Prove that if $X$ and $Z$ are cobordant in $Y$, then for every closed manifold $C$ in $Y$ with dimension complementary to $X$ and $Z$, $I_2(X,C)=I_2(Z,C)$. [HINT: Let $f$ be the restriction to $W$ of the projection map $Y\times I\to Y$, and use the Boundary Theorem.]

Boundary Theorem. Suppose that $X$ is the boundary of some compact manifold $W$ and $g:X\to Y$ is a smooth map. If $g$ may be extended to all of $W$, then $I_2(g,Z)=0$ for any closed submanifold $Z$ in $Y$ of complementary dimension.

My progress:

Since $X$ and $Z$ are cobordant in $Y$, there is a compact manifold with boundary $W\subset Y\times I$ such that $\partial W=X\times \{0\}\cup Z\times \{1\}$. To apply the boundary theorem, I need to involve some boundary. Let $f$ be the restriction of the projection $Y\times I\to Y$ to $W$ and consider $\partial f=f\restriction_{\partial W}: \partial W\to Y$. Now $\partial f$ satisfies the hypotheses of the boundary theorem, so $I_2(\partial f,C)=0$ for any closed submanifold $C$ in $Y$ of complementary dimension.

How do I conclude that $I_2(X,C)=I_2(Z,C)$? I guess I need to use that $\partial W=X\times \{0\}\cup Z\times \{1\}$ but I don't know how.

  • $\begingroup$ What is the definition of $I_2(X,C)$ for a submanifold $X$? Isn't it simply the intersection number of the inclusion $\iota:X\to Y$? $\endgroup$ – cjackal Mar 17 '18 at 2:32
  • $\begingroup$ @cjackal, Yes it is. $\endgroup$ – user500094 Mar 17 '18 at 15:36
  • $\begingroup$ Well, then what is $\partial f$? Isn't it just a disjoint union of the inclusions $X\to Y$ and $Z\to Y$, except that the orientation of one of the inclusions is reversed? Use the additivity of the intersection number to conclude what you desire. $\endgroup$ – cjackal Mar 17 '18 at 18:15
  • $\begingroup$ @cjackal I'm not sure what you mean by union (and orientation) of maps (in this case inclusions), but I don't even see why $\partial f$ is an inclusion (strictly speaking, $X\times \{0\}$ isn't a subset of $Y$). Also, Guillemin and Pollack don't discuss the additivity of $I_2$ -- is this additivity supposed to be obvious? $\endgroup$ – user500094 Mar 17 '18 at 19:00
  • $\begingroup$ Well, it is obvious that the identification of $X\times \left\{0\right\}$ with $X$ gives the viewpoint of $\partial f|_{X\times 0 }$ as an inclusion. Anyway, the intersection number "counts the (signed) intersection points", so the additivity should be obvious. And come to think of it, calling it a disjoint union is a bit nonsense; just try to exploit the additivity and my claim that one of the orientation in $\partial f$ is reversed. $\endgroup$ – cjackal Mar 17 '18 at 19:47

As you already know from the boundary theorem that $I_2(\partial f,C)=0$, it suffices to show that $I_2(X,C)+I_2(Z,C)=I_2(\partial f,C)$.

Now recall that the intersection number mod 2 is counting the number of intersection points of $\iota_X:X\to Y$ and $C$, after a proper homotopy that makes the intersection transverse. (Thus, $I_2(X,C)=I_2(\iota_X,C)$ by definition, and I will identify throughout this post.)

So let me assume that $H:X\times I\to Y$ is one such homotopy, so that $H|_{X\times 1}:X\to Y$ (up to the canonical identification $X\times \left\{1\right\}\cong X$ of course) is transverse to $C$. Similarly, one can find a homotopy $H':Z\times I \to Y$ from $\iota_Z$ to $H'|_{Z\times 1}:Z\to Y$ where the latter is transverse to $C$. Now observe that $\partial f:X\times \left\{0\right\}\sqcup Z\times \left\{1\right\}\to Y$ becomes $\partial f:X\sqcup Z\to Y$ after the canonical identifications. (Do I even need to elaborate on this?) And the homotopies $H$ and $H'$ give ries to a homotopy $H\sqcup H':(X\sqcup Z)\times I \to Y$ from $\partial f$ to $H|_{X\times 1}\cup H'|_{Z\times 1}:X\sqcup Z\to Y$. And now the transversality condition is local, so a disjoint union (coproduct if you like) of two maps transverse to $C$ is transverse to $C$. And the count of intersection points is clearly additive under this disjoint union as $(f\sqcup g)^{-1}(y)=f^{-1}(y)\sqcup g^{-1}(y)$ for any maps $f,g$ and $y\in Y$. Thus, $I_2(\partial f,C)$, which is the count of the number of intersection points of $H|_{X\times 1}\cup H'|_{Z\times 1}:X\sqcup Z\to Y$ with $C$, coincides with the sum of the count of the number of intersection points of $H|_{X\times 1}:X\to Y$ and $H'|_{Z\times 1}:Z\to Y$, which is simply $I_2(X,C)+I_2(Z,C)$ Done.

  • $\begingroup$ Thank you, this makes it more clear. $\endgroup$ – user500094 Mar 18 '18 at 1:11

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