Timeline for Obstructions for a metric to be conformally equivalent to a product metric
Current License: CC BY-SA 3.0
19 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 30, 2013 at 17:20 | vote | accept | Ali Taghavi | ||
| Dec 29, 2013 at 17:48 | answer | added | Robert Bryant | timeline score: 17 | |
| Dec 29, 2013 at 11:22 | comment | added | Robert Bryant | @AliTaghavi: My apologies! I have thought about this and realize that I misunderstood you. You wrote 'conformally equivalent' but I read that to mean 'conformal to', i.e., I assumed that you wanted your diffeomorphism $F$ to preserve the product structure on $M\times N$, but you didn't say this. However, it still seems to me that your second question would be better as "What obstructions are there for a metric g to be conformally equivalent to a product metric?" since the manifolds $M$ and $N$ are not really that important because, above dimension $2$, there are local obstructions. | |
| Dec 28, 2013 at 19:59 | comment | added | Robert Bryant | @AliTaghavi: Sorry, I have to be elsewhere in about 10 minutes and don't have time to chat. I'll think about what I can write as an answer to explain what you are (apparently) not understanding. | |
| Dec 28, 2013 at 19:51 | comment | added | Ali Taghavi | let us continue this discussion in chat | |
| Dec 28, 2013 at 19:46 | comment | added | Robert Bryant | @AliTaghavi: But do you understand that Eremenko did not answer your second question and that your second question is not the generalization to higher dimensions of your first question? Perhaps you meant for your second question to be simply "What obstructions are there for a metric $g$ to be conformally equivalent to a product metric?", but the second question you asked is not that question! | |
| Dec 28, 2013 at 19:35 | comment | added | Ali Taghavi | I wrote the definite because I think there is a misunderestanding: I think for $M=N=\mathbb{R}$ there is no obstruction, as Alexandre Eremenko pointed out in his answer. Every metric on $\mathbb{R}^{2}$ is conformally equivalent to the standard metric or hyperbolic metric . Both metric are a positive multiple of product metric | |
| Dec 28, 2013 at 19:24 | comment | added | Robert Bryant | @AliTaghavi: I know the definition of conformally equivalent; you don't need to repeat it. However, the question "Is every metric on $M\times N$ conformally equivalent to a product metric?" (which, in dimensions above $2$ clearly has a negative answer) is not equivalent to the question "What obstructions are there for a metric $g$ on $M\times N$ to be conformally equivalent to a product metric for metrics $g_1$ and $g_2$ on $M$ and $N$, respectively?" Obstructions exist for this latter problem even when $\dim M = \dim N = 1$. | |
| Dec 28, 2013 at 19:17 | comment | added | Robert Bryant | @DeaneYang: Of course, if you don't specify the splitting in advance, then it's a different question. In dimensions greater than $2$, there are, of course, obstructions to a metric being conformal to a product. I haven't worked them out, but a function count guarantees that they must be there. Let me think about it and see whether there is anything easy to say. | |
| Dec 28, 2013 at 18:12 | comment | added | Ali Taghavi | A metric g on M is conformal equivalent to another metric h if there is a diffeomorphism F on M such that $F^{*}(g)=\lambda. h$ where $\lambda$ is a positive smooth function. By this definition I ask 'Is every metric on $M\times N$, conformaly equivalent to a product metric"?What is that curvature function which you said as an obstruction? | |
| Dec 28, 2013 at 18:04 | comment | added | Ali Taghavi | By conformal equivalent I mean the following: | |
| Dec 28, 2013 at 17:31 | comment | added | Deane Yang | Robert, what if the splitting is not specified in advance? | |
| Dec 28, 2013 at 10:41 | comment | added | Robert Bryant | Your question is not entirely clear. It seems that you are asking a more specialized question, i.e., how to tell when a metric $g$ on $M_1\times M_2$ is conformally equivalent to some product metric $g_1+g_2$, where $g_i$ is a metric on $M_i$, i.e., you are specifying the underlying product structure in advance. If this is your question, then, even when $M_1=M_2=\mathbb{R}$, there are conditions: First, the obvious condition that the tangent spaces of $M_i$ must be $g$-orthogonal, and, second, that a certain second order curvature function must vanish. | |
| Dec 28, 2013 at 6:59 | vote | accept | Ali Taghavi | ||
| Dec 30, 2013 at 17:20 | |||||
| Dec 28, 2013 at 6:45 | answer | added | Alexandre Eremenko | timeline score: 7 | |
| S Dec 28, 2013 at 6:42 | history | suggested | Ed Dean | CC BY-SA 3.0 |
improved English, formatting
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| Dec 28, 2013 at 6:40 | review | Suggested edits | |||
| S Dec 28, 2013 at 6:42 | |||||
| Dec 28, 2013 at 6:30 | history | edited | Ali Taghavi | CC BY-SA 3.0 |
deleted 1 characters in body
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| Dec 28, 2013 at 6:16 | history | asked | Ali Taghavi | CC BY-SA 3.0 |