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Nov
19 |
awarded | Nice Question |
Nov
5 |
comment |
SU(2) invariant Kahler metrics on products of Riemann surfaces
I am giving the SU(2) action. It acts trivially on X and in the usual manner on $\mathbb{P}^1$. |
Nov
5 |
asked | SU(2) invariant Kahler metrics on products of Riemann surfaces |
Oct
14 |
awarded | Nice Question |
May
29 |
comment |
Most harmful heuristic?
I agree with Scott Aaronson. In fact, the physicist way of defining tensors as things that change correctly under coordinates gives a nice way to define tensor fields on manifolds (Simply a smooth collection of multi-index beasts on different open sets such that on the intersection they are related by an appropriate transformation (the transition functions of the tensor bundle)). I am not sure if this "heuristic" actually gives rise to wrong intuitions. |
May
19 |
awarded | Nice Question |
Mar
9 |
accepted | Convexity of a (non-symmetric) function of matrices |
Mar
9 |
comment |
Convexity of a (non-symmetric) function of matrices
Yes. I forgot to add the assumption of semidefiniteness. |
Mar
9 |
revised |
Convexity of a (non-symmetric) function of matrices
Added the hypothesis of semidefiniteness. |
Mar
9 |
asked | Convexity of a (non-symmetric) function of matrices |
Feb
25 |
accepted | Matrix-convexity of inverse of the cofactor matrix |
Feb
16 |
comment |
Matrix-convexity of inverse of the cofactor matrix
Thanks. It arose in proving uniform estimates for an elliptic Monge-Ampere type PDE. |
Feb
13 |
comment |
Matrix-convexity of inverse of the cofactor matrix
Indeed. I corrected that. |
Feb
13 |
revised |
Matrix-convexity of inverse of the cofactor matrix
added 5 characters in body |
Feb
13 |
asked | Matrix-convexity of inverse of the cofactor matrix |
Feb
9 |
awarded | Yearling |
Feb
9 |
accepted | Convexity of a function of matrices |
Feb
9 |
asked | Convexity of a function of matrices |
Feb
9 |
accepted | Geometric mean of two matrices |
Feb
7 |
comment |
Geometric mean of two matrices
Thank you. Unfortunately, case 1 is the more problematic one of the two. This is because $K \neq 0$ cannot be $\leq 0$ (its trace is $0$). |