For a well-based space $X$ denote by $C(\mathbb{R};X)$ the unordered configuration space of points on the real line with labels in $X$, and a point can vanish if its label reaches the basepoint. (Alternatively, you can think about the free $E_1$-algebra over the based space $X$.) Note that $C(\mathbb{R};X)$ is filtered by subspaces $C_{\le r}(\mathbb{R};X)$ which contain configurations of at most $r$ labelled points.

Now let $X$ and $Y$ be well-based and path-connected spaces. According to Segal, we have homotopy equivalences $$C(\mathbb{R};X\times Y)\to \Omega\Sigma(X\times Y),$$ and we use that $\Sigma(X\times Y)$ splits up to homotopy into $\Sigma X\vee \Sigma Y\vee \Sigma(X\wedge Y)$. We also have a homotopy equivalence $$C(\mathbb{R};X\vee Y\vee (X\wedge Y))\to \Omega \Sigma(X\vee Y\vee (X\wedge Y)).$$ Now my question is: Can we invert the homotopy equivalences in such a way that the resulting equivalence $$C(\mathbb{R};X\vee Y\vee (X\wedge Y)) \to C(\mathbb{R};X\times Y)$$ is filtration-preserving? Or is there even an explicit, geometric description of such a map?


1 Answer 1


The answer to your first question is no. And this can be seen by homology considerations. Note that this equivalence induces an isomorphism of Hopf algebras $$ H_*(C(\mathbb R; X \vee Y \vee (X\wedge Y)); \mathbb F) \simeq H_*(C(\mathbb R; X \times Y); \mathbb F)$$ for any field coefficients $\mathbb F$. Also, one knows that $H_*(C(\mathbb R; Z); \mathbb F)$ is isomorphic to the free tensor algebra generated by $\tilde H_*(Z;\mathbb F)$.

The isomorphism must send primitives to primitives.

Examples show that this will not preserve the filtration on homology induced by the filtration you are asking about. Here is a simple example: let $X=Y=S^2$ and $\mathbb F = \mathbb Q$. Let $x \in H_2(X;\mathbb Q)$ and $y \in H_2(Y;\mathbb Q)$ be generators. Then the filtration 1 primitive one might write as $x \bar{\times} y \in H_4(X \wedge Y;\mathbb Q)$ must map to the primitive $x \times y + x * y \in H_4(X \times Y;\mathbb Q)$. But this element has filtration 2. (Here $*$ is the multiplication induced by the H-space structure.)

(Of course, even more simply, filtration 1 on the one side certainly can't be mapped as an equivalence to filtration 1 on the other side!)

As to your second question, just add the three maps $$X \times Y \rightarrow X \hookrightarrow C(\mathbb R; X \vee Y \vee (X\wedge Y)),$$ $$X \times Y \rightarrow Y \hookrightarrow C(\mathbb R; X \vee Y \vee (X\wedge Y)),$$ $$X \times Y \rightarrow X \wedge Y \hookrightarrow C(\mathbb R; X \vee Y \vee (X\wedge Y)),$$ and then extend to a map of $\mathbb E_1$-algebras.


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