# Metric 1-current decomposition

I've been reading Paolini-Stepanov arcticle and in section 4, at page 6, they define a metric current from a transport: $$T_{\eta}(\omega)=\int_{\Theta}[[\theta]](\omega)d\eta(\theta),$$ which satisfies $$\mu_{T_\eta}\leq\int_{\Theta}\mu_{[[\theta]]}d\eta.$$ Where $$\Theta$$ is the set of Lipschitz curves, $$\mu_T$$ is the mass of the current and $$[[\theta]]$$ is the current associated to a Lipschitz curve.

Then they build an example in which this inequality is strict. In the example we take an $$\eta_1$$ concentrated over horizontal segments in $$Q=[0,1]^2$$ (which define the set $$\Theta_1$$) going from left to right and defined on a Borel set $$e$$ to be the 1-Length of the set of the starting points of the curves in $$e$$ $$\eta_1(e):=\mathcal{H}^1(e_0(e\cap\Theta_1).$$ Then they build $$\eta_2$$ in the same way on the vertical segments ion $$Q$$. Now they show that $$T_{\eta_1+\eta_2}$$ satisfies the inequality above in a strict sense.

The conclusion of the paper is that any acyclic normal current comes from a transport and have equality in that relation.

Since the obtained current cannot not be acyclic or normal, I think I am missing something. What am I missing?

Addendum: during the definition of this current they represent them with the notation $$T_{\eta_1}=\bar{e}_1\wedge\mathcal{L}^2\llcorner Q$$ Which is alien to me. Maybe by understanding it I could see the problem above.

• I think if you start by a metric current $T_\eta$ the conclusion simply says that it may be represented by another $\eta'$ for which the claimed equality holds. – Teri Mar 13 at 23:50
• @Teri the solution is always the easiest. I though that we had uniqueness, but we do not. You are right. – Lolman Mar 14 at 18:23

I think for any $$\bar e \in \mathbb R^2$$ the current $$\bar e \wedge \mathcal{L}^2 \llcorner Q$$ is defined in their paper via $$<\bar e \wedge \mathcal{L}^2 \llcorner Q, \varphi> = \int_Q \bar e\cdot \varphi \, dx$$ for any bounded Borel $$\varphi \colon Q \to \mathbb R^2$$. Therefore $$T_{\eta_1} = \bar e_1 \wedge \mathcal{L}^2 \llcorner Q$$, $$T_{\eta_2} = \bar e_2 \wedge \mathcal{L}^2 \llcorner Q$$ and consequently $$T_{\eta_1+\eta_2} = (\bar e_1 + \bar e_2) \wedge \mathcal{L}^2 \llcorner Q$$. Note that this current actually is normal by theorem 4.2, or more explicitly because $$\partial (T_{\eta_1+\eta_2}) = s \wedge \mathcal H^1 \llcorner \partial Q$$, where $$s(x) = \mathrm{sign}(x_1 + x_2 - 1)$$.
• Evaluating $T$ over $\phi$ defined on $Q$ wouldn't make it a 2-current? Not really, but still, it would not be a 1-current. – Lolman Mar 14 at 21:27
• @Lolman why not a 1-current? 1-current is a functional on 1-forms, which can be written as $\varphi = \sum_i \varphi_i \,dx_i$. Or equivalently we can view them as functionals on the vector fields $\varphi = \sum_{i} \varphi_i \partial_{x_i}$. – Skeeve Mar 15 at 6:43