Https :// wigtonphysics .blogspot .com/2018/02/identity-matrix-in-dirac-notation.html

Shows |n> <n| is the outer product that is the identity matrix = Sum_i  of  |i><i|.

However, our Stokes theorem surface integral is saying the integrand (argument) for Berry’s (7a)

[Berry, Michael Victor. "Quantal phase factors accompanying adiabatic changes." Proceedings of the Royal Society of London. A. Mathematical and Physical Sciences 392, no. 1802 (1984): 45-57.]

curl < n | grad n >  = < grad n | X | grad n >  and I can see this, but the next step is

 < grad n | X | grad n >  =  Sum_{m≠n}  < grad n | m> X <m | grad n > 

Hmmmmm.

First, let’s look at what’s |  Sum_{m,n}  < grad n | m> X <m | grad n > | - | m> < grad n |

officially, A X B = < grad n | X | grad n >  =  (A X B)i = eps_{ijk} Aj Bk = Aj Bk – Ak Bj 

so 

 < grad n | X | grad n >  = (Aj Bk – Ak Bj )i + ( Ak Bi – Ai Bk )j + ( Aj Bi – Ai Bj )k

OK, If Sum over all, with m = n, then 

Sum_{m & n}  < grad n | m> X <m | grad n > = 

 (Aj Bk – Ak Bj )i + ( Ak Bi – Ai Bk )j + ( Aj Bi – Ai Bj )k

+  (Bj Bk – Bk Bj )i + ( Bk Bi – Bi Bk )j + ( Bj Bi – Bi Bj )k

+  (Aj Ak – Ak Aj )i + ( Ak Ai – Ai Ak )j + ( Aj Ai – Ai Aj )k

Finally we can see that this is 

 < grad n | X | grad n >  =  Sum_{m ≠ n}  < grad n | m> X <m | grad n >