# Least Action

Nontrivializing triviality..and vice versa.

## The Yang Mills Field Strength: a motivation for differential forms

Although their importance and applications are widely recognized in theoretical physics, differential forms are not part of the standard curriculum in physics courses, except for the rare mention in a general relativity course, e.g. if you have used the book by Sean Carroll. In this blog post, we describe the construction of the field strength tensor for Yang Mills Theory, using differential forms. The derivation (which is based on the one given by Tony Zee in his QFT book) is elegant, and mathematically less tedious than the more conventional derivation based on the matrix gauge field. It also illustrates the power of the differential forms approach.

To begin with, let me define a normalized matrix gauge field $A_\mu$ which is $-i$ times the usual matrix gauge field. So in the equations that appear below, the covariant derivative has no lurking factors of $-i$. This greatly simplifies the algebra for us, as we don’t have to keep track of conjugations and sign changes associated with $i$. The covariant derivative is thus $D_\mu = \partial_\mu + A_\mu$ in terms of this new gauge field. The matrix 1-form $A$ is

$A = A_\mu dx^\mu$

So, $A^2 = A_\mu A_\nu dx^\mu dx^\nu$. But since $dx^\mu dx^\nu = -dx^\nu dx^\mu$, only the antisymmetric part of the product survives and hence we can write

$A^2 = \frac{1}{2}[A_\mu, A_\nu] dx^\mu dx^\nu$

We want to construct an appropriate 2-form $F = \frac{1}{2}F_{\mu\nu}dx^\mu dx^\nu$ from this 1-form A. Now, if d denotes the exterior derivative, then dA is a 2-form, as is A^2. These are the only two forms we can construct from A. So, $F$ must be a linear combination of these two forms. This is a very simple, and neat argument!

Now, the transformation law for the gauge potential is

$A \rightarrow U A U^\dagger + U dU^\dagger$

where $U$ is a 0 form (so $dU^\dagger = \partial_\mu U^\dagger dx^\mu$). Applying d to the transformation law, we get

$dA \rightarrow U dA U^\dagger + dU A U^\dagger - U A dU^\dagger + dU dU^\dagger$

where the negative sign in the third term comes from moving the 1-form d past the 1-form A. Squaring the transformation law yields

$A^2 \rightarrow UA^2 U^\dagger + U A dU ^\dagger + U dU^\dagger U A U^\dagger + U dU^\dagger U dU^\dagger$

Now, $UU^\dagger = 1$, so applying d again to both sides we get $U dU^\dagger = -dU U^\dagger$. So, we can write the square transformation law as

$A^2 \rightarrow U A^2 U^\dagger + U A dU^\dagger - dU A U^\dagger - dU dU^\dagger$
whereas if we recall the expression for the transformation of $dA$, it was just
$dA \rightarrow U dA U^\dagger + dU A U^\dagger - U A dU^\dagger + dU dU^\dagger$

Clearly if we merely add $A^2$ and $dA$, the last 3 terms on the RHS of each cancel out, and we get

$A^2 + dA \rightarrow U(A^2 + dA)U^\dagger$

which is the expected transformation law for a field strength of the form $F = A^2 + dA$:

$F \rightarrow U F U^{\dagger}$

The differential form approach uses compact notation that suppresses the Lorentz index $\mu$ as well as the group index $a$, and gives us a fleeting glimpse into the connection between gauge theory and fibre bundles.

For a gentle yet semi-rigorous introduction to differential forms, the reader is referred to the book on General Relativity by Sean Carroll.