Exponential of derivation of commutative associative algebra

Question

Let $A$ be a commutative associative algebra with unit 1 and derivation $D$. For any $a\in A$, form the formal power series (vertex operator) $$Y(a,z)=\sum_{n\leq -1} \frac{D^{-n-1}a}{(-n-1)!}z^{-n-1}=e^{zT}a.$$ I am looking to show that $[D,Y(a,z)]=\partial Y(a,z)$. I've shown that $$\partial Y(a,z)=\sum_{n\leq -1} \frac{(-n-1)D^{-n-1}a}{(-n-1)!}z^{-n-2}=\sum_{n\leq 1}\frac{D^{-n}a}{(-n-1)!}z^{-n-1} $$ and $$[D,Y(a,z)]=\sum_{n\leq -1} \frac{D(D^{-n-1}a)-(D^{-n-1}a)D}{(-n-1)!} z^{-n-1}=\sum_{n\leq -1} \frac{D^{-n}a-(D^{-n-1}a)D}{(-n-1)!} z^{-n-1}=\partial Y(a,z)-Y(a,z)D.$$ If the identity I'm looking to show holds then $Y(a,z)D$ must be the zero map, but I don't see why this should be true.

Answer 1

Letting the vertex operator $Y(a,z)$ act on $b\in A$ looks promising (maybe all the sources I've been reading presented the vertex operators in that way for a reason?): we have $$Y(a,z)b=\sum_{n\leq -1} \frac{(D^{-n-1}a)b}{(-n-1)!}z^{-n-1}=(e^{zT}a)b,$$ $$\partial Y(a,z)=\sum_{n\leq -1} \frac{(-n-1)(D^{-n-1}a)b}{(-n-1)!}z^{-n-2}=\sum_{n\leq 1}\frac{(D^{-n}a)b}{(-n-1)!}z^{-n-1}, $$ and $$[D,Y(a,z)]b=\sum_{n\leq -1} \frac{D((D^{-n-1}a)b)-(D^{-n-1}a)Db}{(-n-1)!} z^{-n-1}.$$ Since $D$ is a derivation, we have $$D((D^{-n-1}a)b)=D(D^{-n-1}a)b+(D^{-n-1}a)Db=(D^{-n}a)b+(D^{-n-1}a)Db,$$ hence $$D((D^{-n-1}a)b)-(D^{-n-1}a)Db=(D^{-n}a)b, $$ which shows that the coefficients of the two series coincide.