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I would like to know more about Ehresmann connections in vector bundles and how they relate to the electromagnetic field and the electron in quantum mechanics.

Background: The Schrödinger equation for a free electron is

$$ \frac{(-i\hbar\nabla)^2}{2m} \psi = i\hbar\partial_t \psi $$

Now, to write down the Schrödinger equation for an electron in an electromagnetic field given by the vector potential $A=(c\phi,\mathbf{A})$, we simply replace the momentum and time operator with the following operators

$$\begin{array}{rcl} -i\hbar\nabla &\mapsto& D_i = -i\hbar\nabla + e\mathbf{A} \\ i\hbar\partial_t &\mapsto& D_0 = i\hbar\partial_t - e\phi \end{array}$$

I have heard that this represents a "covariant derivative", and I would like to know more about that.

My questions:

  1. (Delegated to Notation for Sections of Vector Bundles.)

  2. I have heard that a connection is a "Lie-algebra-valued one-form". How can I visualize that? Why does it take values in the Lie-algebra of $U(1)$?

  3. Since a connection is a one-form, how can I apply it to a section $\psi$? I mean, a one-form eats vectors, but I have a section here? What is $D_\mu \psi(x^\mu)$, is it a section, too?

I apologize for my apparent confusion, which is of course the reason for my questions.

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Note that separate questions should be asked separately in general, but your 3 questions actually seem like they could be 3 parts of one question, so it might be fine. Also, perhaps it's just me, but I'm not clear on exactly what your question 1 is getting at. If you could go into a bit more detail there, it might help you get a more precise answer. –  David Z Nov 14 '10 at 0:13
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P.S. No need to apologize for your confusion! These are tricky things to work through properly. –  David Z Nov 14 '10 at 0:14
    
Well, the thing is that my confusion is so great that I don't quite know how the right question to ask. :D I think I got a better idea of what I want with question 1 now. Hm, I should probably split them up, even. –  Greg Graviton Nov 14 '10 at 13:25

3 Answers 3

  1. Here's a short(-ish) answer. A vector bundle is a family of vector spaces over a manifold. The vector spaces can have bases. The manifold can have coordinates. The two concepts are not a priori related (now for the bundle of tangent spaces, a coordinate change happens to induce a change of basis; this fact often sows confusion). Once you pick a basis for your vector space, you do define a vector by its components -- but someone else may be describing the same vector in a different basis. To translate to physics: change of basis = gauge transformation. In the case of a charged particle, the wave function is the component of a one-vector section; in a new basis, this number changes by a non-zero complex number (which can vary from point to point). Again, the wave function is a section, and a section means one vector for each point in the manifold. How do we differentiate a function which takes values in different vector spaces over different points? We need a way of connecting the vector spaces. The connection does this; pragmatically, it is just a rule for doing this differentiation. Of course, differentiation will look different in different vector spaces, so the form of the connection will depend on the basis and change under gauge transformations (just as the form of a linear transformation changes under change of bases). That's what various messy formulas about how things "transform" are trying to tell you.
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Nice answer. And only now I realized that I completely forgot to mention vector bundles, so it's good it got mentioned. Just to make it a little bit more precise: vector bundle $V$ is not just a family of vector spaces (except for the trivial case of cartesian product) but rather a fiber bundle with vector space fiber. The point about $G$-structure $P$ is that it acts on this vector bundle: the fiber of $P$ (which is diffeomorphic to $G$) over the point $x$ acts on the fiber of $V$ over the same point. We then say that $V$ transforms in some representation of $G$ (e.g. scalar, spinor, etc.). –  Marek Nov 14 '10 at 3:17
    
