[Physics] Question on connections in general relativity and particle physics

differential-geometryfield-theorygauge-theorygeneral-relativitygroup-theory

$1$ Introduction

It seems that when you learn General Relativity all the technology of bundles are irrelevant (at least in elementary discussions as $[1]$, $[2]$, $[3]$ and others). But, even the simplest field theory (electromagnetism) in a flat background like Minkowski spacetime, $\mathcal{L} = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu}$, do require a (curved) principal fiber bundle. Moreover, we can think in gluons $\mathcal{L} = -\frac{1}{4}G^{a}_{\mu\nu}G_{a}^{\mu\nu}$, with $a$ assuming values on the lie algebra, turning out the usage of fiber bundles even more necessary.But my question lies on the conceptual leap on the usage and definition of "connections in general relativity and particle physics".

$2$ Connections in Spacetimes

On in one hand, it seems that in general relativity we just need the knowledge of: the manifold $\mathcal{M}$ and the metric tensor $g$, $(\mathcal{M},g)$; the tangent spaces $T_{p}\mathcal{M}$, and the set of all tangent vectors $\mathfrak{X}(\mathcal{M})$ to settle down the formal concept of a (affine) connection:

$\nabla: \mathfrak{X}(\mathcal{M}) \times \mathfrak{X}(\mathcal{M}) \to \mathfrak{X}(\mathcal{M}),$

and then construct the "covariant derivative (of general relativity)":

Given a chart:
$$\nabla_{X}Y = [X^{\mu}\nabla_{\mu}Y^{\nu}]\frac{\partial}{\partial x^{\nu}} = [X^{\mu}(\partial_{\mu}Y^{\nu}+\Gamma ^{\nu}\hspace{0.1mm}_{\mu \delta}Y^{\delta})]\frac{\partial}{\partial x^{\nu}}\tag{1}$$

$2.1$ Connections on Principal Fiber Bundles

On the other hand, it seems that in particle physics you need to:

  1. Set your favorite principle fiber bundle: $\mathcal{P}= \big(\mathcal{E},\mathcal{M},\pi,G,G\big)$
  2. Construct the tangent principal fibre bundle $\mathcal{TP}$
  3. Divide the tangent bundle as $\mathcal{TP} = \mathcal{HP} \oplus \mathcal{VP} \implies T_{p}(\mathcal{P}) = H_{p}(\mathcal{P}) \oplus V_{p}(\mathcal{P})$
  4. Use the push-forward $(\Phi_{g})_{*}$ to "push" the horizontal subspaces $H_{p}(\mathcal{P})$ along the fiber: $(\Phi_{g})_{*}[H_{p}(\mathcal{P})] = H_{(\Phi_{g})p}(\mathcal{P}) := H_{q}(\mathcal{P})$ creating a "connection of subspaces" in the bundle.

after all that,

  1. Define (convince yourself/realize) the connection on $\mathcal{P}$ as a "horizontal distribution".

Equivalently:

  1. Set your favorite principle fiber bundle: $\mathcal{P}= \big(\mathcal{E},\mathcal{M},\pi,G,G\big)$
  2. Construct the tangent principal fibre bundle $\mathcal{TP}$
  3. Divide the tangent bundle as $\mathcal{TP} = \mathcal{HP} \oplus \mathcal{VP} \implies T_{p}(\mathcal{P}) = H_{p}(\mathcal{P}) \oplus V_{p}(\mathcal{P})$
  4. Construct the tangent spaces of $G$, $T_{q}G$
  5. Use the fact that $T_{q}G \approx \mathfrak{g}$.
  6. Use the $H_{p}(\mathcal{P})$ to project a $v \in T_{p}(\mathcal{P})$ in $V_{p}(\mathcal{P})$
  7. Use the fact that $V_{p}(\mathcal{P}) = T_{p}G$
  8. Realize, by $5)$, $6)$ and $7)$, the existence of a 1-form (or a $\mathfrak{g}$-valued differential form), $\omega_{p}: T_{p}\mathcal{P} \to \mathfrak{g}$, on $\mathcal{P}$.
  9. Take a particular tangent space in the identity $e$ of $G$: $T_{e}G$
  10. Use the fact $5)$, $T_{e}G \approx \mathfrak{g}$.
  11. Take the vector space of "all left-invariant vector fields" $\mathfrak{X}_{L}(G)$
  12. Take a $\xi \in \mathfrak{g}$
  13. Using $11)$ and $12)$ to mount a vector field $X_{\xi} \in \mathfrak{X}_{L}(G)$, such that $X_{\xi}(e) = \xi$ (to be a unique flow).
  14. Use the technology of flows in manifolds $[4]$, to describe a unique integral curve of $X_{\xi}$ passing at $t=0$ through $e \in G$: $g_{\xi}(t)$.
  15. Set the famous exponential map to define the fundamental vector field: $\textbf{A} =: \xi_{\mathcal{P}}(p) = \frac{\mathrm{d}}{\mathrm{d}t}(p\cdot \mathrm{exp}[t\xi])|_{t=0} = p\xi$
  16. Take a vector $v \in T_{p}(\mathcal{P})$, and then push (forward) it thorough the fiber $(\Phi_{g})_{*}|_{p}(v) = w$.
  17. Then take the defined 1-form of $8)$ and form the adjoint action: $\omega_{\Phi_{g}(p)}[w] = \mathrm{Ad}_{g} [\omega_{p}(v)] := \Phi_{g}^{-1}\omega_{p}(v)\Phi_{g}$

after all that,

  1. Define (convince yourself/realize) the (Ehresmann) connection on $\mathcal{P}$ as a $\mathfrak{g}$-valued 1-form $\omega_{p}: T_{p}\mathcal{P} \to \mathfrak{g}$ which satisfies, for each $p\in \mathcal{P}$,

$$\omega_{p}(\xi_{\mathcal{P}}(p)) = \xi \hspace{5mm} \forall \xi \in \mathfrak{g}$$

$$\mathrm{Ad}_{g} [\omega_{p}(v)] := \Phi_{g}^{-1}\omega_{p}(v)\Phi_{g} \hspace{5mm} \forall v\in T_{p}(\mathcal{P}), \forall g \in G $$

My Question

Well, my question is: why the texts books of general relativity don't define a connection in a principal fibre bundle? (or, why in general relativity we need "less math" to define a connection?).

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$[1]$ Carrol.S. Spacetime and Geometry

$[2]$ Weinberg. S. Gravitation and Cosmology

$[3]$ d'Inverno.R. Introducing Einstein's Relativity

$[4]$ Nakahara.M. Geometry, Topology and Physics

Best Answer

This is because its takes a substantial amount of mathematics to get there as you have just outlined. They need to learn about topology, smooth structures, manifolds, connections, bundles, frame bundles, principal bundles and so on.

Instead, they take the traditional physics language of components which has its advantages although eschewed by mathematicians.

This allows them to get to the physics quickly. It's also sanctified as the method used by Einstein himself.

Undoubtedly this will change in the future but not anytime soon. It's a question of physical pedagogy changing to take into account all the mathematical technology that allows us to think about physics in a more invariant manner. This will take time.

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