I came across the following introduction to a paper I was reading:
"Let $L$ be a second-order, elliptic, differential operator, with smooth coefficients, on a Riemannian manifold $M$. Assume $-L$ is positive and self-adjoint, such as perhaps $-L=-\Delta,$ and let $A=(-L)^{\frac{1}{2}}$ be its positive square root. Thus, $A$ is a self-adjoint operator on the Hilbert space $H=L^2(M)$."
I have the following questions based on this sentence:
$1.$ Why is the operator $L$ defined as a negative? Why not just define it as $L=\Delta$ and $A=L^{\frac{1}{2}}$?
$2.$ Why do we take the square root of $A$? Is it to ensure that $A$ can be defined on the Hilbert space $L^2$ (which requires a square root by definition)?
This is from the paper "Pointwise Fourier Inversion: a Wave Equation Approach" by M. Pinsky and M. Taylor (I can provide a pdf if it will help).
Thank you in advance for your comments.
For (1), depending on whether you are a geometer or an analyst you may have a different sign on your Laplacian. However, if you locally write
$$ \Delta f = \sum \frac{\partial^{2} f}{\partial x_{j}^{2}} $$
then this is negative definite. This being the classic example of the sort of operator $L$ you are interested in, the author might be assuming $L$ to be negative definite and hence $-L$ is positive, for example the well-known positive operator $-\Delta$. Along those lines, we negate $L$ to take its square root.
As for (2), I think your suggestion is on point.