# 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!

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