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The Yang Mills Field Strength: a motivation for differential forms

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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.

Written by Vivek

June 11, 2014 at 10:53