An alternative representation of the principal symbol of the Laplace operator

Question

Assume that $(M,g)$ is a $n$ dimensional Riemannian manifold. We denote by $\Delta$, the Laplacian associated to this Riemannian structure. Are the following two conditions equivalent?

First condition 1) There exist globally defined vector fields $X_i's,\quad i=1,2,\ldots,n$ such that $$\Delta=\sum_{i=1}^n \partial^2/\partial{X_i^2}$$

Second condition 2) There exsits a first order differential operator $D$ on $\chi^{\infty}(M)$ such that at each cotangent vector in $T^*(M)$ we have $$(*)\;\;\;{Det(P(D))}^2= {P(\Delta)}^n$$ where $"P"$ stand for the principal symbol of the corresponding operator.

If the answer to the question is negative, what kind of obstructions would appear to have the "Second condition 2"? Does every Riemannian manifold admit such a differential operator $D$ on $\chi^{\infty}(M)$? Apart from obvious examples in dimension $1,2,4,8$, are there some other examples(in other dimension)?

What about if we replace $(*)$ with $$trace (P(D)P(D)^{tr})=nP(\Delta)$$ Note that $P(D)^{tr}$, the transpos of $P(D)$, is uniquely well defined since each fiber of the bundle $q^*(TM)$, the pull back of $TM$ over $M$ under the natural projection $q:T^*M \to M$ is equiped with an inner product via vector bundle pull back of the metric of fibers of $TM$.

Answer 1

The answer to the first question is negative on the Euclidean sphere $\mathbb S^2$. It is possible to prove that the Laplace operator on the sphere $\mathbb S^2$ is NOT the sum of two squares of smooth vector fields.

However, you can always consider on the sphere $\mathbb S^d$ the $d(d-1)/2$ vector fields $$ \Omega_{jk}=x_j\partial_{x_k}-x_k\partial_{x_j}, \quad 1\le j<k\le d, $$ and you have indeed $ ∆_{\mathbb S^{d-1}}=\sum_{1\le j<k\le d} \Omega_{jk}^2. $

For your second question, assuming say that your manifold $(\mathcal M, g)$ is compact, the positive Laplace operator $A$ is a self-adjoint operator and by the functional calculus, you can define $B=A^{1/2}$ which is in fact a pseudo-differential operator of order 1, whose principal symbol is $b$ and such that $b^2=a$, where $a$ is the principal symbol of $A$. The downside of that approach is that the operator $A^{1/2}$ is never a differential operator.