From self-duality equations to Nahm equations

This page uses the Einstein summation convention: \( a_jb_j = \Sigma_{j} a_jb_j\).

The matrices (Nahm or gauge field) on this page are skew-hermitian.

In this section we discover the Nahm equations by looking at the self-duality equations of Yang-Mills theory.

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Differential forms and Yang-Mills

From studying electromagnetism we know that the electromagnetic potential can be written as a four-vector \( A^{\mu} \), while the electromagnetism field strength is an antisymmetric tensor of rank 2:

\( \quad \displaystyle F^{\mu \nu } = \partial^{\mu}A^{\nu} – \partial^{\nu}A^{\mu} \)

We use these in the special relativistic treatment of Maxwell’s equations.

The physics of the system comes from writing down an action for the free electromagnetic field;

\( \quad \displaystyle S = \frac{1}{2} \int F_{\mu \nu} F^{\mu \nu } d^4x \)

and varying it to find two of Maxwell’s equations,\( \partial_{\mu}F^{\mu \nu} = 0 \). The other two come from the Bianchi identity

\( \displaystyle \quad \partial_{\mu} \epsilon^{\mu \nu \rho \sigma} F_{\rho \sigma} = 0 \)

Exercise: Vary the action (or skip ahead to the Euler-Lagrange equations) to derive the first Maxwell equation, \( \partial_{\mu}F^{\mu \nu} = 0 \), and also verify the Bianchi identity (hint: symmetry/antisymmetry).

A nice geometric way of viewing these objects is to use the language of differential forms. The potential A^μ is a one-form

\( \quad A = A_{\mu}dx^{\mu} \)

and the field strength is a two-form

\( \displaystyle \quad F = \frac{1}{2}F_{\mu \nu}dx^{\mu} \wedge dx^{\nu} \)

The object \( \frac{1}{2}\epsilon^{\mu \nu \rho \sigma}F_{\rho \sigma} \) gives the components of a dual two-form

\( \quad \star F = \frac{1}{2}(\star F)_{\mu \nu} dx^{\mu} \wedge dx^{\nu} \qquad (\star F)_{\mu \nu} = \frac{1}{2}\epsilon^{\mu \nu \rho \sigma}F_{\rho \sigma} \)

Exercise: Compute the components of the dual tensor \( \star F \) in terms of the components of F.

What exactly is this duality? We can think of the \( \star \) appearing here as being a linear operator which maps differential forms to differential forms. In 4-dimensions, p-forms are mapped to 4 − p forms – in fact \( \star \) gives an isomorphism between the space of p-forms and the space of 4 − p forms. The most simple way to view \( \star \) is as a permutation: so for instance on two-forms, \( \star \) maps \( dx^{\mu} \wedge dx^{\nu} \: \text{to} \: dx^{\rho} \wedge dx^{\sigma}\), where μνρσ forms an even permutation of 0,1,2,3. This is neatly encapsulated using the Levi-Civita symbol above.

In the language of differential forms the action becomes

\( \displaystyle \quad S = \int trF \wedge \star F \)

and the Maxwell equations are

\( \quad d \star F = 0 \)
\( \quad dF = 0 \)

What do you notice about the structure of these equations? Is there an indirect way to find solutions?

Yang-Mills theory

In Yang-Mills theory we generalise electromagnetism by allowing the components A^μ of the potential to be square matrices and not just ordinary numbers. In this context the field strength is

\( \quad F^{\mu \nu } = \partial^{\mu}A^{\nu} – \partial^{\nu}A^{\mu} + [A^{\mu}, A^{\nu}] \)

The action must be written in terms of a scalar, so we have to take a trace:

\( \quad \displaystyle S = \frac{1}{2} \int trF_{\mu \nu} F^{\mu \nu }d^4x \)

The equations of motion are

\( \quad \partial_{\mu}F^{\mu \nu} + [A_{\mu}, F^{\mu \nu}] = 0 \)

while the Bianchi identity is

\( \quad \displaystyle \partial_{\mu}\epsilon^{\mu \nu \rho \sigma} F_{\rho \sigma} + [A_{\mu}, \epsilon^{\mu \nu \rho \sigma} F_{\rho \sigma}] = 0 \)

Everything we said about differential forms is still valid though. We can in fact write these equations as

\( \quad d \star F + A \wedge \star F = 0 \qquad \text{(eom)} \)
\( \quad dF + A \wedge F = 0 \qquad \text{(Bianchi identity)} \)

These are non-linear second-order partial differential equations (for A). Instead of trying to solve them directly, we notice the simple but powerful fact that if F is self-dual

\( \quad F = \star F \)

or anti-self-dual

\( \quad F = -\star F \)

then F will solve the equations of motion automatically as a consequence of the Bianchi identity!

Exercise: Why are these the only possible solutions of \( F = \lambda \star F \) ?

Reduction of the self-duality equations

The self-duality equations as written above are hugely important as they stand, and have been extensively studied for many years (their solutions are known as instantons). A remarkable observation is that they in fact contain the equations describing magnetic monopoles (the Bogomolny equations) and also the Nahm equations as dimensional reductions.

By a dimensional reduction we mean the following: supposing our potential A^μ is invariant in a certain direction or directions.

Exercise: Find the reduced self-duality equations that result from supposing x⁰-invariance.

Exercise: Find the reduced self-duality equations that result from supposing x¹,x²,x³-invariance.

So we have obtained again the Nahm equations, as well as the Bogomolny equations related to them via the Nahm transform, as dimensional reductions of the same four-dimensional set of equations.

The two-dimensional reduction is also interesting; the system of equations obtained in this case is known as the Hitchin equations.

Exercise: Derive the Hitchin equations.