Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

(Reformulation of part 1 of Electromagnetic Field as a Connection in a Vector Bundle)

I am looking for a good notation for sections of vector bundles that is both invariant and references bundle coordinates. Is there a standard notation for this?

Background:

In quantum mechanics, the wave function $\psi(x,t)$ of an electron is usually introduced as a function $\psi : M \to \mathbb{C}$ where $M$ is the space-time, usually $M=\mathbb{R}^3\times\mathbb{R}$.

However, when modeling the electron in an electromagnetic field, it is best to think of $\psi(x,t)$ as a section in a $U(1)$-vector bundle $\pi : P \to M$. Actually, $\psi(x,t)$ *itself* is not a section, it's just the image of a section in one particular local trivialization $\pi^{-1}(U) \cong U\times\mathbb{C}$ of the vector bundle. In a different local trivialization (= a different gauge), the image will be $e^{i\chi(x,t)}\psi(x,t)$ with a different phase factor.

Unfortunately, I feel uncomfortable with this notation. Namely, I would prefer an invariant notation, like for the tangent bundle. For a section $\vec v$ of the tangent bundle (= a vector field), I can write $\vec v = v^\mu \frac{\partial}{\partial x^\mu}$. This expression mentions the coordinates $v^\mu$ in a particular coordinate system, but it is also invariant, because I also write down the basis vector $\frac{\partial}{\partial x^\mu}$ of the coordinate system.

The great benefit of the vector notation is that it automatically deals with coordinate changes: $\frac{\partial}{\partial x^\mu} = \frac{\partial}{\partial y^\nu}\frac{\partial y^\nu}{\partial x^\mu}$.

My question:

Is there a notation for sections of vector bundles that is similar to the notation $\vec v = v^\mu \frac{\partial}{\partial x^\mu}$ for the tangent bundle? What does it look like for our particular example $\psi$?

If no, what are the usual/standard notations for this? How do they keep track of the bundle coordinates?

share|improve this question
add comment

2 Answers 2

up vote 2 down vote accepted

Edit: I realized that what I've written wasn't really correct, so let me change the text a little. I'll mark the additions by italics, so that the old text stays as a reference.


I gave a (partial) answer to this in the update of my answer to your previous question so let me copy&paste that answer (with some modifications):

The answer to the first question is no but for a different reason than I stated in the above link and in the text below!. There is no such notation and to see why, we first have to understand where "coordinate-free" vector invariance comes from. The $v$ in you question is a section of a tangent bundle $TM$ and we are decomposing it with respect to some section of the canonical tangent frame bundle $FM$, which also carries a natural action of the group $GL_k({\mathbb{R}})$ (the action is a local change of basis and $k$ is the rank of $TM$). In other words, we have a $GL_k({\mathbb{R}})$-structure here.

The situation is superficially similar with $\psi$: it is a section of a vector bundle $\pi: V \to M$ which carries an $U(1)$-structure. At this point it should also be clear where the difference between the two cases is: in the former you have two bundles $TM$ and $FM$ while in the latter there is only $\pi: V \to M$. So it doesn't really make sense to ask for $\psi$ to be any more invariant than it already is: you have nothing with respect to which you could decompose it. So instead of thinking about $\psi$ as an analogue of section of $TM$, think of it instead as an analogue of a section of $FM$.

I made a mistake in the above reasoning because in the case of one-dimensional $U(1)$ the concepts of $V$ and $F(V)$ (associated frame bundle) coincide. So you also have two bundles in the second case. But the difference comes from the fact that $TM$ is a very special vector bundle: it's structure comes from the manifold $M$ itself, whereas $V$ is an extrinsic structure. So you certainly cannot get a decomposition with respect to coordinate derivates on $M$ as is the case of $TM$.

As for the second question: in gauge theories one usually fixes gauge before-hand (think of Lorenz or Coulomb gauge) and work in that forever. You don't really get anything interesting here by working in some "gauge-free" way (or at least I don't know about it). So, these things really aren't an issue at least until the point when you come across QFT and start wondering how to account for all this huge gauge freedom. And there it actually is a big problem that must be dealt with and it can be dealt with in various ways (including gauge-fixing). But none of this is relevant for you at this point, I guess.

share|improve this answer
    
Thanks! Pondering you answer, I think I get it now. Obviously, I cannot expect that a coordinate change $x^\mu \to y^\mu$ in the base manifold $M$ will induce a change on $\psi$; after all, the coordinates in the fibers are "new" or "independent" of the base manifold. (In this sense, the tangent bundle "is not a proper" vector bundle; the only coordinate changes allowed on the fibers are differentials $Df$ of maps $f: M \to M$). –  Greg Graviton Nov 14 '10 at 20:07
    
