How to make sense of complex coordinates on manifolds?

Solution 1:

This is going to be a long answer, but this is a good question that brings up some rather cool ideas that are often not clearly explained in elementary treatments.

There are two things going on here. First, we need the notion of a complex manifold -- this is a $2n$-dimensional real manifold, together with a maximal atlas of charts whose transition maps are all holomorphic, when we consider the codomain of the charts to be $\mathbb C^n$ (identified with $\mathbb R^{2n}$ in the standard way). Given such a manifold $M$, the complex-valued coordinate functions $(z^1,\dots,z^n)$ of any such chart are called complex coordinates or holomorphic coordinates on $M$. You're correct that all the information about a point is already contained in these complex coordinates.

The second thing going on is that once we have a complex manifold, we can start to ask which smooth complex-valued functions on (open subsets of) the manifold are actually holomorphic. In simplest terms, these are just the ones whose coordinate representations in any holomorphic coordinate chart are holomorphic in the usual sense; the fact that the various charts overlap holomorphically ensures that this is independent of the choice of chart.

But we can also approach this question from the point of view of differential forms. If $f = u+iv$ is a complex-valued smooth function on an open subset of $M$, we can consider its differential as a complex-valued $1$-form: $df = du + idv$. Technically, this is a section of the complexified cotangent bundle, whose fiber at $p\in M$ is the $2n$-dimensional complex vector space $T^*_p M\otimes \mathbb C$, or equivalently the space of real-linear functions from $T_pM$ to $\mathbb C$. The question is, how can we tell from $df$ whether $f$ is holomorphic or not?

If we choose local holomorphic coordinates $(z^j)$ and write their real and imaginary parts as $z^j = x^j + i y^j$, then we can think of any smooth complex-valued function $f$ locally as a function of the $2n$ real variables $(x^1,\dots,x^n,y^1,\dots y^n)$, and write its differential as the following complex-valued $1$-form $$ df = \frac{\partial f}{\partial x^1} dx^1 + \dots + \frac{\partial f}{\partial x^n} dx^n + \frac{\partial f}{\partial y^1} dy^1 + \dots + \frac{\partial f}{\partial y^n} dy^n. $$ On the other hand, consider the complex-valued $1$-forms $dz^1,\dots,dz^n,d\bar z^1,\dots d\bar z^n$, where $dz^j = dx^j + i dy^j$ and $d\bar z^j = dx^j - i dy^j$. They are linearly independent over $\mathbb C$, and therefore also provide a local coframe for the complexified cotangent bundle. A little linear algebra shows that $df$ can also be written in terms of this coframe as follows: $$ df = \sum_{j=1}^n \frac12 \left( \frac{\partial f}{\partial x^j} - i\frac{\partial f}{\partial y^j} \right) dz^j + \sum_{j=1}^n \frac12 \left( \frac{\partial f}{\partial x^j} +i \frac{\partial f}{\partial y^j} \right) d\bar z^j . $$

With real coordinates, the components of $df$ are just the partial derivatives of $f$ with respect to the coordinate functions. Motivated by analogy with that case, let us define the expressions $\partial f/\partial z^j$ and $\partial f/\partial \bar z^j$ as follows: $$ \frac{\partial f}{\partial z^j} = \frac12 \left( \frac{\partial f}{\partial x^j} - i\frac{\partial f}{\partial y^j} \right) , \qquad \frac{\partial f}{\partial \bar z^j} = \frac12 \left( \frac{\partial f}{\partial x^j} +i \frac{\partial f}{\partial y^j} \right). $$ Then $$ df = \sum_{j=1}^n \frac{\partial f}{\partial z^j} dz^j + \sum_{j=1}^n \frac{\partial f}{\partial \bar z^j} d\bar z^j . $$ The cool thing about this formula is that the equations $\partial f/\partial \bar z^j = 0$ for $j=1,\dots,n$ are exactly the Cauchy-Riemann equations, so $f$ is holomorphic if and only if $df$ can be expressed as a linear combination of $dz^1,\dots,dz^n$.

So far, these are just notations, chosen to make that last formula for $df$ look nice. Based on that formula, it's tempting to think of the $2n$ variables $(z^1,\dots,z^n,\bar z^1,\dots, \bar z^n)$ as if they were independent complex variables; and think of $\partial f/\partial z^j$ and $\partial f/\partial \bar z^j$ as if they were partial derivatives, obtained by holding some of those variables fixed and differentiating with respect to the others; and think of a function as being holomorphic if and only if it's "independent of the $\bar z^j$'s." But as you correctly pointed out, this makes no sense, because if you hold $z^j$ fixed, then $\bar z^j$ stays fixed as well.

However, there is more here than meets the eye. First of all, let's think about a complex-valued polynomial function of the real variables $x^j,y^j$. It might or might not be holomorphic, depending on whether it satisfies the Cauchy-Riemann equations. For example, the polynomial $(x^1)^2 - (y^1)^2 + 2ix^1y^1 = (z^1)^2$ is holomorphic, while $(x^1)^2 + (y^1)^2 = z^1 \bar z^1$ is not.

But we can also rewrite $f$ as a polynomial in $(z^j,\bar z^j)$, simply by making the substitutions $x^j =(z^j + \bar z^j)/2$, $y^j =(z^j-\bar z^j)/(2i)$. Then you can check that $f$ is holomorphic if and only if none of the $\bar z^j$'s show up in this representation of $f$.

We can extend this idea to real-analytic functions as well -- a complex-valued real-analytic function can be expressed locally by a convergent power series in $(x^j,y^j)$, and then by making the same substitutions as above, we can write it as a convergent power series in $(z^j,\bar z^j)$. Such a function is holomorphic if and only if no $\bar z^j$'s appear in the power series.

In this context, you can almost make sense of the idea of the $z^j$'s and $\bar z^j$'s being independent variables. Given a power series in the $z^j$'s and $\bar z^j$'s, we can just make the formal substitution $\bar z^j \mapsto w^j$, and obtain a convergent power series in $2n$ complex variables $(z^j,w^j)$. Let's call that function $g(z,w)$. Then the original function $f$ can be recovered as $f(z) = g(z,\bar z)$, and $f$ is holomorphic if and only if $g$ was independent of all the $w^j$'s.

This whole approach makes sense only for real-analytic functions -- if $f$ is merely a smooth function of $(x^j,y^j)$, then it makes no sense to try to evaluate it with $z^j$ and $\bar z^j$ varying independently. But still, people tend to think informally of a function for which $\partial f/\partial \bar z^j =0$ as somehow being "independent of the $\bar z^j$'s."