Weyl algebra and its nontriviality

Question

The Weyl algebra (say, over $\mathbb{C}$) is an universal unital algebra with two generators $x,y$ subject to the relation $xy-yx=1$. This algebra can be constructed in the following way: take two dimensional vector space with basis $\{x,y\}$ and construct the tensor algebra $T(V)$. Then take an ideal $I=(xy-yx-1)$ generated by $xy-yx-1$ and form a quotient $T(V)/I$. It is possible to find a norm on $T(V)$ in which $T(V)$ becomes a normed algebra, therefore we can take the closure $\overline{I}$ of $I$ and form a quotient $T(V)/\overline{I}$. This is again a normed algebra but generators of this algebra satisfy the relation $xy-yx=1$ but this is impossible in any normed unital algebra. This apparent contradiction actually shows that $I$ must be dense in $A$. This leads me to the question: it is really obvious and immediate that $I$ is not the whole $T(V)$? I'm pretty sure that this must be true but it seems to me that it would be hard to give a "two line" proof of this fact. Forgive me if this question is to elementary: but in the literature no one seems to see any problem with the existence (and nontriviality) of Weyl algebra.

Answer 1

Generally the strategy for showing that some syntactic construction is nontrivial is to find a semantic model of it. E.g. the strategy for showing that some relations don't force a group to be a trivial group is to find, say, some matrices satisfying the relations, and the strategy for showing that some axioms in some theory don't contradict each other is to find a model satisfying all the axioms.

In this case the strategy for showing that some relations don't force a ring to be nontrivial is to find a module. An alternative definition of the Weyl algebra, at least in characteristic zero, is that it is algebra of differential operators on $k[x]$, where one generator acts as multiplication by $x$ and the other acts as $\frac{\partial}{\partial x}$. In characteristic zero, this action can be used to prove the stronger statement that the Weyl algebra has a basis given by the monomials $x^i \frac{\partial}{\partial x^j}$.

In positive characteristic $k[x]$ actually fails to be a faithful module of the Weyl algebra because $\frac{\partial}{\partial x^p}$ acts by zero. A module which is faithful regardless of the characteristic can be constructed using the lattice of Young tableaux; see this math.SE answer for details.

The basis above should remind you of a PBW basis, and in fact the Weyl algebra is almost the universal enveloping algebra of the Heisenberg Lie algebra spanned by $1, x, \frac{\partial}{\partial x}$. It is precisely the quotient of this universal enveloping algebra by the relation that $1$ acts by the identity, and either the PBW theorem or a slight modification of it can be used to prove the stronger statement above. (A less coordinate-dependent way of saying this is that the Weyl algebra has a natural filtration and you can show that its associated graded is $k[x, y]$.)

Answer 2

A generic argument that can be used to show that an algebra is non-trivial is Bergman's Diamond Lemma. In this case, it immediately applies (there are no ambiguities) so for example $x$ is not in the ideal.

This method is useful even in situations in which you cannot find a representation to implement the idea in the other two answers, or when the characteristic gets in the way.

Answer 3

One way to prove non-triviality is by modeling the formal $x,y$ as $d/dx$ and (multiplication by) $x$ on (e.g., test-) functions on $\mathbb R$.