I am trying to convince myself that a naïve definition of the Wall self intersection number should not work for odd-dimensional manifolds. Namely, let $X^{2n-1}$ be a smooth oriented closed manifold and let $f : S^n \to X$ be a map from a sphere. Also, let's assume that $X$ and $S^n$ are based and $f$ preserves base-points and $n \geq 2$.

For now, suppose that $f$ is a smooth immersion with double point curves of intersection, and then by orienting the double point curves and attaching them to the basepoint along arcs contained in the image of $f$ to obtain $\mu(f) \in \mathbb{Z} \pi_1(X)$. To remove the dependence on the orientations of the curves, I better mod out by $\{ g = g^{-1} : g \in \pi_1(X)\}$. I would like for $\mu(f)$ to be invariant under changing $f$ by a homotopy, so I better also mod out by setting $1=0$.

Is $\mu(f) \in \mathbb{Z}\pi_1(X) /\langle\{ g = g^{-1} \}, 1=0\rangle$ an invariant of the based homotopy class of $f$?

I am guessing that the answer is no so ignoring everything above perhaps I should just ask: Is there a map $f : S^3 \to T^5$ that has exactly one double point curve in the image that is nontrivial in $\mathbb{Z}^5 = \pi_1(T^5)$?

  • $\begingroup$ Last paragraph: no, there is a smooth lift $\tilde f: S^3 \to \Bbb R^5$, and scaling gives an isotopy through immersions with generic double points; after this isotopy of immersions one has that the corresponding map $f_1: S^3 \to T^5$ factors through a chart, and hence any loop in its image is null-homotopic. $\endgroup$
    – mme
    Jun 18, 2019 at 0:18
  • 2
    $\begingroup$ In some sense the Wall self-intersection form in this case should involve both $\pi_1(X)$ and $\pi_2(X)$. I had a go at defining it (as a complete obstruction to removing double circles) in the final section of my article arxiv.org/abs/1111.3248. I think some of Tobias Eckholm's earlier papers might also be relevant. $\endgroup$
    – Mark Grant
    Jun 18, 2019 at 6:34
  • $\begingroup$ But for your question: Let $g=(f,0):S^n\to X\times\mathbb{R}$. It could be that what you are defining agrees with the Wall invariant of $g$. $\endgroup$
    – Mark Grant
    Jun 18, 2019 at 6:35
  • $\begingroup$ @MarkGrant Thanks for the link - your Prop 6.2 is exactly the sort of thing I had in mind. $\endgroup$
    – user101010
    Jun 18, 2019 at 18:55
  • $\begingroup$ @LSpice Whoops typo thanks $\endgroup$
    – user101010
    Jun 19, 2019 at 19:56

1 Answer 1


I doubt that what you are proposing as the receptacle for the obstruction is the correct abelian group. For one thing, you are not taking into account the involution on the canonical double cover of the double point manifold, i.e., $\{(x,y)\in S^n\times S^n\setminus \Delta_{S^n} |f(x) = f(y)\}$ (where we have assumed $f$ has been made self-transverse). This involution is a crucial aspect of the answer.

I think the answer to the question is going to be complicated.

Let me write a few words as to what might constitute a "non-naive" approach.

1) If $X$ is $2$-connected and $n > 2$, then there is no obstruction to deforming a map $f: S^n \to X$ to an embedding. This follows from the work of Haefliger, Hirsch, et. al.

So assume for the rest of this note that $X$ is not $2$-connected.

2) Consider the problem deforming, through regular homotopy, an immersion $f:S^n \to X^{2n-1}$ to an embedding, where $X$ is not assumed to be $2$-connected.

According to the work of Hatcher and Quinn (see also Corollary G my paper with Bruce Williams, "Homotopical Intersection Theory, II: Equivariance"), the obstruction, given by the double point manifold, lies in a certain twisted Bordism group $$ \Omega_0(E'(f,f)_{h\Bbb Z_2};\xi_{\Bbb Z_2}) \, , $$ where $E'(f,f)$ is the homotopy pullback of the diagram $S^n \to X \times X \overset{\Delta}\leftarrow X$ (the map $S^n \to X \times X$ is given by $x\mapsto (f(x),f(-x))$, and $\Delta$ is the diagonal). The space $E'(f,f)$ has a $\Bbb Z_2$-action and $E'(f,f)_{\Bbb Z_2} = E\Bbb Z_2 \times_{\Bbb Z_2} E'(f,f)$ is the Borel construction. Here $\xi$, defined over $E'(f,f)$, is a certain virtual vector bundle of rank $-1$, which is too technical to describe here.

