# M721: Index Theory

## A local view of the global definition of a differential operator

Recall the global definition of a differential operator (of order m) $D: \mathcal C^\infty(E) \to \mathcal C^\infty(F), \ \text{with} \ E,F \ \text{being smooth vector bundles over a manifold} \ X:\\ DO_{-1}(E,F) = \{ \mathbf 0\}, \text{and inductively}, \ D \in DO_m(E,F) \!\iff\! [D, f] \in DO_{m-1}(E,F).$

Also recall that differential operators are local, so that in particular, they induce (linear) operators between the spaces of sections over chart neighbourhoods in X.  This allows us to reduce to the Euclidean case, and for the remainder of this discussion, we will assume $D \in \mathrm{Hom}_\mathbb R\left(\mathcal C^\infty(\mathbb R^n, \mathbb R^N) \to \mathcal C^\infty(\mathbb R^n, \mathbb R^M)\right)$ (noting that the space of (smooth) sections of $\mathbb R^{n + N} \to \mathbb R^n$ is naturally isomorphic to the space of (smooth) maps from $\mathbb R^n \ \text{to} \ \mathbb R^N$, etc.)

Proposition: If D has the form of a “local differential operator” of order m, then $D \in DO_m\left(\mathcal C^\infty(\mathbb R^n\times \mathbb R^N), \mathcal C^\infty(\mathbb R^n\times \mathbb R^M)\right) \quad \text{(which we henceforth refer to as simply}\ DO_m).$

Proof:

Set $D = \sum_{|\alpha| \le m} A_\alpha D^\alpha, \ \text{with} \ \alpha \ \text{a multi-index,}\ A_\alpha \in M_{M \times N}(\mathcal C^\infty(\mathbb R^n, \mathbb R)),\ \text{and} \ D^\alpha$ as in the definition of a local differential operator.  If $f \in \mathcal C^\infty(\mathbb R^n, \mathbb R) \ \text{and}\ s \in \mathcal C^\infty(\mathbb R^n, \mathbb R^N),$ then Leibniz’s differentiation rule yields $D^\alpha(fs) = \sum_{\mu + \nu = \alpha}\binom{\alpha}{\nu} D^\mu fD^\nu s, \ \text{where}\ \mu,\nu$ are multi-indices, and sums of mulit-indices are taken term-wise, and $\binom{\alpha}{\nu} = \frac{\alpha!}{\nu! (\alpha - \nu)!} = \frac{\alpha_1!\cdots \alpha_n!}{\nu_1!\cdots \nu_n!(\alpha_1 - \nu_1)!\cdots (\alpha_n - \nu_n)!}$.

We proceed by induction on m, the order of D (as a local D.O.).  By the linearity of $[\cdot , f]$, it suffices to prove $D^\alpha \in DO_{|\alpha|}$.  Now, $D^0 s = s, \ \forall s \in \mathcal C^\infty(\mathbb R^n, \mathbb R^N), \ \text{and so}\ [D^0,f](s) = D^0(fs) - f D^0s = 0$, which gives $[D^0,f] = \mathbf 0 \in DO_{-1}, \ \forall f\in \mathcal C^\infty(\mathbb R^n, \mathbb R) \implies D^0 \in DO_0$, establishing the base case.  Now assume that all local differential operators of order at most$m-1$ are (global) differential operators of the same order.  The Leibniz rule above gives:
$[D^\alpha, f](s) = D^\alpha(fs) - fD^\alpha s = \left(\sum_{\mu + \nu = \alpha} \binom{\alpha}{\nu} D^\mu f D^\nu s\right) - fD^\alpha s = \sum_{\mu + \nu = \alpha, \; |\mu| > 0} \binom{\alpha}{\nu} D^\mu f D^\nu s.$

Since $|\mu| > 0, \ |\nu| = |\alpha| - |\mu| < |\alpha| = m$—i.e., each $D^\nu$ in the sum has order at most $m-1$, and so by the inductive hypothesis (and noticing that $DO_{k-1} \subset DO_k, \forall k \in \mathbb N$), we find that $[D^\alpha, f] \in DO_{m-1} \implies D^\alpha \in DO_m.$   QED.

Now we begin to prove the converse to the above proposition.  For the purposes of the next lemma, we may relax the assumption that D is an operator between sections of vector bundles over Euclidean space.

Lemma 1: Let $D \in DO_m(E,F)$, then for every $s \in \mathcal C^\infty(E), \ \text{and} f_1, ..., f_{m+1} \in \mathcal C^\infty(X,\mathbb R).$  If every $f_i$ vanishes at a point $x_0 \in X, \ \text{then} \ (D(f_1 \cdots f_{m+1})s)(x_0) = 0$.

Proof:

Again, by induction on m.  If $D \in DO_{-1}(E,F), \ \text{then}\ D = \mathbf 0, \ \text{whence}\ (Ds)(x_0) = 0, \ \forall x_0 \in X$, and thus the base case is established.  Now suppose the result for $DO_{m-1}$, and let $D \in DO_m$.  Now $(D(f_1 \cdots f_{m+1})s)(x_0) = [D, f_{m+1}]((f_1 \cdots f_m)s)(x_0) + f_{m + 1}(x_0)(D(f_1\cdots f_m)s)(x_0)$.  By assumption, $[D, f_{m+1}] \in DO_{m-1},$ and so $[D, f_{m+1}]((f_1 \cdots f_m)s)(x_0) = 0$.  On the other hand, $f_{m+1}(x_0) = 0$ by hypothesis, and so $(D(f_1 \cdots f_{m+1})s)(x_0) =0.$   QED.

