# show that subset of complex projective space is a submanifold

Let n, m $\in \mathbb{N}$.

I'm trying to show that

$M(n,m) = \{[z_0 : z_1 : … : z_n] \in \mathbb{C}P^n | \sum^{n}_{i=0} z^m_i = 0\}$

is a submanifold and codim(M(n,m))=2.

My idea was to use the regular value theorem.

I guess f should be $f(z) =\sum^{n}_{i=0} z^m_i$, where $z \in \mathbb{C}P^n$. At first: How can I show that f is differentiable? – thomas.en Nov 1 '12 at 18:31
Well, the same $f$ is differentiabe from $\Bbb C^n$, as all its partial derivatives exist and are continuous. – Berci Nov 1 '12 at 19:03
The codimension is meant over $\Bbb R$, isn't it? – Berci Nov 1 '12 at 19:33
Yes, that's right, codimension over $\mathbb{R}$. Ok, so if f is differentiable, I have to show that 0 is a regular value. How do I see that? And if I take $z=0$ then, f(z)=0, so 0 has to be a regular point. But isn't the rank of the Jacobian matrix 0 for z=0? – thomas.en Nov 1 '12 at 21:11