Timeline for Left invariant metric on ${\rm SL}_n(\mathbb{R})$
Current License: CC BY-SA 3.0
22 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 29, 2017 at 8:32 | comment | added | Robert Bryant | @HeeKwonLee: I am not sure exactly what you are asking in these comments. It seems that you are asking about the behavior of the cut locus, which is much more complicated than the conjugate locus. Also, you seem to be asking about the geodesic formula for more general Lie groups and left invariant metrics. I explain this in the following file, which I wrote in response to another person's questions about this. You might find it useful: services.math.duke.edu/~bryant/GroupGeodesicsNotes.pdf | |
| Mar 29, 2017 at 1:46 | comment | added | Hee Kwon Lee | But how do you find the equation ? On $G_1$ $e$ is not surjective. However, for any $A\in G_1$, either $A$ or $A^{-1}$ is of the form $e^V$ for some $V$. So for surjective, multiple is considered. But I can not proceed more. Thank you for your attention. | |
| Mar 29, 2017 at 1:46 | comment | added | Hee Kwon Lee | I have a second question. If $G$ has a left invariant metric $g$ that is an $H$-biinvariant metric for some $H\subset G$, then the following is considered : Which a left invariant vector field $X$ has an integral curve $e^{tX}$ that is a geodesic. Since you give a formula $e^{v^T}e^{v-v^T}$ on $G_1:={\rm SL}(2,\mathbb{R})$, I can check directly that this is a geodesic equation on $G_1$ by using Levi-Civita connection formula. | |
| Mar 29, 2017 at 1:38 | comment | added | Hee Kwon Lee | Thank you for your file. But I showed that ${\rm SL}(2,\mathbb{R})$ has a conjugate point by using a map $e^{v^T}e^{v-v^T}$ you give. However, your file contains a perfect computation so that I know that $SO(2)$ is an image of hyperboloid. | |
| Mar 28, 2017 at 20:51 | comment | added | Robert Bryant | @AsafShachar: I just posted on my home web site a file that computes the conjugate locus. See the link in the above comment to Hee Kwon Lee. My memory of the computation from 5 years ago wasn't perfect, for I forgot another sequence of surfaces that are also part of the conjugate locus, but you can read about it in the file above. The description in my answer of the first conjugate point was, nevertheless, corrrect. | |
| Mar 28, 2017 at 20:50 | comment | added | Robert Bryant | @HeeKwonLee: Here is a short file that explains the computation of the conjugate locus. My memory was not perfect. It turns out that I left out a second countable sequence of surfaces that are also conjugate loci in addition to the ones that I remembered, but you can see how the computation goes. The file can be found at services.math.duke.edu/~bryant/ConjugateLocusofSL2R.pdf. | |
| Mar 28, 2017 at 11:12 | comment | added | Hee Kwon Lee | @Robert Bryant : Thank you for your comment. | |
| Mar 28, 2017 at 8:55 | comment | added | Asaf Shachar | @RobertBryant , which is harder to get a grip on. (In the case of $\text{SL}(2,\mathbb{R})/\text{SO}(2)$ we need to calculate solutions to quadratic equation (for the $\lambda_i$, for $n=3$ this becomes a quintic equation...). So I tried to use both your approaches, and hoped that the Riemannian one would turn out to be easier, but perhaps I was wrong. (The Riemannian distance is of independent interest however, and I am very interested to see how you obtained some information on the cut locus...Thanks again for the effort.) | |
