Is the equation $[\nabla_{\mu},\nabla_{\nu}]=F_{\mu\nu}$ correct? If yes, how does it have to be interpreted?

Question

It seems like simply using the equation \begin{equation} \nabla_{\mu}=\partial_{\mu}+A_{\mu} \end{equation} isn't enough: One obtains \begin{equation} [\nabla_{\mu},\nabla_{\nu}]=\underbrace{[\partial_{\mu},\partial_{\nu}]}_{=0}+\underbrace{\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}+[A_{\mu},A_{\nu}]}_{=F_{\mu\nu}}+A_{\mu}\partial_{\nu}-A_{\nu}\partial_{\mu} \end{equation} and I don't see why $A_{\mu}\partial_{\nu}-A_{\nu}\partial_{\mu}=0$.

Thus, it seems like this naive approach doesn't work and we need to be more rigorous:

Consider the following setting (please see the section "Notation" for more details): A principal $G$-bundle $P\to M$ and a representation $\rho\colon G\to\mathrm{GL}(V)$, $\rho_{*}\colon g\to\mathrm{End}(V)$. Let $E\to M$ be the associated vector bundle, $A\in\Omega^1(P,g)$ a connection $1$-form and $\nabla\colon \Gamma(M,E)\to\Omega^1(M,E)$ the induced covariant derivative.

Obviously, the equation \begin{equation} \mathrm{End}(C^{\infty}(U,V))\ni[\nabla_{\mu},\nabla_{\nu}]=F_{\mu\nu}:=(s^*F)(\partial_{\mu},\partial_{\nu})\in C^{\infty}(U,g) \end{equation} doesn't make sense, as two totally different objects are equated. My guess would be that in this equation, $F_{\mu\nu}$ has to be interpreted as the linear operator $\widetilde{F_{\mu\nu}}\colon C^{\infty}(U,V)\to C^{\infty}(U,V)$ defined by \begin{equation} (\widetilde{F_{\mu\nu}}(\phi))(m)=(\rho_{*}\circ F_{\mu\nu})(m)\phi(m)\in V. \end{equation} What do you think?

Answer 1

Not that much notation, but probably not less true either: $$[\partial_\mu+A_\mu,\partial_\nu+A_\nu]\psi=(\partial_\mu+A_\mu)(\partial_\nu+A_\nu)\psi-(\partial_\nu+A_\nu)(\partial_\mu+A_\mu)\psi=$$ $$=\partial_\mu\partial_\nu\psi+A_\mu\partial_\nu\psi+\partial_\mu (A_\nu\psi)+A_\mu A_\nu\psi-\partial_\nu\partial_\mu\psi-A_\nu\partial_\mu\psi-\partial_\nu (A_\mu\psi)-A_\nu A_\mu\psi=$$ $$=A_\mu\partial_\nu\psi+(\partial_\mu A_\nu)\psi+ A_\nu\partial_\mu\psi-A_\nu\partial_\mu\psi-(\partial_\nu A_\mu)\psi- A_\mu\partial_\nu\psi=$$ $$=(\partial_\mu A_\nu-\partial_\nu A_\mu)\psi=F_{\mu\nu}\psi$$ and therefore $$[\partial_\mu+A_\mu,\partial_\nu+A_\nu]=F_{\mu\nu}$$ The interpretation is that if you parallel-transport the field $\psi$ along a closed loop in the electromagnetic 4-potential/connection $A_\mu$, then $\psi$ is not the same, but differs by a phase factor (more generally, a gauge group element, for SU(2) or SU(3) gauge theories) that depends on the electromagnetic field tensor, which can be interpreted as a kind of curvature (just like the Einstein-tensor is a measure of curvature due to gravitation).

In lattice gauge theory, these are called Wilson loops.

I guess, someone else will find a more bundle-ish version of this explanation.

PS: I have silently assumed you are talking about electromagnetism. For other gauge theories the derivation is almost the same, only that the commutator of $A_\mu$ with $A_\nu$ does not vanish (and consequently enters into the field tensor) like it does for EM.

