 How do I express this vector field, given in terms of an element of elements of $\mathbb{C}$, in the standard notation for derivations?

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In a set of lecture notes (not available online) that I'm currently working through, one is given the following set-up:

Consider rotational vector fields on the plane $\mathbb{R}^{2}\cong\mathbb{C}$ with coordinates $(x,y)\cong z=x+\imath y$. Let $X$ be the vector field given by $X(z)=\imath z=-y+\imath x\in\mathbb{C}\cong T_{z}\mathbb{C}$, i.e. $X(x,y)=-y\partial_{x}+x\partial_{y}$.

This is all fine by me, but there is one thing that I don't understand: How do I express $X(z)$ in terms of the standard notation for tangent vectors (i.e. I'm looking for the complex equivalent of $X(x,y)=-y\partial_x+x\partial_y$). There are two approaches that seem sensible, but they give irreconcilable results---at least so it seems:

1. Use $z=x+\imath y$ and $\partial_z=\partial_x-\imath\partial_y$ and directly "translate" the expression in real coordinates. This yields $$X(z)=-\frac{z-\bar z}{2}(\partial_z+\partial_{\bar z}) +\frac{z+\bar z}{2}(\partial_{\bar z}-\partial_z) =\bar z\partial_{\bar z}-z\partial_z$$

2. Assume that the isomorphism $\mathbb C\cong T_z\mathbb C$ identifies $X(z)\in \mathbb{C}$ with $X(z)\partial_z+0\cdot\partial_{\bar z}$ so that we obtain $$X(z)=\imath z \partial_z$$ But when "translating back" to real variables one then obtains $$X(z)=\frac{1}{2}\imath(x+\imath y)(\partial_x-\imath \partial_y) =\frac{1}{2}\Big((\imath x-y)\partial_x+(x+\imath y)\partial_y\Big)$$

At first sight, it seems reasonable to simply say that the second method is based on a wrong assumption, and to simply dismiss it (even though I don't know why it's wrong). However, continuing in the lecture notes I find

Its trajectories are given by $c_{z}(t)=e^{\imath t}z$. Indeed, $\dot{c}_{z}(t)=\imath e^{\imath t}z=\imath c_{z}(t)$.

Assuming the first approach is correct we indeed obtain something different: It seems like the "translation" to the tangent space notation was not only useless (since the solution simply seems to solve the equation $c'(t)=\imath c$ with initial condition $c(0)=z$) but leads to the wrong solution! Of course, the second approach here does give the correct result... My question is therefore twofold:

What does the vector field $X$ correspond to in terms of the notation for derivations of smooth functions on $\mathbb C$, and is this at all relevant for finding its integral curves (trajectories)? If the answer to the latter is "no", please elaborate!

This post imported from StackExchange Mathematics at 2015-09-13 11:44 (UTC), posted by SE-user Danu
asked Sep 13, 2015

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Note that $\partial_z$ and $\partial_{\overline z}$ do not span the tangent space in a coordinate chart, but the complexified tangent space. A basis for the tangent space, as you know, is formed by $\partial_x$ and $\partial_y$, and of course these also span the complexified tangent space, which has an alternative basis that is generally more useful in complex geometry, consisting of $\partial_z$ and $\partial_{\overline z}$.

The tangent space as a real subbundle of the complexified tangent bundle is isomorphic to the holomorphic  tangent space, which has a real basis $\partial_z$, $i\partial_z$ corresponding to $\partial_x$ and $\partial_y$ (obtained from the inclusion followed by projection onto the holomorphic part). In other words, $a\partial_x + b\partial_y$ maps to $(a + bi)\partial_z$. In 1 I think you lost a factor $i$ by the way; also I think you followed the (common) convention that $\partial_z = \frac12(\partial_x - i\partial_y)$ though you wrote something different.

Note by the way that $\partial_{\overline z}$ maps holomorphic (complex differentiable) functions to 0, so as a derivation on holomorphic functions the two are the same.

answered Oct 3, 2015 by (125 points)

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