Mass of the push forward of a k-current with fixed orientation

Question

$\DeclareMathOperator{\Mass}{Mass}$Let $f: \mathbb{R}^n \to \mathbb{R}^n$ be a smoth map. Given a $2$-vector (in general a $k$-vector but let's stick to $2$) $v_1 \wedge v_2 \in \Lambda_2 (\mathbb{R}^n)$ you can consider the push forward of such vector through $f$, defined as

$$ \langle f_\# (v_1 \wedge v_2 ) , \omega \rangle =\langle v_1 \wedge v_2 , f^\#\omega \rangle$$

where $\langle \cdot, \cdot \rangle$ is the duality pairing and $\omega $ is a $2$-form. Now, for simplicity assume that $v_1 ,v_2 = e_1 , e_2$, elements of the canonical basis for $\mathbb{R}^n$, and let $T$ be a $2$-current defined such that its direction is, at every point, $e_1 \wedge e_2$. I can define its push forward just like before, and its mass will be

$$ \sup_{\|\omega\|^* \leq 1} \langle T, f^\# \omega \rangle $$

where $\|\omega \|^*$ denotes the comass norm. The question is: how does the particular structure of $T$ (i.e. the fact that it is always directed as $e_1 \wedge e_2 $) give me a better mass estimate? What can easily be seen is

$$ \Mass [ (f_\#) T] \leq \|d f\|_\infty ^2 \Mass (T),$$

but for example if $T$ were a $1$-current always in the $e_1$ direction, then it's easy to see that the estimate above can be improved with

$$ \Mass [ (f_\#) T] \leq \|\partial_1 f\|_\infty \Mass (T).$$

Now, maybe this is trivial if one is well-versed in tensor calculus (which I am not), but pulling back a $2$-form (or a $k$-form) totally messes me up, I don't know how to derive a similar estimate, or if there's even one. In case of rectiafiable currents you can improve the estimate by considering the tangential differential, is this true even in this case?

Answer 1

Your assumption that $T$ has direction $e^1 \wedge e^2$ means that for some distribution $u$, $T = u e^1 \wedge e^2$ (actually, since $T$ has finite mass, $u$ is a Radon measure). We're trying to compute $$\langle f_\# (u e^1 \wedge e^2), \omega\rangle = \langle u e^1 \wedge e^2, f^\# \omega\rangle = u \langle e^1 \wedge e^2, f^\# \omega\rangle.$$ The pullback is given by $$\langle e^1 \wedge e^2, f^\# \omega\rangle = \sum_{i < j} f^\# (\langle e^i \wedge e^j, \omega\rangle) \langle e^1 \wedge e^2, df_i \wedge df_j\rangle$$ where $df_i$ is the differential of $f_i$. Now $e^i \wedge e^j$ is a $2$-blade, so by the definition of the comass and the fact that $\omega$ has comass $\leq 1$, $$|f^\#(\langle e^i \wedge e^j, \omega\rangle)| \leq 1.$$ Meanwhile $$\langle e^1 \wedge e^2, df_i \wedge df_j \rangle = \partial_1 f_i \partial_2 f_j.$$ In particular $\langle e^1 \wedge e^2, f^\# \omega\rangle$ is a weighted sum, with weights in $[-1, 1]$, of the components of $\partial_1 f \otimes \partial_2 f$ and we conclude that $$\langle e^1 \wedge e^2, f^\# \omega\rangle \leq C \|\partial_1 f \otimes \partial_2 f\|_{L^\infty} \leq C \|\partial_1 f\|_{L^\infty} \|\partial_2 f\|_{L^\infty}$$ where $C$ is the number of pairs $(i, j) \in \{1, \dots, n\}^2$ with $i < j$. You might be able to improve this $C$.

Anyways we have $$\langle f_\# T, \omega \rangle \leq C\|\partial_1 f\|_{L^\infty} \|\partial_2 f\|_{L^\infty} \|u\|$$ where the norm of $u$ is its total measure, hence $\|u\|$ is the total mass of $T$.