differential geometry - How does a vector field transform under an infinitesimal coordinate transformation?

If I have a vector $X^{\mu}(x)$, and then I consider an infinitesimal coordinate transformation of the form $x^{\mu} \to x^{\mu} + v^{\mu}(x)$, then how does my vector $X^{\mu}(x)$ transform?

From some reading online, it seems like the answer is along the lines of:

$$X^{\mu}(x) \ \to \ X^{\mu}(x) + v^{\sigma}(x) \partial_{\sigma} X^{\mu}(x) - X^{\sigma}(x) \partial_{\sigma} v^{\mu}(x)$$

I'm not really understanding where this is coming from though...is it because we're taking a Taylor expansion of $X^{\mu}(x+v)$? The minus sign is particularly unsettling for me.

Answer

If $X^\mu(x)$ is a vector field, in terms of coordinates $\{x^\mu\}$ and we consider an infinitesimal shift,

$$x^\mu \to x^\mu + v^\mu(x)$$

then the vector field $X^\mu$ changes according to the Lie derivative w.r.t. to the vector $v^\mu$, that is, we have that,

$$\delta X^\mu = \mathcal L_v X^\mu = v^\nu \nabla_\nu X^\mu - X^\nu \nabla_\nu v^\mu$$

by simplying applying the rules of Lie differentiation of a tensor. If the manifold is entirely flat, then covariant derivatives are demoted to partial derivatives,

$$\delta X^\mu = v^\nu \partial_\nu X^\mu - X^\nu \partial_\nu v^\mu$$

recovering the expression given by the OP. Notice that $v^\nu \partial_\nu$ is the same as the vector $v^\mu$ expressed as a derivation.$^\dagger$ As such, we can write,

$$\mathcal L_v X = [v,X]$$

with the Lie bracket, where $v$ and $X$ are the fields expressed as derivations, i.e. operators.

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$\dagger$ A vector as a derivation can be thought of as a directional derivative. Specifically, for a vector $v$ and a map $f : \mathbb R^n \to \mathbb R$, we have that,

$$D_v f(x) = \frac{d}{d\lambda} f(x + \lambda v) \bigg\rvert_{\lambda = 0} = v^\mu \partial_\mu f (x).$$

A case of interest is when $v$ is a vector along a curve, that is, it is the vector tangent along a path, in which case one may define differentiation of a map along a path on the manifold.