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1. • CommentRowNumber1.
   • CommentAuthorUrs
   • CommentTimeOct 29th 2019
   • (edited Oct 29th 2019)
   • PermaLink
   Author: Urs
   Format: MarkdownItexSome naive ramblings, just thinking out loud, in case anyone feels inspired to offer a comment: I am trying to see how close to an ordered configuration space of points one can get with mapping spaces, which a priori give un-ordered configuration spaces of points. The idea I have is -- in words -- the following: An ordered configuration of points in $\mathbb{R}^3$ (say) is, up to homotopy, the same as a) An un-ordered configuration of points in $\mathbb{R}^3 \times \mathbb{R}^1$, such that this b) projects to an $\mathbb{R}^1$-labeled un-ordered configuration in $\mathbb{R}^3$; and c) projects to an $\mathbb{R}^3$-labeled un-ordered configuration in $\mathbb{R}^1$. Meaning that the points in the configuration are distinct not only as points in $\mathbb{R}^3 \times \mathbb{R}^1$, but also after projection as points in $\mathbb{R}^3$ and as points in $\mathbb{R}^1$. Here condition c) is what imposes an ordering on the "labels" in $\mathbb{R}^1$, since an arrangement of distict points on the real line puts these points into linear order. The formal statement of this idea should be that * the ordered un-labeled configuration space $\underset{n \in \mathbb{N}}{\sqcup} \underset{ {}^{ \{1,\cdots, n \} } }{Conf}(\mathbb{R}^3)$ is a fiber product (in the 1-category of topological spaces) of * the unordered $X$-labeled configuration spaces $Conf( \mathbb{R}^k, X )$ with points disappearing when labeled by the base-point of $X$ as follows: $$ \underset{n \in \mathbb{N}}{\sqcup} \underset{ {}^{ \{1,\cdots, n \} } }{Conf}(\mathbb{R}^3) \;\simeq_{homeo}\; Conf\big( \mathbb{R}^3, S^1 \big) \underset{ Conf\big( \mathbb{D}^{3+1} \big)_{rel} }{\times} Conf\big( \mathbb{R}^1, S^3 \big) $$ That this is the case should essentially come down to observing that this fiber product encodes the above "in words" description. First I thought that this 1-categorical fiber product is a homotopy-retraction of the corresponding homotopy fiber product, but now I think this can't be. This is because all items in the above are homotopy equivalent to based mapping spaces as $$ Maps^{\!\ast/\!}\big( S^3, S^4\big) \underset{ Maps^{\!\ast/\!}\big( S^0, S^4\big) }{\times^h} Maps^{\!\ast/\!}\big( S^1, S^4\big) $$ and of these I know the rational models, and there is no way for any homotopy fiber product of these to be equivalent to the rational model for the ordered configuration space. So now I am thinking that maybe I should regard the configuration spaces above as smooth manifolds, and as such as smooth stacks, and then think of them as differential refinements of the homotopy types of these mapping spaces (which are Cohomotopy cocycle spaces), to find that the ordered configuration space, as a smooth manifold/stack, is a homotopy fiber product of differentially refined Cohomotopy cocycle spaces. But not sure if that's a fruitful picture...

   Some naive ramblings, just thinking out loud, in case anyone feels inspired to offer a comment:

   I am trying to see how close to an ordered configuration space of points one can get with mapping spaces, which a priori give un-ordered configuration spaces of points.

   The idea I have is – in words – the following:

   An ordered configuration of points in $\mathbb{R}^3$ (say) is, up to homotopy, the same as

   a) An un-ordered configuration of points in $\mathbb{R}^3 \times \mathbb{R}^1$,

   such that this

   b) projects to an $\mathbb{R}^1$-labeled un-ordered configuration in $\mathbb{R}^3$;

   and

   c) projects to an $\mathbb{R}^3$-labeled un-ordered configuration in $\mathbb{R}^1$.

   Meaning that the points in the configuration are distinct not only as points in $\mathbb{R}^3 \times \mathbb{R}^1$, but also after projection as points in $\mathbb{R}^3$ and as points in $\mathbb{R}^1$.

   Here condition c) is what imposes an ordering on the “labels” in $\mathbb{R}^1$, since an arrangement of distict points on the real line puts these points into linear order.

   The formal statement of this idea should be that

   • the ordered un-labeled configuration space $\underset{n \in \mathbb{N}}{\sqcup} \underset{ {}^{ \{1,\cdots, n \} } }{Conf}(\mathbb{R}^3)$

   is a fiber product (in the 1-category of topological spaces) of

   • the unordered $X$-labeled configuration spaces $Conf( \mathbb{R}^k, X )$ with points disappearing when labeled by the base-point of $X$

   as follows:

   $$\underset{n \in \mathbb{N}}{\sqcup} \underset{ {}^{ \{1,\cdots, n \} } }{Conf}(\mathbb{R}^3) \;\simeq_{homeo}\; Conf\big( \mathbb{R}^3, S^1 \big) \underset{ Conf\big( \mathbb{D}^{3+1} \big)_{rel} }{\times} Conf\big( \mathbb{R}^1, S^3 \big)$$

   That this is the case should essentially come down to observing that this fiber product encodes the above “in words” description.

   First I thought that this 1-categorical fiber product is a homotopy-retraction of the corresponding homotopy fiber product, but now I think this can’t be.

   This is because all items in the above are homotopy equivalent to based mapping spaces as

   $$Maps^{\!\ast/\!}\big( S^3, S^4\big) \underset{ Maps^{\!\ast/\!}\big( S^0, S^4\big) }{\times^h} Maps^{\!\ast/\!}\big( S^1, S^4\big)$$

   and of these I know the rational models, and there is no way for any homotopy fiber product of these to be equivalent to the rational model for the ordered configuration space.

   So now I am thinking that maybe I should regard the configuration spaces above as smooth manifolds, and as such as smooth stacks, and then think of them as differential refinements of the homotopy types of these mapping spaces (which are Cohomotopy cocycle spaces), to find that the ordered configuration space, as a smooth manifold/stack, is a homotopy fiber product of differentially refined Cohomotopy cocycle spaces.

   But not sure if that’s a fruitful picture…

2. • CommentRowNumber2.
   • CommentAuthorUrs
   • CommentTimeOct 30th 2019
   • PermaLink
   Author: Urs
   Format: MarkdownItexFor what it's worth, I have uploaded a graphics that illustrates this construction of the ordered unlabeled configuration space as a fiber product of unordered labeled configuration spaces: [here](https://ncatlab.org/nlab/show/configuration+space+of+points#OrderedUnlabeledConfigurationsFromUnorderedLabeledConfigurations)

   For what it’s worth, I have uploaded a graphics that illustrates this construction of the ordered unlabeled configuration space as a fiber product of unordered labeled configuration spaces: here