# Covariant derivative equivalence

## Main Question or Discussion Point

If we are representing the basis vectors as partial derivatives, then $\frac{\partial}{\partial x^\nu + \Delta x^\nu} = \frac{\partial}{\partial x^\nu} + \Gamma^\sigma{}_{\mu \nu} \Delta x^\mu \frac{\partial}{\partial x^\sigma}$ to first order in $\Delta x$, correct? But in the same manner we could write $\frac{\partial}{\partial x^\nu + \Delta x^\nu} = \frac{\partial x^\kappa}{\partial x^\nu + \Delta x^\nu} \frac{\partial}{\partial x^\kappa} \approx \frac{\partial}{\partial x^\nu}(x^\kappa - \Delta x^\kappa) \frac{\partial}{\partial x^\kappa}$.

But these two equalities seem to lead to a weird result, namely that if $\Delta x$ are not function of $x$ then the $\Gamma$ vanish. But I'm even not sure about the validity of the equality. Could someone shed some light on this?

(If someone is curious about why I'm asking this, it's just something that I was thinking about, out of curiosity.)

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haushofer
What does that derivative with that Delta mean mathematically? Personally, I have no idea to be honest, so I can't help you.

What does that derivative with that Delta mean mathematically? Personally, I have no idea to be honest, so I can't help you.
Well in the same way that $\frac{d}{d g(x) + f(x)}$. Think of the $\Delta x$ just as a different symbol.

Orodruin
Staff Emeritus
Well in the same way that $\frac{d}{d g(x) + f(x)}$. Think of the $\Delta x$ just as a different symbol.