3
$\begingroup$

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

  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!

EDIT: After correcting some elementary mistakes pointed out in the answers, some of the confusion is resolved, but part of the question still stands.

$\endgroup$
0
3
$\begingroup$

First of all, it should be pointed out that the literature is full with different definitions and different relations between real tangent vectors and complex tangent vectors. Having said that, let us try to understand the situation you are in.

The real tangent space at a point, is the real $2$-dimensional space spanned by the vectors $\partial_x$ and $\partial_y$. The complex tangent space is the complexification of the real tangent space. Namely, it is the complex $2$-dimensional space spanned by $\partial_x$ and $\partial_y$. However, when working with complex tangent vectors, it is more common to use the two independent vectors $\partial_z$ and $\partial_{\overline{z}}$ as a basis. The translation formula is$$\partial_z=\frac{1}{2}(\partial_x-i\partial_y),\quad\partial_{\overline{z}}=\frac{1}{2}(\partial_x+i\partial_y),$$or alternatively,$$\partial_x=\partial_z+\partial_{\overline{z}},\quad\partial_y=i(\partial_z-\partial_{\overline{z}}).$$Hence, for the vector field in question we have$$X=-y\partial_x+x\partial_y=-y(\partial_z+\partial_{\overline{z}})+ix(\partial_z-\partial_{\overline{z}})=iz\partial_z+\overline{iz}\partial_{\overline{z}}.$$

However, there is another tangent space in the picture, and this is the holomorphic tangent space, which is the complex $1$-dimensional space spanned by $\partial_z$. It comes with a projection from the complex tangent space to the holomorphic tangent space, given by$$\partial_z\mapsto\partial_z,\quad\partial_{\overline{z}}\mapsto0.$$This projection, when restricted to the real tangent space, yields an isomorphism of real vector spaces from the real tangent space to the holomorphic one, given by$$\partial_x\mapsto\partial_z,\quad\partial_y\mapsto i\partial_z.$$It is customary to identify the real and holomorphic spaces via the above isomorphism. Doing so, our vector field is expressed as$$X=iz\partial_z.$$

As for trajectories, we should notice first that it is a different world to begin with. We are looking for a curve $c:\mathbb{R}\to\mathbb{C}$, and this involves no complex derivatives, since the domain of the curve is real, not complex. Hence, I'd say the answer to your second question is no. The way to find the trajectories is exactly as written in your post.

$\endgroup$
0
1
$\begingroup$

Comments to the question (v2):

  1. There seems to be a factor half missing from some of OP's formulas. Note that $$z~=~x+iy\quad\text{and}\quad\bar{z}~=~x-iy$$ yield via the chain rule $$ \partial_z~=~\frac{1}{2}\left(\partial_x-i\partial_y \right)\quad\text{and}\quad\partial_{\bar{z}}~=~\frac{1}{2}\left(\partial_x+i\partial_y \right).$$

  2. Note that in polar coordinates $z=re^{i\theta}$, the vector field becomes $$ X~=~-y\partial_x+x\partial_y~=~\partial_{\theta}.$$

  3. Using the chain rule, the vector field becomes $$ X~=~i\left(z\partial_z-\bar{z}\partial_{\bar{z}}\right). $$

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.