Notation for Sections of Vector Bundles

Question

(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?

Answer 1

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.

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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.

Answer 2

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.