6
$\begingroup$

$\DeclareMathOperator\Sym{Sym}$Recently, I have been thinking about the space $\Sym^2(S^n)$ of pairs of points in the sphere (including the repeated pairs $(p,p)$ for each $p\in S^n$) and possible ways to describe it. This is topologized as the quotient of $S^n\times S^n$ under the involution that switches components. I am looking (somewhat broadly and ill-definedly) for descriptions of this space and related results. I will describe below some of the results that I have encountered and I would really appreciate any pointers towards descriptions that I have missed in my literature search.

(Side note: If the case of $n=1$ is something you have thought about / want to think about / enjoy, also consider taking a look at this related question that I asked on StackExchange about different ways to prove that $\Sym^2(S^1)$ is a Möbius strip.)

Primarily, I am interested in the homeomorphism type of $\Sym^2(S^n)$, of which I have found two descriptions:

  • On the topology of cyclic products of spheres by Liao cites a description due to Steenrod, where $\Sym^2(S^n)$ is formed by attaching a $2n$-cell to the (unreduced) suspension of $\Sym^{2}(S^{n-1})$. This seems quite useful, since it gives an inductive view of $\Sym^2(S^n)$ as a CW-complex.
  • On the symmetric square of a sphere by James, Thomas, Toda and Whitehead gives a second description, as the mapping cone of a certain map $\Sigma^n(\mathbb R\mathbb P^{n-1})\rightarrow S^n$ (here $\Sigma$ denotes the unreduced suspension). This seems more specialized than the above (their calculations actually use both descriptions), but it also seems to be the more-cited description (perhaps this is because it appears in Hatcher's Algebraic Topology, on page 482).

These are the only two ways I know of to describe the homeomorphism type and I would be very interested if there are others that I have missed.

In terms of computations, it seems that a lot is known about the homology, homotopy and $K$-theory of the space $X_n=\Sym^2(S^n)$. However, these results are widely scattered throughout the literature, so there's a very reasonable chance that I have missed things. Here is what I have been able to find:

  • It seems that the homology groups $H_i(X_n;\mathbb Z)$ are well-understood and should follow from Dold's purely algebraic recipe in Homology of symmetric products and other functors of complexes (though it was known before this too). I have no idea what is known about the cup product for this space.
  • In Homotopy of two-fold symmetric products of spheres, Nakaoka describes the stable homotopy groups $\pi_{n+i}(X_n)$ for $0<i\leq \min(2n-2,9)$. For $n>1$, the space $X_n$ is simply-connected, so we also get $\pi_n(X_n)=\mathbb Z$ and $\pi_i(X_n)=0$ for $0<i<n$, by the Hurewicz theorem. I don't know if any other homotopy groups of these spaces are known.
  • For the real $K$-theory of $X_n$, we have $K^{n+1}(X_n)=0$ and the inclusion $S^n\rightarrow X_n$ induces an injection $K^n(X_n)\rightarrow K^n(S^n)=\mathbb Z$ of index $2^{k}$, where $k$ is the number of integers $0<s<n$ such that $s\cong 1,2\text{ or }4\bmod8$. (See the James–Thomas–Toda–Whitehead paper mentioned above.)

And that's all I know about these spaces! I would be very happy to hear any details that I have missed here.

