Gauge-Field Connections vs. Connections in General Relativity [Higher and More Elegant Electrodynamics · Extra 1]
HaifeiIn a previous article, we said that the electromagnetic field is a connection. This led many readers to think of the connection in general relativity. What are the differences and commonalities between these two kinds of connections?
The electromagnetic connection \(A\) is a connection on a principal bundle, whereas the general-relativistic connection \(\Gamma\) is a connection on a vector bundle. Their definitions appear to be quite different. Is there a way to relate them?
The answer is: yes! We can absolutely regard the connection in general relativity as a connection on a principal bundle as well. In this way, the two kinds of connections are unified as the same mathematical object.
In fact, \(\Gamma\) can be regarded as a connection on the frame bundle (a type of principal bundle). Then, the transformation rule for the Christoffel symbols
\[\boxed{ \begin{aligned} \bar{\Gamma}^i_{\mu j} = \frac{\partial \bar{x}^i}{\partial x^k}\frac{\partial x^l}{\partial \bar{x}^j} \frac{\partial x^\nu}{\partial \bar{x}^\mu} \Gamma^k_{\nu l} \color{red}{ + \frac{\partial \bar{x}^i}{\partial x^k} \frac{\partial^2 x^k}{\partial x^\mu \partial \bar{x}^j}} \end{aligned} }\]is nothing more than a gauge transformation:
\[\boxed{ \bar{A} = g^{-1}A g \color{red}{+ g^{-1} \mathrm{d}g} }\]The extra inhomogeneous term \( g^{-1} \mathrm{d}g\) is inherently part of a gauge transformation.
In addition, as mentioned previously, the curvature of a non-Abelian gauge field is \(\boxed{ F=\mathrm{d} A + A \wedge A }\). Where does this extra \(A \wedge A \) come from? After reading this article, you will know the answer as well.
I. Gauge Transformations
When performing a gauge transformation \(g\) on a principal bundle, the connection \(A\) transforms as follows:
\[A \mapsto g^{-1}A g \color{red}{+ g^{-1} \mathrm{d}g}\]As for where this formula comes from, we will explain it in Section V.
For example, the simplest gauge transformation is the gauge transformation \(U(1)\) on a \(g=\mathrm{e}^{\mathrm{i}\lambda}\)-fiber bundle:
\[\begin{aligned} A &\mapsto g^{-1}A g \color{red}{+ g^{-1} \mathrm{d}g} \\ &= \mathrm{e}^{-\mathrm{i}\lambda} A \mathrm{e}^{\mathrm{i}\lambda} \color{red}{+ \mathrm{e}^{-\mathrm{i}\lambda}\mathrm{d} \mathrm{e}^{\mathrm{i}\lambda}} \\ &= A \color{red}{+\mathrm{i}\ \mathrm{d}\lambda} \end{aligned}\]In physics, the imaginary unit \(\mathrm{i}\) is usually factored out of the connection \(A\). Thus, the above equation is written in physics as:
\[A \mapsto A \color{red}{+ \mathrm{d}\lambda}\]Let \(A^\sharp=(\phi,\mathbf{A})\) and adopt the \((+,-,-,-)\) metric; then the above equation becomes:
\[\boxed{ \begin{aligned} \phi &\mapsto \phi \color{red}{+ \partial_t \lambda} \\ \mathbf{A} &\mapsto \mathbf{A} \color{red}{- \nabla \lambda} \end{aligned} }\]This is the electromagnetic gauge transformation. Note that \(\nabla\) here denotes the gradient, not the covariant derivative.
The \(A\) above is a Lie-algebra-valued 1-form: \(A\in \Omega^1(M, \mathfrak{u}(1))\), meaning that it takes a tangent vector and returns an element of \(\mathfrak{u}(1) \cong \mathrm{i}\mathbb{R}\).
If \(\mathrm{i}\) is factored out, then it is no different from an ordinary real-valued 1-form.
This is why we regard the electromagnetic potential \(A\) as a vector field.
The same thing happens with the Christoffel symbols.
II. Connections on the Frame Bundle
Now, let us directly apply the electromagnetic gauge transformation to the case of the Christoffel symbols. The only difference is that we replace the \(U(1)\)-principal bundle with the frame bundle \(FM\), namely, the \(GL(n)\)-principal bundle.
A frame is defined as an ordered set of basis vectors \(f:=(e_1, e_2, ...,e_n)\). Transformations between frames are described by \(GL(n)\), and the group action of \(GL(n)\) is free and transitive; therefore, the frame bundle is a \(GL(n)\)-principal bundle.
The connection \(\Gamma\) at this point is a Lie-algebra-valued 1-form on a principal bundle:
\[\Gamma_{\text{global}} \in \Omega^1(FM, \mathfrak{gl}(n))\]Choosing a local frame \(s\), it can be locally trivialized as \(\Gamma = s^*\Gamma_{\text{global}} \in \Omega^1(M, \mathfrak{gl}(n))\).
