Timeline for Relation between mean curvature and conformal metric
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
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| Mar 10, 2021 at 20:11 | vote | accept | K2-liz | ||
| Feb 1, 2021 at 12:56 | vote | accept | K2-liz | ||
| Mar 10, 2021 at 20:11 | |||||
| Feb 1, 2021 at 2:25 | comment | added | User | I edit the typo on the conformal factor and (hopefully) find the correct formula. As for the definition of the mean curvature, I have also seen two different definitions. | |
| Feb 1, 2021 at 2:24 | history | edited | User | CC BY-SA 4.0 |
added 9 characters in body
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| Jan 31, 2021 at 19:34 | comment | added | Leo Moos | @RobertBryant I agree that calling the sum of the principal curvatures the mean curvature is slightly absurd, but this convention is not infrequent. In addition to the people I mention above, the mean curvature is also defined as the sum in Mantegazza's book on mean curvature flow. Mantegazza even adds the following remark 'Despite its name, [the mean curvature] is the sum of the eigenvalues of the second fundamental form, not their average mean'. As I said, it's a very minor point, that I'm mainly raising for students who might not be aware of the different conventions. | |
| Jan 31, 2021 at 17:14 | comment | added | Robert Bryant | @LeoMoos: There are two things: There's the definition of mean curvature, which is the average of the principal curvatures. There's also a convention about the notation $H$, which was classically used for the mean curvature, but nowadays is used by some to mean the sum of the principal curvatures, which is fine as long as they explicitly specify it. I suppose some people don't mind calling the sum the mean, but that seems ill advised to me. In any case, the OP explicitly specified that they mean $H$ to be the mean curvature, so I assumed they really meant it. | |
| Jan 31, 2021 at 16:25 | comment | added | Leo Moos | @RobertBryant I have a minor remark on your first comment, where you write that 'the mean curvature is the average of the principal curvatures, not the sum'. In my opinion that is largely a matter of convention. For example, on the one hand do Carmo and Lee define it as the average, on the other hand Simon or Colding-Minicozzi use the convention where $H$ is the sum of the principal curvatures. Just a small point... | |
| Jan 31, 2021 at 14:03 | comment | added | Robert Bryant | Yes, $\bar g = e^{2f} g$ means what you say, and also $\bar g^{ij} = -e^{2f} g^{ij}$, but that's not relevant. The point is that a vector field in local coordinates is $X = X^i\,\partial_i$, with $ |X|_g^2 = g_{ij}X^iX^j$. | |
| Jan 31, 2021 at 12:16 | comment | added | User | I mean $\overline g^{ij} = e^{-2f}g^{ij}$ in the previous comment. | |
| Jan 31, 2021 at 12:00 | comment | added | User | @RobertBryant So when people write $\overline g = e^{2f} g,$ it doesn't mean that locally $\overline g_{ij} = e^{2f}g_{ij}$ (which implies $\overline g^{ij} = e^{2f}g^{ij}$), is it? | |
| Jan 31, 2021 at 11:52 | comment | added | Robert Bryant | No, that is not correct. $g$ is a a tensor of type $(0,2)$ while vector fields are tensors of type $(1,0)$. Maybe you are confused by the notation $\langle X,Y \rangle_g$, which means $g(X,Y)$, so $\bar g(X,Y) = e^{2f} g(X,Y) = g(e^fX,e^fY)$. In particular, if $g(X,X)=1$, then $\bar g(e^{-f}X,e^{-f}X) = 1$. | |
| Jan 31, 2021 at 11:37 | comment | added | User | @Robert Bryant Hello Robert. Thanks for your comment! Just one question. If $\overline g = e^{2f}g$ and $\overline e=e^{-f}e,$ isn't it correct that $\langle \overline e,\overline e\rangle_{\overline g} = e^{-2f}\langle e^{-f} e, e^{-f} e\rangle_{g} = e^{-4f}|e|^2_g?$ | |
| Jan 31, 2021 at 11:25 | comment | added | Robert Bryant | A couple of comments to get a correct formula: First, if $\bar g = \mathrm{e}^{2f} g$ and $\mathbf{e}_i$ is a $g$-orthonormal frame near $p$, then $\mathrm{e}^{-f} \mathbf{e}_i$ is a $\bar g$-orthonormal frame. (Note the minus sign.) Second, the mean curvature is the average of the principal curvatures, not the sum, so you are missing a normalization factor. For example, check your formula for the unit sphere $M=S^{n-1}\subset\mathbb{E}^n$ with $g = |dx|^2$, $\mathrm{e}^{2f} = |x|^{-2}$, and $\eta$ the outward normal. You should get $H=-1$ and $\bar H = 0$. | |
| Jan 31, 2021 at 7:48 | history | answered | User | CC BY-SA 4.0 |