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1. • CommentRowNumber1.
   • CommentAuthorUrs
   • CommentTimeOct 12th 2019
   • (edited Oct 12th 2019)
   • PermaLink
   Author: Urs
   Format: MarkdownItexI am stuck with understanding the rational cohomology of the group-completed configuration space of points in Euclidean 3-space $$ \Omega B_{{}_{\sqcup}} Conf\big( \mathbb{R}^3 \big) $$ where $Conf\big( \mathbb{R}^3 \big)$ is a topological monoid under disjoint union of configurations of points (after translating them a little; as on p. 1-2 of [Segal 73](https://ncatlab.org/nlab/show/configuration%20space%20of%20points#Segal73)). I know the rational cohomology abstractly, but I am stuck interpreting the generators: Namely, by Theorem 1 in [Segal 73](https://ncatlab.org/nlab/show/configuration%20space%20of%20points#Segal73) and by Example 2.5 in [Møller-Raussen 85](https://ncatlab.org/nlab/show/rational%20model%20of%20mapping%20space#MollerRaussen85) we have a rational equivalence $$ \Omega B_{{}_{\sqcup}} Conf\big( \mathbb{R}^3 \big) \underoverset{\simeq}{Segal 73}{\longrightarrow} \Omega^3 S^3 \underoverset{\simeq_{\mathbb{Q}}}{Moller-Raussen 85}{\longrightarrow} \underset{ n \neq 0 \in \mathbb{N} }{\sqcup} S^3 $$ So the rational cohomology of each connected component of $\Omega B_{{}_{\sqcup}} Conf\big( \mathbb{R}^3 \big)$ is free on a degree 3 generator, except in the connected component of the base point, where it is trivial. I am stuck understanding what these degree-3 generators are in terms of configurations of (virtual) points. One idea was to use the [[group completion theorem]] to express $$ H_\bullet \big( \Omega B_{{}_{\sqcup}} Conf\big( \mathbb{R}^3 \big), \mathbb{Q} \big) \overset{??}{\simeq} H_\bullet \big( Conf \big( \mathbb{R}^3 \big), \mathbb{Q} \big) \big[ (\pi_0)^{-1} \big] $$ I am unsure if the assumption of the group completion theorem is met (that $\pi_0$ is in the center of the Pontrjagin ring). It seems intuitively obvious, but there might be a pitfall. The author of the first lines of [arXiv:math/0511645](https://arxiv.org/abs/math/0511645) thinks that it applies. But assuming the group completion theorem applies, I next run into a contradiction, since $$ H_\bullet \big( Conf_n\big( \mathbb{R}^3 \big), \mathbb{Q} \big) \;=\; \mathbb{Q} $$ is just trivial. (One quick argument is that it's the $Sym(n)$-invariants in [this space](https://ncatlab.org/nlab/show/configuration+space+of+points#eq:RealCohomologyOfUnorderedConfigurationSpaceInEuclideanSpace), which are trivial. It's also a special case of Theorem 4 in [arXiv:math/0311323](https://arxiv.org/abs/math/0311323).) So I guess this means that the assumptions of the McDuff-Segal group completion theorem are not actually met here for the $\sqcup$-monoid of configurations in $\mathbb{R}^3$. (??) Or else I am making some silly mistake elsewhere.

   I am stuck with understanding the rational cohomology of the group-completed configuration space of points in Euclidean 3-space

   $$\Omega B_{{}_{\sqcup}} Conf\big( \mathbb{R}^3 \big)$$

   where $Conf\big( \mathbb{R}^3 \big)$ is a topological monoid under disjoint union of configurations of points (after translating them a little; as on p. 1-2 of Segal 73).

