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Oct 28, 2010 at 15:48 vote accept Niel de Beaudrap
Oct 27, 2010 at 10:35 answer added Niel de Beaudrap timeline score: 1
Oct 27, 2010 at 5:58 comment added Niel de Beaudrap @Jon Bannon: close... it maps any rank-1 projector to 1. That is to say, it is the first-year-university trace operation, concieved as a positive map $\mathop{tr}:M_{∣C∣}(\mathbb C) \to M_1(\mathbb C)$.
Oct 26, 2010 at 19:12 history edited Yemon Choi
tweaked the tags
Oct 26, 2010 at 16:14 comment added Jon Bannon To clarify, what precisely is the trace operator on $\mathbb{C}^{C}$? Do you mean the trace on $M_{|C|}(\mathbb{C})$? If so, is this trace normalized so that the trace of the identity in $M_{|C|}(\mathbb{C})$ is 1?
Oct 26, 2010 at 14:54 history asked Niel de Beaudrap CC BY-SA 2.5