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!

  • $\begingroup$ (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. $\endgroup$
    – LéKitty
    Aug 15, 2023 at 0:58

1 Answer 1


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.)

  • $\begingroup$ 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. $\endgroup$
    – LéKitty
    Aug 15, 2023 at 0:37
  • $\begingroup$ 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. $\endgroup$
    – LéKitty
    Aug 15, 2023 at 0:52
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    $\begingroup$ G&P were positing that $X$ is orientable. For $X$ non-orientable and $Y$ orientable, try $X=\Bbb RP^2 \subset \Bbb RP^3\times S^1$. $\endgroup$ Aug 15, 2023 at 0:52
  • $\begingroup$ 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 $I_2(X,X) = 1$? $\endgroup$
    – LéKitty
    Aug 15, 2023 at 1:03

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