Definition of a complex fiber

Question

We define a real hypersurface as a subset $M\subset\Bbb C^n$ which is locally defined as the zero-locus of some $r\in\mathcal C^2(\Omega,\Bbb R)$ ($\Omega\subseteq\Bbb C^n$ open). Then let $z_0\in M$.

If $\Bbb C^n$ was $\Bbb R^n$, we can easily see that such an $M$ is a $(n-1)$-dimensional variety. I hope in complex case it works as well.

Allowing us to think that real facts works also here (and I'm well aware that this could be strongly wrong, but the only source I have is hyper-cryptic, so I hope in the good heart of someone of you!), we know the definition of tangent space to $M$ at $z_0$ (or the fiber of $z_0$): in the real case it was $\ker \Delta r(z_0)$, now my book says that it is "the space of vectors othogonal to $\partial r(z_0)$ under hermitian product", and denotes this with $T_{z_0}^{\Bbb C}M$. I obviously deduced that $$ T_{z_0}^{\Bbb C}M:=\{z\in\Bbb C^n\;:\;z\cdot\overline{\partial r(z_0)}=0\} $$ where $\partial r(z_0)=(\partial_{z_1}r(z_0),\dots,\partial_{z_n}r(z_0))$.

But this can't be right because of something I found after (it deals with the well definition of the signature of the Levi form of $M$, and it would be really long to write: I think the important is to say that at this point I reached a contradiction). The way to avoid contradictions, is to define $T_{z_0}^{\Bbb C}M$ as $\{z\in\Bbb C^n\;:\;z\cdot\partial r(z_0)=0\}$. But this doesn't seem "the orthogonal to $\partial r(z_0)$ under hermitian product"!

Can someone help me? Many thanks!

Answer 1

First, let me emphasize that the introduction of a hermitian structure on $\mathbb C^n$ in this question is ugly, absolutely unnecessary and only obscures the situation.
That said, given your $r\in\mathcal C^2(\Omega,\Bbb R)$ the subset $M\subset \Omega$ defined by $r=0$ is a real manifold at $z_0\in \Omega$ if and only if the real differential form $dr (z_0)\neq 0\in L_\mathbb R(\mathbb C^n,\mathbb R)$.
If this is the case the tangent space to $M$ at $z_0$ is the real hyperplane $$T_{\mathbb z_0}(M)=\ker dr (z_0)\subset \mathbb C^n $$of real dimension $2n-1$.
Inside this real hyperplane lies a unique complex hyperplane of complex dimension $n-1$ (and thus of real dimension $2n-2$) $$ T_{z_0}^\mathbb C (M)= T_{ z_0}(M)\cap i T_{z_0}(M) \subset \mathbb C^n $$ This complex hyperplane is also characterized by $$T_{z_0}^\mathbb C (M)=ker(\partial r(z_0):\mathbb C^n\to \mathbb C)\subset \mathbb C^n$$
That's all there is to it but this topic is one of the worst explained in complex analysis.

Reminder of the formulas:
$$dr(z_0)=\sum \frac {\partial r} {\partial z_j}(z_0)dz_j+\sum \frac {\partial r} {\partial \bar {z_j}}(z_0)d\bar {z_j}=\sum \frac {\partial r} {\partial x_j}(z_0)dx_j+\sum \frac {\partial r} {\partial y_j}(z_0)dy_j\in L_\mathbb R(\mathbb C^n,\mathbb R)$$ and $$\partial r(z_0)=\sum \frac {\partial r} {\partial z_j}(z_0)dz_j \in L_\mathbb C(\mathbb C^n,\mathbb C)$$

Answer 2

Something's wrong with the statements you paraphrased from your source. Let me explain.

When you're talking about real submanifolds, you should treat $\mathbb C^n$ as being identical with $\mathbb R^{2n}$. If $0$ is a regular value of $r\in \mathcal C^2(\Omega,\mathbb R)$ and $M=r^{-1}(0)$, then $M$ is a real $(2n-1)$-dimensional submanifold of $\Omega$. The tangent space $T_{z_0}\mathbb C^n$ is a $2n$-dimensional real vector space, which can be naturally identified with $\mathbb R^{2n}$ (or, alternatively, with $\mathbb C^n$ viewed as a $2n$-dimensional vector space over $\mathbb R$).

In this context, $T_{z_0}M$ is, as usual, the space of vectors in $T_{z_0}\mathbb C^n$ that are in the kernel of the (real-valued) $1$-form $dr$; this is a real $(2n-1)$-dimensional subspace of $T_{z_0}\mathbb C^n$. In terms of standard coordinates $z^j = x^j + i y^J$, we can write an arbitrary $v\in T_{z_0}\mathbb C^n$ in the form $$ v = \sum_{j=1}^n \left(a^j\frac{\partial}{\partial x^j} + b^j\frac{\partial}{\partial y^j}\right), $$ and then the condition $dr(v)=0$ becomes $$ 0 = \sum_{j=1}^n \left(a^j\frac{\partial r}{\partial x^j} + b^j\frac{\partial r}{\partial y^j}\right).\tag{$*$} $$

It doesn't make sense to say that $T_{z_0}M$ is "the space of vectors othogonal to $\partial r(z_0)$ under hermitian product," because a Hermitian inner product can only defined on a complex vector space. If we think of $(\partial_{z_1}r(z_0),\dots,\partial_{z_n}r(z_0))$ as a vector in $\mathbb C^n$, its orthogonal complement under the standard Hermitian inner product is an $(n-1)$-dimensional complex subspace of $\mathbb C^n$, which is wrong.

If you want to write the condition $v\in T_{z_0}M$ in complex notation, you can use the complexified tangent vectors $\partial/\partial z^j = \frac 1 2 (\partial/\partial x^j - i \partial/\partial y^j)$ and $\partial/\partial {\bar z}^j = \frac 1 2 (\partial/\partial x^j + i \partial/\partial y^j)$, and can identify $v$ with the complex vector $(a^1+ib^1,\dots, a^n+ib^n)\in \mathbb C^n$. Then $(*)$ is equivalent to $$ 0 = \sum_{j=1}^n \left( (a^j + i b^j )\frac{\partial r}{\partial z^j} + i (a^j - i b^j) \frac{\partial r}{\partial \bar{z}^j} \right)= 2 \operatorname{Re} (v\centerdot \partial r(z_0)), $$ which is not the same as the equation you quoted.