$\endgroup$
5
  • $\begingroup$ It would seem that this space ${\rm Sym}^2(S^n)$ is a bundle over ${\bf P}^n$ whose fibers are closed $n$-balls minus an antipodal pair of points $-$ and thus in particular that ${\rm Sym}^2(S^n)$ retracts to ${\bf P}^n$, which should give the cohomology ring, and as many of the homotopy groups as are known for ${\bf P}^n$. To get the map to ${\bf P}^n$, identify a pair ${p,q}$ of points in $S^n$ with the line $\ell \subset {\bf R}^{n+1}$ that connects $p$ with $q$, and then map $\ell$ to the diameter of $S^n$ parallel to $\ell$. $\endgroup$ Feb 24, 2021 at 23:59
  • $\begingroup$ Sorry, I should have been clearer in my question. What you are describing is the configuration space of pairs of points in $S^n$, which excludes the diagonal pairs $(p,p)$. But the symmetric square includes these pairs and hence we can't pick a unique line through $p$ and $p$. So the homotopy type is different from $\mathbb R\mathbb P^n$ when $n>1$, which we can also see because $\pi_1(\mathbb R\mathbb P^n)=\mathbb Z/2$ but $\pi_1\big(Sym^2(S^n)\big)=0$. $\endgroup$ Feb 25, 2021 at 0:07
  • 1
    $\begingroup$ Sorry, you're right. Moreover, for the configuration space I should have excluded the entire boundary of $B^n$, not just one antipodal pair of boundary points. Having done that, for the full symmetric square we must restore these boundaries but then identify all the lines $\ell$ that are tangent to $S^n$ at the same point $p$; I can imagine that this will make the topology much more complicated. $\endgroup$ Feb 25, 2021 at 0:22
  • $\begingroup$ Thanks. Yeah, I was thinking of a description along similar lines, but the topology definitely does get complicated. (Although the mapping cone description already sort of gives away that the space is complicated, and Clifford Wagner's thesis shows that a symmetric square of an $n$-manifold is never a manifold for $n>2$. So we can't really expect something too simple.) $\endgroup$ Feb 25, 2021 at 0:33
  • 2
    $\begingroup$ I think this is completely orthogonal to the geometry you actually want, but there are maps $\Sigma \mathrm{Sym}^k S^n\rightarrow \mathrm{Sym}^k S^{n+1}$ that give a spectrum $\mathrm{Sym}^k S$ which turns out to pop up in various places in stable homotopy theory. A good exposition with references is in people.math.harvard.edu/~lurie/ThursdayFall2017/… . Maybe another comment is that Bott has an interesting paper (On symmetric products and the Steenrod squares) that uses $\mathrm{Sym}^2$ to construct Steenrod squares. $\endgroup$ Feb 25, 2021 at 4:30

1 Answer 1

8
$\begingroup$

That cofiber description in the old short paper of James and other famous folks tells you a lot.

The map you have turns out to be adjoint to the standard map $\mathbb RP^{n-1} \rightarrow \Omega^n S^n$.

Continuing the cofibration sequence one place to the right gives a cofibration sequence $$ S^n \rightarrow SP^2(S^n) \rightarrow \Sigma^{n+1}\mathbb RP^{n-1}$$ that ends up being short exact in (co)homology with $\mathbb Z/2$ coefficients.

Furthermore, all possible Steenrod operations acting on the class in dimension $n$ are nonzero: with $x_k$ denoting the nonzero class in degree $k$, for $k=n$ or $n+2\leq k \leq 2n$, one has $Sq^i(x_n) = x_{n+i}$ for $i=2, \dots, n$. This tells you that there is one interesting cup product: $x_n^2 = x_{2n}$.

The $K$-theory can be worked out with the Atiyah-Hirzebruch spectral sequences.

More fun happens when $n=\infty$, so we are looking at $SP^2(S)$, where $S$ is the sphere spectrum. Now one has a cofibration sequence of spectra $$ \Sigma^{-1} SP^2(S) \rightarrow \Sigma^{\infty} \mathbb R P^{\infty} \rightarrow S$$ that, quite remarkably, induces a short exact sequence in homotopy groups in positive degrees: this is a consequence of the Kahn-Priddy Theorem. This also is short exact in all Morava K-theories. Fun, fun, fun!

$\endgroup$
2
  • 1
    $\begingroup$ Thanks, this is great! Could you clarify for me: what is the standard map $\mathbb RP^{n-1}\rightarrow \Omega^nS^n$? Does that take a line $\ell$ to negation in the $\ell$-direction? Do you have a reference for this adjunction, or do you think it's something I should work out myself? $\endgroup$ Feb 25, 2021 at 5:44
  • 5
    $\begingroup$ The standard map is studied by Kahn and Priddy whom are quoted at the end of Nick’s answer. Given a one dimensional subspace in a finite dimensional Euclidean space, consider the reflection map with respect to the hyper plane perpendicular to your chosen line and then compactify. This would give a self map on a sphere of degree -1. You may take loop sum with a map of degree 1to get a map of degree 0. This leads the o the important Kahn-Priddy theorem, etc. $\endgroup$
    – user51223
    Feb 25, 2021 at 11:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.