Functional Derivation of Holonomy

Question

I would like to know how to take the functional derivative of the holonomy, or Wilson line. I have tried it and I will show what I have done below, but before I wanted to say that I also have seen and done this with the characteristic deifferential equation for the holonomy $$ \frac{\partial U}{\partial s}+\dot{\gamma}^a A_{a} U=0 $$ with $\dot{\gamma}$ a tangent vector to the curve and $A$ the connection. By varying this equation I can find what $\frac{\delta U}{\delta A}$ is, but I would like to know how to do it from the expression for $U$ $$ U=\mathcal{P}\exp \left[ -\int_{\gamma} \dot{\gamma}^a(s) A_a(\gamma(s)) ds \right] $$ with $\dot{\gamma}^a=\frac{dx^a}{ds}$ as before. Now I have tried to directly vary this with respect to $A_b$ $$ \frac{\delta U}{\delta A_b(x)}=\mathcal{P} \exp \left[ -\int_{\gamma} \dot{\gamma}^a A_a ds \right] \cdot \frac{\delta}{\delta A_b}\left[ -\int_{\gamma} \dot{\gamma}^a A_a ds \right]. $$ Now if $A_a=A_{a}^{i}\tau^i$ then $$ \frac{\delta}{\delta A_{b}^i }\left[ -\int_{\gamma} \dot{\gamma}^a A_{a}^j \tau^j ds \right]=-\int_{\gamma} \dot{\gamma}^a \delta _{ab}\delta_{ij} \delta^3(\gamma(s)-x) \tau^j ds=-\int_{\gamma}\dot{\gamma}^b \delta^3(\gamma(s)-x) \tau^j ds. $$ So I end with $$ \frac{\delta U}{\delta A_{b}^j}=U(\gamma)\left[ -\int_{\gamma}\dot{\gamma}^b \delta^3(\gamma(s)-x) \tau^j ds \right] $$ Which isn't right. Can someone point me in a better direction.

Answer 1

1. Let us write the Wilson-line of a simple open curve $\gamma: [s_i,s_f]\to \mathbb{R}^4$ as $$ U(s_f,s_i) ~=~ \mathcal{P}\exp \left[ i\int_{\gamma} A_{\mu}~ dx^{\mu} \right].\tag{1} $$

2. The path-ordering $\mathcal{P}$ becomes important if the gauge potential $$A_{\mu}~=~A^a_{\mu} T_a\tag{2}$$ is non-abelian. Here $T_a$ are the generators of the corresponding Lie algebra.

3. The Wilson-line has groupoid properties, e.g., $$U(s_3,s_2)U(s_2,s_1)~=~ U(s_3,s_1), \qquad U(s,s) ~=~ {\bf 1}.\tag{3}$$

4. If one differentiates wrt. the final point $s_f$, one gets $$\frac {\mathrm{d}U(s_f,s_i)}{\mathrm{d}s_f} ~=~ i\dot{\gamma}^{\mu}(s_f)~A_{\mu}(\gamma(s_f)) ~U(s_f,s_i). \tag{4}$$

5. If one differentiates wrt. the initial point $s_i$, one gets $$ \frac {\mathrm{d}U(s_f,s_i)}{\mathrm{d}s_i} ~=~ -U(s_f,s_i)~i\dot{\gamma}^{\mu}(s_i)~A_{\mu}(\gamma(s_i)) . \tag{5}$$

6. OP wants to differentiate the Wilson-line $U(s_f,s_i)$ functionally wrt. the gauge potential components $A^a_{\mu}(x)$. One gets $$ \frac {\delta U(s_f,s_i)}{\delta A^a_{\mu}(x)} ~=~\int_{s_i}^{s_f}\! \mathrm{d}s~ U(s_f,s)~ i\dot{\gamma}^{\mu}(s)\delta^4(x-\gamma(s))T_a~U(s,s_i). \tag{6}\label{eq:6}$$

7. Heuristic proof of $\eqref{eq:6}$. Since we have already used the letter $x\in\mathbb{R}^4$ in $\eqref{eq:6}$ as a fixed space-time point, let us call an arbitrary spacetime point for $y\in\mathbb{R}^4$.