   I know the rational cohomology abstractly, but I am stuck interpreting the generators:

   Namely, by Theorem 1 in Segal 73 and by Example 2.5 in Møller-Raussen 85 we have a rational equivalence

   $$\Omega B_{{}_{\sqcup}} Conf\big( \mathbb{R}^3 \big) \underoverset{\simeq}{Segal 73}{\longrightarrow} \Omega^3 S^3 \underoverset{\simeq_{\mathbb{Q}}}{Moller-Raussen 85}{\longrightarrow} \underset{ n \neq 0 \in \mathbb{N} }{\sqcup} S^3$$

   So the rational cohomology of each connected component of $\Omega B_{{}_{\sqcup}} Conf\big( \mathbb{R}^3 \big)$ is free on a degree 3 generator, except in the connected component of the base point, where it is trivial.

   I am stuck understanding what these degree-3 generators are in terms of configurations of (virtual) points.

   One idea was to use the group completion theorem to express

   $$H_\bullet \big( \Omega B_{{}_{\sqcup}} Conf\big( \mathbb{R}^3 \big), \mathbb{Q} \big) \overset{??}{\simeq} H_\bullet \big( Conf \big( \mathbb{R}^3 \big), \mathbb{Q} \big) \big[ (\pi_0)^{-1} \big]$$

   I am unsure if the assumption of the group completion theorem is met (that $\pi_0$ is in the center of the Pontrjagin ring). It seems intuitively obvious, but there might be a pitfall. The author of the first lines of arXiv:math/0511645 thinks that it applies.

   But assuming the group completion theorem applies, I next run into a contradiction, since

   $$H_\bullet \big( Conf_n\big( \mathbb{R}^3 \big), \mathbb{Q} \big) \;=\; \mathbb{Q}$$

   is just trivial. (One quick argument is that it’s the $Sym(n)$-invariants in this space, which are trivial. It’s also a special case of Theorem 4 in arXiv:math/0311323.)

   So I guess this means that the assumptions of the McDuff-Segal group completion theorem are not actually met here for the $\sqcup$-monoid of configurations in $\mathbb{R}^3$. (??)

   Or else I am making some silly mistake elsewhere.

2. • CommentRowNumber2.
   • CommentAuthorUrs
   • CommentTimeOct 12th 2019
   • (edited Oct 12th 2019)
   • PermaLink
   Author: Urs
   Format: MarkdownItexNow I see that Theorem 6.3 in "Configuration spaces with summable labels" [arXiv:math/9907073](https://arxiv.org/abs/math/9907073) (with the remark above 4.20) gives an explicit description of the group completion of the configuration space in terms of the iterated loop space of another configuration space. That should help. Let's see...

   Now I see that Theorem 6.3 in

   “Configuration spaces with summable labels” arXiv:math/9907073

   (with the remark above 4.20) gives an explicit description of the group completion of the configuration space in terms of the iterated loop space of another configuration space.

   That should help. Let’s see…

3. • CommentRowNumber3.
   • CommentAuthorUrs
   • CommentTimeOct 12th 2019
   • (edited Oct 12th 2019)
   • PermaLink
   Author: Urs
   Format: MarkdownItexAh, no that theorem 3.6 does not help here: By [Boedigheimer 87, Example 11](https://ncatlab.org/nlab/show/configuration+space+of+points#Boedigheimer87) it just reduces to a tautology in the present case. Hm...

   Ah, no that theorem 3.6 does not help here: By Boedigheimer 87, Example 11 it just reduces to a tautology in the present case.

   Hm…

4. • CommentRowNumber4.
   • CommentAuthorUrs
   • CommentTimeOct 12th 2019
   • (edited Oct 12th 2019)
   • PermaLink
   Author: Urs
   Format: MarkdownItexMaybe another strategy: Is there any chance that the free loop space construction commutes with the delooping $B_{\sqcup}$ with respect to taking disjoint union of configurations? I.e. could there be a weak homotopy equivalence $$ Maps(S^1, B_{\sqcup} Conf(\mathbb{R}^3)) \overset{?? }{\simeq} B_{\sqcup} Maps( S^1, Conf(\mathbb{R}^3)) $$ ?? From geometric intuition this seems plausible, but otherwise I have no idea why this could work.

   Maybe another strategy:

   Is there any chance that the free loop space construction commutes with the delooping $B_{\sqcup}$ with respect to taking disjoint union of configurations?

