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The Weyl algebra is $$ A=k\langle x,y\rangle/\langle yx-xy-1\rangle $$ where $k$ is a field, $k\langle x,y\rangle$ is the free algebra, and $yx-xy-1$ is the defining relation. A basis for $A$ is $\{ x^i y^j|i,j\geq0\}$. What I am concerned about here is how to prove that elements of this set are linearly independent. Can we reason by contradiction that if $\sum c_{ij}x^i y^j=0$, then (after rewriting $\sum c_{ij}x^i y^j=\sum f_j(x)y^j$), $f_j(x)=0,\forall j$, thus $c_{ij}=0,\forall i,j$. This is a bit weird, I feel.

Another related question is about the $q$-Weyl algebra $$ B=k\langle x,y\rangle/\langle xx^{-1}-1,x^{-1}x-1,yy^{-1}-1,y^{-1}y-1,yx-qxy \rangle $$ where $q\in k$. How to show that that a basis for $B$ is $\{x^i y^j|i,j\in \mathbb{Z}\}$? Can someone give me some hints on these two problems? Thank you very much!

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I edited for proper use of \langle and \rangle in TeX, i.e. where it said $<x,y>$, it now says $\langle x,y \rangle$. –  Michael Hardy Sep 19 '11 at 13:42
    
Can you use Poincaré-Birkhoff-Witt theorem to reach the conclusion of your first question? The Lie algebra would be the span of $\{1,x,y\}$, but we need a method for identifying the generator $1$ with the multiplicative identity. Moding out a suitable ideal out of the universal enveloping algebra should do it? –  Jyrki Lahtonen Sep 19 '11 at 14:39
    
I tidied up the question a bit, because I found it hard to read at first. Hope you don't mind. –  George Lowther Sep 20 '11 at 0:05
    
@GeorgeLowther:Thank you very much for the editing work!By the way,can you tell me how to edit this?I am not very familiar with using Latex here.:) –  user14242 Sep 20 '11 at 15:40
    
@user14242: You should see "edit" below your post, which you can click on and you can see the source. –  George Lowther Sep 20 '11 at 19:41

3 Answers 3

You can use Poincaré-Birkhoff-Witt, but it's not necessary. Simply observe that the Weyl algebra acts on $k[x]$ (by multiplication by $x$ and differentiation w.r.t. $x$), and all of the basis elements you describe act linearly independently (edit: in characteristic $0$!). The proof for the $q$-Weyl algebra should be similar except that $y$ acts by $q$-differentiation.

Edit: The above proof doesn't work in positive characteristic since there we have the pesky extra relation $\left( \frac{d}{dx} \right)^p = 0$. Here's one that does: it is a folk observation (see Stanley's Differential Posets for details) that the Weyl algebra acts on the free $k$-module $Y$ spanned by the set of Young diagrams as follows:

  • $x$ acts by sending a diagram $\lambda$ to the set of all diagrams obtained by adding a box to $\lambda$ (such that the result is still a Young diagram),
  • $y$ acts by sending a diagram $\lambda$ to the set of all diagrams obtained by removing a box from $\lambda$ (such that the result is still a Young diagram).

(The proof that this defines a representation of the Weyl algebra is left as a pleasant exercise.)

Now suppose that there is a nontrivial linear dependence of the form

$$\sum_n c_n x^{i_n} y^{j_n} = 0.$$

Let $J = \text{max}_n(j_n)$, let $I = \text{max}_{j_n=J}(i_n)$, and consider the action of the LHS on the Young diagram $\lambda_{J+1} = (111...)$ consisting of $J+1$ boxes in a single row. Among the terms in

$$\sum_n c_n x^{i_n} y^{j_n} \lambda_{J+1}$$

the only ones which consist of a single column come from the terms $x^{i_n} y^{j_n}$ with $j_n = J$, and of these the tallest one comes from $x^I y^J$ and has a nonzero coefficient. Hence this sum is nonzero; contradiction.