Its value, as an element of the Lie algebra \(\mathfrak{gl}(n)\), can act on the frame \(f \) to yield \(\Gamma f \), representing an infinitesimal \(GL(n)\) transformation of the frame.
Note that the 1-form \(\Gamma\) here has not yet taken a vector as its argument. More rigorously, we should write the vector to be taken as a placeholder:
\[\Gamma(\cdot) f \]We can also write the vector \(X\) it takes as the subscript \(\Gamma_X\) of \(\Gamma\):
\[\Gamma_X f := \Gamma(X)f \]Written this way, it is very clear: \(\Gamma\) is a Lie-algebra-valued 1-form. Acting on the vector \(X\) gives the Lie-algebra element \(\Gamma_X \in \mathfrak{gl}(n)\), which can further act on \(f\) to give \(\Gamma_X f\).
Up to this point, there is no problem: \(\Gamma\) is a 1-form, that is, a tensor of type (0,1), except that its value after acting on a vector is a matrix in \(\mathfrak{gl}(n)\).
Then why do people always say that the Christoffel symbols \(\Gamma_{\mu i}^j\) are not tensors? Where does the problem lie? The problem arises when we write them in index notation, because we fail to distinguish between the action of \(\Gamma\) as a 1-form and the action of \(\Gamma\) as a Lie-algebra element.
As a 1-form, \(\Gamma\) can be expanded as:
\[\Gamma =\Gamma_\mu\mathrm{d}x^\mu\]where \(\Gamma_\mu\) is an element of the Lie algebra \(\mathfrak{gl}(n)\). Here, there is no issue in saying that \(\Gamma\) is a tensor of type (0,1). Moreover, the components \(\Gamma_\mu\) indeed transform according to the transformation rule for the components of a tensor field of type (0,1).
But what if we further expand \(\Gamma_\mu \in \mathfrak{gl}(n)\) in a local frame? Then we have
\[\Gamma_\mu = (\Gamma_\mu)_{j}^i\partial_i \otimes \mathrm{d}x^j = :\Gamma_{\mu j}^i \partial_i \otimes \mathrm{d}x^j\]If we now write \(\Gamma^i_j := (\Gamma_\mu)^i_j \mathrm{d}x^\mu\), then this \(\Gamma^i_j\) is no longer a Lie-algebra-valued 1-form, but a real-valued 1-form. This is equivalent to taking each matrix element \(\Gamma^i_j\) of the \(\mathfrak{gl}(n)\)-valued \(\Gamma\) as a 1-form.
This is where the problem lies: although \(\Gamma^i_j\) has upper and lower indices, it is itself a 1-form rather than a component of any tensor field. Furthermore, the definition of \(\Gamma^i_j\) as a 1-form depends on the choice of coordinate system (\(\partial_i \otimes \mathrm{d}x^j\)). This is why some people say that it is a coordinate-dependent tensor field.
III. Gauge Transformations on the Frame Bundle
Finally, starting from the gauge transformation
\[\bar{A} = g^{-1}A g \color{red}{+ g^{-1} \mathrm{d}g}\]we derive the transformation rule for the Christoffel symbols:
\[\begin{aligned} \bar{\Gamma}^i_{\mu j} = \frac{\partial \bar{x}^i}{\partial x^k}\frac{\partial x^l}{\partial \bar{x}^j} \frac{\partial x^\nu}{\partial \bar{x}^\mu} \Gamma^k_{\nu l} \color{red}{ + \frac{\partial \bar{x}^i}{\partial x^k} \frac{\partial^2 x^k}{\partial x^\mu \partial \bar{x}^j}} \end{aligned} \]Note that \(g\) here does not refer to the metric tensor, but to a gauge transformation.
First, let us replace \(A\) with \(\Gamma\):
\[\bar{\Gamma} = g^{-1} \Gamma g \color{red}{+ g^{-1} \mathrm{d}g}\]Then expand the 1-forms \(\bar{\Gamma}\) and \(\mathrm{d}g\) in a local frame as \(\bar{\Gamma}_\mu \mathrm{d}\bar{x}^\mu\) and \((\bar{\partial}_\nu g)\mathrm{d}\bar{x}^\nu\):
\[\begin{aligned} \bar{\Gamma}_\mu \mathrm{d}\bar{x}^\mu &\mapsto g^{-1} (\Gamma_\nu \mathrm{d}x^\nu) g \color{red}{+ g^{-1} (\bar{\partial}_\mu g)\mathrm{d}\bar{x}^\mu} \\ &= g^\nu_\mu \ g^{-1} \left(\Gamma_\nu \mathrm{d}\bar{x}^\mu \right) g \color{red}{+ g^{-1} (\bar{\partial}_\mu g)\mathrm{d}\bar{x}^\mu} \end{aligned}\]where \(\bar{\partial}_\mu :=\frac{\partial}{\partial \bar{x}^\mu}\), \(g^\nu_\mu = \frac{\partial x^\nu}{\partial \bar{x}^\mu}\).