Lemma 2: Let $D \in DO_m\left(\mathcal C^\infty(\mathbb R^n\times \mathbb R^N), \mathcal C^\infty(\mathbb R^n\times \mathbb R^M)\right).$  Now, for every $k \in \mathbb N, \ \exists!\{A_\alpha\}_{|\alpha| \le k}$ with the property that $\forall |\beta| \le k, \ D(x^\beta) = \sum_{|\alpha| \le k} A_\alpha(x)D^\alpha(x^\beta).$

The proof is left as an exercise.  We give an outline of the inductive construction.  To find $A_0$, allow $D$ to act on the constant co-ordinate sections $e_i, \ i = 1, ..., N$, and expand each $De_i$ in terms of the basis $f_1, ..., f_M\ \text{for}\ \mathbb R^M$.  Take $A_0$ to be the matrix thus determined—since D maps smooth vector fields to smooth vector fields, the co-ordinates of $A_0$ will be given by smooth functions of x.  This gives a $0^\mathrm{th}$ order approximation of D.

Now we take these sections and begin to multiply them by functions on X to obtain new sections—e.g., take $s_{j,i}: x\mapsto x_je_i\ \text{and let the}\ i^\mathrm{th}\ \text{column of}\ \ A_\beta\ \text{be}\ Ds_{i,j} - De_i = D(x_j-1)e_i,$ where $\beta$ is the multi-index with only a single 1 in position j.  In general, define the  $i^\mathrm{th}\ \text{column of }\ A_\beta$ by $D\left(\left(\frac{x^\beta}{\beta!} - \sum_{\mu + \nu = \beta} \frac{x^\nu}{\nu!}\right)e_i\right)$ for some multi-index $\mu$.  It is clear (or an exercise) that these matrices produce the desired result, and moreover, Lemma 1 shows that at most $2^m$ of these matrices will be non-zero (i.e., that the process terminates).  Uniqueness is, of course, automatic.

Theorem: Differential operators are precisely local operators whose local form is that of a local differential operator.

Proof:

Because differential operators are local, we can immediately reduce to the Euclidean case.  Showing “$\supseteq$” was the content of the proposition, and so it suffices to show “$\subseteq$“.  Invoking the construction from lemma 2 (with k = m), let $\tilde D = \sum_{|\alpha| \le m}A_\alpha D^\alpha$.

Consider the section $s: x \mapsto s_i(x)e_i, \ s_1, ..., s_N \in \mathcal C^\infty(\mathbb R^n, \mathbb R)\ \text{of}\ \mathbb R^{N + n} \to \mathbb R^n$.  Recall Taylor’s Theorem: given $f \in \mathcal C^\infty(\mathbb R^n, \mathbb R), \ \forall x_0 \in \mathbb R^n, \ \text{and}\ \forall m \in \mathbb N, \exists p$, a polynomial of degree m, a neighbourhood $U\ \text{of}\ x_0$, and for each multi-index $|\alpha| = m + 1, \exists h_\alpha \in \mathcal C^\infty(\mathbb R^n, \mathbb R)$ such that $f(x) = p(x) + \sum_{|\alpha| = m + 1}h_\alpha(x) (x - x_0)^\alpha$ for all $x \in U$.  Choose $x_0 \in X$ take m to be the order of D, and apply Taylor’s Theorem to each $s_i$ individually to obtain $s_i(x) = p_i(x) + \sum_{|\alpha| = m + 1} h_{i,\alpha}(x)(x - x_0)^\alpha = p_i(x) + h_i(x)$ on a neighbourhood $U_i\ \text{of}\ x_0$.  Take $U = \bigcap_{i = 1}^N U_i$.

Now by locality, $(Ds)\bigl|_U \ \text{depends only on}\ s\bigl|_U,$ and so $(D - \!\tilde D)(s)(x_0) \!=\! (D - \!\tilde D)\!\!\left(\sum_{i = 1}^N(p_i + h_i)e_i\right)\!\!(x_0) \!=\!\! \sum_{i = 1}^N ((D - \!\tilde D)(p_i + h_i)e_i)(x_0).$  Hence, $\tilde D(p_ie_i) = D(p_ie_i)$ by construction, and so $(D - \tilde D)(s) = \sum_{i = 1}^N (D - \tilde D)(h_ie_i)(x_0)$.  On the other hand, $h_i(x) = \sum_{|\alpha| = M + 1} h_{i,\alpha}(x)(x - x_0)^\alpha$, and so linearity reduces the question to showing $(D - \tilde D)((x - x_0)^\alpha h_{i,\alpha}(x)e_i)(x_0) = 0$.

Finally, $D,\tilde D$ are differential operators of order m (use the proposition to get this for $\tilde D$), and since $(x - x_0)^\alpha\ \text{decomposes into}\ m+1$ smooth functions which all vanish at $x_0\ \text{for}\ |\alpha| = m + 1$, we find by lemma 1, that $(D(x - x_0)^\alpha h_{i,\alpha}(x)e_i)(x_0) = (\tilde D(x - x_0)^\alpha h_{i,\alpha}(x)e_i)(x_0) = 0$, whence $Ds(x_0) = \tilde Ds(x_0) \implies D = \tilde D,\ s,x_0$ having been arbitrary.    QED.