Covariant Derivative Equivalence: Exploring an Intriguing Result

[kent davidge]

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.)

 

[haushofer]

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

 

[kent davidge]

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]

Where are you getting this notation from? It seems non-standard and it is unclear what you mean.

 

[haushofer]

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

I've never seen such a notation.