Hence, we wish to identify the above bordism group as an abelian group.

Let us make a further simplifying assumption that the manifold $X$ is stably parallelized. In this case the virtual bundle is trivial. So the bordism group in this case is then $\pi_0$ of the homotopy orbits of a certain action of $\Bbb Z_2$ on the suspension spectrum $\Sigma^{-1} (E'(f,f)_+)$. After choosing a basepoint in $X$, there is an evident equivariant map $\Omega X \to E(f,f)$ which is $(n-1)$-connected (where $\Bbb Z_2$ acts by reversing loops). So the map $\Sigma^{-1}(\Omega X_+)\to \Sigma^{-1} (E'(f,f))_+$ is $(n-2)$-connected.

Assuming that $n \ge 3$, the latter map will then be at least $1$-connected, and the obstruction be regarded as residing in $\pi_0$ of the homotopy orbits of a certain action of $\Bbb Z_2$ on the suspension spectrum $\Sigma^{-1} (\Omega X_+)$. Here $\Bbb Z_2$ is also acting on the suspension coordinate.

I don't know how to make this computation.

However, if we apply the "transfer" $\Omega_0(E'(f,f)_{h\Bbb Z_2};\xi_{\Bbb Z_2}) \to \Omega_0(E'(f,f);\xi)$ to this problem (see my papers with Williams; this philosophically has the effect of thinking of the self-intersection question as an ordinary intersection question between two distinct manifolds and so we can then ignore the involution), then we are looking at the abelian group $$ \pi_0(\Sigma^{-1} (\Omega X_+)) = \pi_{1}^{\text{st}}((\Omega X)_+) \cong \Bbb Z_2 \oplus \pi_1^{\text{st}}(\Omega X)\, . $$ If we make yet another simplifying assumption that $X$ is $1$-connected, then the displayed group is isomorphic to $\Bbb Z_2 \oplus \pi_2(X)$. (But as noted above this group is solving a different question.)

3) There is the related question of deforming a map $f: S^n \to X$ to an embedding. The first step in solving the question is to find an immersion in that homotopy class. The obstructions can be studied using Smale-Hirsch theory; this essentially a question about find a section of a certain fiber bundle over $S^n$ whose fibers are Stiefel manifolds.

Assuming the immersion exists we can look at all immersions in that homotopy class. Then we can employ the approach of (2) to each such immersion to create a "torsor" of indeterminacies. If we divide out by this indeterminacy, we get the answer. I will not detail this approach here, but a related problem of this kind was solved in a paper of mine, "Poincare complex diagonals."

Addendum: I just looked at the last section of Mark Grant's paper mentioned in the comments to the question. He is doing something similar to what I did at the end of (2) above, in that he is trying to avoid the problem of computing the involution.

  • 1
    $\begingroup$ Nice answer! I certainly had fun thinking about this stuff back then. Regarding your first display, should the Hatcher Quinn invariant live in an $\Omega_1$ (since it is the bordism class of the double circles)? My approach was to try to deal with this obstruction in two steps, using the Gysin sequence of the double cover of the involution. The first obstruction is that $g=(f,0):S^n\to X\times \mathbb{R}$ is regularly homotopic to an embedding. But the secondary obstruction has some indeterminacy (presumably coming from choosing the homotopy). It would be good to nail this down geometrically. $\endgroup$
    – Mark Grant
    Jun 19, 2019 at 15:18
  • $\begingroup$ @MarkGrant Regarding "should the Hatcher Quinn invariant live in an $\Omega_1$ ?" The answer has to do with the indexing convention for the bordism group. The virtual bundle that we use has rank $-1$. If our convention instead were to always shift virtual bundles so that they have virtual rank $0$, then it would be $\Omega_1$. But that is not our convention. In other words $\Omega_j(X;\xi) = \Omega_{j+1}(X;\xi\oplus \Bbb R)$, $\endgroup$
    – John Klein
    Jun 19, 2019 at 17:06
  • $\begingroup$ "But the secondary obstruction has some indeterminacy (presumably coming from choosing the homotopy). It would be good to nail this down geometrically." I did something like this in my paper "Poincare complex diagonals," where I managed to nail down the indeterminacies in trying to find Poincare embeddings of the diagonal of a Poincare complex. This paper is based on another work of mine called "Embedding, compression and fiberwise homotopy theory," where I studied the problem of compressing a codimension zero embedding $P\times [0,1] \to N\times [0,1]$ down to an embedding $P \to N$. $\endgroup$
    – John Klein
    Jun 19, 2019 at 19:31

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