Resently I'm reading Bott-Tu's differential forms in algebraic topology, and comparing it with some differential topology textbook.

While proving the Poincare-Hopf theorem, it defines the intersection number $I(M,N)$, where M, N are submanifold of another smooth manifold K with condition $\dim M+\dim N=\dim K$, to be $I(M,N)=\int_K \eta_M \wedge\eta_N.$ Here $\eta_M$ is the Poincare dual of M, so $\eta_M \wedge\eta_N$ is a top form in K, assuming the dimension condition.

My question is why this definition is coincide with the definition appeared in differential topology, which using the orientable intersection number that "sum up" $\pm 1$ locally according to the orientation.

If it's possible, I perfer an answer without using much algebraic geometry. Any links to webpage or other reference are welcomed as well. Great Thanks!

  • $\begingroup$ Possibly reading this will answer your question, though this is written in terms of cup products rather than wedge products: math.berkeley.edu/~hutching/teach/215b-2011/cup.pdf $\endgroup$ – Lorenzo May 16 '18 at 5:01
  • $\begingroup$ @AreaMan Yeah I know the theorem in this artical, it's a more topological viewpoint of my question, and reading it is really helpful. Yet I still perfer a more "differential geometry" way I guass? I've been hinted to consider a special metric on K to have M locally totally geodesic, if I figure it out later, I would love to post an answer. Anyway thanks. $\endgroup$ – Ricanry May 16 '18 at 9:21
  • $\begingroup$ I'd love to see a differential geometric answer also, if you figure it out. $\endgroup$ – Lorenzo May 16 '18 at 17:39
  • $\begingroup$ I guess the idea is to assume $M$ and $N$ intersect transversely, and to pick the forms $\eta_M$ and $\eta_N$ be supported in a small tubular neighborhood of $M$ and $N$, respectively; then $\eta_M \wedge \eta_N$ will be supported in the intersection of these two neigbhborhoods; this can be chosen to be some local chart in which $M$ and $N$ are transversely intersecting $\Bbb R^k$'s. Why the mass near these points should be $\pm 1$ is not so clear: maybe an explicit and careful choice of the $\eta$ will allow you to force the integrals in these regions to be close to $\pm 1$... $\endgroup$ – user98602 May 16 '18 at 20:19
  • $\begingroup$ My gut says you might get something useful out of the Mathai-Quillen formalism for forms representing Thom classes, as something like Gaussians on the normal bundle, but I've never really thought about this stuff so this might be sending you on a wild goose chase. See here. $\endgroup$ – user98602 May 16 '18 at 20:23

Imagine nonnegative function $f:\mathbb{R}\rightarrow \mathbb{R}$ that is supported in the interval $[-\epsilon,\epsilon]$ with $$\int_{\mathbb{R}} f dx=1.$$ If $\gamma:[0,1]\rightarrow \mathbb{R}$ is any path with $\gamma(0)$ and $\gamma(1)$ outside of $[-\epsilon,\epsilon]$ then $\int_{[0,1]}\gamma^*(fdx)$ is equal to the algebraic intersection number relative to its boundary, of $\gamma$ with $0$.

We can play the same game in $\mathbb{R}^n$, defining an $n$-form $\omega$ supported in an $\epsilon$-ball of $\vec{0}$, so that $$\int_{\mathbb{R}^n}\omega=1.$$ Once again if $\gamma:B\rightarrow \mathbb{R}^n$ is a smooth map of the ball with $\gamma(\partial B)$ outside the $\epsilon$-ball where $\omega$ is supported, then $$ \int_{B}\gamma^*(\omega) $$ will be the intersection number of $B$ relative to its boundary, or alternately the winding number of $\partial B$ about $\vec{0}$.

The reason for this is the full change of variables formula for integrating $n$-forms of compact support on an oriented $n$-manifold. If $f:X\rightarrow Y$ is a smooth, proper map of oriented $n$-manifolds, with $Y$ connected, define the degree of $f$ as follows. Let $y\in Y$ be a regular value of $f$. Since $f$ is proper $f^{-1}(y)$ is compact. Since it is a zero manifold, it is a finite set of points. At each $x\in f^{-1}(y)$ define the sign of $x$ to be $\pm 1$ depending on whether $df_x:T_xX\rightarrow T_yY$ is orientation preserving or orientation reversing. The degree of $f$, denoted $deg(f)$ is $$\sum_{x\in f^{-1}(y)}sgn(x).$$ Standard elementary arguments show that $deg(f)$ is independent of the point $y$, and unchanged by homotopies of $f$ through smooth proper maps.

