I'm trying to understand one part of the proof over here.
After some rearranging using the product rule, we arrive at the representation $$\displaystyle L(v) = \partial_\beta ( \partial_\alpha u \cdot v) + \partial_\alpha( \partial_\beta u \cdot v ) - 2 \partial_{\alpha \beta} u \cdot v.$$ Among other things, this allows for a way to right-invert the underdetermined linear operator ${L}$. As ${u}$ is free, we can use Cramer’s rule to find smooth maps ${w_{\alpha \beta}: ({\bf R}/{\bf Z})^n \rightarrow {\bf R}^d}$ for ${\alpha,\beta=1,\dots,n}$ (with ${w_{\alpha \beta} = w_{\beta \alpha}})$ that is dual to ${\partial_{\alpha \beta} u}$ in the sense that
$$\partial_\alpha u \cdot w_{\alpha' \beta'} = 0$$
$$\partial_{\alpha \beta} u \cdot w_{\alpha' \beta'} = \delta_{\alpha \alpha'} \delta_{\beta \beta'} + \delta_{\alpha \beta'} \delta_{\beta \alpha'} $$ where ${\delta}$ denotes the Kronecker delta.
I believe the relevant part for $u$ being free is over here
there are no dependencies whatsoever between the ${n + \frac{n(n+1)}{2}}$ scalar functions ${\partial_\alpha u \cdot v}, {\alpha=1,\dots,n}$ and ${\partial_{\alpha \beta} u \cdot v}, {1 \leq \alpha \leq \beta \leq n}$.
I am not able to see how Cramer's rule is applied over here to obtain the two equations involved and how does it guarantee the smoothness of the map (inverse of a matrix being smooth?).
The ideas over here seems to be very relevant but I have not been successful in trying to reframe it to the problem here and figuring out how Cramer's rule fit into the picture. Any help would be greatly appreciated!