For a continuous map $f:(M,g)\to (N,h)$, between Riemannian manifolds $(M,g)$ and $(N,h)$ we can pullback the $h$ by $f$. Most of experts take trace from this new tensor and work with it. i.e. $\mathrm{tr_g}(f^*h)$ which I think is equals to $|df|^2$. I think there is a simple reason from Linear Algebra that perhaps I missed it that
Question: why they use trace (e.g. see this post) and not determinant or any other operator?
One primary reason is that it is similar to $\mathrm{tr }A^tB$ that is an inner product over $n\times n$ matrices.
In the case of energy density of harmonic maps, $e(f):=\frac{1}{2}|df|^2$ is very natural operator because it is similar to (up to a constant $m$) the kinetic energy formula $E=\frac{1}{2}mv^2$ in physics.
But these are not sufficient to not consider the determinant (or any other operator) case. I want to know Is the following expression meaningful and can it reveals nice properties of the space as well as trace case? or that is same as trace case? $$K(f):=\int_M\det_g(f^*h)dvol_g.$$
Update: It is also helpful remember that the trace is $\sum_i\lambda_i$ and determinant is $\prod_i\lambda_i$.