# Why do derivatives act on vector fields on a worldsheet?

The covariant derivative of a vector $A^{\mu}$ at a point $x$ is defined as

$$D_z A^{\mu}=\partial_zA^{\mu}+\Gamma^{\mu}_{\rho\sigma}(x)\partial_{z}x^{\rho}A^{\sigma}$$

where Greek symbols are spacetime (dimension $d$) and $z$ is the index of a worldsheet coordinate $\sigma$.

Why doesn't the vector $A^{\mu}$ act as a scalar with respect to the $z$ derivative? How can I prove this?

Assume $A^{\mu}$ depends on $x^\mu(\sigma^b)$, i.e. $A^{\mu}(x^\mu(\sigma^b))$.

• I edited your question to clarify it, but I'm not 100% sure I got it right. If I changed what you meant to ask, please edit it again to fix it. By the way, we'd really rather you not delete your question and post a new one. This time it's fine, but in the future you should edit the old question instead. – David Z Nov 26 '14 at 8:12
• Possible duplicate by OP: physics.stackexchange.com/q/148384/2451 – Qmechanic Nov 26 '14 at 9:05
• Thanks for the edit. I apologize for the mess with the previous post. – sol0invictus Nov 26 '14 at 9:20

It's the chain rule: $$\partial_x f(y(x)) = \partial_y f(y(x)) \cdot \partial_x y(x)$$ Your vector field $A^\mu$ depends on the worldsheet coordinates only through the worldsheet coordinates $x^\mu$. Thus, when how $A^\mu$ behaves under an infinitesimal shift on the world-sheet you need to take into account how $A^\mu$ depends on $z$ - that's $\partial_z A^\mu$. But you also need to take into account how the worldsheet coordinates change - which is the connection piece.
$$D_z A^\mu = \underbrace{\partial_z A^\mu}_{\text{change in } A^\mu} + \underbrace{\Gamma^\mu_{\rho \sigma}(x) \partial_z x^\rho A^\sigma}_{\text{change of what "} \mu \text{" means.}}$$
• So if I had a quantity $D_zX^{\mu}(\sigma)$ would it be just partial derivative with no connection. This X is the same x where my previous covariant derivative was taken – sol0invictus Nov 26 '14 at 9:20