Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

146 submissions , 123 unreviewed
3,953 questions , 1,403 unanswered
4,889 answers , 20,762 comments
1,470 users with positive rep
507 active unimported users
More ...

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

+ 2 like - 0 dislike
417 views

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 in Mathematics by Danu (175 points) [ no revision ]

1 Answer

+ 2 like - 0 dislike

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 doetoe (110 points) [ no revision ]

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ysicsOverfl$\varnothing$w
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
To avoid this verification in future, please log in or register.




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...