Timeline for Understanding zeta function regularization

7 events

when toggle format	what		by	license	comment
Jul 7, 2018 at 15:45	comment	added	Elle Najt		Since Jacobi's identity is obvious for diagonalizable matrices, and since the equation is invariant under the action $A \to S A S^{-1}$, then it holds for all matrices because the functions on both sides of the equation are continuous and the subspace of diagonalizable (over the complex numbers) matrices is dense in the space of matrices. This shows the identity in $M_n(\mathbb{C})$.
Apr 24, 2012 at 13:10	vote	accept	Daniel Moskovich
Apr 20, 2012 at 14:45	comment	added	Qiaochu Yuan		@Tom: done. (More characters.)
Apr 20, 2012 at 14:45	history	edited	Qiaochu Yuan	CC BY-SA 3.0	added 3 characters in body
Apr 20, 2012 at 1:26	comment	added	Daniel Moskovich		I very much like this answer. Thank you! I'm more-or-less happy with Jacobi's identity popping up in this context, because it is how the Alexander polynomial (equivalent to Ray-Singer torsion for a knot complement) indeed arises as a quantum invariant (Aarhus integral gives rise to exp tr of the equivariant linking matrix, whatever that means), so it fits well with my preconceptions.
Apr 19, 2012 at 15:17	history	edited	Qiaochu Yuan	CC BY-SA 3.0	added 42 characters in body
Apr 19, 2012 at 15:10	history	answered	Qiaochu Yuan	CC BY-SA 3.0