# Self intersection and cohomology of boundary of tubular neighbourhood

Let $$X$$ a compact orientable manifold of dimension $$2n$$ and $$Y$$ a compact submanifold of dimension $$n$$. Further let $$U$$ a tubular neighbourhood of $$Y$$. When I did some calculations I got the conjecture:

$$Y$$ has self-intersection $$\pm 1$$ $$\Leftrightarrow$$ $$H^n(U,\partial U;\mathbb R)\simeq H^{n}(U;\mathbb R)$$

Is this really true?

It seems logical that there is some connection between both sides because the information about how $$Y$$ is embedded in $$X$$ is encoded in $$\partial U$$.

• Just a quick comment in case it helps anyone: If you take $Y = \mathbb{R}P^2$ and $X = \mathbb{C}P^2$, then $Y$ has self intersection number $\pm 1$, but $H^2(U,\partial U)$ has order $8$ while $H^2(U)$ has order $2$. So, the conjecture is false with integral coefficients. Of course, this is not a counterexample to the actual conjecture, since both these homology groups groups vanish with real coefficients. – Jason DeVito Feb 11 at 16:55

## 1 Answer

In addition to what you said above $$Y$$ should be connected (for convenience) and oriented (by necessity). This is all about the Euler class and the Gysin sequence.

You're effectively asking whether the maps $$H^n(U) \to H^n(\partial U)$$ and $$H^{n-1}(U) \to H^{n-1}(\partial U)$$ are zero and surjective, respectively.

Because $$U$$ is the total space of a vector bundle over $$Y$$, we may as well replace $$U$$ with $$Y$$. The maps above are the pullbacks induced by the projection $$\pi: \partial U \to Y$$; note that this is a sphere bundle over $$Y$$.

In particular, these maps fit into the Gysin sequence

$$\cdots \to H^j(\partial U) \to H^{j-n+1}(Y) \xrightarrow{e \smile } H^{j+1}(Y) \xrightarrow{\pi^*} H^{j+1}(\partial U) \to H^{j-n+2}(Y) \to \cdots .$$

The class $$e \in H^{n}(Y;\Bbb Z)$$ is called the Euler class, and following the isomorphism $$H^n(Y;\Bbb Z) \cong \Bbb Z$$ coming from the orientation of $$Y$$, agrees with the self-intersection number of $$Y$$. (This follows quickly from the definition of Euler class, which would be too long a discussion to be appropriate for this post.) In particular, it is nonzero in real cohomology if and only if the self-intersection number is nonzero.

First set $$j = n-2$$. We want the map $$H^{n-1}(\partial U) \to H^0(Y)$$ to be zero so that the map $$H^{n-1}(Y) \to H^{n-1}(\partial U)$$ is surjective. Equivalently, we want the map $$H^0(Y) \xrightarrow{e \smile} H^n(Y)$$ given by the cup product with Euler class to be injective. This map is is either an isomorphism or zero (depending on whether or not the self-intersection number is zero). In particular, if $$e$$ is nonzero, then indeed $$H^{n-1}(Y) \to H^{n-1}(\partial U)$$ is surjective.

Now set $$j = n-1$$. Then again the map $$H^0(Y) \to H^n(Y)$$ given by the cup product with Euler class is either an isomorphism or zero, and so if $$e$$ is nontrivial, the map $$H^n(Y) \to H^n(\partial U)$$ is zero.

Therefore if $$e$$ is nontrivial (which is true if and only if $$Y$$ has nonzero self-intersection), the map $$H^n(U, \partial U) \to H^n(U)$$ is an isomorphism. This is true more generally than just self-intersection $$\pm 1$$.

Conversely, the map $$\pi^*: H^n(Y) \to H^n(\partial U)$$ is injective if $$e$$ is zero, in which case the natural map $$H^n(U, \partial U) \to H^n(U)$$ is zero.

• If I consider integral coefficients and only demand that the cohomology groups on the right hand side have equal rank. Then the LHS is still self intersection $\pm 1$, right? – klirk Feb 12 at 22:13
• @klirk I don't know how one could ever argue just from knowing that they have equal rank, and in fact I'm skeptical. I'd rather not try to find a counterexample, though. The above argument does imply that if the natural map $H^n(U, \partial U) \to H^n(U)$ is an isomorphism with integral coefficients, then the self-intersection is $\pm 1$. – user98602 Feb 12 at 22:22