Proof a $(2n-1)$-compact manifold I have no idea how prove that
$$\{(z_0,\ldots,z_n)\in\mathbb{C}^{n+1} \quad| \quad z_0^d+z_1^2\ldots+z_n^2=0, \quad |z_0|^2+|z_1|^2\ldots+|z_n|^2=2\}$$ is a $(2n-1)$-compact manifold.
How give the charts. For $n=1$ and $d=2$ not distinguish which 1- manifdold is?
 A: You do not want to give charts. Here are some hints to get you started.
Compactness should follow immediately from the fact that a sphere in $\Bbb R^{2n+1}$ is compact and level sets of continuous functions are closed.
In general, any time you have a set $X\subset\Bbb R^N$, say, that is defined by equations, you'd like to use those equations to define a smooth mapping $f\colon \Bbb R^N\to \Bbb R^s$ so that $X=f^{-1}(0)$. If you check that the rank of $df_x$ is $s$ for every $x\in X$, then it follows from the implicit function theorem that $X$ is an $(N-s)$-dimensional manifold. In your problem you have a mix of holomorphic and non-holomorphic, so you should either convert everything to real coordinates or, if you're familiar with doing calculus with $\partial/\partial z$ and $\partial/\partial\bar z$, you can write the functions in terms of $z$ and $\bar z$.
A: Let $(f, g):\mathbb{C^{n+1}}\to\mathbb{C}\times\mathbb{R}$ by
$(f, g)(z_0,\ldots, z_n)=(f(z_0,\ldots, z_n), g(z_0,\ldots, z_n))=(z_0^d+z_1^2\ldots+z_n^2, |z_0|^2+|z_1|^2\ldots+|z_n|^2)$
then $(0, 2)\in\mathbb{C}\times\mathbb{R}$ is a regular value, $f$ is holomorphic and the jacobian $J$ have rank 3 on $\mathbb{C}$.
Therefore $(f, g)^{-1}(0, 2)$ is a $2n-1$ manifold, (Brieskorn Manifold).