Factoring out \(\mathrm{d}\bar{x}^\mu\) gives:
\[\begin{aligned} \bar{\Gamma}_\mu &= g^\nu_\mu \ g^{-1} \left(\Gamma_\nu \right) g \color{red}{+ g^{-1} (\bar{\partial}_\mu g)} \end{aligned}\]Then substitute the gauge transformation
\[\begin{aligned} g = g^i_j \partial_i \otimes \mathrm{d}x^j= \frac{\partial x^i}{\partial \bar{x}^j} \partial_i \otimes \mathrm{d}x^j \end{aligned}\]and take components to obtain:
\[\begin{aligned} (\bar{\Gamma}_\mu )^i_j &= g^\nu_\mu(g^{-1})^i_k \left(\Gamma_\nu \right)^k_l g^l_j \color{red}{+ (g^{-1})^i_k (\bar{\partial}_\mu g^k_l)} \end{aligned}\]Finally, substituting the components of the gauge transformation
\[\begin{aligned} g^i_j = \frac{\partial x^i}{\partial \bar{x}^j} ,\quad (g^{-1})^i_j = \frac{\partial \bar{x}^i}{\partial x^j} \end{aligned}\]gives:
\[\begin{aligned} (\bar{\Gamma}_\mu )^i_j &= \frac{\partial x^\nu}{\partial \bar{x}^\mu}\frac{\partial \bar{x}^i}{\partial x^k} \frac{\partial x^l}{\partial \bar{x}^j} \left(\Gamma_\nu \right)^k_l \color{red}{+ \frac{\partial \bar{x}^i}{\partial x^k} \frac{\partial^2 x^k}{\partial \bar{x}^\mu \partial \bar{x}^j}} \end{aligned}\]Let \(\Gamma_{\mu j}^i := (\Gamma_{\mu})^{i}_{j}\); then we obtain the transformation formula for the Christoffel symbols:
\[\boxed{ \begin{aligned} \bar{\Gamma}^i_{\mu j} = \frac{\partial \bar{x}^i}{\partial x^k}\frac{\partial x^l}{\partial \bar{x}^j} \frac{\partial x^\nu}{\partial \bar{x}^\mu} \Gamma^k_{\nu l} \color{red}{ + \frac{\partial \bar{x}^i}{\partial x^k} \frac{\partial^2 x^k}{\partial x^\mu \partial \bar{x}^j}} \end{aligned} }\]Is it necessary to memorize this formula? No; it is enough to remember \(\boxed{ \bar{A} = g^{-1}A g \color{red}{+ g^{-1} \mathrm{d}g} }\).
IV. Gravity
In particular, spacetime, the base manifold, is equipped with a metric. The frames are then orthonormal bases, so the structure group changes from \(GL(4)\) to the metric-preserving subgroup, namely the Lorentz group \(SO(1,3)\). Thus, the gravitational field can be regarded as the gauge field of the Lorentz group \(SO(1,3)\)[1], just as the electromagnetic field is a \(U(1)\) gauge field and the gluon field is a \(SU(3)\) gauge field, and so on.
V: Derivation of Gauge Transformations
So where does the gauge transformation formula \(\boxed{ \bar{A} = g^{-1}A g \color{red}{+ g^{-1} \mathrm{d}g} }\) come from? It can likewise be derived from a more fundamental relation.
The more fundamental relation referred to here is \(\boxed{ \mathrm{D} = \mathrm{d} + A \ \wedge }\), namely, the covariant exterior derivative equals the exterior derivative plus the connection.
On the frame bundle, \(\mathrm{D}\) is the covariant derivative \(\nabla\).
Under the gauge transformation \(g\), the field transforms according to \(\bar{\psi} = g^{-1} \psi\). We want the covariant exterior derivative \(\mathrm{D} \psi = \mathrm{d} \psi + A \wedge \psi\) of the field to transform in the same way, namely:
\(\mathrm{\bar{D}}\bar{\psi} = g^{-1}(\mathrm{D}\psi )\).
This is equivalent to requiring \(\boxed{ \bar{\mathrm{D}} = g^{-1} \mathrm{D} g }\).
