Definition of a tensor field Could anybody explain to me the following:

If $$T_{ij}=\nabla_i V_j-\nabla_j V_i$$
  where $V_i$ is a covector field and $\nabla_i$ is the covariant derivative,
  then $T_{ij}$ is a tensor field.

I am new to all hese covector, covariant, etc. and am very confused.
I believe the definition of a tensor is one which transforms according to the tensor transformation rule... right? But I don't see how that applies here. 
Thank you in advance.
 A: Alternatively to Neal, you can approach this in a coordinate-free way using the Jacobian and some wedge products.  
In particular, if $f(x) = x'$ defines a coordinate transformation, then $\underline f(a) = a \cdot \partial f$ is the Jacobian, and $\overline f(a)$ denotes the transpose.  Covectors transform according to the transpose.  That is,  
$$V = \overline f(V'), \quad \nabla = \overline f(\nabla')$$  
where the primes refer to objects in the new coordinates.  If $T$ is a tensor, it must follow $T = \overline f(T')$.  We know that $T = \nabla \wedge V$, so $T' = \nabla' \wedge V'$ if $T$ is a tensor.  Let's check that:  
$$T = \nabla \wedge V = \overline f(\nabla') \wedge \overline f(V') = \overline f(\nabla ' \wedge V') = \overline f(T')$$  
If you can convert that approach to index notation, you deserve credit for the problem.
A: Here are two paths you can take toward solving the problem.


*

*You can take your definition of "tensor" as a set of indexed quantities which transform according to a certain rule.  You need to show that $T_{ij} = \nabla_iT_j - \nabla_jT_i$ transforms according to these rules.  You know the expression for the covariant derivative in terms of derivatives and Christoffel symbols, so expand out $\nabla_i T_j - \nabla_j T_i$ and simplify until you get a linear combination of quantities you already know are tensors.  Then you're done, since a linear combination of tensors is again a tensor.

*To work from a coordinate-free vantage point, you can use a slightly different definition of "tensor": a $(p,q)$-tensor is a pointwise linear map which takes as input $p$ vector fields and $q$ covector fields and returns a function.  From this perspective, you have defined the function $T(X,Y) = \nabla_XY - \nabla_YX$.  You would like to show it is a tensor.  So show that it is pointwise linear in both inputs, i.e., that for any smooth functions $f,g$, $T(fX,gY) = fgT(X,Y)$.
For more, you might check out the Wikipedia page on the torsion tensor.
A: Here's how to check it if your connection is torsion-free (e.g. you're using the Levi-Civita connection).
First, note that
\begin{align}
T_{ij} & = \nabla_i V_j - \nabla_j V_i \\
& = (\partial_i V_j - \Gamma^k_{\phantom{k}ji} V_k) - (\partial_j V_i - \Gamma^k_{\phantom{k}ij} V_k) \\
& = \partial_i V_j - \partial_j V_i - \Gamma^k_{\phantom{k}ij}V_k + \Gamma^k_{\phantom{k}ij} V_k \\
& = \partial_i V_j - \partial_j V_i,
\end{align}
where the $\Gamma^k_{\phantom{k}ij}$ are the Christoffel symbols of the connection and $\Gamma^k_{\phantom{k}ij} = \Gamma^k_{\phantom{k}ji}$ by the assumption that the connection is torsion-free.
Now suppose we make a change of coordinates $x_i \mapsto x_{i'}$. Then
\begin{align}
T_{i'j'} & = \partial_{i'} V_{j'} - \partial_{j'} V_{i'} \\
& = \partial_{i'} \left( \frac{\partial x_k}{\partial x_{j'}} V_k \right) - \partial_{j'} \left( \frac{\partial x_l}{\partial x_{i'}} V_l \right) \\
& = \frac{\partial^2 x_k}{\partial x_{i'} \partial x_{j'}} V_k + \frac{\partial x_k}{\partial x_{j'}} \partial_{i'} V_k - \frac{\partial^2 x_l}{\partial x_{j'} \partial x_{i'}} V_l - \frac{\partial x_l}{\partial x_{i'}} \partial_{j'} V_l \\
& = \frac{\partial x_k}{\partial x_{j'}} \partial_{i'} V_k - \frac{\partial x_l}{\partial x_{i'}} \partial_{j'} V_l \\
& = \frac{\partial x_k}{\partial x_{j'}} \frac{\partial x_l}{\partial x_{i'}} \partial_{l} V_k - \frac{\partial x_l}{\partial x_{i'}} \frac{\partial x_k}{\partial x_{j'}} \partial_{k} V_l \\
& = \frac{\partial x_k}{\partial x_{j'}} \frac{\partial x_l}{\partial x_{i'}} (\partial_l V_k - \partial_k V_l) \\
& = \frac{\partial x_k}{\partial x_{j'}} \frac{\partial x_l}{\partial x_{i'}} T_{kl}.
\end{align}
Then we see that the $T_{ij}$ form the components of a type $(0,2)$ tensor.
