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I have been browsing "Topological Degree Theory and Applications" by Cho, Chen and O'Regan as well as "Mapping Degree Theory" by Outerelo and Ruiz, but I have not been able to quite answer myself the following question:

Let $\gamma:\mathbb{S}^1\to\mathbb{R}^n$, $n\geq 3$, be a closed (rectifiable, piece-wise smooth, smooth, if necessary) curve and let $p\in\mathbb{R}^n\setminus\gamma(\mathbb{S}^1)$ be a point not lying on $\gamma$. Do we have a good notion of a winding number $w(\gamma,a)$ that generalizes the case $n=2$ ?

Remarks:

  1. If $\gamma$ is smooth with $D\gamma$ of rank $1$ everywhere, then $\gamma(\mathbb{S}^1)$ is a submanifold ($\gamma$ is injective by definition and $\mathbb{S}^1$ is compact). In particular, it is a connected manifold, and we have a well-defined topological degree $\deg(\gamma)$ since it does not depend on a choice of a regular value of $\gamma$ on the image. This is one straight-forward generalization of the case $n=2$ for curves in higher dimensions. There are probably other approaches as well.
  2. Outerelo and Ruiz introduce $w(f,a)$ for continuous $f$ defined only on hypersurfaces $X:=\partial\bar{U}\subseteq\mathbb{R}^n$ (for some $U$ open and bounded). In particular, the construction does not seem to work for curves for $n\geq 3$. More generally, the topological degree appears to be defined only for maps whose domain and codomain are of equal dimension. (In this interpretation and modulo technical details, the winding number in the plane is the degree of a map defined on the unit disk which restricts on the boundary to our curve.)

I am hoping that (1.) could somehow be "relativized" to include an external vantage point $p$.

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    $\begingroup$ Any two closed curves not touching $p$ are homotopic in $\mathbb R^3 \setminus \{p\},$ so I don't see how a useful definition is possible. $\endgroup$ – Anthony Carapetis Apr 10 '18 at 2:46
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    $\begingroup$ @AnthonyCarapetis being homotopic is irrelevant here - you want the curves to be isotopic, in which case your statement is false, as proved by the existence of the subject of "knot theory". $\endgroup$ – Igor Rivin Apr 10 '18 at 3:12
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    $\begingroup$ Well, I guess it depends what properties of the winding number you're trying to generalize; such a soft question perhaps should not be answered with a definite yes or no. To me the ambient isotopy class doesn't seem to be a good answer to this question, especially given its insensitivity to any "vantage point" $p$. $\endgroup$ – Anthony Carapetis Apr 10 '18 at 3:28
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    $\begingroup$ See also the earlier question, Generalization of winding number to higher dimensions. $\endgroup$ – Joseph O'Rourke Apr 10 '18 at 14:43
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    $\begingroup$ @IgorRivin Why do you want the curves to be isotopic? The winding number of curves in the plane is a homotopy invariant. Moreover "isotopic", to my knowledge, refers to embedded submanifolds. Embedded curves in $\mathbb{R}^2$ are somewhat pointless (and the winding number is either zero or one). $\endgroup$ – Najib Idrissi Apr 10 '18 at 14:56
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No, you cannot define a winding number if $n\geq 3$ since, as pointed out in a comment by Anthony Carapetis, any two curves in $\mathbb{R}^n\setminus\{ p\}$ are homotopic, and a winding number should be invariant under homotopies. You can however, define a linking number between two continuous disjoint images of spheres in $\mathbb{R}^n$ of dimensions $k$ and $n-k-1$. If $k=1$, the other sphere needs to have dimension $n-2$. In the case of a planar curve, $n=2$ and $k=1$, and the other sphere has dimension $0$ i.e., consists of two points. If one of the points is in the unbounded component of the image of the curve, the corresponding linking number will be equal to the winding number.

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  • $\begingroup$ You have not justified the initial "No". $\endgroup$ – Igor Rivin Apr 10 '18 at 3:08
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    $\begingroup$ @IgorRivin I modified my answer. $\endgroup$ – Piotr Hajlasz Apr 10 '18 at 3:17
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It's natural to assume that the winding number is an integer. Nevertheless, this number should behave in a continuous way (see what follows). Consider a curve in $\ \mathbb R^n,\ (n\ge3).\ $ Consider points $p\ q\in\mathbb R^n\ $ outside the curve. Then there exists continuous $ f:[0;1]\rightarrow\mathbb R^n\ $ which is outside the curve, and is connecting $\ p\ q.\ $ The winding number should be constant for points $\ f(t),\ $ hence it should be the same for $\ p\ $ and $\ q.\ $ Thus, under the given understanding, a winding number would be useless.

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    $\begingroup$ Very nice, intuitive argument! $\endgroup$ – M.G. Apr 10 '18 at 13:26

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