Prove that the diagonal $∆$ in $S^2×S^2$ is not globally definable by $2$ independent function.

Question

Prove that the diagonal $∆$ in $S^2×S^2$ is not globally definable by $2$ independent function. In contrast, show that the other standard copies of $S^2$ in $S^2×S^2$ – ie, $S^2×\{a\}$ for $a∈S^2$ are definable.

Definition: We say $Z$ is globally definable if there exists a smooth function $\theta$ on neighborhood of $Z$ such that $\theta ^{-1}=Z$ and $d\theta$ is non zero at every point $z \in Z$

Here is what I got so far,( or I think I got so far)

Let $X=S^2$ then $\Delta \subset X\times X$. I know that $I_2 (\Delta,\Delta)=I_2(i_\Delta,\Delta)$ where $i_\Delta: X \to X\times X$ ie: $i_\Delta(x)=(x,x)$.

(Note: $I_2(\Delta,\Delta)$ is the mod 2 intersection number of the diagonal to itself.)

I know a theorem that say if $Z$ is globally definable by independent functions then $I_2(Z,Z)=0$. So I think I can use the contrapositive of this theorem, show that $I_2 (\Delta,\Delta) \not = 0$ and this will guarantee that the diagonal $∆$ in $S^2×S^2$ is not globally definable by $2$ independent function.

But $L(i_\Delta)= I_2 (\Delta,\Delta)= \chi (S^2)=2$

Question: By the hairy ball theorem, the tangent bundle of $S^2$ is non-trivial. Let $Z$ be submanifold of $S^2 \times S^2$. Because the tangent bundle is non trivial, can I say the normal bundle $N(Z,S^2 \times S^2)$ is non-trivial as well?

If there is a way to show that $\Delta$ is not orientable, this will work too, because if $\Delta$ is not orientable then $N(\Delta, S^2 \times S^2)$ are not trivial.

Answer 1

No, it won't be an odd number of times. Recall that $I(\Delta,\Delta) = \chi(X)$, and $\chi(S^2) =2$. Mod-$2$ intersection numbers aren't going to cut it here.

You need to update your exercise relating mod-$2$ self-intersection and cutting out by independent functions to oriented self-intersection.

Answer 2

An equivalent question would be: does there exist a neighborhood $U\supset \Delta$ and a function $f:U \to \mathbb R^2$ such that $\Delta = f^{-1}(z)$ for some $z$, where $df$ is surjective.

The answer is no.

First off, some impressions concerning your ideas:

1. Your question about non orientable vector bundles over $\Delta \cong S^2$ has a negative answer: Every bundle over $S^2$ is orientable since it is simply connected.

2. The normal bundle of a submanifold which has non-zero intersection number is never trivial, nor has it direct sum decopmosition with a trivial summand. Otherwise you could (by tubular neighborhood theorem) push the submanifold away along a given parametrization (given by a trivial line bundle), which would yield intersection number $0$.

3. You can show that $N(\Delta, M\times M) \cong TM$, which is related to $1.$ and $2.$ and answers the other questions in your question.

4. Of course for every non empty manifold $M$ there exist submanifolds (e.g. points) with trivial normal bundle.

Now to prove my answer, observe that a point $z$ has trivial normal bundle in $\mathbb R^2$ (this is, since points just don't admit too many bundles at all...). If you can obtain $\Delta$ as a pullback $f^{-1}(z)$ ($z$ being a regular value) you also pull back the normal bundle $$N(\Delta,S^2\times S^2) \cong N(\Delta,U) \cong f^*N(z,\mathbb R^2) \cong \epsilon^2,$$ where the latter denotes the trivial plane bundle over $\Delta$. But we have seen in multiple ways that this is not true, hence a contradiction.