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Let $M$ be a non-orientable three dimensional manifold. I am interested in knowing how to characterize the topological properties of the map $$M\to S^3$$ where $S^3$ is parameterized by $(\phi_1, \phi_2, \phi_3, \phi_4)$ with the constraint $$\sum_{i=1}^{4} \phi_i^2=1$$ More concretely, I want to know how the integral $$\mathcal{I}[M]\equiv \frac{2}{\pi^2} \int_{M} (\epsilon^{abc} \phi_1 \partial_a \phi_2 \partial_b \phi_3 \partial_c \phi_4) \;dx dy dz\;$$ is quantized.

In particular, it is known that the integral $\mathcal{I}[M]$ is quantized to be an integer for any oriented $M$. See the post An integral map from 3-torus $\mathbb{T}^3$ to 3-sphere $S^3$ for an explanation when $M=T^3$. Here I am interested to know how $\mathcal{I}[M]$ is quantized when $M$ is non-orientable.

Presumably, I expect $\mathcal{I}[M]$ is quantized to be $$\mathcal{I}[M]\in \frac{1}{p_M}\mathbb{Z}$$ for some integer $p_M$ that depends on $M$, and I would like to know what $p_M$ is. More concretely,

1) When $M=RP^2 \times S^1$, what is $p_{RP^2 \times S^1}$?

2) When $M=KB\times S^1$, $KB$ is klein bottle, what is $p_{KB}$?

3) What is the largest possible value of $p_{M}$ for all possible non-orientable manifold $M$?

Thanks for your help!

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    $\begingroup$ How do you integrate over a non-orientable manifold? $\endgroup$
    – Angina Seng
    Oct 5, 2018 at 3:11
  • $\begingroup$ @LordSharktheUnknown I just evaluate the density $\frac{2}{\pi}(\epsilon^{abc} \phi_1 \partial_a \phi_2 \partial_b \phi_3 \partial_c \phi_4)$ and integrate over the whole manifold (or sum over the discretized manifold if you would like to think about a discrete summation). $\endgroup$
    – user34104
    Oct 5, 2018 at 3:27
  • $\begingroup$ I don't really understand the question, but one always has the mod-2 degree of a map between closed unoriented manifolds of the same dimension, an element of $\Bbb Z/2$. In fact, the Hopf degree theorem states that if $M$ is 3-dimensional, $[M, S^3] \cong H^3(M;\Bbb Z)$. When $M$ is oriented the latter group is $\Bbb Z$ (determined by the degree), but when $M$ is unorientable it is $\Bbb Z/2$. $\endgroup$
    – user98602
    Oct 5, 2018 at 5:58
  • $\begingroup$ @MikeMiller Thanks! Can I understand the Hopf degree theorem this way? Let M be non-orientable, and N be its double cover which is orientable. One can first calculate the above integral on N, and it should be an integer $\mathcal{I}[N]\in \mathbb{Z}$, which is the degree. Then we compute $\mathcal{I}[M]$. $\mathcal{I}[M]= \frac{1}{2}\mathcal{I}[N]$. And different lifts $M \to N$ and $M\to N'$ of the same $M$ may differ by $2$ in $\mathcal{I}[N]$, so only the fractional part of $\mathcal{I}[M]$ is well defined. This suggests $p_M=2$. Is the above understanding correct? $\endgroup$
    – user34104
    Oct 5, 2018 at 14:04

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