This question concerns Eq. (2.10) of the paper https://arxiv.org/abs/hep-th/0305116 by Bena, Polchinski and Roiban.

In section 2.1 they are showing that the infinite number of conserved quantities for the principal chiral model

\begin{equation} L = \frac{1}{2\alpha_0} \mathrm{Tr}(\partial_\mu g^{-1}\partial_\mu g) \end{equation}

are given by the fixed-time Wilson lines $U(\infty,t;-\infty,t)$ where

\begin{equation} U(x;x_0) = \mathrm{P}\, e^{-\int_{\mathcal{C}}a} \end{equation}

and $a$ is a 1-parameter family of flat connections given by Eq. (2.3).

My question is what becomes of the last two terms (i.e. $-a_0a_1 +a_1a_0$) in the second line of Eq. (2.10). Do they cancel? I don't see why the should because the $a$'s are non-commuting (Lie algebra-valued).


2 Answers 2


To simplify, take the notation : $U_y(x)= U(y,t;x,t)$, $U^{-1}_z(x)= U(x,t;z,t)$, $a_i(x) = a_i(x,t)$

Note that you have (on the spatial choosen path $ C = \int dx^1 = \int dx $) :

$\partial_x U^{-1}_z(x)=-a_1(x) U^{-1}_z(x)$, and $\partial_x U_y(x)= U_y(x)a_1(x)$

The minus sign difference can be understood because $\partial_x (U_y U^{-1}_y)=0$

Now, the last line of $2.10$ is :

$a_0(y,t)U(y,t;z,t) − U(y,t;z,t)a_0(z,t)$

With our notations, we have :

$a_0(y) U^{-1}_z(y) - U_y(z)a_0(z) \\= U_y(y)a_0(y) U^{-1}_z(y)- U_y(z)a_0(z)U^{-1}_z(z) \\ =[U_y(x)a_0(x) U^{-1}_z(x)]_z^y \\ = \int_z^y ~dx~\partial_x(U_y(x)a_0(x) U^{-1}_z(x)) \\ = \int_z^y ~dx~(\partial_x U_y(x))a_0(x) U^{-1}_z(x) + \int_z^y ~dx~U_y(x)(\partial_x a_0(x)) U^{-1}_z(x) + \int_z^y ~dx~ U_y(x)a_0(x) (\partial_x U^{-1}_z(x)) \\ = \int_z^y ~dx~U_y(x) (a_1(x) a_0(x) + a'_0(x) -a_0(x)a_1(x)) U^{-1}_z(x)$

So, this is the second line of $2.10$. There is a global minus sign difference, I think it is because here the integration is going from $z$ to $y$, while the first line of $2.10$ uses the integration from $y$ to $z$

  • $\begingroup$ Thanks for your great explanation. By the way, is there any advantage to including a minus sign in the exponent of the Wilson line? It seems like it would be easier to define it without the minus sign in which case the derivation would go as below. $\endgroup$
    – user11881
    Commented Jan 4, 2014 at 18:17
  • $\begingroup$ No, the sign in the exponential is just some convention. And it is not clear, in the second line, what are $x_1$ and $x_2$. If it is $x_1=y, x_2=z$, then there is no problem of sign, and I think it is the case. So, no problem of sign. $\endgroup$
    – Trimok
    Commented Jan 4, 2014 at 18:31

Defining the Wilson loop without the minus sign in the exponent gives \begin{align} \partial_t U(y,t;z,t) & = \partial_t \mathrm{P} \, e^{\int_{(z,t)}^{(y,z)} dx^\mu a_\mu} \\ & = \partial_t \mathrm{P} \, e^{\int_z^y dx a_1} \\ & = \int_z^y dx \, U(x,t;z,t)\dot{a}_1(x,t)U(y,t;x,t) \\ & = \int_z^y dx \, U(x,t;z,t)[a_0' - a_0a_1 + a_1a_0]_{(x,t)}U(y,t;x,t) \\ & = \int_z^y dx \partial_x \left[ U(x,t;z,t) a_0(x,t)U(y,t;x,t) \right] \\ \end{align}


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