What does it mean for the leading symbol of a differential operator to be scalar?

I would like to better understand what it means for the leading symbol of a differential operator to be scalar.

Concretely, I am currently looking at the Laplace - Beltrami operator on an n-dimensional Riemannian manifold $(M,g)$.

Locally, $$\triangle_g = \frac{1}{\sqrt{|g|}} \sum^n_{i,j = 1} \partial_i g^{ij} \sqrt{|g|} \partial_j$$ where $|g|$ denotes the determinant and $(g^{ij})$ the inverse matrix to $(g_{ij})$.

Expanding the above expression we obtain $$\triangle_g = \sum_{i,j = 1}^n g^{ij} \partial_i \partial_j + \sum_{i,j = 1}^n (g^{ij} \frac{\partial_i \sqrt{|g|}}{\sqrt{|g|}} + \partial_ig^{ij})\partial_j$$

and so the leading symbol is given by $$p_2(x,\xi) = \sum_{i,j = 1}^n g^{ij} \xi_i \xi_j$$

(hope I am correct so far)

Now, there are two questions that I am trying to look up but cannot find an answer to that helps me fill in all the gaps that I currently have in my knowledge:

(1) what does it mean when it is pointed out that the leading symbol is scalar ? Does it mean that I have $g^{ij} = \lambda \delta_{ij}$ for some number $\lambda$ ?

(2) since $\triangle_g$ is elliptic, does this mean that the leading symbol is scalar ? I don't think that would be true, but can I diagonalize the matrix $g$ so that it the leading symbol becomes scalar ? But what does it mean for a non-constant matrix $g$ to be diagonalized ?

I hope the questions are not too confused, please let me know in case more clarification is necessary.

Is there a book that you'd recommend me looking at, given my above questions ?

Many thanks!

-

1 Answer

(1) Either it means that (i) the leading symbol does not depend on the point x or (ii) that it is a scalar function, i.e. the functions on which the differential operator acts are defined only on one variable. Both do not make any sense in your case, because (i) is not invariant under coordinate transformations and (ii) forces n=1.

(2) Elliptic means, per definition, that the leading symbol is invertible for all $\xi \not= 0$.

-
Hm .. ok I probably need to look into this to produce a more precise question. But thanks for your help! – harlekin Jan 22 '12 at 18:56