Algebra of pseudo-differential operators

Question

The class of pseudod-ifferential operators form an associative algebra of Fourier integral operators. Moreover, given symbols $a,b,c\in C^\infty$ (each associated to some pseudo differential operator), for the composition of symbols (#) there holds:

$$\text{op}(a)\circ \text{op}(b) = \text{op}(a\# b),$$

so, obviously, $\#$ should be an associative operation as well.

The latter is equivalent to prove for

$$(a \# b)(x,y)=\sum_{|\alpha|\geq 0} \frac 1 {\alpha!} \partial_y^\alpha a(x,y) D_x^\alpha b(x,y)\quad (1)$$

there holds $(a\#b)\# c = a\#(b\# c)$, where $D=-i\partial$.

To check this, I first defined $\circ_N$ which considers in $(1)$ only the multi-indexes up to length $N$, i.e.

$$(a \circ_N b)(x,y)=\sum_{|\alpha|\leq N} \frac 1 {\alpha!} \partial_y^\alpha a(x,y) D_x^\alpha b(x,y).$$

Now, using the general Leibniz product rule I get

$$LS := (a\circ_N b) \circ_N c = \sum_{|\alpha|\leq N} \frac 1 {\alpha!} \partial_\xi^\alpha (\sum_{|\beta|\leq N}) \partial_\xi^\beta a D_x^\beta b) D_x^\alpha c = \sum_{|\alpha|\leq N,\ |\beta|\leq N,\ \alpha_1+\alpha_2=\alpha} \frac{1} {\beta!\alpha_1!\alpha_2!} (\partial_\xi^{\alpha_1+\beta} a) (\partial_\xi^{\alpha_2}D_x^\beta b) (D_x^\alpha c).$$

Similarly, we see

$$RS := a \circ_N (b\circ_N c) = \sum_{|\alpha|\leq N,\ |\beta|\leq N,\ \alpha_1+\alpha_2=\alpha} \frac{1} {\beta!\alpha_1!\alpha_2!} (\partial_\xi^\alpha a) (\partial_\xi^{\beta}D_x^{\alpha_2} b) (D_x^{\alpha_1+\beta} c).$$

But unfortunately it is $LS\neq RS$; the larger $N$, the more different terms arise on each side and they will never cancel.

What's my mistake?

Answer 1

Why do you think the two are different?

If you don't limit the sum to length $N$ you have (I use $\gamma$ and $\delta$ for the multiindices in the summation for $RS$ to avoid confusion):

$$ LS := (a\circ b) \circ c = \sum_{\alpha_1,\alpha_2,\beta} \frac{1} {\beta!\alpha_1!\alpha_2!} (\partial_\xi^{\alpha_1+\beta} a) (\partial_\xi^{\alpha_2}D_x^\beta b) (D_x^{\alpha_1+\alpha_2} c).$$

$$ RS := a \circ (b\circ c) = \sum_{\gamma_1,\gamma_2,\delta} \frac{1} {\delta!\gamma_1!\gamma_2!} (\partial_\xi^{\gamma_1 + \gamma_2} a) (\partial_\xi^{\delta}D_x^{\gamma_2} b) (D_x^{\gamma_1+\delta} c). $$

and one sees that replacing $\alpha_1 = \gamma_1$, $\alpha_2 = \delta$ and $\beta = \gamma_2$ the two expressions are exactly identical.

So your mistake is a mistake of notation. Dummy variables in summations can be renamed.

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You are correct, however, in saying that truncated to order $N$ the operation is not necessarily associative. But even there you should be able to see partial cancellations.

Answer 2

Certainly @WillieWong's answer directly responds to the question.

It may be worth adding that, yes, indeed, there are complications in proving associativity of composition of pseudo-differential operators in the Kohn-Nirenberg model of them, as being on $\mathbb R^n$. Keyword for the composition of of the K-N model of pseudo-differential operators: "Moyal product" (well-known to physicists, I gather).

That is, I have to say that it was a great relief to me when I learned (from some marvelous writing of G. Folland) that Weyl's presentation of pseudo-differential operators, as related to Heisenberg groups (as opposed to just $\mathbb R^n$) made the composition simply be the convolution of operators attached to functions on the group. So, from general principles, we have associativity.