Why trace is more natural than (preferred to) determinant for smooth map $f:M\to N$?

Question

Cross-post from MSE.

For a continuous map $f:(M,g)\to (N,h)$, between Riemannian manifolds $(M,g)$ and $(N,h)$ we can pullback $h$ by $f$. Most experts take the trace from this new tensor and work with it, i.e. $\operatorname{tr}_g(f^*h)$ which I think is equal to $\lvert df\rvert^2$. 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, this and this posts) and not determinant or any other operator?

One primary reason is that it is similar to $\operatorname{tr} A^tB$ that is an inner product over $n\times n$ matrices.

In the case of energy density of harmonic maps, $e(f)\mathrel{:=}\frac{1}{2}\lvert df\rvert^2$ 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 reveal nice properties of the space as well as trace case? or that is same as trace case? $$K(f)\mathrel{:=}\int_M\det_g(f^*h)d\mathrm{vol}_g.$$

It is also helpful remember that the trace is $\sum_i\lambda_i$ and determinant is $\prod_i\lambda_i$.

Answer 1

It suffices to understand the special case of a linear map $T:U\to V$ where $U,V$ are Euclidean vector spaces. (Think $U=T_pM$, $V=T_{f(p)}N$, $T=df(p)$.)

Suppose first that $n=\dim V\leq \dim U=m$. Let $\lambda_1,\dotsc, \lambda_n, $ be the eigenvalues of the symmetric nonnegative operators $TT^*:V\to V$, multiplicities included. Then (see Lemma 1.1 in The co-area formula) there exist Euclidean coordinates $x^1,\dotsc, x^{m}$ on $U$ and Euclidean coordinates $y^1, \dotsc, y^n$ on $V$ such that $T$ is described un these coordinates by $$ y^i=\sqrt{\lambda_i} x^i,\;\;\forall i=1,\dotsc, n. $$ Denote by $g_V$ the inner product on $V$ and by $g_U$ the inner product. Then $$ g_V=\sum_{i=1}^n (dy^i)^2,\;\;T^* g_V= \sum_{i=1}^n \lambda_i (dx^i)^2. $$ We deduce $\DeclareMathOperator{\tr}{tr}$ \begin{align*} \tr_{g_U} T^*g_V&{}=\sum_i \lambda_i=\tr TT^*, \\ \det\limits_{g_U} T^*g_V&{}=0. \end{align*} In this case, more useful in applications is the Jacobian of $T$, $\DeclareMathOperator{\Jac}{Jac}$ $$ \Jac(T)\mathrel{:=}\prod_{i=1}^n\lambda_i =\det T T^*. $$ The Jacobian of $T$ plays an important role in the coarea formula, $$ \int_M \Jac_x(f) u(x) dV_g(x)=\int_N\left(\int_{f^{-1}(y)} u(x) dV_{f^{-1}(y)}(x)\right) dV_h(y),\;\;\forall u\in C_0(M). $$ This contains the change-in-variables formula as a special case.

If $m=\dim U<\dim V$, then we can find Euclidean coordinates $x^1,\dotsc, x^m$ and Euclidean coordinates $y^1,\dotsc, y^n$ on $V$ such that, in these coordinates $T$ is described by $$ y^i=\sqrt{\mu_i}x^i,\;\;\forall i=1,\dotsc, m,\;\;y_j=0,\;\;j>m, $$ where $\mu_1,\dotsc, \mu_m$ are the eigenvalues of the symmetric nonnegative operator $T^*T:U\to U$. (To see this apply the previous result to the map $T^*:V\to U$.)

In this case \begin{align*} \tr_{g_U} T^*g_V&{}=\sum_{i=1}^m \mu_i=\tr T^*T, \\ \det\nolimits_{g_U} T^*g_V&{}= \prod_{i=1}^m \mu_i=\det T^*T. \end{align*}