   • Imagine that $\tilde{A}(y)=A(y)+\delta A(y)$ is an infinitesimal variation of the gauge potential $A(y)$.

   • Imagine that $\delta A(y)$ only differs from zero in an infinitesimally small neighborhood $\Omega$ of the fixed space-time point $x$.

   • Assume that the curve $\gamma$ intersects the neighborhood $\Omega$ at the parametervalue interval $[s_x-\varepsilon,s_x+\varepsilon]\subseteq [s_i,s_f]$. (If the curve $\gamma$ does not intersect the neighborhood $\Omega$, then the equation $\eqref{eq:6}$ becomes trivially correct: $0=0$.)

   On one hand, such infinitesimal variation of the gauge potential yields $$\delta U(s_f,s_i)~=~U(s_f,s_x+\varepsilon)~\delta U(s_x+\varepsilon,s_x-\varepsilon)~U(s_x-\varepsilon,s_i), \tag{7}\label{eq:7}$$ and $$\begin{align}\delta U(s_x+\varepsilon,s_x-\varepsilon)~\approx~&2\varepsilon i~ \dot{\gamma}^{\mu}(s_x)~\delta A_{\mu}(\gamma(s_x)) \cr ~=~&\int_{\Omega} \!\mathrm{d}^4y~\delta^4(y-\gamma(s_x))~2\varepsilon i\dot{\gamma}^{\mu}(s_x)~\delta A_{\mu}(y)\cr ~\approx~& \int_{\Omega} \!\mathrm{d}^4y~\int_{s_x-\varepsilon}^{s_x+\varepsilon}\! \mathrm{d}s~\delta^4(y-\gamma(s))~i\dot{\gamma}^{\mu}(s)~\delta A_{\mu}(y).\end{align}\tag{8}\label{eq:8}$$ On the other hand, the defining property of a functional derivative yields $$\begin{align}\delta U(s_f,s_i) ~=~&\int_{\mathbb{R}^4} \!\mathrm{d}^4y~ \frac {\delta U(s_f,s_i)}{\delta A^a_{\mu}(y)} ~\delta A^a_{\mu}(y)\cr &~=~\int_{\Omega} \!\mathrm{d}^4y~ \frac {\delta U(s_f,s_i)}{\delta A^a_{\mu}(y)} ~\delta A^a_{\mu}(y).\end{align}\tag{9}\label{eq:9}$$ A comparison of eqs. $\eqref{eq:7}$, $\eqref{eq:8}$ and $\eqref{eq:9}$ yields eq. $\eqref{eq:6}$.

Answer 2

Lewandowski, Newman and Rovelli gave all the details in a 1993 paper "Variations of the parallel propagator and holonomy operator and the Gauss law constraint" We have $$d U(x(s),x_i)/ds + \dot \gamma (s)A(s) U(x(s),x_i) = 0 \ (1)$$ Differentiating yields $$d \delta U(x(s),x_i)/ds + \dot \gamma (s)A(s) \delta U(x(s),x_i) = -\delta(\dot \gamma (s)A(s))U(x(s),x_i)$$ Now the ansatz is to write $\delta U = U \Lambda$ using (1) we have: $$U(x(s),x_i) \dot \Lambda = -\delta(\dot \gamma (s)A(s))U(x(s),x_i)$$ As the inverse of U(a,b) is U(b,a) we have to solve $$\dot \Lambda = - U(x_i,x(s))\delta(\dot \gamma (s)A(s))U(x(s'),x_i)$$ then $\delta U(x(s),x_i) = U(x(s),x_i) \Lambda = $ $$= U(x(s),x_i) \int^s_{s_i}U(x_i,x')\delta(\dot \gamma (s')A(s'))U(x(s'),x_i) ds'$$ $$= \int^s_{s_i}U(x(s),x')\delta(\dot \gamma (s')A(s'))U(x(s'),x_i) ds'$$ We get a formula which enables us to vary the connection A, or the curve (a loop is a peculiar case). the result will be seen as a Pauli matrix or something else sandwiched between the two Us. It looks like the derivation product rule. (abcd..)' = a'bcd.. +ab'cd.. + abc'd.. + abcd'.. + ...