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.


2 Answers 2

  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 $$\frac {dU(s_f,s_i)}{ds_f} ~=~ i\dot{\gamma}^{\mu}(s_f)~A_{\mu}(\gamma(s_f)) ~U(s_f,s_i). \tag{4}$$

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

  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 $$ \frac {\delta U(s_f,s_i)}{\delta A^a_{\mu}(x)} ~=~\int_{s_i}^{s_f}\! ds~ U(s_f,s)~ i\dot{\gamma}^{\mu}(s)\delta^4(x-\gamma(s))T_a~U(s,s_i). \tag{6}$$

  7. Heuristic proof of (6). Since we have already used the letter $x\in\mathbb{R}^4$ in (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 intersects the neighborhood $\Omega$, then the equation (6) becomes trivially correct: $0=0$.)

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

  • $\begingroup$ I'm working on how to see clearly that the holonomy breaks into two at the point where the functional derivative acts, I'm checking the definition of the ordered exponential, do you have any advise or hint on how we can see this breakdown from that definition? Thanks. $\endgroup$
    – Saoirse
    Commented Jun 30, 2022 at 23:33

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'.. + ...


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