Answer 2

The field strength $F$ is $\mathfrak{g}$-valued. In that case if $X^a$ is a basis of the Lie algebra we can write $$F=\dfrac{1}{2}F_{\mu\nu}^a dx^\mu\wedge dx^\nu \otimes X^a.$$

In particular, given any representation $R:G\to {\rm GL}(V)$ of $G$ on the vector space $V$, we have the derived representation $dR:\mathfrak{g}\to {\rm End}(V)$ of the Lie algebra, and therefore we have the representative of $F$ in this representation.

Denoting $dR(X^a)=T^a$ the generators of the representation $R$ we have that $F$ is represented by $$F_R=\dfrac{1}{2}F_{\mu\nu}^a dx^\mu\wedge dx^\nu \otimes T^a.$$

In particular $(F_R)_{\mu\nu}:= (F_R)_{\mu\nu}^a T^a$ is a linear operator on $V$ and can act on any $V$-valued object.

When we write $[\nabla_\mu,\nabla_\nu]=F_{\mu\nu}$ what we mean by $F_{\mu\nu}$ is really $(F_R)_{\mu\nu}$ because there one representation being understood there. After all the covariant derivative is induced from the principal connection on each associated vector bundle, which are constructed from representations of $G$! The thing is that people abuse notation and left the representation be understood implicitly.

In that case $[\nabla_\mu,\nabla_\nu]$ is a map that can act on sections of the associated bundle and so is $F_{\mu\nu}$ by the reasons I have outlined above. You are not equating objects of different nature here.

Answer 3

Here's what should be the mathematically rigorous statement and proof:

Let $\rho_*\colon g\to\mathrm{End}(V)$ be the Lie algebra homomorphism induced by the Lie group homomorphism $\rho\colon G\to\mathrm{GL}(V)$. Consider the right action \begin{equation*} \mathrm{End}(V)\times V\ni(A,v)\mapsto A\cdot v:=A(v)\in V. \end{equation*} If $A\in C^{\infty}(U,g)$, \begin{align} \rho_*A\colon C^{\infty}(U,V)&\to C^{\infty}(U,V)\\ \phi&\mapsto(\rho_*\circ A)\cdot\phi \end{align} is $C^{\infty}(U)$-linear. $\rho_*(A+B)=\rho_*A+\rho_*B$ and $[\rho_*A,\rho_*B]=\rho_*[A,B]$.

If $x\colon U\to\mathbf{R}^n$ is a chart, each vector field \begin{equation*} \partial_\mu=\frac{\partial}{\partial x^\mu}\in\Gamma(U,TM) \end{equation*} induces an endomorphism \begin{align*} \partial_{\mu}\colon C^{\infty}(U,V)&\to C^{\infty}(U,V)\\ \phi&\mapsto\mathrm{d}\phi(\partial_{\mu})=\partial_{\mu}(\phi\circ x^{-1})\circ x \end{align*} and \begin{equation} \nabla_{\mu}=\partial_{\mu}+\rho_*A_{\mu}. \end{equation} Corollary: \begin{equation} [\nabla_\mu,\nabla_\nu]=\rho_*F_{\mu\nu} \end{equation}

Proof:

The equation \begin{equation*}\tag{1} \partial_{\mu}\circ(\rho_*A_{\nu})=\rho_*(\partial_{\mu}A_{\nu})+\rho_*A_{\nu}\circ\partial_{\mu} \end{equation*} implies \begin{equation*} [\partial_{\mu},\rho_*A_{\nu}]+[\partial_{\nu},\rho_*A_{\mu}]=\rho_*(\partial_{\mu}A_{\nu})-\rho_*(\partial_{\nu}A_{\mu}). \end{equation*} Thus, using the structure equation, we obtain \begin{align*} [\nabla_\mu,\nabla_\nu]=[\partial_{\mu}+\rho_*A_{\mu},\partial_{\nu}+\rho_*A_{\nu}]=[\partial_\mu,\partial_\nu]+[\partial_{\mu},\rho_*A_{\nu}]+[\partial_{\nu},\rho_*A_{\mu}]+[\rho_*A_\mu,\rho_*A_\nu]\\=\rho_*\partial_{\mu}A_{\nu}-\rho_*\partial_{\nu}A_{\mu}+\rho_*[A_{\mu},A_{\nu}]=\rho_*(\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}+[A_{\mu},A_{\nu}])=\rho_*F_{\mu\nu}, \end{align*} (I used the structure equation in the last step.)

