How can I show components of something transform as a (0,2) tensor? Suppose that $B_i$ are the components of a covariant vector. Show that $\frac{\partial B_j}{\partial x^k} - \frac{\partial B_k}{\partial x^j}$ are the components of a (0,2) tensor. 
I know that the components of a (0,2) tensor transform like this:
$\frac{\partial \bar{B_j}}{\partial \bar{x^k}} -\frac{\partial \bar{B_k}}{\partial \bar{x^j}}  = \sum_{\alpha,\beta=1}^n \left ( \frac{\partial B_{\alpha}}{\partial x^{\beta}} - \frac{\partial B_{\beta}}{\partial x^{\alpha}}\right)\frac{\partial x^{\alpha}}{\partial \bar{x}^j} \frac{\partial x^{\beta}}{\partial \bar{x}^k}$  
So $$\bar{B}_j = \sum_{\alpha=1}^n B_{\alpha} \frac{\partial x^{\alpha}}{\partial\bar{x}^j}$$
and $$\bar{B}_k = \sum_{\beta=1}^n B_{\beta} \frac{\partial x^{\beta}}{\partial\bar{x}^k}$$
Can someone please help me solve this question? I am preparing for my exam and my professor said this question is important to know and understand. 
 A: You are already in the half way. The next step is just differentiate $\bar{B}_j =  B_{\alpha} \frac{\partial x^{\alpha}}{\partial\bar{x}^j}$ and $\bar{B}_k = B_{\beta} \frac{\partial x^{\beta}}{\partial\bar{x}^k}$ as
$$\frac{\partial\bar{B}_j}{\partial \bar{x}^k } = \frac{\partial B_{\alpha}}{\partial \bar{x}^k } \frac{\partial x^{\alpha}}{\partial\bar{x}^j}+B_{\alpha} \frac{\partial^2 x^{\alpha}}{\partial\bar{x}^k\partial\bar{x}^j} = \frac{\partial B_{\alpha}}{\partial x^{\gamma}} \frac{\partial x^{\gamma}}{\partial\bar{x}^k} \frac{\partial x^{\alpha}}{\partial\bar{x}^j}+B_{\alpha} \frac{\partial^2 x^{\alpha}}{\partial\bar{x}^k\partial\bar{x}^j}$$
and 
$$\frac{\partial\bar{B}_k}{\partial \bar{x}^j } = \frac{\partial B_{\beta}}{\partial \bar{x}^j } \frac{\partial x^{\beta}}{\partial\bar{x}^k}+B_{\beta} \frac{\partial^2 x^{\beta}}{\partial\bar{x}^j\partial\bar{x}^k} = \frac{\partial B_{\beta}}{\partial x^{\gamma}} \frac{\partial x^{\gamma}}{\partial\bar{x}^j} \frac{\partial x^{\beta}}{\partial\bar{x}^k}+B_{\beta} \frac{\partial^2 x^{\beta}}{\partial\bar{x}^j\partial\bar{x}^k}$$
changing dummy indices from first equation $\gamma \rightarrow \beta$ (for the first term) and the second equation $\gamma \rightarrow \alpha$ (for the first term) and $\beta \rightarrow \alpha$ (for the second term) and substract them, you'll have the desired result. Note that the last terms cancel each other because we always assume the transformation is smooth so that the mixed partial derivatives is commute.
