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If $\gamma $ is a smooth closed curve in $\mathbb R^2-\{0\}$, I want to know whether the winding number of $\gamma $ about $0$, i.e., $\frac{1}{{2\pi i}}\int\limits_\gamma {\frac{1}{z}} dz$ is equal to the index of $\gamma $ about $0$, that is, the degree of ${S^1} \to {S^1}$ through $t \to \frac{{\gamma (t)}}{{\left| {\gamma (t)} \right|}}$?

If so, it gives us a convenient way to calculate the index of curve. Is there any analogy in higher dimension?

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Yes, the index of a curve around $0$ is equal to the winding number $\displaystyle\frac{1}{2\pi i}\oint_\gamma \frac{dz}{z}$. Note that this can also be written as a line integral: $\displaystyle\frac{1}{2\pi}\oint_\gamma \frac{-y\,dx+x\,dy}{r^2}$, where $r^2=x^2+y^2$.

There are two simple generalizations of this formula to three dimensions. First, if $S$ is any closed paramterized surface, then we can compute the index of $S$ around the origin using the surface integral $$\frac{1}{4\pi}\oint_S \frac{\textbf{r}\cdot\textbf{n}}{r^3}\;dA, $$ where $\textbf{n}$ is the normal vector to the surface, $\textbf{r}$ is the vector $(x,y,z)$, and $r=\sqrt{x^2+y^2+z^2}$. This formula generalizes to $n$ dimensions, with $4\pi$ replaced by the $(n-1)$-dimensional area of the unit sphere in $\mathbb{R}^n$, and $r^3$ denominator replaced by $r^n$.

The other generalization is the integral formula for the linking number of two closed curves. If $\gamma_1$ and $\gamma_2$ are disjoint closed curves in $\mathbb{R}^3$, their linking number is given by the following double line integral: $$ \frac{1}{4\pi}\oint_{\gamma_1}\oint_{\gamma_2} \frac{\textbf{r}_1-\textbf{r}_2}{\|\textbf{r}_1-\textbf{r}_2\|^3}\cdot(d\textbf{r}_1\times d\textbf{r}_2) $$ In the case where one of the curves is the $z$-axis, this presumably simplifies to the line integral for winding number given above.

More broadly, the idea of computing topological invariants using integrals leads to De Rham cohomology. Winding number is really a cohomology class on the punctured plane, and you are computing the value of this cohomology class on a closed curve by integrating a certain differential form over the curve. Using De Rham cohomology, this can be generalized to virtually any real-valued cohomology class on any differentiable manifold.

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Thanks for the great answer! –  Samuel Reid Apr 14 '12 at 16:48
    
+1 This is a great answer. (And I didn't know that the linking number generalized the winding number in that way. Very interesting.) –  Jesse Madnick Apr 14 '12 at 17:02

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