I am a maths student taking a module in (the mathematics of) Relativity so I get quite confused when looking for stuff that may help me understand where I go wrong in certain questions as I'm not fully aware what most symbols and notation are.

I have been asked to show, when given a first order covariant tensor $T_{a}$ that $A_{ab}$ is a second order covariant tensor where $$A_{ab}=T_{a,b}-T_{b,a}$$ ( ,b denotes the partial derivative with respect to $y^b$, and Einstein's Summation Convention is assumed).

I have the following; By definition under a change in co-ordinates $\left\{y^a\right\} \rightarrow \left\{\overline{y}^a\right\}$, $$\overline{A_{ab}}=\overline{T}_{a,b}-\overline{T}_{b,a}=\frac{\partial}{\partial \overline{y}^b}\left(\frac{\partial y^p}{\partial \overline{y}^a} A_p\right)-\frac{\partial}{\partial \overline{y}^a}\left(\frac{\partial y^q}{\partial \overline{y}^b} A_p\right)$$ and upon working it all through using the product rule and chain rule I keep coming up with $\overline{A}_{ab}=0$ which is not the definition of a second order covariant tensor. I really do not understand where I am going wrong and I can't put the rest of my working out in here as I'm only just starting using LaTeX so it would take me forever.

I'm not sure if the notation I've used is understandable to physicists but its all I know. Any help would be much appreciated, thanks .

[Also - is this question OK on this stackexchange rather than the maths one? I thought it had more to do with physics so I posted it here.]

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    $\begingroup$ You forgot to transform the derivative! You must start with $\frac{\partial}{\partial y^b}$ and transform it into a derivative in $\bar{y}$, too - it looks as if you have just replaced the ${}_{,b}$ with the derivative in $\bar{y}$, which is wrong. $\endgroup$ – ACuriousMind Nov 22 '16 at 14:09
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    $\begingroup$ Another way of saying what @ACuriousMind is saying is that the $,b $ notation means the covariant derivative, not just the partial derivative. $\endgroup$ – Lewis Miller Nov 22 '16 at 15:19
  • $\begingroup$ That looks like a comma separating the $a,b$ and $b,a$. If the notation is being followed correctly this would imply a partial derivative whereas the notation for covariant derivative is usually a semicolon as in $a;b$ or $b;a$. $\endgroup$ – K7PEH Nov 22 '16 at 15:59
  • $\begingroup$ OP meant partial derivative; I think he just wants to show that exterior derivative is covariant, unlike most expression with tensor partial derivatives. I am writing an answer now. $\endgroup$ – Brian Moths Nov 22 '16 at 16:00
  • $\begingroup$ I guess I just dated myself. I studied tensors 50 years ago from a 35 year old text. The notation then (Applications of Tensor Analysis, A J McConnell) used a comma to denote the covariant derivative. I still have the text and just checked. $\endgroup$ – Lewis Miller Nov 23 '16 at 4:26

So to recap, you are starting with a covector $T_a$, and then you take was is called the "exterior derivative" to get a rank (0,2) tensor $A_{ab}$ defined by $A_{\mu \nu} = T_{\mu,\nu} - T_{\nu,\mu}$. Here the comma in the subscript denotes partial derivative so that $T_{\mu,\nu}=\dfrac{\partial}{\partial y^\nu} T_\mu$. Notice also that we are considering the partial derivative and not the covariant derivative (we may not even have a metric). I am using greek indices as coordinate indices and roman indices as abstract inidices.

Now suppose we have a prime coordinate system, related to the unprimed system by the covector transformation law $T'_{\nu'} =J^\nu_{\nu'} T_\nu$, where $T'$ is the coordinates of the tensor in the transformed coordinate system. Then we would expect a rank (0,2) tensor to transform according to the law $A'_{\mu'\nu'} =J^\mu_{\mu'}J^\nu_{\nu'} A_{\mu \nu}$. But is this true for $A$ as we have defined it. Let's find out.

