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Let $\phi_1, \phi_2, \phi_3, \phi_4 \in \mathbb{R}$ be real valued functions, such that

$$\phi_j(x,y,z):(x,y,z) \in \mathbb{T}^3 \to \phi_j(x,y,z) \in \mathbb{R}.$$

Here $\mathbb{T}^3$ is a 3-torus, with $j=1,2,3,4$.

The $\phi_j(x,y,z)$ satisfies a constraint $$\sum_{j=1}^4 (\phi_j)^2=1,$$ which means that $(\phi_1, \phi_2, \phi_3, \phi_4)$ is a vector on a 3-sphere $S^3$.

Consider the integral computed from the domain $(x,y,z) \in\mathbb{T}^3$ to the target of $(\phi_1, \phi_2, \phi_3, \phi_4) \in S^3$. We can choose the $\mathbb{T}^3$ has a unit length 1, and the $S^3$ has a unit radius 1.

Question 1:

Can we show that $$(2/\pi^2) \int_{T^3} (\epsilon^{abc} \phi_1 \partial_a \phi_2 \partial_b \phi_3 \partial_c \phi_4) \;dx dy dz\;\in \mathbb{Z}?$$ is integer valued? (Or up to a front factor to be fixed.) Is this true or is it wrong? (At least for certain function $\phi_j(x,y,z)$, I find the integral can be integer valued.


(Bonus, but you can skip this one below to claim the answer.)

Question 2: More generally, is there some homotopy type of constraint, such that the integral map from the domain $\mathbb{T}^d$ to the sphere $S^d$, certain integral of the similar form $$\# \int_{T^d} (\epsilon^{\mu_1 \mu_2 \mu_3 \dots \mu_d} \phi_1 \partial_1 \phi_2 \dots \partial_{\mu_{d-1}} \phi_{d} \partial_{\mu_d} \phi_{d+1}) \;d^dx \;\in \mathbb{Z}?$$ where $$\sum_{j=1}^d (\phi_j)^2=1,$$ Up to a proper normalization $\#$?

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  • $\begingroup$ Why do you suspect this to be true? $\endgroup$ Sep 3, 2018 at 23:43
  • $\begingroup$ I take some examples and plug in -- I also thought it is related to some homotopy constraint, like $\pi_3(S^3)=\mathbb{Z}$ or $\pi_3(T^3)=0?$ $\endgroup$
    – wonderich
    Sep 3, 2018 at 23:46
  • $\begingroup$ What are these $\phi_j$? Not clear to me how they are defined given that there are three coordinates for $T^3$. $\endgroup$ Sep 3, 2018 at 23:46
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    $\begingroup$ @Andres Mejia, then you get a 0 which is an integer. $\endgroup$
    – wonderich
    Sep 3, 2018 at 23:47
  • $\begingroup$ You can try to plug in more complicated functions, and it still works $\endgroup$
    – wonderich
    Sep 3, 2018 at 23:47

1 Answer 1

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Consider the three forms $\psi = x_1 dx_2\wedge dx_3 \wedge dx_4$. Write $\phi : \mathbb T^3 \to \mathbb R^4$, $\phi = (\phi_1, \cdots, \phi_4)$. Then

\begin{align} \int_{\mathbb T^3} \phi^* \psi &= \int_{\mathbb T^3} \phi_1 d\phi_2 \wedge d\phi_3 \wedge d\phi_4 \\ &= \int_{\mathbb T^3} \epsilon^{abc} \phi_1 \partial_b \phi_2 \partial _c \phi_3 \partial _c\phi_4 \ \mathrm d x\ \mathrm d y\ \mathrm d z. \end{align}

On the other hand,

$$\int_{\mathbb T^3} \phi^* \psi = \operatorname{deg} (\phi) \int_{\mathbb S^3} \psi, $$

where $\operatorname{deg}$ is the deg of the map $\phi$, which is an integer. Finally, by Stokes theorem,

$$\int_{\mathbb S^3} \psi = \int_B d\psi = \int_B dx_1 \wedge dx_2 \wedge dx_3 \wedge dx_4.$$

The last term is the volume of the unit ball in $\mathbb R^4$ and is $\pi^2/2$. Thus your term equals $\operatorname{deg}(\phi)$.

The generalization to the higher dimensional case should be easy.

Edit To clarify, in general for two compact orientable $n$-dimensional manifold $M, N$, the degree of a smooth map $\phi : M\to N$ defined as $$ \int_M \phi^* \alpha = \operatorname{deg}(\phi) \int_N \alpha, \ \ \ \forall \alpha $$ is always an integer. I am following section 4 in Bott and Tu here. The above equality depends only on the cohomology class $[\alpha]$ instead of $\alpha$ itself. Thus we can assume $\alpha$ is a bump form support in a small open set around any point $q\in N$. Given a smooth $\phi$, let $q\in N$ be a regular value for $\phi$ (which exists by Sard's theorem). Then $\phi^{-1}(q)$ is a compact smooth submanifold of dimension $0$: that is, a finite set of points. Also there are open neighborhood of $q\in N$ so that $\phi : \phi^{-1}(B) \to B$ is a covering. Thus

$$ \int_M \phi^* \alpha = \int_{\phi^{-1}(B)} \phi^* \alpha = \sum (\pm 1) \int_B\alpha $$

This $\sum (\pm 1)$ is the degree of $\phi$, you have $\pm 1$ since $\phi$ is a local diffeomorphism.

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  • $\begingroup$ thanks, this is very useful, I accept it as an answer for now $\endgroup$
    – wonderich
    Sep 7, 2018 at 1:47

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