Change of variables says that if $\omega$ is an $n$-form with compact support on $Y$, then $f^*\omega$ will have compact support. Both integrals are well defined and $$ \int_Xf^*\omega= deg(f)\int_y\omega.$$

Here is a slightly flawed but intuitive proof. The set of singular points of $f$ has measure zero by Sard's theorem. Remove the singular points of $f$, $S(f)\subset Y$ from $Y$ and $f^{-1}(S(f))$ from $X$. You now have a map $$f:X-f^{-1}(S(f))\rightarrow Y-S(f)$$ that is a local diffeomorphism. Since you removed a set of measure zero, $$ \int_{Y-S(f)} \omega =\int_Y\omega.$$ Using the stack of records theorem, $$\int_{X-f^{-1}(S(f))}\omega=deg(f)\int_{Y-S(f)}\omega.$$ Since the pull back of an $n$-form at points where $df$ is singular is zero, we have $$\int_{X}f^*\omega=\int_{X-f^{-1}(S(f))}f^*\omega.$$ Completing the proof.

If $M\subset N$ is a compact regular submanifold, you can cover $M$ with coordinate patches $(U_{\alpha},\phi_{\alpha})$ of $N$ where $M\cap U_{\alpha}$ is a slice for all $\alpha$. That is in local coordinates $\phi_{\alpha}$, $M$ is the set of points with $x_{m+1}=x_{m+2}\ldots=x_n=0$. We call $(x_{m+1},\ldots,x_n)$ the complementary coordinates to $M$.

Complete this to a cover of $N$ by adding $N-M$. Choose a partition of unity subordinate to the open cover of $N$ so that $\rho_{\alpha}:M\rightarrow \mathbb{R}_{\geq 0}$ has support inside $U_{\alpha}$. Choose $\epsilon>0$ so that inside the support of each $\rho_{\alpha}$ if the complementary coordinates $M$ have norm less than $\epsilon$ are contained in $U_{\alpha}$. Let $\omega_{\alpha}$ be an $n-m$ form with support in the points in $U_{\alpha}$ whose complementary coordinates have norm less than $\epsilon$ and $\int_{\mathbb{R}^{n-m}}\omega_{\alpha}=1$. Finally let $$\omega=\sum_{\alpha}\rho_{\alpha}\omega_{\alpha}.$$ The form is smooth. With a little more care you can make sure its closed.

The point is if $K$ is a regular manifold that is transverse to $M$ of dimension $n-m$, then the pull back under inclusion $i:K\rightarrow N$, $$i^*(\omega)$$ has its support near the points where $K$ intersects $M$, and if $\epsilon$ is small enough, the integral near each point is $\pm 1$ depending on the sign of the intersection of $K$ and $M$ at that point.

That is it.

  • 1
    $\begingroup$ That's really a fundamental proof! I got a neater way to do the same thing, using Thom class which coincides with Poincare duality. $\endgroup$ – Ricanry May 18 '18 at 14:32
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    $\begingroup$ I really like this. $\endgroup$ – user98602 May 18 '18 at 16:13

Assuming readers are familiar with Bott-Tu's book, so I'll use the language in their book. Set $\dim M=m,\ \dim N=n$.

It is well known that Thom isomophism gives $\mathcal{T}:H^*(M)\cong H^{*+k}_{cv}(E)$, where E is a vector bundle over M with rank k, given by intergral along the fiber.

Now $M$ is a submanifold of $K$, so there's a tubular neighborhood $T$, which is diffeomorphism to the normal bundle $S$ in $K$. Let$\Phi =\mathcal{T}(1) \in H^{m}(T)$ be the Thom class,$\ j:T \to M$ be the inclusion, then it's proved in Bott-Tu that the Poincare dual of M $\eta_M=j^*\Phi$.

When computing $\int_K \eta_M \wedge\eta_N$, using Poincare duality and our discussion above, we got $$\int_K \eta_M \wedge\eta_n=\int_N\eta_M|_N= \sum_{p\in M \cap N} \int_{B_{\epsilon}\ (p)\cap N}\eta_M .$$ By choosing a Riemannian metric on K, we might assume that, at every point $p\in M \cap N$, $T_p M \bot T_pN$. (The metric is clearly exist, since such $p$ is finite, so we could take some alart, choosing metric on it, then using partition of unity.) So at point $p$, pull back of Thom class $j^*\Phi_p$ lies in $T_pN$. Besides, Thom isomorphism gives by intergral along the fiber, so intergral of Thom class along fiber $S_p = T_pN$ is $\pm 1$, determinded by orientation.

Thus we only have to chech the signal of each intergral. Let $\omega_M$ be the volumn form on N, then $\omega_M \wedge \eta_M$ is a positive orientation on $T_p K =T_pS \oplus T_pM$. Therefore, if $\omega_M \wedge \omega_N$ is a positive orientation, then the orientation on $T_pS$ and $T_pN$ should be same, leading to the integral of$j^*\Phi_p \in T_pN$ shouldn't change its signal, and the negative part is exactly the same.


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