| Mar 28, 2017 at 8:55 | comment | added | Asaf Shachar | @RobertBryant Thanks. I do not really need the Riemannian distance. Indeed, I also tried to realize your quotient method explicitly (as you suspected there are always $3$ elements in the quotient space, whose simultaneous stabilizer is the identity, and we can obtain an explicit description of them. However, like you mentioned we must first embed $\mathrm{SL}(2,\mathbb{R})$ in $\mathrm{SL}(3,\mathbb{R}$), so the distance on $\mathrm{SL}(2,\mathbb{R})$ unfortunately goes through the distance in $M=\text{SL}(3,\mathbb{R})/\text{SO}(3)$... | |
| Mar 28, 2017 at 8:36 | comment | added | Robert Bryant | @AsafShachar: I'll look for my notes about how to do the conjugate locus computation. I remember that it was not trivial, but not too hard. (The method I used did not extend to $\mathrm{SL}(n,\mathbb{R})$ for $n>2$.) Of course, describing the cut locus is a much more subtle question, and the (partial?) answer that I found showed that, indeed, the cut points occur before the first conjugate point. Do you need the Riemannian distance? I'm sure that using the quotient space method above (which gives a left-invariant metric that is not Riemannian) will yield a nicer metric. | |
| Mar 28, 2017 at 8:02 | comment | added | Asaf Shachar | @RobertBryant as can be seen in this question. Actually, I am afraid that proving your claim about the conjugate points by itself is non-trivial (You can see my modest attempt here). I guess this follow-up question caused HeeKwonLee to ask for an elaboration here... | |
| Mar 28, 2017 at 8:02 | comment | added | Asaf Shachar | @RobertBryant This answer is great, I have some further questions though: Do you know how to compute the cut locus of $I_2$ in ${\frak{sl}}(2,\mathbb{R})$? (even when assuming your claim about the conjugate points, a geodesic can stop minimizing before it reaches a conjugate point of course). The reason I am asking is that I am trying to compute explicitly this Riemannian distance on $\text{SL}(2,\mathbb{R})$. (My reason for that is that this will easily induce a left invariant distance on $\text{GL}(2,\mathbb{R})$, ... | |
| Mar 26, 2017 at 16:37 | comment | added | Robert Bryant | @HeeKwonLee: What I recall is that it's not hard to compute the differential of the mapping $E:{\frak{sl}}(2,\mathbb{R})\to\mathrm{SL}(2,\mathbb{R})$ defined by $E(v) = e^{v^T}e^{v-v^T}$. If I remember correctly, the locus in ${\frak{sl}}(2,\mathbb{R})$ where $E$ fails to be a local diffeomorphism (i.e., the determinant of the differential of $E$ vanishes) is described by the set of hyperboloids $\det(v) = k^2\pi^2$ where $k = 1,2,\ldots$. The statement about conjugate points follows from this. | |
| Mar 26, 2017 at 15:14 | comment | added | Hee Kwon Lee | I have a question. How can we prove that a conjugate point of $I_2$ is at $t=\pi /\sqrt{{\rm det}\ v}$ ? | |
| Oct 28, 2012 at 18:03 | history | edited | Robert Bryant | CC BY-SA 3.0 |
Corrected the even-odd problem for metrics on SL(n,R)
|
| Oct 28, 2012 at 17:13 | history | edited | Robert Bryant | CC BY-SA 3.0 |
Added a construction of a left-invariant metric on SL(n,R)
|
| Oct 24, 2012 at 1:21 | history | edited | Robert Bryant | CC BY-SA 3.0 |
Removed an erroneous metric construction
|
| Oct 23, 2012 at 22:00 | history | edited | Robert Bryant | CC BY-SA 3.0 |
Added information about an 'ordinary' metric on SL(2,R)
|
| Oct 22, 2012 at 11:56 | history | edited | Robert Bryant | CC BY-SA 3.0 |
fixed a typo and reworded final paragraph
|
| Oct 20, 2012 at 15:42 | history | edited | Robert Bryant | CC BY-SA 3.0 |
added informatin about geodesics
|
| Oct 19, 2012 at 15:35 | history | edited | Robert Bryant | CC BY-SA 3.0 |
fixed typos and grammar
|
| Oct 17, 2012 at 21:18 | history | answered | Robert Bryant | CC BY-SA 3.0 |