Timeline for Volume of submanifold as integral of delta-function
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Sep 24, 2022 at 15:35 | history | edited | dennis | CC BY-SA 4.0 |
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| Sep 22, 2022 at 15:43 | comment | added | Ryan Budney | What is your definition of "volume" of the manifold $M$? I think if you are taking the induced Riemann metric from being a submanifold of Euclidean space, this integral won't agree. Your integral it something like the dual of the Thom class, so you have a normalization problem. | |
| Sep 22, 2022 at 12:44 | comment | added | dennis | @RyanBudney Actually I don't think that's true. If each $f_i$ is scaled by a constant $k_i$, then $\prod_i \delta(f_i)\to \prod_i \frac{\delta(f_i)}{|k_i|}$ and $\det(JJ^T)\to\det(JJ^T)\prod_i k_i^2$ such that all $k_i$ cancel. | |
| Sep 22, 2022 at 9:46 | answer | added | Robert Bryant | timeline score: 4 | |
| Sep 22, 2022 at 7:44 | comment | added | Ryan Budney | That integral does not transform appropriately. If you scale $f$, the integral changes, but the volume of the manifold does not. | |
| Sep 21, 2022 at 19:02 | history | edited | dennis | CC BY-SA 4.0 |
edited title
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| Sep 21, 2022 at 18:51 | comment | added | dennis | @VladimirZolotov No $\det(J J^T)=1$ because $(J J^T)_{ij}=\sum_\mu J_{i\mu}J_{j\mu}$ and $J_{i\mu}\equiv J_{1\mu}=\delta_{1\mu}$. | |
| Sep 21, 2022 at 17:34 | comment | added | Vladimir Zolotov | What if we only have one $f_1 = x_1$ but $n$ is big? It looks like $\det(JJ^T) = 0$ so the whole integral is $0$. | |
| Sep 21, 2022 at 16:50 | history | edited | dennis | CC BY-SA 4.0 |
added 45 characters in body
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| Sep 21, 2022 at 16:35 | history | asked | dennis | CC BY-SA 4.0 |