Let $X \subset Y$ be a compact embedded manifold of odd dimension such that $\dim X = \frac12 \dim Y = n$. At the end of the section titled "Oriented Intersection Theory", Guilleman and Pollack claim the following (paraphrasing, see bottom of page 115 of Differential Topology for the exact wording):
If $I_2(X,X) \neq 0$, then $Y$ cannot be oriented.
Why is this true?
If $X$ is orientable then it must be the case that $I(X,X) = 0$, and thus if $I_2(X,X) \neq 0$ when $\dim X$ is odd, I understand "something about orientation" has gone awry. It seems to imply that $X$ is not orientable as a submanifold of $Y$, for instance.
I don't see why this says anything about the orientation of $Y$. My guess is that I've overlooked something quite basic, but I'm quite stuck! Here's what I've tried so far:
If $Y$ were orientable and we could induce an orientation on $X$, perhaps by pulling back an orientation form on $Y$, then we would have the contradiction $I_2(X,X) \neq 0 = I(X,X) \mod 2$. However, it's easy to cook up examples of an orientable manifold $M$ with a nonorientable embedded submanifold $S\subseteq N$; take the Möbius band embedded in $\mathbb R^3$ for instance. I think you can fix this using interior multiplication if you additionally have $k = \operatorname{codim} S$ linearly independent vector fields $v_1,...,v_k$ on $M$ which are nowhere tangent to $S$, but I don't see how to obtain such a collection of vector fields in our case. It might be possible to do this locally by writing $X\cap U$ as the level set of a submersion $\varphi:U\to \mathbb R^{n}$, but I don't think this can be extended to a global collection of linearly independent vector fields due to the aforementioned examples.
I also tried to work out an example where $M$ is the Möbius band, $X = M\times S^1$ and $Y = \mathbb R^6$. The vector field $v = \frac{\partial}{\partial \theta}$ where $\theta$ is the coordinate for $S^1$ is nowhere tangent to $M\times \{pt\}$, and thus any orientation form $\omega$ on $X$ will induce an orientation on $M$ by pulling back $v\lrcorner \omega$ along an embedding $M\to X$. The Möbius band is not orientable and therefore neither is $X$. Furthermore, by taking the product of the embeddings $M\to \mathbb R^3$ and $S^1\to \mathbb R^2$ and then composing with $\mathbb R^5 \to \mathbb R^6$, we obtain an embedding $\iota: X \to \mathbb R^6$. In effect, this is the opposite of the "circle embedded in the Möbius band" example Guilleman and Pollack discuss, for $Y$ is orientable while $X$ is not. However, I don't know how to compute $I_2(X,X)$ in this case and thus can't verify whether $I_2(X,X) = 0$ as we expect.
Thank you!
If $Y$ is orientable, then $I(X,Z)=(-1)^{\dim X\dim Z}I(Z,X)$ for oriented submanifolds $X,Z$ of complementary dimension, and so $I(X,X)=0$ when $X$ is odd-dimensional of half the dimension of $Y$. Therefore, when $Y$ is orientable, we must have $I_2(X,X) = I(X,X) \pmod2 = 0$ when $X$ is orientable.
No, there is no way to “induce” an orientation on a submanifold. Of course, if its normal bundle is orientable — in particular, if its normal bundle is trivial — then you can. And no, I don’t see why you want to infer that the submanifold must also be non-orientable when the normal bundle is non-trivial. Just consider $S^1 = \Bbb RP^1 \subset \Bbb RP^2$. (This is the obvious compactification of G&P’s standard example of the central circle on the Möbius strip.)