most recent 30 from http://mathoverflow.net 2013-05-22T10:26:08Z http://mathoverflow.net/feeds/question/19924 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/19924/why-is-it-called-spectral-triple Rasmus 2010-03-31T09:04:14Z 2010-03-31T09:32:50Z

<p>I know the definition a spectral triple and that it is some kind of non-commutative generalisation of (the ring of functions on) a compact spin manifold.</p> <p>But, why is it called <em>spectral</em> triple?</p>

http://mathoverflow.net/questions/19924/why-is-it-called-spectral-triple/19926#19926 Jan Jitse Venselaar 2010-03-31T09:32:50Z 2010-03-31T09:32:50Z

<p>Well, it uses the spectral properties of the Dirac operator $D$ in the spectral triple quite extensively. Also, in the article where he (essentially) introduces the notion of spectral triples ( <a href="http://www.alainconnes.org/docs/reality.pdf" rel="nofollow">http://www.alainconnes.org/docs/reality.pdf</a> ) Alain Connes writes about the naming:</p> <p><em>"We shall need for that purpose to adapt the tools of the differential and integral calculus to our new framework. This will be done by building a long dictionary which relates the usual calculus (done with local differentiation of functions) with the new calculus which will be done with operators in Hilbert space and spectral analysis, commutators.... The first two lines of the dictionary give the usual interpretation of variable quantities in quantum mechanics as operators in Hilbert space. For this reason and many others (which include integrality results) the new calculus can be called the quantized calculus’ but the reader who has seen the word “quantized” overused so many times may as well drop it and use “spectral calculus” instead."</em></p>