In the unprimed coordinate system, the coordinates of $A$ are given by $$A_{\mu \nu}=T_{\mu,\nu}-T_{\nu,\mu}$$

Now in the primed coordinate system the coordinates of $A$ are given by $$A'_{\mu' \nu'}=T'_{\mu',\nu'}-T'_{\nu',\mu'}= \partial_{\nu'} T'_{\mu'} - \partial_{\mu'} T'_{\nu'} = \partial_{\nu'} J^{\mu}_{\mu'} T_\mu-\partial_{\mu'} J^{\nu}_{\nu'} T_\nu.$$

Now the partial derivative operator transforms according to the rule $\partial_{\mu'}= J^{\mu}_{\mu'} \partial_\mu$, so we get

$$A'_{\mu' \nu'}= J^{\nu}_{\nu'}\partial_{\nu} J^{\mu}_{\mu'} T_\mu-J^{\mu}_{\mu'}\partial_{\mu} J^{\nu}_{\nu'} T_\nu =J^{\nu}_{\nu'} J^{\mu}_{\mu'} \partial_{\nu} T_\mu + T_\mu J^{\nu}_{\nu'}\partial_{\nu} J^{\mu}_{\mu'} - J^{\mu}_{\mu'} J^{\nu}_{\nu'} \partial_{\mu} T_\nu-T_\nu J^{\mu}_{\mu'}\partial_{\mu} J^{\nu}_{\nu'} \\ = J^{\mu}_{\mu'} J^{\nu}_{\nu'} A_{\mu \nu} + T_\mu J^{\nu}_{\nu'}\partial_{\nu} J^{\mu}_{\mu'}- T_\nu J^{\mu}_{\mu'}\partial_{\mu} J^{\nu}_{\nu'},$$

where the second to last equality is by the product rule, and the last equality groups the first and third terms of the second last expression using the definition of $A_{\mu \nu}$. But the first term in the last expression is all we are supposed to get according to our expectation $A'_{\mu'\nu'} =J^\mu_{\mu'}J^\nu_{\nu'} A_{\mu \nu}$. Therefore, we must show that $T_\mu J^{\nu}_{\nu'}\partial_{\nu} J^{\mu}_{\mu'}- T_\nu J^{\mu}_{\mu'}\partial_{\mu} J^{\nu}_{\nu'}=0$. This is the crucial step which only works because $A$ is an exterior derivative. Moving back to primed derivatives and changing a dummy index, we have

$$T_\mu J^{\nu}_{\nu'}\partial_{\nu} J^{\mu}_{\mu'}- T_\nu J^{\mu}_{\mu'}\partial_{\mu} J^{\nu}_{\nu'} =T_\mu \partial_{\nu'} J^{\mu}_{\mu'}- T_\nu \partial_{\mu'} J^{\nu}_{\nu'} = T_\mu \partial_{\nu'} J^{\mu}_{\mu'}- T_\mu \partial_{\mu'} J^{\mu}_{\nu'}=T_\mu \left( \partial_{\nu'} J^{\mu}_{\mu'} - \partial_{\mu'} J^{\mu}_{\nu'}\right),$$

so all that remains is to show that $\partial_{\nu'} J^{\mu}_{\mu'} = \partial_{\mu'} J^{\mu}_{\nu'}$. But remember that the coordinate transformation Jacobian matrix $J$ is given by $J^{\mu}_{\mu'}= \dfrac{\partial y^\mu}{\partial y'^{\mu'}},$ so that $\partial_{\nu'} J^{\mu}_{\mu'} = \dfrac{\partial^2 y^\mu}{\partial y'^{\nu'} \partial y'^{\mu'}} = \dfrac{\partial^2 y^\mu}{\partial y'^{\mu'} \partial y'^{\nu'}} = \partial_{\mu'} J^{\mu}_{\nu'},$ and we are done.

Going forward you will learn that the exterior derivative can operate on tensors of any rank, and will always give a result that transforms correctly under changes of coordinates, but that other expressions involving partial derivatives do not have cancellations of the extra terms (involving derivatives of the jacobian matrix), and so these expressions do not transform correctly so they aren't really tensors, and you must instead use the covariant derivative instead of partial derivatives.


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