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I've come across two slightly different conventions for the symbol of a differential operator $D$ (let's say on $\mathbb{R}^n$) and haven't thought much about the motivation behind them until now.

The first convention is to replace each instance of $\frac{\partial}{\partial x_j}$ by $i\xi_j$, where $\xi_j$ is a real variable for each $j=1,\dotsc,n$.

The second is to replace $\frac{\partial}{\partial x_j}$ by $\xi_j$.

Some of the advantages of the first convention seem to be:

  • It fits exactly with the way differentiation and multiplication correspond under the Fourier transform;
  • Self-adjoint operators have self-adjoint principal symbols;
  • Positive operators have positive principal symbols.

On the other hand, I've sometimes come across the second convention as well. For example it is used in Notes on the Atiyah–Singer Index Theorem by Liviu Nicolaescu; see Example 2.1.6 (p. 49) and Cor 2.1.15 (p. 52). I've also seen other people use this convention as well.

Are there certain advantages to using the second convention also?

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    $\begingroup$ In your first definition, you might want to use a different index since you also use $i$ to mean the imaginary unit. $\endgroup$
    – cmk
    Feb 25, 2022 at 14:52
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    $\begingroup$ The first notation was used primarily when using the Fourier transform and pseudodifferential operators to study linear PDOs.It made sense, because otherwise you had to keep track of the $i$'s. A related debate was whether $\Delta$ denoted a positive operator or the "physicists' Laplacian". But when attention shifted away from linear PDEs towards nonlines PDEs, the Fourier transform was no longer used much and the symbol was usually treated as a real-valued function on the cotangent bundle, it no longer made sense to use the first convention. $\endgroup$
    – Deane Yang
    Feb 25, 2022 at 15:09
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    $\begingroup$ see also the answer at math.stackexchange.com/a/706263/87355 $\endgroup$ Feb 25, 2022 at 15:17
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    $\begingroup$ If one wants to play with algebraic differential operators over an arbitrary base field, there is no distinguished $\sqrt{-1}$. $\endgroup$
    – Z. M
    Feb 25, 2022 at 15:17

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