   I.e. could there be a weak homotopy equivalence

   $$Maps(S^1, B_{\sqcup} Conf(\mathbb{R}^3)) \overset{?? }{\simeq} B_{\sqcup} Maps( S^1, Conf(\mathbb{R}^3))$$

   ??

   From geometric intuition this seems plausible, but otherwise I have no idea why this could work.

5. • CommentRowNumber5.
   • CommentAuthorDavid_Corfield
   • CommentTimeOct 13th 2019
   • PermaLink
   Author: David_Corfield
   Format: MarkdownItexThis [MO question](https://mathoverflow.net/q/211378/447) of any interest?

   This MO question of any interest?

6. • CommentRowNumber6.
   • CommentAuthorUrs
   • CommentTimeOct 13th 2019
   • PermaLink
   Author: Urs
   Format: MarkdownItexAs the single reply there says, the readily known results don't help here. But it looks like today with Vincent we found the full statement and proof.

   As the single reply there says, the readily known results don’t help here.

   But it looks like today with Vincent we found the full statement and proof.

7. • CommentRowNumber7.
   • CommentAuthorDavid_Corfield
   • CommentTimeOct 13th 2019
   • PermaLink
   Author: David_Corfield
   Format: MarkdownItexTo appear in the next one of the series of outputs?

   To appear in the next one of the series of outputs?

8. • CommentRowNumber8.
   • CommentAuthorUrs
   • CommentTimeOct 13th 2019
   • PermaLink
   Author: Urs
   Format: MarkdownItexNeed to find a snappy title. This one is maybe too long: "Equivariant Cohomotopy cocycle spaces generate non-abelian quantum gauge field theory on Horava-Witten boundaries".

   Need to find a snappy title. This one is maybe too long: “Equivariant Cohomotopy cocycle spaces generate non-abelian quantum gauge field theory on Horava-Witten boundaries”.

9. • CommentRowNumber9.
   • CommentAuthorUrs
   • CommentTimeOct 17th 2019
   • (edited Oct 17th 2019)
   • PermaLink
   Author: Urs
   Format: MarkdownItexI was wrong in #6: That fact you pointed to in #5 ([here](https://mathoverflow.net/a/211484/381)) is what I need: > The scanning map > $ Conf_n\big( \mathbb{R}^k \big) \longrightarrow \Omega^k S^k {}_{\vert deg = n}$ > induces an isomorphism in cohomoloigy in degrees $\leq n/2$. (I was previously conflating $n$ with $k$, since I have this urge to confuse myself by sticking to a different convention than everyone else uses...) Where is this actually proven? On the first page of [Segal 73](https://ncatlab.org/nlab/show/configuration%20space%20of%20points#Segal73) this is attributed to [Barratt-Priddy 72](https://ncatlab.org/nlab/show/group+completion+theorem#BarrattPriddy72) and [Quillen 94](https://ncatlab.org/nlab/show/group+completion+theorem#Quillen94). But I haven't yet spotted it in either (of the second I only have a GoogleBooks-copy, which only shows the first few pages)

   I was wrong in #6: That fact you pointed to in #5 (here) is what I need:

   The scanning map

   $Conf_n\big( \mathbb{R}^k \big) \longrightarrow \Omega^k S^k {}_{\vert deg = n}$

   induces an isomorphism in cohomoloigy in degrees $\leq n/2$.

   (I was previously conflating $n$ with $k$, since I have this urge to confuse myself by sticking to a different convention than everyone else uses…)

   Where is this actually proven?

   On the first page of Segal 73 this is attributed to Barratt-Priddy 72 and Quillen 94.

   But I haven’t yet spotted it in either (of the second I only have a GoogleBooks-copy, which only shows the first few pages)

10. • CommentRowNumber10.
    • CommentAuthorUrs
    • CommentTimeOct 18th 2019
    • (edited Oct 18th 2019)
    • PermaLink
    Author: Urs
    Format: MarkdownItex[edit: I see now the apparent contradiction in #1 came from me not properly feeding the Moller-Raussen formula through the homotopy fiber computing the based mapping space...]

