BRST Quantization of the Point Particle: Sign of Structure Functions (Polchinski) Currently I am studying string theory and I encountered a bunch of interrelated problems in the context of BRST quantization which I can't solve for myself although I tried hard for some days.
The question concerns eq. (4.2.20) of Polchinski's textbook. In eq. (4.2.1) he defines the structure functions as follows
$$[\delta_\alpha,\delta_\beta]=f^\gamma_{\alpha\beta}\delta_\gamma\tag{4.2.1} $$
In the following calculation I use precisely this definition. In the subsection about the BRST quantization of the point-particle, Polchinski computes the commutator of two gauge transformations. I tried to verify eq. (4.2.20) using 
$$\delta_{\tau_1}X^\mu(\tau)~=~-\delta(\tau-\tau_1)\partial_\tau X^\mu(\tau) \tag{4.2.19}$$ 
from eq. (4.2.19) as follows
$$\begin{align}
[\delta_{\tau_1},\delta_{\tau_2}]X^\mu(\tau)&=\delta_{\tau_1}(\delta_{\tau_2}X^\mu(\tau))-\delta_{\tau_2}(\delta_{\tau_1}X^\mu(\tau))\\[2pt]
&=\left[-\delta(\tau-\tau_1)\partial_\tau(\delta_{\tau_2}X^\mu(\tau))\right]-\left[-\delta(\tau-\tau_2)\partial_\tau(\delta_{\tau_1}X^\mu(\tau))\right]\\[2pt]
&=\left[-\delta(\tau-\tau_1)\partial_\tau(-\delta(\tau-\tau_2)\partial_\tau X^\mu(\tau))\right]-\left[-\delta(\tau-\tau_2)\partial_\tau(-\delta(\tau-\tau_1)\partial_\tau X^\mu(\tau))\right]\\[2pt]
&=\left[\delta(\tau-\tau_1)\partial_\tau\delta(\tau-\tau_2)-\delta(\tau-\tau_2)\partial_\tau\delta(\tau-\tau_1)\right]\partial_\tau X^\mu(\tau)\\[2pt]
&=\int\mathrm{d}\tau_3\,(-1)\left[\delta(\tau_3-\tau_1)\partial_{\tau_3}\delta(\tau_3-\tau_2)-\delta(\tau_3-\tau_2)\partial_{\tau_3}\delta(\tau_3-\tau_1)\right]\left(-\delta(\tau-\tau_3)\partial_\tau X^\mu(\tau)\right)\\[2pt]
&\equiv\int\mathrm{d}\tau_3\,f^{\tau_3}_{\tau_1\tau_2}\delta_{\tau_3}X^\mu(\tau)
\end{align}$$
where the structure functions are given by
$$f^{\tau_3}_{\tau_1\tau_2}=(-1)\left[\delta(\tau_3-\tau_1)\partial_{\tau_3}\delta(\tau_3-\tau_2)-\delta(\tau_3-\tau_2)\partial_{\tau_3}\delta(\tau_3-\tau_1)\right]$$
In going from the third to the fourth line, the terms of the form $\partial_\tau\partial_\tau X^\mu(\tau)$ cancel. In the next-to-last line I introduced a $\delta$-distribution in order to get $\delta_{\tau_3}X^\mu(\tau)$. Comparing my structure functions with eq. (4.2.21), there is a different sign. In the second line of eq. (4.2.20) the minus sign seemingly appears out of nothing. Can you make sense of this? Or is there a mistake in my calculations? 
Obviously, the sign of the structure functions directly influences the sign of the BRST variation of the c-ghost in eq. (4.2.22e).
It may well be that I miss the wood for the trees since I am trying for a few days. I guess I made some silly mistake, but I can't see where. It would be nice if someone could help me with these problems!
 A: The issue is essentially the same as OP's other Phys.SE question, namely the order of operations: The second BRST transformation "acts inside" the first BRST transformation, so
\begin{align}
[\delta_{\tau_1},\delta_{\tau_2}]X^\mu(\tau)&~~~=~~~\delta_{\tau_1}(\delta_{\tau_2}X^\mu(\tau))-\delta_{\tau_2}(\delta_{\tau_1}X^\mu(\tau))\\[2pt]
&\stackrel{(4.2.19)}{=}\left[-\delta(\tau-\tau_2)\partial_\tau(\delta_{\tau_1}X^\mu(\tau))\right]-\left[-\delta(\tau-\tau_1)\partial_\tau(\delta_{\tau_2}X^\mu(\tau))\right]\\[2pt]
&~~~=~~~\ldots
\end{align}
OP's second line has the opposite order, which causes an overall sign.
A: Let us start from the beginning: we know that if $\delta\tau=\xi$, then we have $\delta X^\mu(\tau)=-\xi\overset{\cdot}{X}^\mu(\tau)$. Furthermore, $$\delta\delta X^\mu(\tau)=-\xi\overset{\cdot}{\xi}\overset{\cdot}{X}^\mu(\tau)-\xi^2\overset{\cdot\cdot}{X}^\mu=-\xi\overset{\cdot}{\xi}\overset{\cdot}{X}^\mu(\tau)$$ where in the last step I used the equations of motion.
The condensed notation used in Polchinski must reproduce the very same result: \begin{align}
\delta\delta X^\mu(\tau)&=\xi^\alpha\delta_\alpha(\xi^\beta\delta_\beta X^\mu)=\xi^\alpha\delta_\alpha\left(\int d\tau_2\xi(\tau_2)\delta_{\tau_2}X^\mu(\tau)\right)=-\xi^\alpha\delta_\alpha\left(\int d\tau_2\xi(\tau_2)\delta(\tau-\tau_2)\overset{\cdot}{X}^\mu(\tau)\right)\\
&=-\int d\tau_1\delta_{\tau_1}\left(\int d\tau_2\xi(\tau_2)\delta(\tau-\tau_2)\overset{\cdot}{X}^\mu(\tau)\right)
\end{align}
Suppose now that $\delta_{\tau_1}(...)=-\delta(\tau-\tau_1)\partial_\tau(...)$, then we see that
\begin{align}
\delta\delta X^\mu(\tau)&=\int d\tau_1d\tau_2\xi(\tau_1)\xi(\tau_2)\delta(\tau-\tau_1)\partial_\tau\left(\delta(\tau-\tau_2)\overset{\cdot}{X}^\mu\right)=\int d\tau_2\xi(\tau)\xi(\tau_2)\partial_\tau\left(\delta(\tau-\tau_2)\overset{\cdot}{X}^\mu\right)\\
&=-\int d\tau_2\overset{\cdot}{\xi}(\tau)\xi(\tau_2)\delta(\tau-\tau_2)\overset{\cdot}{X}^\mu(\tau)=-\xi(\tau)\overset{\cdot}{\xi}(\tau)\overset{\cdot}{X}^\mu(\tau),
\end{align}
which is exactly what we obtained proviously. This computation agrees with your result and reproduces the result that one obtains in the usual way (without condensed notation).
From the above result, we see that $\delta_{\tau_1}\delta_{\tau_2}X^\mu(\tau)=+\delta(\tau-\tau_1)\partial_\tau\left(\delta(\tau-\tau_2)\partial_\tau X^\mu(\tau)\right)$, exactly as you wrote.
This results seem to contradict the answer given by Qmechanic, but they apparently reproduce the correct results.