First of all, thanks for answering! Indeed, $\psi(x)$ itself is not a section in the vector bundle, it's the image of a section in a local trivialization $\pi^{-1}(U) \cong U\times\mathbb{C}$. In a different local trivialization, i.e. after a gauge change, the image is $e^{i\chi(x)}\psi(x)$. So far so good. –  Greg Graviton Nov 14 '10 at 13:15
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Now, what I am wondering is this: for a vector field $\vec v$ (= a section of the tangent bundle), I can write $\vec v = v^\mu \frac{\partial}{\partial x^\mu}$. This expression is invariant because it explicitly mentions the coordinate system $x^\mu$. In contrast, "$\psi(x)$" is not an invariant notation for the section of a $U(1)$ vector bundle, because I didn't specify the local trivialization. Is there a notation to make it explicit as well, like for $\vec v$? (I'm asking because I have a hard time thinking in coordinates if I can't write them down in an invariant way.) –  Greg Graviton Nov 14 '10 at 13:16
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Hi all. The reason I gave my answer the way I did is that the precision, especially the notation and language, can be obfuscating. The questioner seemed most puzzled by the general structure of the story. That said, it is true that a variety of answers can certainly help a variety of readers. To Greg: if you really want, you can put an index on \psi indicating that it is a one-vector. –  Eric Zaslow Nov 14 '10 at 13:27
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@Greg: I updated my answer to address this question (hope it helps). @Eric: you are certainly correct and I would give your explanation as an introduction to subject to some complete newcomer (which doesn't mean it's bad; on the contrary, I like it very much). But it seemed to me that Greg already knows about these things a little and is also conversant in the math of differential geometry, so I decided to make my answer a little bit more technical (I know I would have liked to see this mathematical point of view when I was learning this stuff myself). –  Marek Nov 14 '10 at 14:13
  1. Not really sure about this question. Are you perhaps just asking about notation? You can choose whatever you like. But usually you choose one coordinate system and just work in that. It's certainly a non-issue if you are working in flat spacetime: there you have nice global coordinates for everything.

    • First sentence is correct. For arbitrary Lie group $G$ with Lie algebra $\mathfrak{g}$ you might get a so called G-structure (e.g. $O(n)$-structure for Riemannian manifolds) and you can define a connection on that. It's rather heavy-duty mathematics, but at the end of the day you obtain a $\mathfrak{g} $-valued (more precisely, it takes values in the adjoint representation of $\mathfrak{g} $) one-form $ \omega $ and covariant derivative $D_{\mu}$ such as the one you've written.

    • In general I visualize connections like this: you put in a vector and the connection gives you back an element of the Lie algebra that is a generator of some transformation in the Lie group. E.g. on a Riemannian manifold you insert into the connection the direction you want to go to and you'll get back (very roughly said) information on how much the space is curved in that direction. More preciselly, if you integrate $\oint_{\gamma} \exp(\omega(\dot{\gamma}(t))) dt$ you obtain some element $T \in G$ that tells you how any vector gets parallel-transported along the closed curve $\gamma$ (this is called holonomy).

    • As for the last question of why $ U(1) $: well, because it's electromagnetism. Various groups will give you various interactions (e.g. $SU(3)$ gives you QCD). These groups arise because the theories contain something called gauge symmetry. I am sure you know that Maxwell equations, when written out in $A$ and $\phi$, are invariant to certain transformations. More concretely and clearly, as ${\bf F} = d{\bf A}$, where ${\bf F}$ is the electromagnetic tensor and ${\bf A} $ is the four-potential. In this form it's obvious that equations for $ {\bf F}$ don't change upon the transformation ${\bf A'} = {\bf A} + d\chi$. Now, let us step back to $\psi$ and note that if you want to make the theory locally invariant (why would you do that? because it's nice to have local properties instead of global ones. And the theory is clearly invariant with respect to global phase change, so let's just try this) with respect to the change of phase, you'll have to introduce new degree of freedom ${\bf A}$ that transforms precisely like the four-potential in Maxwell theory! Actually, what we just did is that we recovered photons. So, to conclude: if you choose $ U(1)$ as a group of symmetries, electromagnetism pops out. But note that to incorporate all of this consistently, you need to work in the framework of quantum field theory because Schrödinger's equation clearly isn't relativistically invariant, which is a feature we'd surely like to have in the theory of electromagnetism.

  2. This can be a little confusing, especially in the physical literature and I doubt I can really make it clearer. Probably this will just confuse you even more but here goes: there exist various spaces of sections and $ G$-structures and representations of both $ G $ and $ \mathfrak{g}$ that one should really discern between but which are identified in physics. To make this rigorous would take too much space, so I suggest you look at some books on gauge theory. I'll just tell you that vector (more precisely vector field) is also a section and vice-versa. But note that another two terms are getting mixed together here: abstract vector as a concept from linear algebra (this is our $ \psi $: you know it's this kind of vector because it has no spacetime index $\mu $) and vector as an element of Minkowski spacetime, say $V^{\mu}$. As for the $ D_{\mu} $, it acts (by extension) also on tensor algebra of the space of sections $ \psi $ (in this simple case where $\psi $ has no index, the tensor algebra is isomorphic to normal tangent tensor algebra) and gives you a one-form living in this tensor algebra, so yeah: it is a section, but in a totally different (altough isomorphic) space!

