I'm going through Frankel's 'The Geometry of Physics', and I'm stumped on problem 2.4(3) (iii) where we're asked to investigate whether $\partial_i v^j - \partial_j v^i$ transforms like a tensor on a change of coordinates. Okay, so setting the necessary tools up, we have:
Contravariant vectors (and hence indices on a tensor) transform like:
$$v^i \boldsymbol \partial_i = v^i \left(\frac{\partial x'^j}{\partial x^i}\right) \boldsymbol \partial'_j = v'^j \boldsymbol \partial'_j \rightarrow v'^j = v^i\left(\frac{\partial x'^j}{\partial x^i}\right)$$
And covariant vectors (and hence indices on a tensor) transform like:
$$a_i\boldsymbol {dx}^i = a_i\left(\frac{\partial x^i}{\partial x'^j}\right)\boldsymbol{dx'}^j = a'_j\boldsymbol{dx'}^j \rightarrow a'_j = a_i\left(\frac{\partial x^i}{\partial x'^j}\right)$$
Okay so I define $T_i^j$:
$$T_i^j = \partial_i v^j - \partial_j v^i = \frac{\partial v^j}{\partial x^i} - \frac{\partial v^i}{\partial x^j}$$
And then, I start in the primed coordinate system so I can use the above identities.
$${T'}_i^j = \partial'_i v'^j-\partial'_j v'^i = \partial'_i\left\{v^l\left(\frac{\partial x'^j}{\partial x^l}\right)\right\}- \partial'_j\left\{ v^l \left( \frac{\partial x'^i}{\partial x^l} \right)\right\}$$
$$= \partial'_i v^l\left(\frac{\partial x'^j}{\partial x^l}\right) + v^l \partial'_i\left\{\left(\frac{\partial x'^j}{\partial x^l}\right)\right\}- \partial'_j v^l \left( \frac{\partial x'^i}{\partial x^l} \right) - v^l\partial'_j \left\{ \left( \frac{\partial x'^i}{\partial x^l} \right)\right\}$$
$$= \left[ \partial'_i v^l\left(\frac{\partial x'^j}{\partial x^l}\right) - \partial'_j v^l \left( \frac{\partial x'^i}{\partial x^l} \right) \right]^{(1)} + \left[v^l\left( \partial'_i\left\{\left(\frac{\partial x'^j}{\partial x^l}\right)\right\} - \partial'_j \left\{ \left( \frac{\partial x'^i}{\partial x^l} \right)\right\}\right) \right]^{(2)}$$ $$ = \left[ \partial_k v^l \left( \frac{\partial x^k}{\partial x'^i} \right) \left(\frac{\partial x'^j}{\partial x^l}\right) - \partial_k v^l \left( \frac{\partial x^k}{\partial x'^j} \right) \left( \frac{\partial x'^i}{\partial x^l} \right) \right]^{(1)} + \left[v^l\left( \partial'_i\left\{\left(\frac{\partial x'^j}{\partial x^l}\right)\right\} - \partial'_j \left\{ \left( \frac{\partial x'^i}{\partial x^l} \right)\right\}\right) \right]^{(2)} $$
And that is where I get stuck.
The problem is , I can't seem to pull a common factor of $\left( \frac{\partial x_k}{\partial x'^i}\right) \left( \frac{\partial x'^j}{\partial x^l}\right)$ out of (1). The $i$ index and the $j$ index transform differently between the left and right parts. This isn't a problem in the book when $F_{ij} = \partial_i A_j - \partial_j A_i$ is investigated because both indices are down. When that's true, you can change the dummy indices on one term of $\partial_k v^l$, and pull the necessary factor out.
It's the same problem in (2) as well. Normally I'd say those terms would cancel, but the $i$ and $j$ indices are again swapped in the two terms.
So my gut says that it doesn't transform as a tensor, however when I look at solutions like this: https://kupdf.net/download/the-geometry-of-physics-frankel-solutions_58b5601d6454a7647cb1e95d_pdf This person claims it is a tensor and just seems to hand-wave away the swapping index issue.
Additionally, the way this problem is set up leads me to believe that $T_i^j$ should in fact be a tensor, as the previous part you're asked whether $\partial_i v^j$ is a tensor, and it is not.
I also tried this with $v^j = g^{ji}v_i$ out of hopes that 'lowering' $v$'s index might make this a little easier, but it doesn't seem to.
Anyway, can someone convince me that this handwave is okay? How can I deal with the swapped indices and pull a common term out of (1)?
Jackozee Hakkiuz has pointed out that in fact, the link I referred to actually says the object isn't a tensor. I guess I convinced myself it was, and was glossing over the $\neq$. So I was right that $\partial_i v^j - \partial_j v^i$ isn't a tensor.