# How to show in a clean way that $z^4 + (x^2 + y^2 - 1)(2x^2 + 3y^2-1) = 0$ is a torus?

How to show in a clean way that the zero-locus of $$z^4 + (x^2 + y^2 - 1)(2x^2 + 3y^2-1) = 0$$ is a torus?

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For what definition of torus? – Myself May 20 '13 at 22:03

This took me longer than it should have, given the simplicity of the answer.

First, analyze the equation for when it has solutions: clearly, we require $z^4 \geq 0$, and thus, $(x^2 + y^2 - 1)(2x^2 + 3y^2 - 1) \leq 0$. Thus, one should be positive and the other negative. Each factor vanishes on a plane curve that's a conic section, respectively the unit circle at $(0,0)$ and the ellipse centered at $(0,0)$ with its major axis along the $x$-axis of length $1/\sqrt{2}$ and its minor axis along the $y$-axis of length $1/\sqrt{3}$. The points $(x,y)$ with a corresponding $z$ above them are exactly those in one of those figures and outside the other; since the ellipse is completely contained in the circle, the corresponding set of points is the circle with the ellipse removed, an annular region.

Now, for any $(x,y)$ in the annulus, the corresponding $z$ is a fourth root of some associated nonnegative number; any positive real number has exactly two real fourth roots, one positive and one negative, while zero of course has only one, which is zero.

Putting this together, literally, we see that the implicit equation describes a shape that's obtained from the annulus as follows:

• Make two copies, superimposed;
• Glue their edges to the $xy$-plane;
• With one copy, "inflate upwards", and with the other, "inflate downwards". The edges remain fixed but the interiors flex away from the $xy$-plane.

The result is, in short, two annuli glued in the natural way along their boundaries, which is a topological torus.

Of course, I didn't use any details from the question other than to verify that the "shadow" of the torus was an annulus. In fact, that's all that matters: that the equation is of the form

$$z^2 + f(x,y) g(x,y) = 0$$

where the zero sets of $f$ and $g$ are closed plane curves one contained in the interior of the other. The resulting surface will always be a torus (and you won't always be able to use polar coordinates to figure it out).

(edit: Just noticed I wrote $z^2$ there rather than $z^4$. Actually, that would work too, as would any even power of $z$.)

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GMTA. You were 20 seconds before me... +1. – yohBS May 20 '13 at 22:58

I don't know how "clean" this is, but here's my try.

Write $x=r \cos \alpha$, $y=r\sin\alpha$. The equation becomes $$z^4+\frac{5- \cos2 \alpha }{2}r^4+ \frac{\cos2 \alpha -7}{2}r^2+1=0$$ The details don't matter too much as long as nothing changes sign, which is indeed the case. So we'll write the above as $$z^4+f(\alpha)r^4-g(\alpha)r^2+1=0$$ where $\forall\alpha \ f>0,g>0$. By rescaling $r$ we can further simplify that to $$z^4+\tilde{r}^4-\tilde g(\alpha)\tilde{r}^2+1=0$$ where $\tilde r=f^{1/4} r$ and $\tilde g=\frac{g}{\sqrt f}>0$.

This is an equation for an ellipse in the $z^2$-$r^2$ plane. To see this you can complete the square to get $$\left(z^2\right)^2+\left(\tilde{r}^2-\frac{\tilde g(\alpha)}{2}\right)^2=\left(\frac{\tilde g(\alpha)}{2}\right)^2-1$$ It is easy to check that $(\tilde g/2)^2-1>0$ for all $\alpha$.

We therefore get that for any fixed $\alpha$, $z$ and $r$ are related by some closed curve $z(\beta), r(\beta)$ which is homeomorphic to $S_1$. In other words, the surface is homeomorphic to $S_1\times S_1$. QED.

Generalization

This was a bit ugly. What about that was essential and what was details? The crux of the matter is the term $f(x,y)=\left(x^2+y^2-1\right)\left(3x^2+2y^2-1\right)$. The contour $f(x,y)=0$ looks like this:

and any other term with the same topology (two copies of $S_1$ where one is in the interior of the other) would do the trick (also, replacing $z^4$ with any other even power).

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Looks nice! At the end, I guess you would need the implicit function theorem to actually prove that the bijection you give is continuous. – Damien L May 21 '13 at 9:53

Not an answer, but here's a picture. Cheers.

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If one happens to know some Morse theory the following is another possible solution:

Consider the function $\Phi:\mathbb{R^3}\rightarrow\mathbb{R}$ defined by $$\Phi(x,y,z)=z^4+(x^2+y^2-1)(2x^2+3y^2-1).$$ You want to prove that $\Phi^{-1}(0)=:S$ is a torus. Computing $d_{(x,y,z)}\Phi$ you can check that $0$ is a regular value and hence $S$ is indeed a smooth surface embedded in $\mathbb{R}^3$. In particular $S$ is orientable.

Define now the function $$f:S\rightarrow\mathbb{R},\ (x,y,z)\mapsto y.$$ Using that $T_{(x,y,z)}S=\ker d_{(x,y,z)}\Phi$ it is not difficult to check that $f$ is a Morse function with critical points $(0,-1,0),(0,-\frac{1}{\sqrt{3}},0),(0,\frac{1}{\sqrt{3}},0),(0,1,0)$. Now use the Hessian of $f$ at these points to show that their indices are $0,1,1$ and $2$, respectively. Using the Morse inequalities you get that $\chi(S)=0$.

Since $S$ is an orientable surface with Euler characteristic equal to $0$, it must be a torus.

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I missed this answer when the question was first posed. I like it! I first wanted to do a Morse theory answer but unfortunately, I don't really know how to do Morse theory. – Ryan Reich Sep 14 '13 at 5:21