No need to apologize. These things are pretty hard and it takes a lot of time to grok all of it and I guess lot of the physicists would just hand-wave most of the mathematical content away and go ahead to calculate something. Which is possible here because $ U(1) $ is only one-dimensional, its Lie algebra is one-dimensional, and so is $ \psi $. So you just work with numbers all the time and there arises no need for abstract concepts. Except that this doesn't help you one bit when you'll try to generalize this (either to QCD with non-abelian group $ SU(3) $, or to the curved spacetime), or even if you try to think about some concepts that are rather basic from the point of view of geometry (e.g. try thinking about what is the meaning of the above-mentioned holonomy with respect to $\gamma $ in the case of $ U(1 ) $). So it's good you're trying to understand the core principles already.


Update in regard to Greg's question in the comments:

I am not sure I understand you completely but I got a feeling (probably wrong) that you are mixing up various notions of invariance here. There are at least two notions of physical invariance (under the action of Lorentz group and under the action of $U(1)$) and also a notion of invariance in the sense of being "coordinate-free". Now, if my feeling is correct, you are asking for an analogue of $v = v^{\mu}{{\rm \partial} \over {\rm \partial} x^{\mu}}$ for $\psi$. These two things are indeed very similar, but in a little disguised way. More precisely: $v$ 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 $G$-structure here and this is where "coordinate-free" invariance comes from. The situation is 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 $FM$.

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Just to nitpick: a connection is specified by a one-form with values in the Lie algebra (not the adjoint bundle); but this is a one-form on the associated principal bundle. you only get a form description on the base manifold locally. You might be confused with the curvature which can be seen as a form on the base manifold with values in the adjoint bundle. –  Eric Dec 12 '10 at 21:34
    
@Eric: thank you for nitpicking! What I meant is that it is a one-form in the adjoint representation of the Lie algebra (right?). But true, this part is quite confusing. I'll think about it and try to rewrite it. –  Marek Dec 13 '10 at 9:58
    
@Eric: as for the curvature: AFAIK locally curvature is a two-form with values in adjoint rep. of Lie algebra. So in this regard it's the same as the connection. Both can be represented also on adjoint bundle. Or is this not so? –  Marek Dec 13 '10 at 10:00
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Well, if you're working on a vector bundle then the connection forms are really only defined relative to a local frame. If you patch them together to get something global then it doesn't take values in any nice bundle. However, the curvature form is a global two form with values in the adjoint rep. But if you go from a vector bundle to its associated principal bundle, then the connection form is a global 1 form on the principal bundle with values in the Lie algebra (not ad rep) and the curvature form is a 2 form on the principal bundle with values in the Lie algebra. –  Eric Dec 13 '10 at 13:43
    
@Eric: I see, thanks for the clarification! I definitely need to think more about these things (I'd certainly need one purely mathematical lecture in fiber bundles and principal bundles; right now I am patching my mathematical knowledge from the bits I learned in physical literature) and will update my answer then. –  Marek Dec 13 '10 at 15:14

I can fully understand your confusion since it is natural that you feel overwhelmed by this new viewpoint on the theory.

The answers given by Eric and Marek are just fine and I will not directly talk about principal bundles, local trivialization and the like. I want to present a very intuitive approach here.

I would suggest that you go one or two steps back and try to understand the notion of a covariant derivative in classical differential geometry. There, the covariant derivative $D$ assures that if you derive some quantity $F$ on a manifold, say some surface, this new quantity $DF$ will also lie "on the manifold" (actually, something related to it like the tangent space).

The following example will hopefully illustrate the issue what it means that something has to "stay on the manifold".

Mass-point on a surface

Ok, lets do the most simple example one could think of, the motion of a free mass point on a surface in Newtonian mechanics. As you know, the Lagrangian in this case is just the kinetic energy,

$$L = T = \frac{m}{2}\mathbf{v}^2$$

So, what is now $\mathbf{v}^2$? We have to assume that in every point, the velocity is tangential to the surface. Then, we know that $\mathbf{v}^2 = g_{ab}\dot{x}^a\dot{x}^b$ where we sum over the indices and the surface is described by some metric $g_{ab}(x)dx^adx^b$ and we have replaced the velocity by the time derivative of the position of the particle.

