Where does the minus sign come from in this expression? I'm trying to solve a problem from a course on General Relativity. Consider the change of coordinates $x^\mu \to x'^\mu  = x^\mu + \varepsilon \xi^{\mu} (x)$ and for a tensor $T$ define $\delta T(x) = T'(x) - T(x)$. 
I want to show that for a vector $V^\mu$ 
$$\delta V^\mu = - \varepsilon \left[ \xi^a \partial_a V^\mu - V^a\partial_a \xi^\mu \right] + O(\varepsilon^2)$$
I understand that this is to gain insight in the concept of Lie derivative but I don't understand where the minus sign inside the parenthesis comes from.
I've performed the calculations and I always get the same expression but with a plus sign.
$$\delta V^\mu = - \varepsilon \left[ \xi^a \partial_a V^\mu + V^a\partial_a \xi^\mu \right] + O(\varepsilon^2)$$
Am I wrong? In case that I am, where does the sign come from?
 A: Your standard vector transformation is:$$V^{\prime \mu} (x^\prime) = \frac{\partial x^{\prime \mu}}{\partial x^\nu} V^\nu(x).$$
For the given coordinate transformation, we have $$ \frac{\partial x^{\prime \mu}}{\partial x^\nu} = \delta^\mu_\nu + \varepsilon \partial_\nu \xi^\mu.$$
So we get $$V^{\prime \mu} (x^\prime) = (\delta^\mu_\nu + \varepsilon \partial_\nu \xi^\mu) V^\nu(x).$$
However, we want both sides in the same coordinates in order to calculate $\delta V^\mu$. So write $$ V^\nu(x)=V^\nu(x^\prime - \varepsilon \xi)=V^\nu(x^\prime)-\varepsilon\xi^a\partial_a V^\nu (x^\prime)+O(\varepsilon^2).$$
Substituting this into the above gives
$$V^{\prime \mu} (x^\prime) = V^\mu(x^\prime) + \varepsilon V^\nu(x^\prime) \partial_\nu \xi^\mu - \varepsilon\xi^a\partial_aV^\mu(x^\prime) + O(\varepsilon^2).$$
Finally, subtract $V^\mu(x^\prime)$ from both sides and drop the explicit coordinate notation, since everything is in the same coordinates now, and we get
$$ \delta V^\mu = -\varepsilon \left[\xi^a\partial_aV^\mu - V^a \partial_a \xi^\mu \right] + O(\varepsilon^2), $$
which is what you wanted.
I'm not sure where your minus sign has come from. Maybe you used the transformation law for covectors instead of vectors.
