What I'm trying to show is that, if $v_a$ is a covector field, $\partial_{[a, v_b]} = \frac{1}{2}(\partial_a v_b – \partial_b v_a)$ transforms like a type $(0,2)$ tensor.
First of all, a type $(0,2)$ tensor transforms according to:
$$T'_{ab} = \frac{\partial x_a}{\partial x_\mu'} T_a + ~\frac{\partial x_b}{\partial x'_\nu}T_b$$
We have:
$$\partial'_{[a,v_b]} = \frac{1}{2}\frac{\partial x_\mu}{\partial x'_\nu}(\partial_a v_b – \partial_b v_a)$$
Working with one term:
$$\frac{1}{2} \frac{\partial x_\mu}{\partial x'_\nu}\left(\frac{\partial}{\partial x_a}v_b \right) = \frac{1}{2} \left[ \frac{\partial{x_\mu}}{\partial x'_\nu \partial x_a} v_b + \frac{\partial}{\partial x_a} \frac{\partial x_\mu}{\partial x'_\nu} v_b\right]$$
In the second term on the right, $x_\mu = x_a$, right? Is the partial derivative in the correct place, or do I have things switched around (that is, in the first term, $x_a = x_\mu$ instead of turning into a partial?)
What I think should be a simple, mechanical homework is pretty difficult for me.
Best Answer
You probably know that $\partial_\mu v_\nu$ is not a tensor. The curl $\partial_\mu v_\nu-\partial_\nu v_\mu$ is however. The transformation law for the derivative of a covector is $$(\partial_\mu v_\nu)'=\frac{\partial x^\rho}{\partial x'^\mu}\frac{\partial}{\partial x^\rho}\left(\frac{\partial x^\sigma}{\partial x'^\nu}v_\sigma\right)=\frac{\partial x^\rho}{\partial x'^\mu}\frac{\partial^2 x^\sigma}{\partial x^\rho x'^\nu}v_\sigma+\frac{\partial x^\rho}{\partial x'^\mu}\frac{\partial x^\sigma}{\partial x'^\nu}\partial_\rho v_\sigma$$ Now calculate the transformation law for $\partial_\nu v_\mu$ to find the transformation law of the curl. And for the record, a rank two covariant tensor transforms as $$t'_{\mu\nu}=\frac{\partial x^\rho}{\partial x'^\mu}\frac{\partial x^\sigma}{\partial x'^\nu}t_{\rho\sigma}$$ EDIT: What text are you using? I've never seen the notation you use for tensor transformations. It looks wrong to me.
EDIT 2: Derivation of the transformation law for an (r, s) tensor. We can derive the laws from the definition of a tensor. I actually had this here typed up already for a different purpose: We will start with a scalar field in one dimension, aka a function, $\phi(x)$. A differential change in the function is given by elementary calculus as \begin{equation} d\phi=\frac{d\phi}{dx}dx. \end{equation} Say we now have an actual scalar field, defined in multiple dimensions as $\phi=\phi(x,y)$. Then a differential change in the field is given by elementary multivariable calculus as \begin{equation} d\phi=\frac{\partial\phi}{\partial x}dx+\frac{\partial\phi}{\partial y}dy. \end{equation} In general, in $n$ dimensions, with $n$ variables $x^1,\dotsc,x^n$, the change in the field $\phi(x)$ (where the single $x$ is understood to represent the $n$-tuple variables) is given by \begin{equation} d\phi=\sum_\mu \frac{\partial \phi}{\partial x^\mu}dx^\mu=\frac{\partial \phi}{\partial x^\mu}dx^\mu, \end{equation} where in the last step I have employed the Einstein summation convention. Summation signs will no longer appear from this point on. Suppose that we also have $n$ scalar fields, $\phi_1,\dotsc,\phi_n$, which are all functions of our $n$ variables (The subscripts on the fields are not tensor indices, they merely denote which one of our $n$ fields we are talking about. Tensor indices always refer to a particular coordinate and scalar fields never have tensor indices). And suppose further that these scalar fields give us the equations for a global coordinate transformation. An example would be the equations $x=r\cos\theta$ and $y=r\sin\theta$, which tell us how to convert from polar coordinates to cartesian coordinates and vice-versa. In this example, $x$ and $y$ are functions, scalar fields in fact, of the polar coordinates $r$ and $\theta$. So let us now define our $\mu$th scalar field to give us the transformation equation for the $\mu$th transformed coordinate. \begin{equation} \phi_\mu=x'^\mu(x) \end{equation} Once again, the subscript on $\phi$ is not an index, it is simply lets us know that it is the $\mu$th scalar field. If we place (179) into (178), we have our first transformation law \begin{equation} dx'^\mu=\frac{\partial x'^\mu}{\partial x^\nu}dx^\nu. \end{equation} We define $dx^\mu$ to be a contravariant tensor. A contravariant tensor always has an upper index. This index is an actual tensor index, as such it can be summed over. A quantity $A^\mu$ is a contravariant vector (remember that vectors have one tensor index) if it transforms like the ur-vector $dx^\mu$ (or ur-tensor; we will later see that multi-indexed tensors are products of vector components) \begin{equation} A'^\mu=\frac{\partial x'^\mu}{\partial x^\nu}A^\nu. \end{equation} A contravariant vector $A^\mu(x)$ that depends on the coordinates $x$ and transforms as \begin{equation} A'^\mu(x')=\frac{\partial x'^\mu}{\partial x^\nu}A^\nu(x) \end{equation} is a contravariant vector field. Since we now have a tensor with an upper index, it would seem logical to look for an object that transforms differently, so that we can give it a lower index.