    [edit: I see now the apparent contradiction in #1 came from me not properly feeding the Moller-Raussen formula through the homotopy fiber computing the based mapping space…]

11. • CommentRowNumber11.
    • CommentAuthorUrs
    • CommentTimeOct 19th 2019
    • (edited Oct 19th 2019)
    • PermaLink
    Author: Urs
    Format: MarkdownItexSo I finally understood my silly mistake that led to the above confusion. The correct rational model of $\Omega^3 S^3$ is instead $ \simeq_{\mathbb{Q}} \underset{n \in \mathbb{N}} {\sqcup} \ast$ ([here](https://ncatlab.org/nlab/show/rational+model+of+mapping+space#RationalModelsForBasedMappingSpaceSDToSD)) and so that matches the rational cohomology of the unordered configuration space $Conf(\mathbb{R}^3)$ being trivial in positive degree. My only remaining confusion is that I keep seeing a non-trivial rational 2-cocycle on the loop space $\Omega Conf_n(\mathbb{R}^3)$ of the unordered configuration space for $n \geq 3$ points: Namely the Sym-invariant part of the respective [[graph complex]] is a dg-model for $Conf_n(\mathbb{R}^3)$ itself, and so its based loop space is modeled by shifting all graphs down in degree by 1, and keeping only the co-unary part of the differential. But the 3-point interaction vertex has co-binary differential, exhibiting the "3-term relation". Hence after passing to based loops, this becomes a closed 2-cocycle. Moreover, it seems immediate that this 2-cocycle is not exact: The only possible form of a trivialization is the graph obtained from it by inserting a single internal edge in the essentially unique way, and a quick explicit check shows that the differential of that is not the 3-point interaction vertex (but is zero). But if this is the case, we found a non-trivial 2 class on $\Omega Conf_3(\mathbb{R}^3)$; and then by adding in more disconnected free vertices also on all $\Omega Conf_{n \geq 3}(\mathbb{R}^3)$. At the same time, all these are loop spaces of rationally contractible spaces by the above. That's a contradiction. I see two possibilities: Either I am making a really basic mistake in unwinding the definition of the graph differential. Or I am unwittingly violating some simply-connectedness assumption that invalidates the model for the looped configuration space by degree-shifted graphs. Hm...

    So I finally understood my silly mistake that led to the above confusion.

    The correct rational model of $\Omega^3 S^3$ is instead $\simeq_{\mathbb{Q}} \underset{n \in \mathbb{N}} {\sqcup} \ast$ (here)

    and so that matches the rational cohomology of the unordered configuration space $Conf(\mathbb{R}^3)$ being trivial in positive degree.

    My only remaining confusion is that I keep seeing a non-trivial rational 2-cocycle on the loop space $\Omega Conf_n(\mathbb{R}^3)$ of the unordered configuration space for $n \geq 3$ points:

    Namely the Sym-invariant part of the respective graph complex is a dg-model for $Conf_n(\mathbb{R}^3)$ itself, and so its based loop space is modeled by shifting all graphs down in degree by 1, and keeping only the co-unary part of the differential.

    But the 3-point interaction vertex has co-binary differential, exhibiting the “3-term relation”. Hence after passing to based loops, this becomes a closed 2-cocycle.

    Moreover, it seems immediate that this 2-cocycle is not exact: The only possible form of a trivialization is the graph obtained from it by inserting a single internal edge in the essentially unique way, and a quick explicit check shows that the differential of that is not the 3-point interaction vertex (but is zero).

    But if this is the case, we found a non-trivial 2 class on $\Omega Conf_3(\mathbb{R}^3)$; and then by adding in more disconnected free vertices also on all $\Omega Conf_{n \geq 3}(\mathbb{R}^3)$.

    At the same time, all these are loop spaces of rationally contractible spaces by the above.

    That’s a contradiction.

    I see two possibilities: Either I am making a really basic mistake in unwinding the definition of the graph differential. Or I am unwittingly violating some simply-connectedness assumption that invalidates the model for the looped configuration space by degree-shifted graphs.

    Hm…