Attempt at proving $-i\ \langle{u_n|\nabla_k u_n}\rangle=-\dfrac{i}{2}tr[v^\dagger(k)\nabla_k v(k) ]$ from Kane and Fu's paper I am trying to prove result (3.4) of the following paper:
https://doi.org/10.1103/PhysRevB.76.045302,
namely, that $$-i\ \langle{u_n|\nabla_k u_n}\rangle=-\dfrac{i}{2}tr[v^\dagger(k)\nabla_k v(k) ]$$
where I am using the Einstein summation convention and $v$ is given by
$$v_{nm}(k)=\langle{u_n|P\Theta|u_m\rangle}$$
where $\Theta$ is the time reversal operator and $P$ is the space inversion operator, and all the $\langle{u_n|}$ are evaluated at the same $k$ (this is a vector).
My attempt is the following:
I first write $-i\langle{u_n|\nabla_k u_n\rangle}=-\langle{u_n|r|u_n\rangle}$ because I know that both $\Theta$ takes $r\mapsto r$ and $P$ takes $r\mapsto -r$. I will also use that $v v^+=v^\dagger v=1$, $P^\dagger P=1$ and $\langle{\Theta \psi |\Theta \phi\rangle}=\langle{\phi|\psi\rangle}=\langle\psi|\phi\rangle^*$.
So, here it goes:
$$-\langle{ u_n |r|u_n  \rangle}=-\langle{ P\Theta\ r\ u_n | P\Theta u_n \rangle}=\langle{ r\ P\Theta\ u_n | P\Theta u_n \rangle}=\langle{ r\ P\Theta\ u_n | u_m \rangle}\langle{u_m |P\Theta u_n \rangle}=v_{mn}\langle{ r\ P\Theta\ u_n | u_m \rangle}$$
I now use the chain rule (recalling that the  $k-$space representation of $r$ is a derivative with respect to $k$) to write for example $r\langle{\psi|\phi\rangle}=\langle{r\ \psi|\phi\rangle}+\langle{\psi|r\ \phi\rangle}$. Doing so, gives:
$$\langle{ u_n |r|u_n  \rangle}=v_{mn}\left(r\  \langle{P\Theta u_n|u_m\rangle}-\langle{P\Theta u_n|r\ u_m\rangle}   \right)$$
I now insert the resolution of the identity in the last term:
$$-\langle{ u_n |r|u_n  \rangle}=v_{mn}\left( r\ v^*_{mn}-\langle{P\Theta u_n|u_s \rangle}\langle{u_s|r\ u_m\rangle} \right)=v_{mn}r\ v^*_{mn}-\underbrace{v_{mn}v^*_{sn}}_{\delta_{ms}}\langle{u_s|r\ u_m\rangle}$$
$$=tr[v\ r\ v^\dagger]-\langle{u_m|r|u_m\rangle}=tr[v\ r\ v^\dagger]-\langle{u_n|r|u_n\rangle}$$
Which is clearly wrong!
I suspect that a mistake is that $$-\langle{ P\Theta\ r\ u_n | P\Theta u_n \rangle}=\langle{ r\ P\Theta\ u_n | P\Theta u_n \rangle},$$ but even if I don't change the sign here, the last result picks up a minus sign, which gives that $$-i\ \langle{u_n|\nabla_k u_n}\rangle=-\dfrac{i}{2}tr[v(k)\nabla_k v^\dagger(k) ]=\dfrac{i}{2}tr[v^\dagger(k)\nabla_k v(k) ]$$
which differs by a sign from the result I should have gotten; so, even if that sign change when I (anti)commute $r$ past $P\Theta$ doesn't fix things.
 A: The Kane-Fu paper seems to have a sign error.
First, we know that $PkP^{-1} = -k$ and $\Theta k \Theta^{-1} = -k$. Hence, we have $P\Theta \nabla_k \Theta^{-1} P^{-1} |\phi\rangle = \nabla_k|\phi\rangle$ (as an operator statement, e.g. acting on a ket!). This is consistent with what you have previously wrote regarding $r$. 
Let us work in reverse to prove the result. I will be using Einstein summation notation.
Let us be very careful about the derivatives. I define $\langle \nabla_k \phi | \equiv \nabla_k \langle \phi|$ and $|\nabla_k \phi\rangle \equiv \nabla_k | \phi\rangle$. 
\begin{align}
(v^\dagger)_{mn} \nabla_k v_{nm}
&= \langle P\Theta u_m|u_n\rangle \nabla_k \langle u_n|P\Theta u_m\rangle \\
&= \langle P\Theta u_m|u_n\rangle[\langle \nabla_k u_n|P\Theta u_m\rangle + \langle u_n|\nabla_k P\Theta u_m\rangle]\\
&= \langle \nabla_k u_n|P\Theta u_m\rangle\langle P\Theta u_m|u_n\rangle + \langle P\Theta u_m|u_n\rangle\langle u_n|\nabla_k P\Theta u_m\rangle \\
&= \langle\nabla_k u_n|u_n\rangle + \langle P\Theta u_m|\nabla_k P\Theta u_m\rangle \\
&= \langle\nabla_k u_n|u_n\rangle + \langle P\Theta u_m| P\Theta \nabla_k u_m\rangle \\
&= \langle\nabla_k u_n|u_n\rangle + \langle u_m| \nabla_k u_m\rangle^* \\
&= -\langle u_n| \nabla_k u_n\rangle - \langle u_m| \nabla_k u_m\rangle \text{$\hspace{3em}$ (since $\langle u_n|u_n\rangle =1$)}\\
&= -2\langle u_n|\nabla_k u_n\rangle
\end{align}
Hence, putting in $-i/2$, we get the desired result, though with opposite sign to the paper. I'm not exactly sure, but I believe the error of getting $0$ in your calculation comes from being a bit too careless about interchanging $r$ as an operator and $r$ as a derivative (eg not between bras and kets). 
