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Jul 24, 2013 at 3:06 comment added Will Sawin @Mariano: I believe my example does satisfy your condition (if you differentiate it to make it a Riemannian metric). The unique Haar volume form is invariant to rotations around $0$, and is scaled by the scaling map at $0$. But there is a unique form with these properties that does not vanish on lines through $0$, since that group action is transitive on a dense subset, and the nonvanishing at $0$ handles the ambiguity of the scaling factor. But my metric's volume form is clearly invariant and scales, so it must be the same.
Jul 24, 2013 at 2:43 comment added Mariano Suárez-Álvarez Good point! IMO that was a natural condition for a metric to be the most natural one :-)
Jul 24, 2013 at 2:37 comment added Will Sawin Since the Grassmanian is correct, any induced metric from it will be finite volume, while the Haar measure is infinite volume.
Jul 24, 2013 at 2:29 comment added Mariano Suárez-Álvarez It'd be nice to have a metric whose volume form is the Haar measure under the Euclidean group. Do you know if this one works?
Jul 24, 2013 at 1:40 history answered Yoav Kallus CC BY-SA 3.0