The partial derivative of a vector $V^\lambda , _\nu$ is not a tensor. Neither is a Christoffel symbol $\Gamma^\lambda _{\mu \nu}$.

Is the addition of these two objects a tensor? If they were tensors, the addition woulnd't even be defined.

Using $\Gamma^\lambda _{\mu \nu}=w^\lambda(\partial_\mu e_\nu)$ how can I prove that $\partial_\nu V^\lambda+w^\lambda(\partial_\mu e_\nu)$ is a tensor? It looks like the covariant derivative, wich is a tensor, but it isn't.

I have tried to find the tranformation rule:

\begin{equation} \partial_\gamma V^\alpha+w^\alpha(\partial_\beta e_\gamma)=\frac{\partial x^\nu}{\partial x^\gamma}\frac{\partial}{\partial x^\nu}\frac{\partial x^\alpha}{\partial x^\lambda}V^\lambda+\frac{\partial x^\alpha}{\partial x^\lambda}w^\lambda\left(\frac{\partial x^\mu}{\partial x^\beta}\frac{\partial}{\partial x^\mu}\frac{\partial x^\nu}{\partial x^\gamma}e_\nu\right) \end{equation}

But I am unsure on how to operate the different derivatives.

  • 2
    $\begingroup$ You seem to be adding terms with different index structures?? $\endgroup$ – Qmechanic Apr 13 '19 at 19:53
  • $\begingroup$ @Qmechanic yes! I thought that that was a result of neither the Christoffel symbol or the derivative of the vector being tensors. Is the addition even possible? $\endgroup$ – IchVerlore Apr 13 '19 at 19:59
  • $\begingroup$ Have you tried to see how it transforms under coordinate transformation? That will tell you if the thing is a tensor or not... eqn's 5-7 here mathworld.wolfram.com/Tensor.html $\endgroup$ – N. Steinle Apr 13 '19 at 21:05
  • $\begingroup$ This seems extremely confused, like your last question. You should read a basic introduction to tensors from the beginning before trying to do anything else... $\endgroup$ – knzhou Apr 13 '19 at 21:05

It's not that since the Christoffel symbol and the partial derivative are not tensors you cannot add them. Think of them as sets of values: the fact that are not tensors means just that they do not transform covariantly (or contravariantly) when you do a change of coordinates; it does not forbid you to sum them and as a matter of fact the covariant derivative maps tensors (of a certain rank) in tensors (of a different rank).

In your post you are not writing the Christoffel symbol as applied to the field you are deriving in the partial derivative. The covariant derivative would be: $$\nabla_{\mu}V_{\nu}:=\partial_{\mu}V_{\nu}-\Gamma_{\mu\nu}^{\lambda}V_{\lambda}$$ Now if I understand correctly you really mean to sum the three index Christoffel symbol with the two index partial derivative right?

In that case you do not have a tensor. You can see it by proceeding as you tried but fixing the fact that you are changing coordinates so $x\rightarrow x'$. The transformation law becomes: \begin{equation} \partial_\gamma V^\alpha+w^\alpha(\partial_\beta e_\gamma)\;\;\rightarrow\;\;\frac{\partial x^\nu}{\partial x'^\gamma}\frac{\partial}{\partial x^\nu}(\frac{\partial x'^\alpha}{\partial x^\lambda}V^\lambda)+\frac{\partial x'^\alpha}{\partial x^\lambda}w^\lambda\frac{\partial x^\mu}{\partial x'^\beta}\frac{\partial}{\partial x^\mu}(\frac{\partial x^\nu}{\partial x'^\gamma}e_\nu) \end{equation} Despite you can still see this as a three index object (the first term is simply multiplied by $I_{\beta}\equiv 1 \;\forall\;\beta$), you can see that now the spurious second derivatives do not cancel each others and therefore you do not get something that transform as a tensor.

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