What's the calculation formula of topological number for mappings of $\pi_{3}(S^2)=\mathbb{Z}$?

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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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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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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.