# Normal and locally normal currents

I am reading the Geometric Measure Theory book by H. Federer and I have some questions about currents:

1. Assuming $$T \in \mathscr{D}_{m}(U),$$ we call $$T$$ locally normal if and only if $$T$$ is representable by integration and either $$\partial T$$ is representable by integration or $$m=0 .$$ Furthermore call $$T$$ normal if and only if $$T$$ is locally normal and $$\operatorname{spt}T$$ is compact.

Now, I want to construct a locally normal current that is not normal! Does the following current work:

$$\mathbf{E}^{n} \llcorner \psi$$ corresponding to all weakly differentiable real-valued functions $$\psi,$$ with \begin{aligned} \left[\partial\left(\mathbf{E}^{n} \llcorner \psi\right)\right](\phi) &=\int\left\langle e_{1} \wedge \cdots \wedge e_{n}, \psi(x) d \phi(x)\right\rangle d\left[\begin{array}{l} n \\ z \end{array}\right) \\ &=(-1)^{n-1} \int \psi(x) \operatorname{div} \xi(x) d \mathfrak{L}^{n} x \end{aligned}

1. $$M \subset U$$ oriented submanifold, then there is a corresponding $$n$$-current $$[M]$$ defined by $$[M](\omega)=\int_{M}\langle\omega(x), \xi(x)\rangle d H^{n}(x), \omega \in D^{n}(U)$$

Now, I want to calculate $$\partial [M]$$! Is it $$\omega$$ on $$M$$??

• For the first one, why not just take an $m$-dimensional plane $P \subset U = \mathbf{R}^n$? Pick an orientation for it, then $[P] \in \mathcal{D}_m(U)$ has non-compact support. For the second, if $M$ has a manifold boundary $\partial M$ in $U$, then $\partial [M] = [\partial M]$. (In general the boundary of $[M]$ need not be representable by integration if $M$ were badly behaved.) – Leo Moos 2 days ago
• Thank you so much @Leo moos. – Zeno cosini 2 days ago