Why take 'complex powers' of pseudo-differential operators? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T19:49:54Z http://mathoverflow.net/feeds/question/105021 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/105021/why-take-complex-powers-of-pseudo-differential-operators Why take 'complex powers' of pseudo-differential operators? Uday 2012-08-19T05:21:52Z 2012-08-19T07:30:08Z <p>Given a pseudo-differential operator $P$ of order zero, <a href="http://cdsweb.cern.ch/record/469079/" rel="nofollow">Seeley</a> showed that the holomorphic family of operators $\lbrace P^{z} : z\in \mathbb{C} \rbrace$ of all complex powers is contained in the class of pseudo-differential operators. </p> <p>Apart from knowing that we can take powers of these operators, is there any application of this theory? I understand the utility of raising operators to fractional powers. But, irrational and complex powers are not clear. </p> http://mathoverflow.net/questions/105021/why-take-complex-powers-of-pseudo-differential-operators/105027#105027 Answer by Rafe Mazzeo for Why take 'complex powers' of pseudo-differential operators? Rafe Mazzeo 2012-08-19T07:30:08Z 2012-08-19T07:30:08Z <p>There are many reasons to do this. A principal one (that was Seeley's motivation) is that the meromorphic operator family $P^{-s}$ is trace-class when the real part of $s$ is greater than $-n/m$ where $n$ is the dimension and $m$ is the order of $P$. This gives a meromorphic continuation of the spectral zeta function $\zeta(s) = \sum_{j=1}^\infty \lambda_j^{-s}$. This is an interesting generalization of the Riemann zeta function. Furthermore, its behaviour sheds light on the distribution of eigenvalues (the Weyl law). This zeta function is equivalent to the heat trace $\sum_{j=1}^\infty e^{-\lambda_j t}$ via the Mellin transform. There are lots of other things that come out of this, however -- take a look at the recent book by Scott Traces and Determinants of Pseudodifferential Operators''. </p>