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May 8, 2017 at 16:37 comment added Moishe Kohan See my answer here for an explicit left-invariant distance function.
S Mar 26, 2017 at 21:20 history suggested evgeny
not algebraic geometry
Mar 26, 2017 at 20:57 review Suggested edits
S Mar 26, 2017 at 21:20
S Mar 26, 2017 at 15:35 history suggested Hee Kwon Lee CC BY-SA 3.0
Math expression need a dollar
Mar 26, 2017 at 15:16 review Suggested edits
S Mar 26, 2017 at 15:35
S Aug 25, 2015 at 21:00 history suggested Ali Taghavi
I add a tag
Aug 25, 2015 at 20:27 review Suggested edits
S Aug 25, 2015 at 21:00
Oct 17, 2012 at 21:18 answer added Robert Bryant timeline score: 25
Sep 28, 2012 at 6:25 answer added Misha timeline score: 15
Sep 28, 2012 at 5:56 comment added Misha @Deane: Yes, of course, provided manifold is connected, otherwise one has to make some ad hoc choices for the distance between points in different components. I was just trying to point out (I guess, unsuccessfully) that not every distance is Riemannian and not every Riemannian metric defines distance. On the other hand, $SL(n,R)$ is, of course connected, so every left-invariant Riemannian metric defines a left-invariant distance function.
Sep 28, 2012 at 2:34 comment added Deane Yang Mischa, it is true, however, that a Riemannian metric does define a distance function, right? So I believe your answer is a full positive answer to the question, right?
Sep 27, 2012 at 22:50 comment added Misha Every Lie group has a left-invariant metric: Start with any positive definite inner product on the Lie algebra and ntranslate it to the rest of the group using left multiplication. Note also that Riemannian metric is not the same thing as a distance function.
Sep 27, 2012 at 21:17 answer added paul garrett timeline score: 12
Sep 27, 2012 at 21:07 comment added Anthony Quas The discrete metric?
Sep 27, 2012 at 20:09 comment added safsaf32 I mean just a distance function.
Sep 27, 2012 at 19:58 comment added Igor Rivin What is an "ordinary metric"?
Sep 27, 2012 at 19:27 history asked safsaf32 CC BY-SA 3.0