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Addendum - proof of equation $(1)$

More explicitely, equation $(1)$ means \begin{equation*} \partial_{\mu}(\rho_*(A_{\nu})\phi)=\rho_*(\partial_{\mu}A_{\nu})\phi+\rho_*(A_{\nu})\partial_{\mu}\phi \end{equation*} for all $\phi\in C^{\infty}(U,V)$ ("product rule"). Even more explicit: \begin{equation*} \partial_{\mu}((\rho_*A_{\nu}\cdot\phi)\circ x^{-1})\circ x=\rho_*(\partial_{\mu}(A_{\nu}\circ x^{-1})\circ x)\cdot\phi+\rho_*A_{\nu}\cdot(\partial_{\mu}(\phi\circ x^{-1})\circ x)\in C^{\infty}(U,V) \end{equation*} If we define $A_\nu:=A_\nu\circ x^{-1}\in C^{\infty}(x(U),g)$ (notice the abuse of notation), this is equivalent to \begin{equation*} \partial_{\mu}(\rho_*A_{\nu}\cdot\phi)=\rho_*\partial_{\mu}A_{\nu}\cdot\phi+\rho_*A_{\nu}\cdot\partial_{\mu}\phi\in C^{\infty}(x(U),V) \end{equation*} for all $\phi\in C^{\infty}(x(U),V)$.

Since the partial derivative is the total derivative of a function defined on an open interval, it suffices to prove the equation for $n=1$:

Let $I\subset\mathbf{R}$ is an open interval, $V$ and $W$ normed vector spaces, $O\colon W\to\mathrm{End}(V)$ a linear and continuous function and $f\colon I\to W$, $g\colon I\to V$ differentiable functions. If $Ow\in\mathrm{End}(V)$ is continuous for all $w\in W$, $Of\cdot g\colon I\to V$ is differentiable and \begin{equation*} (Of\cdot g)'=Of'\cdot g+Of\cdot g'. \end{equation*} Proof: Let $x\in I$. To simplify the notation, we define \begin{equation*} \delta F:=F(x+\delta)-F(x),F(x)=:F \end{equation*} for all functions $F\colon I\to X$. We want to prove that for every $\epsilon>0$ there exists an $r>0$ s.t. \begin{equation*} |\delta(Of\cdot g)-\delta\cdot(Of'\cdot g+Of\cdot g')|<|\delta|\epsilon \end{equation*} for all $\delta\in(-r,r)$. This follows from the following facts:

• For every $\epsilon>0$ there exists an $r>0$ s.t. $|\delta f-\delta\cdot f'|<|\delta|\epsilon$ and $|\delta g-\delta\cdot g'|<|\delta|\epsilon$ for all $\delta\in(-r,r)$.

• $\delta(Of\cdot g)=O\delta f\cdot g+Of\cdot\delta g+O\delta f\cdot\delta g$

• $O\delta f\cdot\delta g=O(\delta f-\delta\cdot f'+\delta\cdot f')\cdot(\delta g-\delta\cdot g'+\delta\cdot g') =O(\delta f-\delta\cdot f')\cdot(\delta g-\delta\cdot g')\\+O(\delta f-\delta\cdot f')\cdot(\delta\cdot g')+O(\delta\cdot f')\cdot(\delta g-\delta\cdot g')+O(\delta\cdot f')\cdot(\delta\cdot g')$

• Triangle inequality

• $|(O(w))(v)|\leq|O||w||v|$ for all $(v,w)\in V\times W$.