MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This is the mirror of previous post.

For an immersion $f\colon N^n\to M^{2n}$ with fixed lift $\tilde{f}\colon N\to \tilde{M}$ and $N,M,\tilde{M}$ are oriented, first we define a unordered double point set $S_2[f]=\lbrace(x_1,x_2)\in N\times N | x_1\neq x_2, f(x_1)=f(x_2)\rbrace/(x_1,x_2)\sim(x_2,x_1)$.

Second, for each $x=[(x_1,x_2)]\in S_2[f]$, we can think an equivariant intersection number $I[x]=a(x)w(x)\in \lbrace \pm g|g\in\pi_1(M)\rbrace\subset \mathbb{Z}[\pi_1(M)]$ as follows:

$a(x)\in \pi_1(M)$ satisfies $a(x)\circ\tilde{f}(x_1)=\tilde{f}(x_2)$, here we are regarding $a(x)\colon \tilde{M}\to \tilde{M}$.

$w(x)$ is defined to be $1$ or $-1$ if the isomorphism between oriented vector spaces $(d(a(x)\tilde{f})_{x_1},d\tilde{f}_{x_2})\colon T_{x_1}N\oplus T_{x_2}N\to T_{y}\tilde{M}$, where $y=a(x)\circ\tilde{f}(x_1)=\tilde{f}(x_2)$.

Eventually, we define the Wall's self intersection $\mu(f)=\sum_{x\in S_2[g]}I[x]\in \mathbb{Z}[\pi_1(M)]/\lbrace\sum a_g(g-(-1)^ng^{-1})\rbrace$ as in Wall's Surgery on compact manifolds Section 5 or Andrew Ranicki's book page 259.

Here, the the equivalence relation $\sum a_g(g-(-1)^ng^{-1})$ is added to eliminate the ambiguity in the choice of representative element in $S_2[f]$, namely $[(x_1,x_2)]=[(x_2,x_1)]\in S_2[f]$.

Now, For $t\in [0,1]$, let $f_t\colon N^n\to M^{2n}$ be an immersion (or regular homotopy by definition) and the regular homotopy $\tilde{f_t}\colon N\to \tilde{M}$ is also given.

I want to prove that in this situation $\mu(f_0)=\mu(f_1)$. i.e. the self-intersection number is an invariant under the regular homotopy.

Here is the argument given in Ranicki's book page 259.

Let $f\colon N\times[0,1]\to M\times [0,1]$ be the trace of $f_t$. i.e. $f(x,t)=(f_t(x),t)$.

Then, $f\times f^{-1}(\Delta(M\times[0,1])-\Delta(N\times[0,1])$ is 1-dimensional manifolds.

i.e. The unordered double point set of $f_1$ is obtained from the unordered double point set of $f_0$ by a sequence of "birth" and "death" of cancelling pair consisting two points in unordered double point set. Hence, such cancelling pair contributes to the self intersection number by $\underline{a-(-1)^nw(a)a=0}\in Q_{(-1)^n}(\mathbb{Z}[\pi_1(M)])$. Hence, $\mu(f_0)=\mu(f_1)$.

I want to prove rigorously that the "birth" or "death" of each cancelling pair contributes to self-intersection number as much as the underlined part. I think that I should compare the orientation of tangent space of $N$ at some double points and tangent space of $M$ at the image of them. But I can't fill in the details.

Are there anybody who can help me?

share|cite|improve this question
up vote 3 down vote accepted

The double points of a generic immersion $f:P^p \to Q^q$ of closed manifolds without triple points constitute a closed $(2p-q)$-dimensional manifold $S_2[f]$ (defined as in the question). This is also true for a generic immersion $(f,\partial f):(P,\partial P) \to (Q,\partial Q)$ of manifolds with boundary: in this case the double points constitute a $(2p-q)$-dimensional manifold with boundary $(S_2[f],S_2[\partial f])$. In particular, this is the case for the trace immersion of the regular homotopy $$(f,\partial f):(P,\partial P)=N^n \times (I,\partial I) \to (Q,\partial Q)=M^{2n}\times (I,\partial I)$$ for $p=n+1$, $q=2n+1$, with $(S_2[f],S_2[\partial f])$ a 1-dimensional manifold with boundary. Following Wall work in the universal cover $\tilde{M}$ : the count of the double points of $\partial f$ is $$\mu(\partial f)=\delta\mu(f)-(-)^n\overline{\delta\mu(f)}\in Z[\pi_1(M)]$$ with $\delta\mu(f)\in Z[\pi_1(M)]$ the count of the double point arcs of $f$. The double point circles of $f$ can be ignored here.

share|cite|improve this answer

Even more is true in the context of surgery theory: let $q \ge 3$ with $q$ odd, assume $M$ is a $1$-connected closed smooth manifold of dimension $2q$. Let $$I^{\text{fr}}_q(M)$$ denote the space of immersions $\phi: S^q \to M$ such that the normal bundle of $\phi$ is trivializable.

Assertion: the mod-2 double point invariant $$ \mu: I^{\text{fr}}_q(M) \to {\mathbb Z_2} $$ is actually an invariant of the underlying homotopy class of $\phi$, i.e., if two framed immersions $S^q \to M$ are merely homotopic then their double point invariants coincide.

share|cite|improve this answer

I don't know about a rigorous proof, but here is a heuristic explanation you may find helpful.

You have a compact $1$-dimensional manifold given by the self-intersections of the trace $f$ of your regular homotopy from $f_0$ to $f_1$. This should be thought of as a cobordism between the compact $0$-dimensional manifolds $S_2[f_0]$ and $S_2[f_1]$. That is to say, it is a $1$-manifold embedded in $N\times N\times[0,1]$ which intersects the boundary in $S_2[f_0]$ at one end of the cylinder and $S_2[f_1]$ at the other end. So it is a disjoint union of embedded arcs and circles (if I could find a good picture I'd link it here; the best I could find is on page 44 of Milnor's book "Topology from the differentiable viewpoint"). This helps to visualise the "birth" and "death" of pairs of double points, as you scan the cobordism along the cylinder. Unfortunately you also have to think about self-intersection numbers and homotopy classes.

In fact, the correct way to think about Wall's invariant (morally, at least) is as a $0$-dimensional $\mathbb{Z}_2$-equivariant normal bordism class in the based loop space $\Omega M$ with inversion of loops. For this approach, see the papers of Hatcher and Quinn or Klein and Williams (Appendix A).

Update: I've just finished writing up a detailed proof of the fact (which seems to be well-known to the experts) that the invariants of Wall and Hatcher-Quinn coincide, see here.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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