In trying to understand precisely how fiber bundle theory maps to physical models, I came across this quotation:

We can think of the elements of the principal bundle as generalized frames for the original fiber bundle. This means they correspond with different ways that we can convert the intrinsic dynamics described abstractly by a section of the fiber bundle to something concrete that we observe.

These generalized frames are often called 'gauges'. The structure group of the principal bundle is called the 'gauge group' and the automorphism of the principal bundle that fixes the base is called a 'gauge transformation.'

In the case of a vector bundle, this means that, by choosing a gauge, or an orthonormal frame for the fiber, we obtain a set of numbers: the coordinates of the section with respect to the frame.

So please tell me if I understand this correctly: we have two logically distinct fiber bundles here: the principal bundle and the associated vector bundle. The associated bundle is the 'matter field' whose sections represent the observable quantities, e.g. phase, and the principal bundle is the bundle of 'generalized frames' whose sections represent the bases we use to describe the sections of the associated bundle numerically?

Is this correct?

Sorry if this is vague. I am trying to get an intuitive 'handle' on how the technical math of fiber bundles maps to what I know from physics.

  • 1
    $\begingroup$ Could you provide a source for that quotation? $\endgroup$ – tparker Nov 3 '18 at 22:13
  1. The Principle fiber bundle can be thought of as an expansion of spacetime: Given a gauge group $G$ and a principal $G$ bundle $\pi: P \to M$ over spacetime $M$, we locally get $$\pi^{-1}(U) \cong U \times G, \ U \subset M$$ since a PFB is just a fiber bundle with fiber $G$. In the case of $U(1)$, one can think of the PFB as a way to keep track of the phase a given particle has at a given point in spacetime, more generally, one can think of it as a way to explicitly encode intrinsic properties of a particle (there are other examples, but let's omit them for the moment).
  2. A representation $\rho: G \to V$ on a vector space $V$ should be thought of as the ''transformation rule'' i.e how a field transforms under a gauge transformation. Examples include rotations in isopin space (even though they are not a ''real'' gauge theory) as well as the well known $U(1)$ gauge transformation, see also this answer of mine.
  3. Given a representation $\rho$ as above, one can form the associated vector bundle $$E = P \times_{\rho} V,$$ which in the principle bundle formalism is used instead of the vector space $V$. More precisely, take any (classical) field, for example the dirac field, which is a smooth function $$ \psi:\mathbb{R}^{1,3} \to \mathbb{C}^4$$ satisfying the Dirac Equation. In our new formalism, the fields would be smooth sections $$\Psi: M \to E.$$ Matter fields would then be thought of as those sections $\Psi$ and not as the bundle.
  4. Choosing a gauge would more commonly be thought of as choosing a local section for $P$. The locality is to be emphasized as it's, in general, not possible to choose global sections. By means of a local section, ''everything'' can be pulled back to $U \subset M$ which then gives gauge invariant (and, in general, ''curved'') differential equations, for example Dirac Equation in curved spacetime.(Which would, however, need the notion of a connection form, which we should ignore for the moment). This would concretly be done as follows: Given a ''matter field''

    $$\Psi:M \to E$$ and a (local) section $$s: U \to P$$ one can (locally) write $$\Psi(x) = [s(x), \psi'(x)]$$ (by the definition of $E$). Now, the section $s$ does define a family of isomorphisms $([s(x)])_{x \in U}$ which, for a fixed $x_0 \in U$, is given by $$\quad [s(x_0)]:V \to E_{x_0}, \quad v \mapsto [s(x_0),v]$$ where I denoted the fibre of $E$ over $x_0$ by $E_{x_0}$. Using this family of isomorphisms, we get a function $$\psi': U \to V, \ x \mapsto \psi'(x)$$ which then can be used for calculations.


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.