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This works well if char$k$=0.Then what about the case char$k$=$p$? –  user14242 Sep 19 '11 at 15:47
    
@user14242: okay, then observe that the Weyl algebra acts on the free $k$-module with basis the set of partitions $\lambda$ as follows: $x$ acts by sending a partition $\lambda$ to the sum over all partitions obtained by adding one box to $\lambda$, and $y$ acts by sending a partition $\lambda$ to the sum over all partitions obtained by removing one box from $\lambda$. See www-math.mit.edu/~rstan/pubs/pubfiles/77.pdf for details. I believe this action is faithful for all $k$. –  Qiaochu Yuan Sep 19 '11 at 15:59
    
+1 A nice argument! –  Jyrki Lahtonen Sep 19 '11 at 18:07
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A remark: the $q$-Weyl algebra mentioned in the question is not the one involving $q$-differentiation: it is the subalgebra of endomorphisms of $k[t^\pm]$ generated by the map given by multiplication by $t$ , by the unique $k$-algebra automorphism whichmaps $t$ to $qt$, and their inverses. –  Mariano Suárez-Alvarez Sep 19 '11 at 22:30

One simple way to do this is to use Bergman's Diamond Lemma, and it works over any commutative ring $k$.

If we order monomials in the free algebra $k\langle x,y\rangle$ first by degree and then lexicographically, for example, then the relation $$yx-xy-1$$ gives a 'rewriting rule' of the form $$yx \leadsto xy + 1.$$ It is immediate that there are no ambiguities (in the sense of Bergman) so that it follows at once that the classes of the monomials in $k\langle x,y\rangle$ which do not contain $yx$ as a subword form a basis of the quotient $A_1=k\langle x,y\rangle/(yx-xy-1)$. This is precisely what you want.

Exactly the same reasoning gives you the quantum case (but now there will be overlapping ambiguities: for example, $xx^{-1}x$... but a little check shows they are all resolvable.)

If you intend to do this kind of things, you must learn about the Diamond Lemma. :)

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You just have to find a representation of the algebra (as Qiaochu discussed). A simple way of doing this is to consider the Weyl algebra acting on the (commutative) polynomial ring $k[u,v]$. Regarding this as a vector space over $k$ with basis $u^iv^j$ ($i,j\ge0$), you can define $x,y$ as linear operators on $k[u,v]$, $$ \begin{align} &x(u^iv^j)=u^{i+1}v^j,\\ &y(u^iv^j)=iu^{i-1}v^j+u^iv^{j+1}. \end{align} $$ You can check that $xy-yx=1$, so $k\langle x,y\rangle$ is a representation of the Weyl algebra. Also, if $\sum_{ij}c_{ij}x^iy^j=0$, then applying this to $1\in k[u,v]$ gives $\sum_{ij}c_{ij}u^iv^j$, so $c_{ij}=0$.

Note: All I did here was to look at the multiplication rules $x\cdot x^iy^j=x^{i+1}y^j$ and $y\cdot x^iy^j=ix^{i-1}y^j+x^iy^{j+1}$, then replace these by operations on $u^iv^j$.

The same thing works for the q-Weyl algebra. Consider $V=k[u,v,u^{-1},v^{-1}]$ as a vector space over $k$ with basis $u^iv^j$ ($i,j\in\mathbb{Z}$). Define $x,y$ acting on $V$ as $$ \begin{align} &x(u^iv^j)=u^{i+1}v^j,\\ &y(u^iv^j)=q^iu^iv^{j+1}. \end{align} $$ Then, $x,y$ are invertible linear operators with $yx=qxy$, so $k\langle x,y\rangle$ is a representation of the q-Weyl algebra. As before, if $\sum_{ij}c_{ij}x^iy^j=0$ then applying this to $1\in V$ gives $\sum_{ij}c_{ij}u^iv^j=0$, and $c_{ij}=0$.

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