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For this question, the Weyl algebra is the algebra of differential operators on $\Bbb C$: $$A_1(\Bbb{C})=\Bbb{C}\langle x,\partial\rangle/(\partial x- x\partial -1)$$ Although for a lot of purposes it seems that people like to use the Bernstein filtration $\deg(x)=\deg(\partial)=1$, we can also impose the "order" filtration $\deg(x)=-1$, $\deg(\partial)=1$. It seems to me that the latter makes $A_1(\Bbb{C})$ into a graded ring, so I am surprised that authors seem allergic to saying this, writing $\text{gr}A_1(\Bbb{C})$ even when considering this filtration.

Am I wrong— is this actually not a grading? And if it is, then what's up with all these gr's?

Eric Nathan Stucky
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  • I have very little experience of non-commutative rings, but I have only ever heard "graded" being used about commutative rings or algebras. And with the Bernstein filtration $\operatorname{gr}_F A_n$ becomes isomorphic to the polynomial ring in $2n$ variables, so commutative (which would distinguish it from $A_n$ itself). Further, the associated graded ring can be used to show properties of $A_n$, so there is a higher purpose with all these $\operatorname{gr}$'s. – LetGBeTheGraph Apr 29 '21 at 09:33
  • I know graded rings are defined in generality, but your cultural observation is appreciated (On reflection, I do agree that it is somewhat disquieting to think about a commutative grading on a noncommutative ring!) Still, I think that with respect to this filtration, gr$A_1(\Bbb C)$ is not the polynomial ring (and is "equal" to $A_1(\Bbb C)$). Of course, I do appreciate that with respect to the Bernstein filtration, the gr's are important ^.^ – Eric Nathan Stucky May 01 '21 at 02:51

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