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Feb 20, 2018 at 12:02 comment added LSpice @AliTaghavi, sorry; I missed the word 'isometric'.
Feb 20, 2018 at 7:53 comment added Ali Taghavi @LSpice but it is not an isometry since the metric of $GL(n,R)$ is not the Euclidean metric.
Feb 20, 2018 at 2:49 comment added LSpice @AliTaghavi, $\mathrm{GL}(n, \mathbb R) \to \mathfrak{gl}(n, \mathbb R)$ does it; its derivative is the identity map.
Feb 19, 2018 at 8:11 comment added Ali Taghavi Is there an isometric embedding of $GL(n,\mathbb{R})$ into some $M^{k\times k}$ whose derivative preserves the Lie algebra structure?
Feb 17, 2018 at 22:56 comment added Ali Taghavi Thanks for your answer. yes but it is not an isometric embedding.
Feb 17, 2018 at 22:53 history answered Igor Rivin CC BY-SA 3.0