Recognizing a quadric surface from its traces

A simple “calculus” (but really analytic geometry) exercise is to find the traces of a given surface in coordinate planes. Just take the surface equation F(x,y,z)=0, set, for example, y=0 in it, maybe simplify and voilà—you have the trace in the xz-plane. Calculus textbook writers rely on quadric surfaces, in which case the intersections are (possibly degenerate) quadrics.

Can the process be reversed? That is, can we determine a quadric surface from its coordinate traces?

Let’s begin in one dimension lower: can we determine a quadric curve from its intersections with coordinate axes?

Of course not!

There is an obvious algebraic reason too: adding xy to the equation has no effect on the traces. The correct way to go one dimension lower is to also go one degree lower: that is, determine a line from its intercepts. This, of course, is a famous problem in precalculus. The solution is known: a line is uniquely determined if and only if it does not pass through the origin.

Going back to the original problem, we notice that there is no hope of identifying the surface if the traces are empty, as they can easily be for an ellipsoid. However, even three nonempty traces do not always determine the surface. Indeed, imagine that the ellipses drawn above lie in the z=1, and use them to build a paraboloid (or a cone) with a vertex at (0,0,0). All three traces are nonempty, and they do not depend on the shape of the ellipse.

Apparently, we must assume more: all three traces are nondegenerate quadrics, namely circles/ellipses, parabolas or hyperbolas. Can we recover the surface now? The answer is yes but the proof is not entirely trivial. It naturally splits into two cases, depending on whether the surface passes through the origin. (We can tell if it does by looking at any of the traces.)

Case 1: the surface does not pass through the origin. Then we look for its equation in the form p(x,y,z)=1 where p is a polynomial with zero constant term. Correspondingly, we write the equations of traces as f(x,y)=1, g(x,z)=1 and h(y,z)=1. Now, p-f vanishes on the xy-trace and at the origin, hence on the entire xy-plane. This means that p-f involves only the terms divisible by z: namely, xz, z^2, and yz. Of these, the first two are found by inspecting g and the last one by inspecting h. The process by which we obtain p from f,g,h can be described by merging the polynomials: we take the union of all monomials they contain. Something similar happens in calculus when students are asked to recover a function from its partial derivatives.

Case 2: the surface passes through the origin. Now we look for its equation in the form p(x,y,z)=0 where p still has zero constant term. The obvious issue is that 0 is not as useful for normalization as 1. Having written the trace equations as f(x,y)=0, g(x,z)=0, and h(y,z)=0, we still don’t know if we got their scaling right. The idea is then to insist that at the origin f_x=g_x, f_y=h_y, and g_z=h_z, which must be the case if f,g,h are scaled to fit together into p.

Case 2a: Two of the equations with derivatives are 0=0. This means that one of three functions, say f, has zero gradient at the origin. However, we ruled out this possibility by assuming that all traces are nondegenerate quadrics.

Case 2b: At most one of the equations is 0=0. The other two give us enough information to determine the ratios of coefficients with which f,g,h appear in p. Actually, we can just say that f appears with coefficient 1, and scale g,h accordingly. Now we merge f, g and h into p as in Case 1. QED

This is not as satisfactory as the identification of a line by its intercept, because the condition (nondegenerate traces) is not necessary. For example, the right circular cylinder x^2+y^2=1 is determined by its traces, even though two of them are degenerate quadrics. Two questions:

  • Is there a clean necessary and sufficient condition for unique determination of a quadric surface by its traces?
  • What happens for hypersurfaces of degree n-1 in \mathbb R^n?

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