1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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