Why is this curve a topological manifold?

Question

Why is $$M=\{(z_1,z_2)\in \mathbb{C}^2 \, |\,\, z_1^3-z_2^4=0 \}$$ a topological manifold?

I understand for example why why $|z|=1$ is a topological manifold, since I can write every point as $z_0=e^{i2\pi k}$ for a unique $k$ and then with $r=e^{i2\pi t}$ write $r$|$_{[k-1/2,k+1/2]}^{-1}$ as a local coordinate chart near $z$. But I can't see how I would do this withe the curve above... What am I missing?

Answer 1

The (complex) curve has only one singular point ($0$) so to check that it is a topological manifold, you need to check that the link of that point (that is, the space where $|z_1|^2 + |z_2|^2 = \epsilon,$ and $z_1^3 - z_2^4 = 0$ is a topological circle. This requires some computation (however, not that difficult, since by the second equation $z_2 = e^{i \theta} z_1,$ where theta is a fourth root of unity, so you can try all the possibilities.

Answer 2

Maybe give a try following approach using implicit function theorem:

Let $z_1 = x_1 + ix_2$, $z_2 = x_3 + ix_4$ then $z_{1}^{3}-z_{2}^{4} = 0$ is equivalent to $$\begin{cases} -x_3^4 + 6x_3^2x_4^2-x_4^4+x_1^3-3x_1x_2^2 = 0\\ -4x_3^3x_4+4x_3x_4^3+3x_1^2x_2-x_2^3= 0 \end{cases}$$

so let $F:\mathbb{R}^4\to\mathbb{R}^2$ such that $$F:\left(\begin{array}{ccc}x_1\\x_2\\x_3\\x_4\end{array}\right) \mapsto \left(\begin{array}{ccc}-x_3^4 + 6x_3^2x_4^2-x_4^4+x_1^3-3x_1x_2^2\\-4x_3^3x_4+4x_3x_4^3+3x_1^2x_2-x_2^3\end{array}\right)$$ $M$ is the zero-locus of $F$ which is obviously continuously differentiable. Now we need to show that $DF$ is epimorphism (equivalently $\forall \textbf{t} \in M$ there exists minor of maximal rank in $DF$) but $$DF = \left(\begin{array}{} 3x_1^2-3x_2^2 & 6x_1x_2 \\ -6x_1x_2 & 3x_1^2-3x_2^2 \\ -4x_3^3 + 12x_3x_4^2 & -12x_3^2x_4+4x_4^3 \\ 12x_3^2x_4-4x_4^3 & -4x_3^3+12x_3x_4^2 \end{array}\right)$$ and now it easy to see that both $\left|\begin{array}{}3x_1^2-3x_2^2 & 6x_1x_2 \\ -6x_1x_2 & 3x_1^2-3x_2^2\end{array}\right|$ and $\left|\begin{array}{}-4x_3^3 + 12x_3x_4^2 & -12x_3^2x_4+4x_4^3 \\ 12x_3^2x_4-4x_4^3 & -4x_3^3+12x_3x_4^2\end{array}\right|$ are equal zero if and only if $x_1=x_2=x_3=x_4=0$ it sufficies to check that singular point (look at Igor Rivin's sketch).