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What is the general approach for showing that given equation does not represent a regular surface. For eg. if I have to prove that $y(x-a)+zx=0$ is not a surface, how do I approach?

ab123
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1 Answers1

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The implicit function theorem tells you that if you consider the level surface $f(x,y,z)=0$, then the surface is a regular surface in a neighborhood of any point $(x_0,y_0,z_0)$ with $\nabla f(x_0,y_0,z_0)\ne 0$ [or use the derivative $df(x_0,y_0),z_0)$].

At points where this criterion fails (if any!), you need to ask yourself: Can I write this surface locally as a graph $z=\phi(x,y)$, $y=\phi(x,z)$, or $x=\phi(y,z)$ for a smooth function $\phi$? I've mentioned this criterion several times before; see this one, for example.

Ted Shifrin
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  • I also thought the same, but one solution I read states - "We consider the function $f(x, y, z) = y(x − a) + zx.$ We have $f_x = y + z$, $f_y = x − a$ and $f_z = x.$ The latter two expressions cannot be simultaneously $0$, so $f^{-1}(0)$ is a regular surface" – ab123 Oct 25 '17 at 21:03
  • Oh, of course, I was stooopid. There was $x-a$ in the first term but $x$ in the second term, so the gradient is everywhere nonzero. But keep my answer in mind if you end up with something like $x^2+y^2=z^2$ or $x^2=y^3$ ... – Ted Shifrin Oct 25 '17 at 21:16
  • But your way also makes sense, I showed that there is no local graph of function around $(a,0,0)$. Every such graph comes out to be not one-one or not differentiable around the point. Maybe you can check and tell me a mistake in it – ab123 Oct 25 '17 at 21:23
  • Hmm, let's see. At the point $(a,0,0)$, $f_z\ne 0$, so we should be able to solve fine for $z$ as a function of $x,y$. Sure: $z=-\frac yx(x-a)$, and this is (well-defined and) smooth (so long as we keep $|x-a|<|a|$). – Ted Shifrin Oct 25 '17 at 21:33
  • But this function is not one-one, as $(x,y) \rightarrow$ $(1,0)$ and $(2,0)$ give same image of $z$ – ab123 Oct 25 '17 at 21:39
  • No one says it has to be one-to-one!! Think about $y=x^2$. That's a perfectly fine function and curve. Projection from the surface to the $xy$-plane (locally) is still one-to-one and that's the chart on your surface. – Ted Shifrin Oct 25 '17 at 21:41