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This happens for operators ```in divergence form'', i.e., Laplace operators associated to a Riemann metric on $\Omega$. If the Riemann metric is described by the tensor $(g_{ij}(x))_{1\leq i,j \leq …

This is mostly enhancing Nik Weaver's comments. Suppose that $M$ is compact of dimension $m$. If $m\geq 2$, then for a generic metric $g$ on $M$ the eigenvalues $\lambda_k$ of the Laplacian $\Delta …

Here is a good reference for this Michael Taylor: Pseudodifferential Operators, Princeton University PRess, 1981 In Chapter 12 it explains how to construct $f(A)$ when $A$ is elliptic selfadjoi …

Let me explain how to canonically produce a family of $\Delta$-invariant finite dimensional subspaces of $C^\infty(M)$ that completely determines the geometry of $M$. First, I need to introduce s …

First, the existence of a spectral projection implies that the index of the family $(D_\alpha)$ is a trivial element of $K^0(B)$. Next, if you fix a spectral projection $\Pi_0^+ :=\Pi^+_{\alpha_0}$. …

First observe that on a compact Riemann manifold $(M, g)$ the operator $1+t \Delta$, $t>0$. $\Delta = d^*d: C^\infty(M)\to C^\infty(M)$ has a unique fundamental solution. Jacques Hadamard has cons …

My answer is rather a question. Suppose that $(M,g)$ is a compact Riemann manifold and for any positive integer $N$ there exists an eigenvalue of the Laplacian that has multiplicity $>N$. Can one co …

The group of isometries of a compact Riemann manifold is a compact Lie group. This group acts on the eigenspaces of the Laplacian. If the group of isometries is non-abelian, then it is natural t …

For any $N\times N$ matrix $A$ we set $\DeclareMathOperator{\tr}{tr}$ $\newcommand{\bE}{\mathbb{E}}$ $$ |A|^2:=\sum_{i,j}|a_{ij}|^2 =\tr(A A^*). $$ There exists a constant $C=C(N)>0$ such that for a …

Two years ago my student Daniel Cibotaru wrote a dissertation entitled Localization formulae in odd K-theory in which he answers precisely this question in great generality. More precisely, g …

See Theorem 3.4.3, page 93, of these notes for a detailed proof of the fact that $T: H\to H$ is Fredholm if and only if there exists $Q:H\to H$ such that $QT-1$ is compact. If $Q=(T-\lambda)^{-1} …

I asume $M$ is a surface and $SM$ is the unit circle bundle. The bundle $SM\to M$ is also a principal $S^1$-bundle and the obvious $S^1$ action on $SM$ that preserves the natural metric on $SM$. …

Here is an argument using a bit of real algebraic geometry. Denote by $\newcommand{\eS}{\mathscr{S}}$ $\eS^n_+$ the cone of positive semi-definite symmetric $n\times n$ real matrices. $\newcommand{\ …

This is a bit too long to fit as a comment. For $d\geq 3$ there is only one spin structure on $S^d$ thus a unique spin Dirac operator $D: C^\infty(S^+)\to C^\infty(S^-)$ where $S^\pm$ denotes the bun …

If we assume a bit more, namely that $1-\Delta$ is (essentially) self-adjoint, and we denote by $A$ its self-adjoint extension, then we can define the Sobolev space $H^s(M)$, $s\geq 0$, as the domain …