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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.

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    $\begingroup$ 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. $\endgroup$ – Teri Mar 13 at 23:50
  • $\begingroup$ @Teri the solution is always the easiest. I though that we had uniqueness, but we do not. You are right. $\endgroup$ – Lolman Mar 14 at 18:23
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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 $s(x) = \mathrm{sign}(x_1 + x_2 - 1)$.

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  • $\begingroup$ 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. $\endgroup$ – Lolman Mar 14 at 21:27
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    $\begingroup$ @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 $\varphi = \sum_{i} \varphi_i \partial_{x_i}$. $\endgroup$ – Skeeve Mar 15 at 6:43
  • $\begingroup$ I had a moment of confusion. Of course it's correct. Thank you, I see it now. $\endgroup$ – Lolman Mar 15 at 22:42

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