In Etingof's Introduction to representation theory there is a bit of notation I don't understand. If $k$ is a field, he defines the Weyl-algebra by the free algebra on $x$ and $y$ quotienting out the relation $$yx-xy=1.$$ Next, Etingof proves the fact that $\{x^iy^j | i, j \in \mathbb{N} \}$ is a basis for this algebra. In the proof, the author says: if $a$ is a variable, let $E=t^a k[a][t,t^{-1}]$, and later on he says $a$ is just a formal symbol so 'reallly' $E=k[a][t,t^{-1}]$. I do not understand what vector space he has in mind. I am interested in this because he specifically says his proof shows we can view the Weyl-algebra as the algebra of differential operators in $t$, which sounds useful.
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- 1$\begingroup$ I guess the point of introducing this formal variable is that if you are in characteristic $p$, possibly the derivative of a polynomial can be zero even if the polynomial itself is not zero, but it can be fixed by introducing a formal variable as exponent. $\endgroup$user171326– user1713262017-06-26 20:26:17 +00:00Commented Jun 26, 2017 at 20:26
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2 Answers
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$t^a k[a][t, t^{-1}]$ is the free $k[a]$-module on a basis whose elements are named $t^{a + n}$ where $n \in \mathbb{Z}$. The point of the $t^a$ factor is as a mnemonic to remind you what the intended action of the Weyl algebra is. Namely, there is a derivation $\frac{\partial}{\partial t}$ which acts on each basis element by
$$\frac{\partial}{\partial t} t^{a+n} = (a+n) t^{a+n-1}$$
and this action is extended $k[a]$-linearly to every element of $t^a k[a][t, t^{-1}]$. Together with the action of multiplication by $t$ we get a module over the Weyl algebra, where $x \mapsto t$ and $y \mapsto \frac{\partial}{\partial t}$. To see that this is actually a module over the Weyl algebra you need to verify the relation
$$\frac{\partial}{\partial t} (t f(t)) - t \frac{\partial}{\partial t} f(t) = f(t)$$
for $f(t) \in t^a k[a] [t, t^{-1}]$ which is straightforward.
The point of having $a$ be a formal variable here is to deal with the possibility that $k$ has positive characteristic. If $k$ has characteristic $p$ then a funny thing happens to $\frac{\partial}{\partial t}$, namely
$$\left( \frac{\partial}{\partial t} \right)^p t^n = n(n-1)(n-2) \dots (n-(p-1)) t^{n-p}$$
which, if $n$ is an ordinary integer, is always equal to zero! So the action of the Weyl algebra on $k[t, t^{-1}]$ (obtained from the above action by setting $a = 0$) is not faithful. The introduction of $a$ fixes this cleanly.
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As I understand it his $E$ consists of elements $t^a f(a,t,t^{-1})$ where $f$ is a polynomial. There is a $k$-vector space isomorphism to $k[a][t,t^{-1}]$ where $t^a f(a,t,t^{-1})$ maps to $ f(a,t,t^{-1})$. As far as I can tell, this remark is just to point out that these vector spaces are isomorphic. So one could define the representation of the Weyl algebra on $f(a,t,t^{-1})$ but the action $t^a f(a,t,t^{-1})$ has a more natural description.
