Can Yang-Mills field strength be defined as covariant derivative squared?

Question

In Yang-Mills theory the field strength tensor $F_{\mu \nu}$ can be calculated as $$ \begin{equation} F_{\mu\nu} \equiv \frac{i}{g} [D_\mu,D_\nu] = \partial_\mu A_\nu - \partial_\nu A_\mu -ig[A_\mu,A_\nu], \end{equation} $$ where covariant derivative is defined as $$ D_\mu = \partial_\mu -ig A_\mu. $$ Can Yang-Mills field strength be defined as covariant derivative squared instead, rather than as the non-commutatibity of the covariant derivatives? The two definitions are essentially the same, but the former affords an alternative way of deriving the covariance properties of field strength (see below).

It's easier to demonstrate the idea in terms of differential forms. The field strength 2-form $F$ (also called curvature 2-form) is defined as (here the coupling constant and $i$ are absorbed into the definition of $A$ and $F$) $$ F = F_{\mu \nu}dx^\mu\wedge dx^\nu = d\wedge A + A\wedge A, $$ gauge field 1-form $A$ is defined as $$ A = A_\mu dx^\mu, $$ and the covariant derivative of the Dirac fermion field $\psi$ (spinor) is defined as $$ D\psi = (d+A)\psi. $$

Now the covariant derivative squared $D\wedge D$ of a spinor is $$ D\wedge D \psi = (d+A)\wedge (d+A)\psi $$ $$ = (d\wedge d + d\wedge A + A \wedge d + A\wedge A) \psi $$ $$ = 0 + d\wedge (A \psi) + A \wedge (d \psi) + (A\wedge A) \psi $$ $$ = (d\wedge A) \psi - A \wedge (d \psi) + A \wedge (d \psi) + (A\wedge A) \psi $$ $$ = (d\wedge A + A\wedge A) \psi $$ $$ = F \psi, $$ or in short $$ F = D\wedge D, $$ where we have used the fact that $d\wedge d = 0$ and 1-forms $d$ and $A$ anti-commute in wedge (outer) $\wedge$ product.

The beauty of the covariant derivative squared definition is that the covariance of field strength 2-form $F$ $$ F \rightarrow F' = gFg^{-1} $$ is an automatic outcome, since by definition of the covariant derivative we have $$ F\psi = D\wedge D \psi \rightarrow g D\wedge D \psi = g F \psi = g F g^{-1}g\psi = F'\psi', $$ where $$ \psi \rightarrow \psi' = g\psi. $$

Answer 1

It is a standard fact in gauge theory that the following definitions for the field strength $F$ of a gauge field $A$ with exterior covariant derivative $\mathrm{d}_A = \mathrm{d} + A\wedge{}$ (what you call $D$) are equivalent:

• $F = \mathrm{d}_A A$ - "The field strength is the covariant derivative of the gauge field."
• $F\wedge \omega = \mathrm{d}_A^2 \omega$ for any form $\omega$ - "Acting twice with the covariant derivative is the same as wedging with the field strength." or "The field strength is the obstruction to the covariant derivative fulfilling $\mathrm{d}^2_A = 0$."
• $F(X,Y) = A([X - A(X), Y - A(Y)])$ for any two vector fields $X,Y$ on the principal bundle. - "The field strength is the obstruction of the Lie bracket of two horizontal vector fields to be horizontal" [You won't usually see this one in physics treatments since it is not particularly useful for our purposes]
• $F$ is the holonomy around an infinitesimal loop, cf. Ambrose-Singer theorem
• ...

In so far as the spirit of your question aims at the second expression, it is therefore correct. However, you should use standard notation: One does not usually use a wedge between the derivative operator and its argument, and the equation $F = D\wedge D$ is malformed: The l.h.s is a Lie-algebra-valued 2-form, the r.h.s. an operator. If you absolutely insist, write $F \wedge{} = D^2$ or $F_{\mu\nu} = [D_\mu,D_\nu]$.

Lastly, there is no difference in the expressions $F\wedge \omega = \mathrm{d}^2_A \omega$ and $F_{\mu\nu} = [D_\mu,D_\nu]$ except that one is coordinate-free and the other isn't. You could equally well deduce the transformation behaviour of the field strength from the commutator expression: $$ F_{\mu\nu}\psi = [D_\mu,D_\nu]\psi\mapsto g[D_\mu,D_\nu]\psi = \dots,$$ precisely in the same way "by definition of the covariant derivative".