Timeline for Is there an Error on pg. 17 of Tromba's "Teichmuller Theory in Riemannian Geometry"?

11 events

when toggle format	what		by	license	comment
Sep 19, 2018 at 20:16	history	edited	Alex M.	CC BY-SA 4.0	The previous AMS link required a username and a password, this Springer one does not
Jan 17, 2011 at 22:00	history	edited	BrainDead	CC BY-SA 2.5	Clarified my question 2); deleted 1 characters in body
Jan 17, 2011 at 15:59	vote	accept	BrainDead
Jan 17, 2011 at 15:57	history	edited	BrainDead	CC BY-SA 2.5	Fixed the defintion of $\mathcal{M}$.
Jan 17, 2011 at 15:04	answer	added	Willie Wong		timeline score: 4
Jan 17, 2011 at 5:07	comment	added	Theo Buehler		Sorry I confused $\mathcal{A}^{s}$ and $\mathcal{M}$. But then Cayley-Hamilton tells you that $J^{-1} = (\text{tr}(J)\text{id} - J)$, so $\text{tr}(J^{-1}H) = \text{tr}(J)\text{tr}(H) - \text{tr}(JH)$.
Jan 17, 2011 at 4:53	comment	added	BrainDead		Why is $tr(J) =0$?
Jan 17, 2011 at 4:48	comment	added	Theo Buehler		Maybe I'm dense but isn't 1) just the Cayley-Hamilton theorem: $J^{2} - \text{tr}(J) \cdot J + \text{det}(J)\cdot\text{id} = 0$, so $J^{2} = - \text{id}$?
Jan 17, 2011 at 4:44	history	edited	BrainDead	CC BY-SA 2.5	added 2 characters in body
Jan 17, 2011 at 4:43	comment	added	BrainDead		I realized that the answer to question 0) is "yes," or at least I see that the map $H \mapsto tr(J^{-1}H)$ is surjective.
Jan 17, 2011 at 4:37	history	asked	BrainDead	CC BY-SA 2.5