A few questions about the covariant derivative

[etotheipi]

Hey everyone, I was trying to learn in an unrigorous way a bit about making derivatives in the general manifold, but I'm getting confused by a few things. Take a vector field $V \in \mathfrak{X}(M): M \rightarrow TM$, then in some arbitrary basis $\{ e_{\mu} \}$ of $\mathfrak{X}(M)$ we have$$\frac{\partial V}{\partial x^\rho} = \frac{\partial}{\partial x^\rho} \left( V^{\mu} e_{\mu} \right) = \frac{\partial V^{\mu}}{\partial x^\rho} e_{\mu} + V^{\mu} \Gamma^{\sigma}_{\rho \mu} e_{\sigma} = \frac{\partial V^{\mu}}{\partial x^\rho} e_{\mu} + V^{\sigma} \Gamma^{\mu}_{\rho \sigma} e_{\mu} = \left(\frac{\partial V^{\mu}}{\partial x^\rho} + V^{\sigma} \Gamma^{\mu}_{\rho \sigma} \right) e_{\mu}$$because of definition of the symbols $\partial_a e_b = \Gamma^{c}_{ab} e_c$. Then we define a "covariant derivative" as a (1,1) tensor field$$\nabla V = \left(\frac{\partial V^{\mu}}{\partial x^\rho} + V^{\sigma} \Gamma^{\mu}_{\rho \sigma} \right) e_{\mu} \otimes e^{\rho} \equiv {(\nabla V)^{\mu}}_{\rho} e_{\mu} \otimes e^{\rho}$$This tensor can naturally be seen as a map from the tangent space to itself, e.g. if we act the tensor upon an arbitrary basis vector $e_{\nu}$$$\nabla V(\, \cdot\,, e_{\nu}) = \left(\frac{\partial V^{\mu}}{\partial x^\nu} + V^{\sigma} \Gamma^{\mu}_{\nu \sigma} \right) e_{\mu} = \frac{\partial V}{\partial x^{\nu}}$$My first question is that sometimes I have seen the notation $\nabla_{\nu} V$ used, but never seen it defined. Is this defined to be $\nabla_{\nu} V \equiv \nabla_{e_{\nu}} V \equiv \partial_{\nu} V$, as above, or something else?

Secondly, say we have some other vector field $U \in \mathfrak{X}(M): M \rightarrow TM$, then how do we define $\nabla_{U} V$, which is a function $\mathfrak{X}(M) \times \mathfrak{X}(M) \rightarrow \mathfrak{X}(M)$? Maybe something like$$\nabla(U,V) = \nabla(U^{\mu} e_{\mu}, V)\equiv \nabla_{U} V = \nabla_{U^{\mu} e_{\mu}} V = U^{\mu} \nabla_{e_{\mu}}V = U^{\mu} \partial_{\mu} V$$and I was wondering if this is correct? Also, if $f$ is some function and $W$ a vector field, then why does $\nabla_{fW} = f \nabla_W$?

Thanks!

 

[strangerep]

Since you said "unrigorous"...
$$\nabla_\rho V^\mu ~:=~ \frac{\partial V^\mu}{\partial x^\rho} ~+~ V^\sigma \Gamma^\mu_{~\rho \sigma} $$ $$U^\rho \nabla_\rho V^\mu ~=~ U^\rho \left( \frac{\partial V^\mu}{\partial x^\rho} ~+~ V^\sigma \Gamma^\mu_{~\rho \sigma} \right)$$

 

[etotheipi]

Right cool thanks, that looks like what I have. Now I feel stupid for writing out so much, oh well

 

[etotheipi]

There's one other aspect I wanted to clarify... do we define the covariant derivative of the components of an arbitrary tensor $T$ with respect to some basis to be$$\nabla_V T^{a_1, \dots, a_m}_{b_1,\dots,b_n} = (\nabla_{V} T)^{a_1, \dots, a_m}_{b_1,\dots,b_n}$$where $\nabla_V = V^{\mu} \nabla_{e_{\mu}}$, as usual? Or, is the notation on the LHS just shorthand for that on the RHS?

The LHS does seem a bit ambiguous, given that a tensor component is a function on a manifold - whose covariant derivative would ordinarily just be a partial derivative.

 

[PeterDonis]

do we define the covariant derivative of the components of an arbitrary tensor $T$ with respect to some basis

No, we don't. There is no such thing as "the covariant derivative of the components". The covariant derivative is an operator that acts on tensors. It doesn't act on components.

Also, the covariant derivative is a different operator from the absolute derivative along a curve, which is what you seem to be talking about.

The covariant derivative operator, $\nabla$, takes an $(m, n)$ tensor (i.e., a tensor with $m$ upper and $n$ lower indexes) to an $(m, n + 1)$ tensor. That's why, in component notation, it is often written as $\nabla_a$, but it would be more correct to write, not $\nabla_a T_{bc}$ but $(\nabla T)_{abc}$.

The absolute derivative operator along a curve, $\nabla_V$, tells how a tensor changes along a curve whose tangent vector is $V$. If all you know is the values of the tensor on the curve, it makes no sense to write $\nabla_V$ as $V^a \nabla_a$; to write it in the latter form, you need to know the components of a vector field $V$ in some neighborhood of the curve, not just the tangent vector of the curve itself.

This Insights article might be helpful:

https://www.physicsforums.com/insights/precession-in-special-and-general-relativity/

 

[etotheipi]

Thanks, that clarifies it! In that case the $\nabla_V T^{a_1, \dots, a_m}_{b_1,\dots,b_n}$ is just shorthand.

And yeah, I should have said that $T$ is a tensor field, otherwise it certainly doesn't make sense

 

[dsaun777]

Since you said "unrigorous"...
$$\nabla_\rho V^\mu ~:=~ \frac{\partial V^\mu}{\partial x^\rho} ~+~ V^\sigma \Gamma^\mu_{~\rho \sigma} $$ $$U^\rho \nabla_\rho V^\mu ~=~ U^\rho \left( \frac{\partial V^\mu}{\partial x^\rho} ~+~ V^\sigma \Gamma^\mu_{~\rho \sigma} \right)$$

Is that second equation a directional derivative?

 

[etotheipi]

Is that second equation a directional derivative?

I believe it's the $\mu$ component of the vector field $\nabla_{U} V$, i.e. the $\mu$ component of the derivative of $V$ along the vector $U$.

Also, if $U = \dot{\gamma}$ is the tangent vector to a curve $\gamma$, then it's the $\mu$ component of the derivative of the vector field $V$ along the curve.