelled out on the nLab. But after I have walked home I can spell it out. You can also see Andre Henriques' "Integrating L-oo algebras" First you show that a closed n-form on the n-sphere extends to a closed n-form on the n+1-ball precisely if its integral over the sphere vanishes: details are in <a href="http://www.math.uni-hamburg.de/home/schreiber/nactwist.pdf#page=89">section 5.2.2. of nactwist</a>. Then use this to show that every n-form on a higher-dimensional sphere can be extended: foliate the sphere by n-balls, notice that the integral of the closed form over threir boundary now vanishes since the boundary is now contractible in the higher dimensional sphere. So extend the forms slice-wise and observe that you can then add a form depending on the direction orthogonal to the foliation to make the result be closed again. (There is probably a more elegant way to achieve this... )

   So over $U$ in degree $k$ the complex consists of closed vertical $n$-forms (with sitting instants) on $U \times \Delta^k$ modulo forms in the image of the degenracy maps (which are forms coming from pullback of forms on $U \times \Delta^{k-1}$ via the degeneracy maps).

   We see that this has no homology groups in higher degree by observing (under Dold-Kan) that the original simplicial set has no homotopy groups there: if you have a closed $n$-form on the $k \gt n$-sphere, it always extends to a closed $n$-form on the $k+1$-ball.

   An explicit proof of this last statement is currently not spelled out on the nLab. But after I have walked home I can spell it out. You can also see Andre Henriques’ “Integrating L-oo algebras”

   First you show that a closed n-form on the n-sphere extends to a closed n-form on the n+1-ball precisely if its integral over the sphere vanishes: details are in section 5.2.2. of nactwist.

   Then use this to show that every n-form on a higher-dimensional sphere can be extended: foliate the sphere by n-balls, notice that the integral of the closed form over threir boundary now vanishes since the boundary is now contractible in the higher dimensional sphere. So extend the forms slice-wise and observe that you can then add a form depending on the direction orthogonal to the foliation to make the result be closed again.

   (There is probably a more elegant way to achieve this… )

7. • CommentRowNumber7.
   • CommentAuthordomenico_fiorenza
   • CommentTimeOct 5th 2010
   • PermaLink
   Author: domenico_fiorenza
   Format: MarkdownItexthe proof at <a href="http://www.math.uni-hamburg.de/home/schreiber/nactwist.pdf#page=89">section 5.2.2. of nactwist</a> is the one I had in mind, and whose details are puzzling me: it works perfectly fine with smooth forms, but I don't see how it works adding the sitting instant constraint: Poincare' lemma should not hold there.

   the proof at section 5.2.2. of nactwist is the one I had in mind, and whose details are puzzling me: it works perfectly fine with smooth forms, but I don’t see how it works adding the sitting instant constraint: Poincare’ lemma should not hold there.

8. • CommentRowNumber8.
   • CommentAuthorUrs
   • CommentTimeOct 5th 2010
   • (edited Oct 5th 2010)
   • PermaLink
   Author: Urs
   Format: MarkdownItexLet me see: the proof invokes the Poincaré-lemma to find from the closed $n$-form $B_n$ on $S^n$ a form $a_{n-1}$ on $S^n$ such that $d a_{n-1} = B_n$ (using that the period of $B$ is assumed to vanish). There is no sense in which $a_{n-1}$ here could have sitting instants, since there are no boundaries, and it is not required to. What is required is that the form on the $(n+1)$-disk filling the $n$-sphere which is defined as $$ d (f \wedge a_{n-1}) $$ where $f$ depends only on the radial coordinates and $a_{n-1}$ only on the spherical ones, becomes constant in the radial direction in a neighbourhood $r \in (1-\epsilon, 1]$. But this is true by the requirement that $f : [0,1] \to [0,1]$ itself has sitting instants. Am I overlooking something?

   Let me see: the proof invokes the Poincaré-lemma to find from the closed $n$-form $B_n$ on $S^n$ a form $a_{n-1}$ on $S^n$ such that $d a_{n-1} = B_n$ (using that the period of $B$ is assumed to vanish).

