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(Problem 13 of Milnor's Topology from the differentiable viewpoint).

Let $M^n,N^m\subset \mathbb{R}^{m+n+1}$ be compact, oriented differential manifolds without boundary of dimension $n$ and $m$.

Consider

$\phi:M^n \times N^m\rightarrow S^{n+m}:(p,q)\mapsto \frac{q-p}{||q-p||}$

and define $l(M,N)=$deg$(\phi)$. I would like to prove $l(M,N)=(-1)^{(m+1)(n+1)}l(N,M)$.

So far I did the following. Define

$\phi_1:M^n \times N^m\rightarrow S^{n+m}:(p,q)\mapsto \frac{q-p}{||q-p||}$

and

$\phi_2:N^m \times M^n\rightarrow S^{n+m}:(q,p)\mapsto -\frac{q-p}{||q-p||}$.

We want deg$(\phi_1)=$deg$(\phi_2)$. Let $z$ be a regular value of $\phi_1$. If I am correct this means that $-z$ is a regular value of $\phi_2$. Moreover $\phi_1^{-1}(z)$ and $\phi_2^{-1}(-z)$ are in one to one correspondence via $(p,q)\mapsto (q,p)$. Therefore we are left to show for each such $(p,q)$:

sgn$(\phi_1,(p,q)) = (-1)^{(m+1)(n+1)}$sgn$(\phi_2,(q,p))$

where sgn$(\phi_1,(p,q))$ is +1 if the derivative map $(d{\phi_1})_{(p,q)}$ preserves orientation and -1 otherwise.

I am stuck proving this last equality. It looks like the antipodal map but theres is some change of coordinates going as well. I think I am messing up working with different orientations. Does anyone know how to proceed? Thank you for your help!

PP123
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  • So $M$ and $N$ are submanifolds of $\Bbb R^{m+n+1}$? – Angina Seng May 30 '17 at 17:44
  • Yes indeed! Sorry for the confusion, I'll edit my question. – PP123 May 30 '17 at 17:46
  • $\phi_2 = \phi_1 \circ c$ where $c : N \times M \to M \times N$ is the map which switches the factors. What is the degree of $c$? – Balarka Sen May 30 '17 at 18:03
  • The degree of c should be just sgn$(c,(q,p))$ for some $(q,p) \in N \times M$. Then for $(v_n,v_m)\in T_{(q,p)}(N \times M)= T_q M \times T_p N$ we get $dc_{(q,p)}(v_n,v_m)=(v_m,v_n)$. But then I get for (n_1,...,n_m) base of $T_qN$ and $(m_1,...,m_n)$ base of $T_pM$ and thus ${(n_1,0),...,(0,m_n)}$ as base for $T_{q,p}(N\times M): \frac{s(dc(n_1,0)),...,dc(0,m_n))}{s((n_1,0),...,(0,m_n))}=\frac{s((0,n_1),...,(m_n,0))}{(n_1,0),...,(0,m_n)}$. But how do I work this further out? – PP123 May 30 '17 at 18:22

1 Answers1

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As Milnor suggests, let the linking map be $$\lambda:M^m\times N^n\to S^k,$$ $$\lambda(p,q)=\dfrac{p-q}{||p-q||},$$ with $m+n=k$, where $M,N\subset\mathbb{R}^{k+1}$ are disjoints, oriented, boundaryless manifolds. Define $$\lambda':N^n\times M^m\to S^k,$$ $$\lambda'(q,p)=\dfrac{q-p}{||q-p||}=-\dfrac{p-q}{||p-q||}$$ the linking map of interest. Let $\phi:S^k\to S^k$ be the antipodal map. We see that we have $\lambda(p,q)=(\phi\circ\lambda')(q,p)$ for all $p\in M$ and $q\in N$. Taking into account the reversal map given by $$ \psi:M^m\times N^n\to N^n\times M^m$$ which maps $(p,q)$ to $(q,p)$ and has degree $(-1)^{mn}$, we can proceed. We have $\lambda=\phi\circ\lambda'\circ\psi$.

Thus, as problem 1 of Milnor suggests, $$\mathrm{deg}(\lambda)=\mathrm{deg}(\phi)\mathrm{deg}(\lambda')\mathrm{deg}(\psi).$$

Hopf theorem of section 7 and the end of section 5 of Milnor can help you now. We know that for $k$ odd, the antipodal map is homotopic to the identity but for $k$ even, it isn't. So, it is easily shown that $$\mathrm{deg}(\phi)=(-1)^{k+1}.$$ It then follows that $$\mathrm{deg}(\lambda)=(-1)^{n+m+1+mn}\mathrm{deg}(\lambda')=(-1)^{(m+1)(n+1)}\mathrm{deg}(\lambda')$$ which is what we wanted to show.

EDIT: I think I need to further explain what is up with this reversal map.

Problem 8 in milnor shows that for two smooth manifolds, we have $$T_{(x,y)}(M\times N)=T_x M\times T_y N.$$ With this in mind, the differential of the reversal map can be formulated as $$(v_1,\dots, v_m,w_1,\dots, w_n)\mapsto (w_1,\dots w_n,v_1,\dots, v_m).$$ You can show that the determinant of this linear application is $(-1)^{mn}$ by counting permutations. I hope this helps.

alerouxlapierre
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  • I don't see why $\lambda = \phi \circ \lambda$ '. $\lambda$ works on $M \times N$ and$ \phi \circ \lambda$ ' on $N \times M$... – PP123 May 30 '17 at 18:28
  • I see exactly you concern. I will edit my post. – alerouxlapierre May 30 '17 at 18:32
  • You also need to take into account the degree of the reversal map from $M\times N$ to $N\times M$ which is $(-1)^{mn}$. – Angina Seng May 30 '17 at 18:33
  • Yes, I am seeing this right now. – alerouxlapierre May 30 '17 at 18:33
  • I corrected it, thank you @LordSharktheUnknown for the comment. – alerouxlapierre May 30 '17 at 18:41
  • Thank you both very much! One final question. How do you work out the degree of reversal map? I got this far: Take $(q,p)\in N×M,(vn,vm)\in T_{(q,p)}(N×M)=T_qM×T_pN$. Remark $(d\phi){(q,p)}(vn,vm)=(vm,vn)$. Thus for $(n_1,...,n_m)$ base of $T_qN$ and $(m_1,...,m_n)$ base of $T_pM$ and thus $((n_1,0),...,(0,m_n))$ base for $T{(q,p)}(N×M)$ I get $\frac{s_{MxN}(d\phi(n_1,0)),...,d\phi(0,m_n))}{s_{NxM}((n1,0),...,(0,mn))}=\frac{s_{MxN}((0,n_1),...,(m_n,0))}{s_{NxM}(n1,0),...,(0,m_n)}$. How do I work this out? Thank you! – PP123 May 30 '17 at 18:55
  • I also edited to add this :) – alerouxlapierre May 30 '17 at 18:56
  • Sorry for reawakening a dead thread, I wanted to ask briefly, why do we need the reversal map? It seems as if left composing $\lambda$ with $\phi$ is enough to transition between the two maps. – rubikscube09 Jan 29 '20 at 17:26