Let $T>0$. If $f:[0,T]\to\mathbb R$, let $$\operatorname{Var}_\varsigma(f):=\sum_{i=1}^k|f(t_i)-f(t_{i-1})|$$ denote the variation of $f$ along a partition $\varsigma=(t_0,\ldots,t_k)$ with $0=t_0<\cdots<t_k=T$ and $k\in\mathbb N$ of $[0,T]$.
Now, let $f_{ij}:[0,T]\to\mathbb R$ be of bounded variation for $i,j=1,2$ and consider $$f:=\left(\begin{array}{cc}f_{11}&f_{12}\\f_{21}&f_{22}\end{array}\right)$$ as taking values in $\mathbb R^2\otimes\mathbb R^2\cong\mathbb R^{2\times2}$. Let $\pi$ denote the projective norm on $\mathbb R^2\otimes\mathbb R^2$. Note that $$\pi(A)=\sup\left\{\sum_{n\:=\:1,\:2}|\langle Af_n,g_n\rangle|:(f_1,f_2)\text{ and }(g_1,g_2)\text{ are orthonormal bases of }\mathbb R^2\right\}$$ for all $A\in\mathbb R^{2\times2}$.
Let $g:=\operatorname{tr}f$. How are $\operatorname{Var}_\varsigma(f)$, where the variation has to be understood wrt $\pi$, and $\operatorname{Var}_\varsigma(g)$ related?
It's clear that $$\operatorname{Var}_\varsigma(f)\ge\operatorname{Var}_\varsigma(g),\tag1$$ but it's not clear to me whether we can show that the inequality is strict or not. If that's not possible in general, is there an example with strict inequality? Or am I on the wrong track and we even got equality in $(1)$?