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I don't know much about currents, but I saw a paper of Smirnov which seems relevant to a problem I am working on. In the very last paragraph of page 848 of the following paper

Smirnov says:

If T is an (n-1)-dimensional current with div(T)=0 ($\partial T=0$), then $T=\partial S$ where $S$ is an n-dimensional current. Moreover $S$ can be identified with a function in $BV$ and $T$ with its gradient.

I would appreciate an explanation of the above in simpler terms that one could understand without much knowledge about currents. Can it be explained in a simpler case only using vector fields, the gradient and divergence operator, BV functions, ... in $R^n$?

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Smirnov claims that his paper can be read and understood without using currents (i.e., geometric measure theory). Perhaps you should try skipping this sentence and see how far you can get with the paper. The paper, by avoiding geometric measure theory, has to derive some of the machinery from scratch. That's not always so easy to follow but in principle it's self-contained. – Deane Yang Nov 4 '13 at 21:03
There's also a beautiful book by Giusti called Minimal surfaces and functions of bounded variation. I don't know if he discusses currents explicitly but he does define and explore the properties of BV functions and their derivatives. – Deane Yang Nov 4 '13 at 21:05
The result I quoted is not proved in Smirnov's paper. He mentions it as a general result with the same "theme". I have a fairly good understanding of BV functions but it doesn't seem to be enough to understand what he says. – MathStudent Nov 4 '13 at 21:31
up vote 3 down vote accepted

Currents are distributions dual to smooth forms. I often get confused about the degrees, so I'm never sure what the standard meaning of an $m$-current. But according to Smirnov (Sec 1.4) an $m$-current is an element of the dual space of smooth $m$-forms. As is usual with distributions, differential operators on forms can be extended to differential operators on currents via the adjoint formula. So, one can define a differential $\partial$ on currents as the adjoint of the de Rham differential $d$ on forms: $\langle \partial T, \alpha \rangle = \langle T, d\alpha \rangle$, up to a choice of sign.

One can embed smooth $(n-m)$-forms in $m$-currents (on an $n$-dimensional space) using the formula $\langle T_\alpha, \beta \rangle = \int \alpha \wedge \beta$. One must restrict the support so that this integral is always defined. That is, if $\beta$ is allowed to be arbitrary, then $\alpha$ must have compact support, and vice versa. Restricted to the embedded form, the $\partial$ operator is just the de Rham operator $d$ (up to sign). This follows from the formula $\int (d \alpha) \wedge \beta = \pm \int \alpha \wedge (d\beta)$.

Since it follows from the definition that $\partial^2 = 0$, one can use $\partial$ to define a cohomology on currents. Turns out that this cohomology on currents is isomorphic to the de Rham cohomology of the embedded forms. So, the cohomology on $m$-currents with compact support is isomorphic to the de Rham cohomology with compact support in degree $n-m$.

So, if I interpret your notation correctly, the claim that for an $(n-1)$-current $T$, the identity $\partial T = 0$ implies $T = \partial S$ for some $n$-current $S$, is simply statement that the de Rham cohomology of $1$-forms (with appropriate support) vanishes. Since, I think, Smirnov is working on $\mathbb{R}^n$, that is only to be expected. If you restrict $T$ and $S$ to be embedded smooth forms, then $T$ is simply a $1$-form and $S$ a $0$-form, with $d S = T$, which is consistent with the interpretation proposed by Smirnov.

I'm not sure that I can help with showing, under relaxed smoothness conditions, that $S$ necessarily belongs to the BV class.

BTW, the details of the above discussion can be found in de Rham's own book. Perhaps also in the Federer reference given by Smirnov.

Update: Above, I gave a formula for embedding smooth $(n-m)$-forms in $m$-currents. If you have a preferred measure $\mu$ on your space (like the Lebesgue measure on $\mathbb{R}^n$, for instance), then you can also embed $m$-polyvectors in $m$-currents. In coordinates, $m$-polyvectors have $m$ contravariant indices, which are fully antisymmetriezed. So a $1$-polyvector is just a vector field. The embedding in currents uses the formula $\langle T_v, \alpha \rangle = \int (v\cdot \alpha) d\mu$, where $\cdot$ denotes the contraction of the $m$ contravariant indices of $v$ and the $m$ covariant indices of $\alpha$. If you work out the adjoint formula for the differential acting on an embedded vector field $v$, you find that $\partial T_v = T_{\operatorname{div} v}$ (up to sign), where in Cartesian coordinates $\operatorname{div} v = \partial_i v^i$.

Note that the above formulas for embedding forms and vector fields and currents are different. In particular, if you use the metric to lower indices and thus identify vector fields and $1$-forms, you also get two different ways of embedding vector fields (and polyvector fields) in currents. So if you start with a vector field $v$ ($1$-polyvector), then you can either embedded it in $(n-1)$-currents (first method above) or in $1$-currents (second method above). You get different actions of $\partial$ on $v$ using different embeddings. I already gave the formula for the second embedding, for the first embedding you find $(\partial v)^{ij} = \partial^{[i}v^{j]}$ (up to sign, and all in Cartesian coordinates). You can see that this is the formula for the exterior differential of forms, just with all indices raised. I believe that it is this last formula that Smirnov has in mind when talking about $(n-1)$-currents.

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hank you for your response Igor. Let say T is a smooth compactly supported vector field in which can be identified as a 1-form. Then what does it mean to have $\partial T=0$? What does the above imply about the vector field T? I wonder if it provides some sort of decomposition for divergence free vector fields. – MathStudent Nov 5 '13 at 1:50
user1026, that's a good question. It appears that $\partial T$ should mean the exterior derivative of a 1-form and not its divergence. Either that or $S$ is really supposed to be an $(n-2)$-form. – Deane Yang Nov 5 '13 at 4:22
See the update in my answer. – Igor Khavkine Nov 5 '13 at 9:00
Thank you Igor. This answers my question, but I have trouble understanding the last formula for $(\partial v)^{i,j}$. What does the condition $\partial T_v=0$ imply if $v$ is a vector field embedded as an $(n-1)$-current? How can it be written in terms of derivatives of $v$? I am looking for conditions under which a vector field $v$ can be written as a gradient of a function in BV, as mentioned in Smirnov's paper. – MathStudent Nov 5 '13 at 15:33
If you're working on $\mathbb{R}^n$, then a vector field $v_1, \dots, v_n)$ can be written as the gradient of a function if and only if $\partial_iv_j - \partial_jv_i = 0$ for all $1 \le i,j \le n$. This is equivalent to what you quote from Smirnov only if the condition $\partial T = 0$ corresponds to this condition and not $\mathrm{div} T = 0$. – Deane Yang Nov 5 '13 at 20:24

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