   There is no sense in which $a_{n-1}$ here could have sitting instants, since there are no boundaries, and it is not required to.

   What is required is that the form on the $(n+1)$-disk filling the $n$-sphere which is defined as

   $$d (f \wedge a_{n-1})$$

   where $f$ depends only on the radial coordinates and $a_{n-1}$ only on the spherical ones, becomes constant in the radial direction in a neighbourhood $r \in (1-\epsilon, 1]$. But this is true by the requirement that $f : [0,1] \to [0,1]$ itself has sitting instants.

   Am I overlooking something?

9. • CommentRowNumber9.
   • CommentAuthordomenico_fiorenza
   • CommentTimeOct 5th 2010
   • PermaLink
   Author: domenico_fiorenza
   Format: MarkdownItexthat's right. it is the passage from spheres and disks to simplices and their boundaries that is puzzling me: what I miss is a way to go from a 1-form (say) with sitting instants on $\partial\Delta^2$ to a smooth form on $S^1$; and to go from a smooth 1-form on $D^2$ which becomes constant in the radial direction to a smooth 1-form on $\Delta^2$ which is constant in the normal direction to the boundary: behaviour at the vertices is puzzling me :(

   that’s right. it is the passage from spheres and disks to simplices and their boundaries that is puzzling me: what I miss is a way to go from a 1-form (say) with sitting instants on $\partial\Delta^2$ to a smooth form on $S^1$; and to go from a smooth 1-form on $D^2$ which becomes constant in the radial direction to a smooth 1-form on $\Delta^2$ which is constant in the normal direction to the boundary: behaviour at the vertices is puzzling me :(

10. • CommentRowNumber10.
    • CommentAuthorDavidRoberts
    • CommentTimeOct 6th 2010
    • PermaLink
    Author: DavidRoberts
    Format: MarkdownItexThat's ok, because a smooth curve can turn a 90 degree corner, if it is flat at the corner, for example if it has sitting instants within an $\varepsilon$-neighbourhood. In other words, the circle can map smoothly onto $\partial \Delta^2$ if one takes the parameterisation to be flat at the corners. Then the smooth form on $\partial \Delta^2$ pulls back to $S^1$ easy ;)

    That’s ok, because a smooth curve can turn a 90 degree corner, if it is flat at the corner, for example if it has sitting instants within an $\varepsilon$-neighbourhood. In other words, the circle can map smoothly onto $\partial \Delta^2$ if one takes the parameterisation to be flat at the corners. Then the smooth form on $\partial \Delta^2$ pulls back to $S^1$ easy ;)

11. • CommentRowNumber11.
    • CommentAuthorUrs
    • CommentTimeOct 6th 2010
    • (edited Oct 6th 2010)
    • PermaLink
    Author: Urs
    Format: MarkdownItexSorry, Domenico and myself have further discussed this by private email. We should move the discussion back here. What Domenico actually spotted was that there had been a clause missing in the <a href="http://ncatlab.org/nlab/show/Lie+integration#SmoothIntegration">definition of forms with sitting instants on a simplex</a>: it only did say that the forms have to become perpendicularly constant towards a face. But it needs to say that they become perpendicularly constant towards a face _on the value they have on that face_ . Meaning: if a $p$-form approaches an $(k \lt p)$-face, it has to become constantly _vanishing_ perpendicular to that face. That clause had been missing on the nLab page, and this is what made Domenico wonder. I have now added this phrase.

    Sorry, Domenico and myself have further discussed this by private email. We should move the discussion back here.

    What Domenico actually spotted was that there had been a clause missing in the definition of forms with sitting instants on a simplex:

    it only did say that the forms have to become perpendicularly constant towards a face. But it needs to say that they become perpendicularly constant towards a face on the value they have on that face . Meaning: if a $p$-form approaches an $(k \lt p)$-face, it has to become constantly vanishing perpendicular to that face.

    That clause had been missing on the nLab page, and this is what made Domenico wonder. I have now added this phrase.