To find the transformation law for the second type of tensor, we'll consider the gradient of a scalar field $\phi(x)$, which is given by \begin{equation} \nabla\phi=\frac{\partial\phi}{\partial x^\mu}\hat{e}^{\mu}=\partial_{\mu}\phi\hat{e}^{\mu}, \end{equation} where we use the convenient notation $\partial_\mu=\frac{\partial}{\partial x^\mu}$, which is introduced in the main paper. Since $\hat{e}^{\mu}$ is a basis vector, the $\mu$th vector component is given by \begin{equation} (\nabla\phi)_\mu=\partial_{\mu}\phi. \end{equation} Our notation already reveals that this is leading directly to the lower-indexed tensor we are looking for. To follow convention, which is to give tensors in primed coordinate in terms of unprimed coordinate versions, like in (181), we'll say that (184) gives us the gradient in the primed coordinate system. We will go back to traditional partial derivative notation and invoke the chain rule to obtain \begin{equation} \partial'_{\mu}\phi=\frac{\partial\phi}{\partial x'^\mu}=\frac{\partial x^\nu}{\partial x'^\mu}\frac{\partial\phi}{\partial x^\nu}=\frac{\partial x^\nu}{\partial x'^\mu}\partial_\nu\phi, \end{equation} which is the transformation law for covariant vectors. We find that the scalar field $\phi$ is mathematically not necessary. It is, however, a good pedagogical tool to start with the familiar gradient. We now have the covariant ur-vector \begin{equation} \partial'_{\mu}=\frac{\partial x^\nu}{\partial x'^\mu}\partial_\nu. \end{equation} In analogy with (181) and (182), we have the covariant vector $C_\mu$ \begin{equation} C'_\mu=\frac{\partial x^\nu}{\partial x'^\mu}C_\nu \end{equation} and the covariant vector field $C_\mu(x)$ \begin{equation} C'_\mu(x')=\frac{\partial x^\nu}{\partial x'^\mu}C_\nu(x). \end{equation}
We see that the difference in the transformation laws is that contravariant vectors transform with $\partial x'/\partial x$ terms and covariant vectors transform with $\partial x/\partial x'$ terms. To sum everything up, contravariant vectors transform like coordinate differentials and covariant vectors transform like partial differential operators.
That was some easy multivariable calc. Now onto differential geometry. This is what I'm typing this up now. Let $M$ be a differentiable manifold. Let $T_p(M)_s^r$ be the set of tensors rank (r, s) defined on $T_p(M)$ (the tangent space of $M$ at the point $p$). If we assign to every $p\in M$ a tensor $t_p\in T_p(M)_s^r$, then the map $t:p\mapsto t_p$ defines a tensor field of type (r, s). Tensors can be expanded as $$t=t_{j_1\dotsc j_s}^{i_1\dotsb i_r}\left(\frac{\partial}{\partial x^{i_1}}\otimes\dotsb\otimes\frac{\partial}{\partial x^{i_r}}\right)\otimes\big( dx^{j_i}\otimes\dotsb\otimes dx^{j_s}\big)$$ where $t_{j_1\dotsc j_s}^{i_1\dotsb i_r}$ are the components. So for a rank two covariant tensor we have $$T=T_{ij}\left(dx^i\otimes dx^j\right)$$ For this to be coordinate invariant $T_{ij}$ must transform oppositely to $\left(dx^i\otimes dx^j\right)$, i.e like $$\left(\frac{\partial}{\partial x^i}\otimes\frac{\partial}{\partial x^j}\right)$$ We derived this transformation above. The justification of the above expansion of $t$ is highly nontrivial and probably not within your reach if you are just learning about tensors for the first time.