For the "extra" coordinates on the fibers, I can always write $\psi = \sum_\alpha\psi_\alpha s_\alpha$ where the $s_\alpha$ are a bunch of linearly independent sections. This is coordinate independent, but it doesn't really buy me anything. [However, I did read the notation $\psi((x,t),g)$ once, where $g\in U(1)$ was an element of the structure group. The intention was probably to carry around the gauge $g$ at the point $(x,t)$, in the spirit of $\psi((x,t),g)=g\psi((x,t),1)$. That's what prompted my question here, but somehow this notation seems bogus to me.] –  Greg Graviton Nov 14 '10 at 20:13
1  
@Greg: Well, I'd say that $TM$ is a special vector bundle rather than "not proper". But I agree that it can cause great confusion. Especially in general relativity, where the connection (very roughly said) works on $TM$ itself (so $TM$ plays both the role of a tangent bundle and of the field bundle). As to the $s_{\alpha}$ decomposition: actually it's not quite true that it doesn't buy you anything. It gives you the same thing as tetrad formalism in general relativity, which is often much easier to work with than coordinates. –  Marek Nov 14 '10 at 20:36
add comment

Sorry I just noticed your comment on a previous question where you stated you would ask a separate question. Here is a response. (Sorry for any overlap with marek's.)

Just as in order to talk about vectors in an n-dimensional vector space as n-tuples of numbers, you need to first choose basis, in order to talk about sections of vector bundles in a concrete way, you must choose a "frame" $\{e_a\}$ (which is just a fancy way of saying a family of basis vectors/sections). Then the notation is exactly as before, a section $s$ looks like $s^a e_a$ (summed over $a$).

In your example of a wave function for a charged particle, the vector bundle is one (complex) dimensional, so the single basis vector $e$ is usually dropped from the notation. But it should be there, morally, as you note.

A gauge transformation by $\chi$ amounts to changing your basis vector $e$ by multipliying it by $\exp(-i\chi)$, so the (invariant) vector $\Psi = \psi e$ has coordinate $\psi \exp(i\chi)$ in the new basis. The gauge field $A_\mu$ is a 1x1 matrix, and has to be changed by adding $\exp(i\chi)\partial_\mu \exp(-i\chi) = -i\partial_\mu \chi$. That way $D_\mu \Psi = \partial_\mu \psi + iqA_\mu \psi$ makes invariant sense, where $q$ is the charge (or representation, telling how $A$ acts on $\psi$ -- i.e. by multiplying with $q$ in front).

Hope that helped.

share|improve this answer
    
Ah, ok. So, I have to choose a (local) frame first, and then I can write down coordinates for any section. (In the case of the tangent bundle, there are natural choices of frames that arise from the coordinates on the manifold, but not the case for general vector bundles.) Coordinate/frame changes are described by the structure group (here $U(1)$). –  Greg Graviton Dec 16 '10 at 10:51
    
I think my original question was this: why is $A_\mu$ an element of the Lie algebra? I.e., why does the formula $D_\mu\Psi = \partial_\mu \psi + iqA_\mu \psi$ work? I mean, you can calculate that it's gauge invariant, but I dislike the partial derivatives $\partial_\mu$, they are no invariant geometric objects. What I would like to see is a geometrically meaningful object from which the representation $D_\mu = ...$ in local coordinates can be derived. This is probably be the mysterious one-form on the principal bundle? –  Greg Graviton Dec 16 '10 at 10:58
    
Your first comment: yes. Your second comment: the Lie algebra is Lie(Aut(G)) = Lie(G), where Aut(G) is the automorphisms of G with its transitive right G action. This is geometric. The derivative is the differential of an automorphism, so its an endomorphism. –  Eric Zaslow Dec 16 '10 at 19:57
    
Yes, there are several cleaner mathematical formulations of covariant derivatives (connections) on vector bundles, but they are all equivalent. You might consult the first chapter of Berline-Geztler-Vergne. Here's a quick take: a connection is the data of which directions (in the tangent space of the whole vector bundle) are horizontal and which are vertical. Parallel transport is then lifting a curve to a horizontal curve starting at your vector. The covariant derivative is the vertical part of the derivative of that lifted curve. (You can encode this as a one-form on the total space.) –  Eric Zaslow Dec 16 '10 at 20:05
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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