Timeline for (n-1)-dimensional normal currents and Smirnov's paper
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 6, 2013 at 21:30 | comment | added | A random mathematician | Thanks Igor and Deane. Now I understand what Smirnov said in his paper. | |
| Nov 5, 2013 at 21:09 | comment | added | Igor Khavkine | @user1026, the first question you need to ask yourself is this: are you comfortable with the statement that, on $\mathbb{R}^n$, a $1$-form $\alpha$ is the exterior derivative of a $0$-form $\beta$, $\alpha_i = (d\beta)_i = \partial_i \beta$, iff $(d\alpha)_{ij} = \partial_i \alpha_j - \partial_j \alpha_i = 0$? If not, you simply need to read up on the Poincaré lemma and de Rham cohomology. If yes, then you need only notice that raising all indices with the Euclidean metric converts these conditions from $1$-forms to vector fields. I'm just expanding slightly on what Deane Yang already said. | |
| Nov 5, 2013 at 20:24 | comment | added | Deane Yang | 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 $\mathrm{div} T = 0$. | |
| Nov 5, 2013 at 15:33 | comment | added | A random mathematician | 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. | |
| Nov 5, 2013 at 13:12 | vote | accept | A random mathematician | ||
| Nov 5, 2013 at 9:00 | comment | added | Igor Khavkine | See the update in my answer. | |
| Nov 5, 2013 at 9:00 | history | edited | Igor Khavkine | CC BY-SA 3.0 |
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| Nov 5, 2013 at 4:22 | comment | added | Deane Yang | 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. | |
| Nov 5, 2013 at 1:50 | comment | added | A random mathematician | 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. | |
| Nov 4, 2013 at 22:12 | history | answered | Igor Khavkine | CC BY-SA 3.0 |