For the solution of the system we will need two terms. First of all, we want to calculate

$$\frac{\partial{L}}{\partial{x^k}} = \frac{m}{2}\partial_k{g_{ab}(x)}\dot{x}^a\dot{x}^b$$

second,

$$\frac{d}{dt}\frac{\partial{L}}{\partial{\dot{x}^k}} = \frac{d}{dt}\frac{m}{2}g_{ab}(x) \left( \delta^a_k\dot{x}^b + \dot{x}^a\delta^b_k \right) = m\frac{d}{dt}g_{kb}(x)\dot{x}^b$$

since $g$ is symmetric. Now,

$$m\frac{d}{dt}g_{kb}(x)\dot{x}^b = m\left( \partial_lg_{kb}(x)\dot{x}^l\dot{x}^b + g_{kb}(x)\ddot{x}^b\right) $$

Finally, the equations of motion are given by

$$\frac{d}{dt}\frac{\partial{L}}{\partial{\dot{x}^k}} - \frac{\partial{L}}{\partial{x^k}} = 0$$

and dropping m, re-using the symmetry in $g$ and renaming some indices, we arrive at

$$g_{kb}\ddot{x}^b+\frac{1}{2}\left( \partial_a g_{kb} + \partial_b g_{ka} - \partial_k g_{ab} \right)\dot{x}^a\dot{x}^b = 0$$

which is exactly by applying $g^{ik}$

$$\ddot{x}^i + \Gamma^i_{ab}\dot{x}^a\dot{x}^b = 0$$

Where the Christoffel symbols can directly seen as

$$\Gamma^i_{ab} = \frac{1}{2}g^{ik}\left( \partial_a g_{kb} + \partial_b g_{ka} - \partial_k g_{ab} \right)$$

and I hope I did not miscalculate anything.

Relation to electrodynamics

Now this equation of motion is already magic since it is precisely the equation of motion for a testparticle in general relativity. But what about (other) gauge field theories?
Here, the curvature is not defined with respect to the manifold directly but to a group somehow "attached" to it. That's why it will have some group indices but one can drop this if the Lie Algebra of the group is one dimensional as in the case of electrodynamics. There, our curvature is $F_{\mu\nu}$ but we could also state $F^a_{\mu b\nu}$ where now $a$ and $b$ are indices of the group. This looks much more familliar to the curvature of general relativity, $R^{\mu}_{\nu\alpha\beta}$ where all indices correspond to the manifold, in some sense, the tangent space is the group of general relativity, roughly speaking.

For historical reasons, the Christoffel symbols that somehow catch (not invariantly!) the, force, acting on the particle because of the curvature are in gauge theories called gauge fields $A$ and rescaled,

$$\beta A^a_{\mu b}dx^{\mu}" = "\Gamma^a_{\mu b}dx^\mu $$

with again group indices $a$ and $b$.

Now if you derive something on the manifold, you will always have to define that derivative with respect to the Christoffel Symbols to stay "in the manifold". On the other hand, the derivation will also have to respect the group character, one could say the result has to stay "in the group". This will be realized by a covariant derivative and here the "Christoffel symbols" are called gauge fields.

Sincerely

Robert

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@Marek: Thanks for the LaTex edit. I will also have to edit formulas for some typos :) –  Robert Filter Dec 15 '10 at 12:09
    
I added formatting to your equations. By the way, just a minor terminology question: is $F^a_{\mu b \nu}$ standard? I am more used to (and also think it is more natural) to put tensor indices together and Lie algebra indices together to emphasize that ${\mathbf F}$ is actually a two-form with values in a Lie algebra. As in ${{F_{\mu \nu}}^a}_b$. –  Marek Dec 15 '10 at 12:09
    
@Marek: You are perfectly right, I just wanted to make the correlation to the general relativistic curvature obvious. The answer is also not to be seen to compete with yours or Eric's, its just another approach to familiarize the term covariant derivative from a different perspective. –  Robert Filter Dec 15 '10 at 12:18
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@Marek: Thank you :) Lagrangian mechanics is the most beautiful theory of all to me. I always try to grasp an idea of new things from its perspective. –  Robert Filter Dec 15 '10 at 12:34
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Your example is just as well thought of as Hamiltonian mechanics, and is part of a general (and simple!) story. On the cotangent bundle of a Riemannian manifold (equivalently, the tangent bundle -- the two are the same when you have a metric), you can take the Hamiltonian to be the energy, or length-squared/2 of the velocity vector, as you have done. Then, Hamiltonian flow equals geodesic flow. This is what you have demonstrated in your example. –  Eric Zaslow Dec 17 '10 at 16:53

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