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I am trying to find my way through Davies' book, and one of the difficult points is his choice of what "elliptic operator" means in his text. This is the first time that I encounter some of the concepts in the text, so I apologize if something below if wrong (any mistake is mine, not Davies').

One begins with a smooth matrix function $x \mapsto(a_{ij} (x))$ (Davies takes it to be measurable, but I don't need this) that is positive-definite and symmetric, from which one constructs the quadratic form

$$Q(f) = \int \limits _U \sum _{i,j} a_{ij} \frac {\partial f} {\partial x_i} \frac {\partial \bar f} {\partial x_j}$$

for $f \in C_c ^\infty (U)$ (the smooth compactly-supported functions on the open subset $U \subset \mathbb R^n$).

Assume that $\lambda > 0$ is a continuous function on $U$ and that $a > \lambda$ (i.e. $a - \lambda I_n$ is a positive-definite matrix function). Theorem 1.2.6 guarantees that $Q$ is closable on $f \in C_c ^\infty (U)$, therefore, according to theorem 1.2.1, there exists a self-adjoint operator $H \ge 0$ such that

$$Q(f) = \begin{cases} \langle \sqrt H f, \sqrt H f \rangle, & f \in \operatorname{Dom} \sqrt H \cap \overline{ C_c ^\infty (U) } \\ \infty, & f \in \overline{ C_c ^\infty (U) } \setminus \operatorname{Dom} \sqrt H \end{cases}$$

where the closure is taken in $L^2(U)$.

Such an operator $H$ is called "elliptic" by Davies.

Davies next notes that if $U$ has smooth boundary then

$$H = - \sum _{i,j} \frac \partial {\partial x_i} \left( a_{ij} \frac {\partial f} {\partial x_j} \right) \ .$$

  1. Why does Davies leave the "other" elliptic operators out?

  2. Should I understand that the Gaussian bounds for the integral kernel of $\mathrm e ^{tH}$ do not hold for general 2nd order elliptic operators, or has the theory been improved since then?

  3. The reason why I am asking is that, in a system of coordinates, the Laplace operator of a Riemannian manifold does not have the form required by Davies, which I find really strange, because this is one of the most important operators on a manifold and it is disappointing to see that Davies' theory does not hold in this simple and important case.

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  • $\begingroup$ What do you mean by "other elliptic operators"? (I know that some other text define an abstract version of ellipticity in terms of sending $H^{ṣ+2}$ onto $H^{s}$) $\endgroup$
    – Adrián González Pérez
    Commented Jun 19, 2018 at 12:32
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    $\begingroup$ @AdriánGonzález-Pérez: Take a look at the general definition and notice that Davies' one is more restrictive. In particular (to give a stupid trivial example) $-\Delta$ is elliptic according to Davies, but $-\Delta + 1$ is not. $\endgroup$
    – Alex M.
    Commented Jun 19, 2018 at 12:44
  • $\begingroup$ The Laplacian does have the form required by Davies. The expression for the Euclidean Laplacian you have in mind is given by setting $a_{ij}=\delta_{ij}$. This reduces then to show that form of your $H$ will be $H=-\sum_{i}\frac{\partial^2}{\partial x_i^2}$. $\endgroup$
    – Hadrian Quan
    Commented Jun 19, 2018 at 13:12
  • $\begingroup$ @HadrianQuan: I am asking about the Laplacian on Riemannian manifolds, expressed in local coordinates. I do not understand your comment. $\endgroup$
    – Alex M.
    Commented Jun 19, 2018 at 13:22
  • $\begingroup$ Hi @AlexM., the portion of Davies you've quoted is defining an elliptic differential operator on $\mathbb{R}^n$. If you'd like to define such an operator on a general manifold then the definition is given a bit differently. In particular, to respond to your question 3), the expression for the Laplacian on a Riemannian manifold does reduce to the definition given by Davies on $\mathbb{R}^n$. It is $$\Delta_g=-\frac{1}{\sqrt{det(g)}}\sum_{i,j}\partial_i(g^{ij}\sqrt{det(g)}\partial_j)$$. Since the Euclidean metric has $g_{ij}:=\delta_{ij}$, and $det(g_{ij})=1$, this reduces to your equation. $\endgroup$
    – Hadrian Quan
    Commented Jun 22, 2018 at 0:27

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