Non-vanishing of mod 2 intersection number for odd-dimensional manifolds implies the ambient manifold is nonorientable.

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!

• (Commenting rather than editing) I should clarify that by "It seems to imply that $X$ is not orientable as a submanifold of $Y$" I was trying to communicate that $I_2(X,X) \neq 0$ together with "$X$ is odd dimensional" seems to indicate some failure of orientability in relation to the choice of embedding $X\to Y$. As I note in my question and as @Ted Shifren says in his answer, it's in general impossible to "induce" an orientation on an embedded submanifold, so my statement of course doesn't have literal meaning. Aug 15, 2023 at 0:58

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.)
• This argument makes sense, but I still don't see why if $X$ is odd dimensional and $I_2(X,X) \neq 0$ then $Y$ is necessarily nonorientable. I'm worried about the case where $Y$ is orientable but $X$ is not. Then $I_2(X,X)$ is still defined but $I(X,X)$ is not. Aug 15, 2023 at 0:37
• Perhaps I've misunderstood the claim that Guillemin and Pollack are making. Are they simply observing that if $X$ is known to be orientable and $Y$'s orientability is in question, then a disagreement between $I_2(X,X)$ and $I(X,X)$ modulo 2 indicates it's impossible for $Y$ to be orientable? If so then great -- I thought this couldn't be the case as the definition of $I(X,X)$ required $Y$ to be orientable from the start. Aug 15, 2023 at 0:52
• G&P were positing that $X$ is orientable. For $X$ non-orientable and $Y$ orientable, try $X=\Bbb RP^2 \subset \Bbb RP^3\times S^1$. Aug 15, 2023 at 0:52
• Gotcha, a simple misunderstanding then, thank you. I am still curious about the more general question however: if $X$ is odd-dimensional and $I_2(X,X)\neq 0$, then is $Y$ non-orientable? I'm looking for examples where $X$ is nonorientable and has odd dimension equal to half $\dim Y$. This is why I asked about $X = M\times S^1$ and $Y = \mathbb R^6$ where $M$ is the Möbius band. Does this (or any other example with $X$ odd dimensional) satisfy $I_2(X,X) = 1$? Aug 15, 2023 at 1:03