Complex powers of elliptic operators

Remarks

The powers satisfy $A^s A^t = A^{s+t}$, $A^0 = \id$ and $A^1 = A$, and $A^{-1}$ is the inverse of $A$. Seeley proved that $A^s$ is a classical pseudo-differential operator of order $ms$. For $\operatorname{Re} s > n/m$, the operator $A^{-s}$ is trace class, and the spectral zeta function $$\zeta_A(s) = \operatorname{tr} A^{-s} = \sum_{j \ge 1} \lambda_j^{-s}$$ is holomorphic there. It extends to a meromorphic function on $\C$ with at most simple poles, and it is holomorphic at $s = 0$. This gives the zeta-regularized determinant $\det A = e^{-\zeta_A'(0)}$. More generally, Seeley defined $A^s$ by a contour integral of $\lambda^s (A - \lambda)^{-1}$ for elliptic $A$ whose principal symbol has no eigenvalues on some ray from $0$, without assuming self-adjointness.

Example: on the circle $S^1 = \R / 2\pi\Z$ with $dx$, let $A = 1 - d^2/dx^2$, which is elliptic of order $2$ and positive. Its eigensections are $e^{ikx}/\sqrt{2\pi}$, $k \in \Z$, with eigenvalues $1 + k^2$, so $A^{-s} e^{ikx} = (1 + k^2)^{-s} e^{ikx}$. Then $\zeta_A(s) = \sum_{k \in \Z} (1 + k^2)^{-s}$ converges for $\operatorname{Re} s > \frac{1}{2} = \frac{n}{m}$ with $n = 1$ and $m = 2$.

Non-example: the operator $-d^2/dx^2$ on $S^1$ is elliptic and self-adjoint but not positive, since it kills the constants. The formula would need $0^s$, which is not defined for all $s$; one first restricts to the orthogonal complement of the kernel.