# Form of $SU(N)$ gauge transformations in $SU(N)$ Yang-Mills theory

For $SU(N)$ Yang-Mills theory, instantons correspond to finite action solutions $A_\mu(x)$ of the Euclidean equation of motion. The requirement of finite action demands that $A_\mu(x)$ is a pure gauge at the boundary of $\mathbb{R}^4$ given by $$A_\mu(x)=-\frac{i}{g}(\partial_\mu U)U^{-1}$$ where $U\in SU(N)$.

For $SU(2)$, therefore, we have, $$U(x)=\exp[i\theta_a(x)T^a]$$ where $T^a=\sigma^a/2$ are the generators of $SU(2)$ in the fundamental representation. However, $U$ is taken to be $$U=\frac{x_4+i\boldsymbol{\sigma}\cdot\textbf{x}}{\tau}$$ where $\tau^2=x_4^2+\textbf{x}\cdot\textbf{x}$ while studying instantons of class $n=1$.

Is the latter expression of $U$ a special case of the former expression? In that case, how is the latter expression derived from the former?

This has nothing to do with instantons or quantum field theory, it is just an elementary fact about $2\times 2$ matrices:

The Pauli matrices $\sigma^i$ together with the identity $\mathbf{1}_2$ form a basis of the vector space of 2-by-2 matrices. Therefore, $U(x)$, as a 2-by-2 matrix-valued function, may be written as $$U(x) = \zeta(x)\mathbf{1}_2 + \omega_i(x)\sigma^i$$ and your expression for $U(x)$ follows from the choices $\zeta(x) = x_4/r,\omega_i(x) = x_i/r$ for $r = \sqrt{x^2}$.

If you want to see how you need to choose the $\theta_a(x)$ in the exponential for $U(x)$ you wrote, simply use the standard relation $$\exp(\mathrm{i}\alpha\vec n\cdot\vec\sigma) = \mathbf{1}_2\cos(\alpha) + \mathrm{i}(\vec n\cdot\vec \sigma)\sin(\alpha)$$ and compare coefficients to get the $\theta_a = \alpha n_a$.

Of course, it is. The only condition one imposes on $U$ is that it is unitary. You can easily check that the latter matrix is.

$$U^{\dagger}=\frac{x_4-\mathrm{i} \vec{\sigma} \cdot \vec{x}}{\sqrt{x_4^2+\vec{x}^2}};$$

$$U^{\dagger} \cdot U=\frac{x_4-\mathrm{i} \vec{\sigma} \cdot \vec{x}}{\sqrt{x_4^2+\vec{x}^2}} \cdot \frac{x_4+\mathrm{i} \vec{\sigma} \cdot \vec{x}}{\sqrt{x_4^2+\vec{x}^2}}=\frac{x_4^2+\vec{x}^2}{x_4^2+\vec{x}^2}=1.$$

• The fact that U is unitary is a trivial part. It would have answered my question if you supplied a choice of $\theta_a(x)$, that enables one to reduce the former expression to the latter. Or expressed the latter as the exponential of the generators $\sigma^a/2$. – SRS Dec 21 '16 at 13:41
• Because $U$ is unitary, it may be diagonalized in the appropriate basis, and one can evaluate $\mathrm{log} U$. Any $2 \times 2$ matrix can be expanded in terms of $\vec{\sigma}$'s, which implies one can find $\theta^a$ in terms of $U$. – Andrey Feldman Dec 21 '16 at 13:50