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It is well-known that, when mapping $|\vec{n}(\vec{x})|=1$, we can use $N=\int{\mathrm{d}x_1\mathrm{d}x_2\vec{n}\cdot(\partial_1\vec{n}\times\partial_2\vec{n})}$ to calculate the topological winding number, which is integer-valued since $\pi_{2}(S^2)=\mathbb{Z}$.

We also have homotopy group $\pi_{3}(S^2)=\mathbb{Z}$. For this case, do we have the calculation formula of the integer-valued topological number $\mathbb{Z}$ ?

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    $\begingroup$ The question is not understandable to me. What exactly is "topological number" (is it the topological degree?) and what do you want to compute? Do you want to identify, for a map $f: S^3\to S^2$, the class $[f]$ within $\pi_3(S^2)\simeq \mathbb{Z}$? $\endgroup$ Commented Jun 16, 2014 at 11:54
  • $\begingroup$ The $N=\dots$ formula shows how to calculate the topological winding number $\mathbb{Z}$. I want to know its counterpart for $\pi_3$ case. $\endgroup$
    – xiaohuamao
    Commented Jun 16, 2014 at 12:25

2 Answers 2

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There is such a formula: the Gauss formula for linking number.

Consider $f : S^3 \to S^2$. Applying Sard's Theorem, after a small homotopy of $f$ we may assume that $f$ is smooth and that all but finitely many points of $S^2$ are regular values of $f$. Given a regular value $p \in S^2$, its inverse image $L_p = f^{-1}(p)$ is a link in $S^3$, and $L_p$ has a natural orientation (defined so that the induced transverse orientation is preserved by $f$). Choose two distinct regular values $p,q \in S^2$. Then Gauss formula applied to $L_p$ and $L_q$ gives the integer invariant for $f$: $$\frac{1}{4 \pi} \int\!\!\int_{L_p \times L_q} \frac{\vec r_1 - \vec r_2}{|\vec r_1 - \vec r_2|^3} \cdot (d \vec r_1 \times d \vec r_2) $$

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  • $\begingroup$ Nice, didn't know that, thanks. If I understand it, this generalizes the fact that any two circles in the standard Hopf fibration have linking number 1, right? $\endgroup$ Commented Jun 16, 2014 at 13:59
  • $\begingroup$ Yes, that's right. $\endgroup$
    – Lee Mosher
    Commented Jun 16, 2014 at 14:34
  • $\begingroup$ But $L_p$, $L_q$, how you defined it, might both be unions of more circles; does the formula still hold, or you need to find such homotopy and such $p,q$ so that $L_p$, $L_q$ are both connected? $\endgroup$ Commented Jun 16, 2014 at 14:47
  • $\begingroup$ It does not matter how many components $L_p,L_q$ have. If $L_p$ has $m$ components and $L_q$ has $n$ components then the space over which you are integrating, $L_p \times L_q$, will have $mn$ components, each of which is a torus. $\endgroup$
    – Lee Mosher
    Commented Jun 16, 2014 at 23:18
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This is not really an answer, just a few remarks:

If I understand the assignment correctly, you want to have an "integral formula" such that, given $f: S^3\to S^2$, the formula would returns $\varphi [f]$, where $\varphi: \pi_3(S^2)\to\mathbb{Z}$ is a given isomorphism. Is it right?

I doubt that there can be such formula; on an oriented $3$-manifold, you naturally integrate $3$-forms and I don't see a way how to convert an $S^2$-valued map to something like that. Moreover, I would be even more sceptical if you wanted a general formula for identifying any homotopy class $\pi_k(S^n)$ by integration, because the problem is complicated and unlikely reducible to simple formulas.

However, in your case, if you can identify the preimage $f^{-1}(y)$ for some regular value $y\in S^2$ such that $T_y S^2\simeq \langle v_1, v_2\rangle$ -- the preimage $f^{-1}(y)$ is a disjoint union of topological circles -- and compute, for each $x\in f^{-1}(y)$, the vectors $f^*(v_i)$ in the normal space $N_x S^3$ that are mapped by $f_*$ to $v_i$ -- then the number you are looking for is the sum of "how many times the frame $(f^*(v_1), f^*(v_2))$ winds around, if you make one revolution around each circle in the preimage $f^{-1}(y)$"; moreover, you should care a bit about the orientation, but I'm skipping this for the moment. This might look technical but is quite geometric in nature.

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  • $\begingroup$ In your last paragraph, are you essentially factoring $f$ into the composition $p\circ \tilde{f}\colon S^3\to S^3\to S^2$ where $p$ is the Hopf fibration, and then associating the 'value' of $f$ with the topological degree of $\tilde{f}$? That seems like the approach I would also take. $\endgroup$
    – Dan Rust
    Commented Jun 16, 2014 at 13:37
  • $\begingroup$ Yes, I think it is equivalent, good point. But you can also describe it in terms of "framing of the preimage of a regular value" which I was trying to do (homotopies on the level of maps correspond to framed cobordisms on the level of framed submanifolds). $\endgroup$ Commented Jun 16, 2014 at 13:40

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