## How the Jacobian can Detect Cusps and Self-intersections

Let be a subset of the plane defined by the vanishing of a smooth (that is, derivatives of all orders exist) function . For simplicity, it is assumed throughout that is not identically zero.

For example, if then the set 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 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 has any cusps or self-intersections. The intuitive principle of driving a car along the curve 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 and such that for all . Finding a self-intersection is equivalent to finding a non-trivial () solution to and . This is a difficult “global” problem. However, the curve has been specified implicitly by , 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 , meaning that knowledge of what looks like anywhere except in a small neighbourhood of a point is irrelevant to determining whether there is a singularity or self-intersection at . This suggests considering a Taylor series of about the point of interest. [Technically, we would also want 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 lying on , that is, . Then where the dots denote higher-order terms. If either or is non-zero then contains a linear term, and the linear term dominates the behaviour of when is sufficiently close to . The linear equation 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 is not analytic: if for every point , either or is non-zero at , then 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 into neighbourhoods which are sufficiently small that it is possible to determine exactly what looks like. To belabour the point, if a point in is not in then it is not in .)

Textbooks generally do not mention explicitly though that if fails this “Jacobian test” then it may still be a manifold. The above makes this clear; if then it is necessary to examine higher-order terms before making any claims. As a simple example, fails the test at because both and are zero at the origin, yet is the straight line given equivalently by .