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How the Jacobian can Detect Cusps and Self-intersections

October 17, 2013 2 comments

Let M be a subset of the plane defined by the vanishing of a smooth (that is, derivatives of all orders exist) function f\colon \mathbb{R}^2 \rightarrow \mathbb{R}.  For simplicity, it is assumed throughout that f is not identically zero.

For example, if f(x,y) = x^2 + y^2 - 1 then the set M = \{ (x,y) \in \mathbb{R}^2 \mid f(x,y) = 0\} is the unit circle. This can be readily visualised using the online Sage Notebook. Sign in to Sage, create a new worksheet, and type: var(‘x,y’); implicit_plot(x^2+y^2-1==0, (x, -2, 2), (y, -2, 2)); NOTE: WordPress automatically converts the first apostrophe in var(‘x,y’); from the correct {}' to the incorrect `.

Next, try: var(‘x,y’); implicit_plot(x^3+y^3-x*y==0, (x, -2, 2), (y, -2, 2));

Note that this second example has a self-intersection; if one considers starting at one end of the line and walking to the other end, there will be a point that will be passed over twice, once when entering the loop and once upon exit.

Thirdly, try: var(‘x,y’); implicit_plot(x^3-y^2==0, (x, -2, 2), (y, -2, 2));

This third example has a cusp; if one endeavours to drive a car from the bottom end of the line to the top end, one will get stuck at the cusp with the car pointing to the left but the “road” continuing off to the right.

Roughly speaking, and as a good introduction to differential geometry, a (concrete) manifold is a subset of \mathbb{R}^n that does not contain cusps (that is, it is smooth) and does not contain self-intersections (that is, it is locally Euclidean). Regardless, it is interesting to ask how difficult it is to determine if the set M = \{ (x,y) \in \mathbb{R}^2 \mid f(x,y) = 0\}  has any cusps or self-intersections.  The intuitive principle of driving a car along the curve M to detect cusps or self-intersections may appear difficult to translate into a mathematical tool, especially for detecting self-intersections.

Note though that driving a car corresponds to a parametrisation of the curve: roughly speaking, constructing functions x(t) and y(t) such that f(x(t),y(t)) = 0 for all t. Finding a self-intersection is equivalent to finding a non-trivial (t \neq t') solution to x(t)=x(t') and y(t)=y(t').  This is a difficult “global” problem. However, the curve has been specified implicitly by f(x,y) =0, and perhaps the problem is not global but only local with respect to this implicit formulation?

Detecting cusps and self-intersections really is only a local problem with respect to f(x,y), meaning that knowledge of what M = \{ (x,y) \in \mathbb{R}^2 \mid f(x,y) = 0\} looks like anywhere except in a small neighbourhood of a point (x,y) is irrelevant to determining whether there is a singularity or self-intersection at (x,y).  This suggests considering a Taylor series of f about the point (x,y) of interest. [Technically, we would also want f to be analytic if we are relying on a Taylor series argument, but we are focusing here on gaining a feel for the problem in the first instance.] It is also worth noting that this is also an example of “unhelpful intuition”; thinking of driving cars is a hindrance, not a help.

Choose a point (x',y') lying on M, that is, f(x',y') = 0.  Then f(x,y) = \alpha (x-x') + \beta (y-y') + \cdots where the dots denote higher-order terms.  If either \alpha or \beta is non-zero then f contains a linear term, and the linear term dominates the behaviour of f when (x,y) is sufficiently close to (x',y').  The linear equation \alpha (x-x') + \beta (y-y') = 0 describes a straight line; there can be no cusps or self-intersections.  This argument can be made rigorous using the implicit function theorem, even when f is not analytic: if for every point (x,y) \in M, either \frac{\partial f}{\partial x} or \frac{\partial f}{\partial y} is non-zero at (x,y), then M = \{ (x,y) \in \mathbb{R}^2 \mid f(x,y) = 0\} contains no self-intersections or cusps.  This result, stated more generally in terms of the rank of the Jacobian matrix, appears near the beginning in most differential geometry textbooks.  Without the above considerations though, it may appear mysterious how the Jacobian can detect cusps and self-intersections. (The key point is not to think in terms of a parametrisation of the curve, but instead, break the space \mathbb{R}^2 into neighbourhoods U_i  which are sufficiently small that it is possible to determine exactly what M \cap U_i looks like. To belabour the point, if a point in U_i is not in M \cap U_i then it is not in M.)

Textbooks generally do not mention explicitly though that if M fails this “Jacobian test” then it may still be a manifold. The above makes this clear; if \alpha=\beta=0 then it is necessary to examine higher-order terms before making any claims. As a simple example, f(x,y) = x^3 fails the test at (0,0) because both \frac{\partial f}{\partial x} and \frac{\partial f}{\partial y} are zero at the origin, yet M = \{ (x,y) \in \mathbb{R}^2 \mid x^3 = 0\} is the straight line given equivalently by x=0.