Expanding the left-hand side of the above equation:
\[\begin{aligned} \mathrm{\bar{D}}\bar{\psi} &= (\mathrm{\mathrm{d}+\bar{A}}\ \wedge)(g^{-1}\psi) \\ &= \mathrm{d}(g^{-1}\psi) + \bar{A} \wedge (g^{-1}\psi) \\ &= (\mathrm{d} g^{-1})\psi + g^{-1} \mathrm{d}\psi + \bar{A}g^{-1} \wedge \psi \end{aligned}\]Expanding the right-hand side:
\[\begin{aligned} g^{-1}(\mathrm{D}\psi ) &= g^{-1} (\mathrm{d} + A\ \wedge )\psi\\ &= g^{-1}\mathrm{d} \psi + g^{-1} A \wedge \psi \end{aligned}\]Comparing both sides gives:
\[\bar{A}g^{-1} \wedge \psi = g^{-1} A \wedge \psi - (\mathrm{d} g^{-1} )\psi\]Factoring out \(\psi\):
\[ \bar{A} g^{-1}= g^{-1} A - \mathrm{d}g^{-1}\]Multiplying on the right by \(g\) and using \(\mathrm{d}(g^{-1}g)=0\) gives:
\[\boxed{ \bar{A} = g^{-1} A g + g^{-1} \mathrm{d} g }\]This yields the transformation formula for the connection under a gauge transformation.
VI. Curvature Tensor
The curvature \(F\) on a principal bundle is defined as the square of the covariant exterior derivative \(\mathrm{D}\), and is a 2-form:
\[\boxed{ F =\mathrm{D}^2 }\]Substituting \(\mathrm{D} = \mathrm{d} + A \ \wedge\) gives:
\[\begin{aligned} \mathrm{D}^2 \psi &= (\mathrm{d} + A\ \wedge)^2 \psi \\ &= \cancel{\mathrm{d}\mathrm{d}} + \mathrm{d} (A \wedge \psi) + A \wedge (\mathrm{d}\psi) + A \wedge A \wedge \psi \\ &=\mathrm{d} A \wedge \psi \cancel{- A \wedge \mathrm{d}\psi} \cancel{+A\wedge \mathrm{d}\psi} + A \wedge A \wedge \psi \\ &= (\mathrm{d} A+ A \wedge A)\wedge \psi \end{aligned}\]Therefore:
\[\boxed{ F=\mathrm{d} A + A \wedge A }\]Expanding \(F\) as \(F = \frac{1}{2}F_{\mu\nu} \mathrm{d}x^\mu \wedge \mathrm{d}x^\nu\) and \(A\) as \(A = A_{\mu} \mathrm{d}x^{\mu}\), then substituting into the above equation, gives:
\[F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu + [A_\mu , A_\nu]\]The derivation above applies to all principal bundles. Now consider the frame bundle as a principal bundle. Its fibers carry a free and transitive \(GL(n)\) action, and its Lie algebra is \(\mathfrak{gl}(n)\). Thus, both sides of the above equation are matrices in \(\mathfrak{gl}(n)\): \(F_{\mu\nu} = \{ (F_{\mu\nu})^i_j\}, \ \ A_{\mu}=\{(A_\mu)^i_j\}\).
Let \(R^i_{j\mu\nu} := (F_{\mu\nu})^i_j\), \(\Gamma^{i}_{\mu j}:= (A_\mu)^i_j\). Substituting into the above equation gives:
\[\boxed{ R^i_{j\mu\nu}= \partial_\mu \Gamma^i_{\nu j} - \partial_\nu \Gamma^i_{\mu j} + \Gamma^{i}_{\mu k} F^{k}_{\nu j} - \Gamma^{i}_{\nu k} F^{k}_{\mu j} }\]This is precisely the curvature tensor.
Wait—why is the connection \(\Gamma^i_{\mu j}\) not a component of a tensor field, whereas \(R^i_{j\mu\nu}\) is a component of a tensor field?
This is because, although the connection \(A\) is not gauge-invariant, the curvature \(F\) is gauge-invariant (or, rather, gauge-covariant). As can be seen from the following equation, the transformation law for curvature contains only a homogeneous term:
\[\begin{aligned} \mathrm{\bar{\nabla}} &= g^{-1} \mathrm{\nabla} g \\ \Rightarrow \bar{F} &= \bar{\nabla}^2 = (g^{-1} \nabla g)^2 = \underbrace{g^{-1} F g}_{\text{homogeneous}} \end{aligned}\]In contrast, the transformation formula for the connection has an inhomogeneous term:
\[A \mapsto \underbrace{g^{-1}A g}_{\text{homogeneous}} \color{red}{+ \underbrace{g^{-1} \mathrm{d}g}_{\text{inhomogeneous}}}\]Summary
This article introduced:
- How to regard the connection in general relativity (a connection on a vector bundle) as the connection of a gauge field (a connection on a principal bundle):
- How to regard the Riemann curvature tensor as the curvature of a non-Abelian gauge field:
References
- ^Utiyama, Ryoyu. “Invariant theoretical interpretation of interaction.” Physical Review 101.5 (1956): 1597. https://doi.org/10.1103%2FPhysRev.101.1597