Can anyone prove that a Weyl Algebra is not isomorphic to a matrix ring over a division ring?


5 Answers 5


Notation: The Weyl algebra is $$k[x_1, x_2, \ldots, x_n, \partial_1, \partial_2, \ldots, \partial_n]$$ with the obvious relations.

The Weyl algebra doesn't contain any division rings larger than $k$, and it is infinite dimensional over $k$. So, assuming you don't allow infinite matrices, that's a proof.

How to see that it doesn't contain any division ring larger than $k$? I just need to show that any nonconstant differential operator is not invertible. One way to see this is to notice that multiplying differential operators multiplies symbols, and the symbol of a nonconstant differential operator is a nonconstant polynomial.

  • $\begingroup$ Whats the base field? $\endgroup$
    – Casebash
    Nov 8, 2009 at 4:42
  • $\begingroup$ The field of constants. I have edited the answer; see if that is clearer. $\endgroup$ Nov 8, 2009 at 13:49
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    $\begingroup$ I guess you could also use the fact (which follows easily by looking at highest order terms) that the Weyl algebra is a domain and not a division ring. So it can't be a matrix ring over a division ring. $\endgroup$
    – GS
    Jan 14, 2010 at 10:40

A different proof would be to show that a Weyl algebra is not semisimple, that is, that it is not a direct sum of simple submodules as a left module over itself. However, note that there is an infinite descending chain of left submodules of a Weyl algebra given by $A_n\supseteq A_nd\supseteq A_nd^2\supseteq A_nd^3\supseteq...$ where $d$ is any non-invertible element. A direct sum of a finite number of simple modules can't have an infinite descending chain of submodules. Then, by the converse of Artin-Wedderburn, $A_n$ is not a direct sum of matrix algebras over a divsion ring.

Of course, showing this sequence of submodules never stabilizes can be done by looking at the associated graded algebra, and noting that the $\overline{A_nd^n}$ are always distinct there. However, then this answer starts getting closer to David's answer, so maybe this wasn't a truly different proof.

  • $\begingroup$ What does the bar mean? $\endgroup$
    – Casebash
    Nov 8, 2009 at 6:50
  • $\begingroup$ Image in the associated graded ring. $\endgroup$ Nov 8, 2009 at 14:51
  • $\begingroup$ Makes sense now $\endgroup$
    – Casebash
    Nov 8, 2009 at 21:15

I would only like to add a simple proof that the Weyl algebra doesn't even HAVE any (non-trivial) finite-dimensional representations. Already in the case n=1, consider the relations $$[\partial_x,x]=1.$$ Now suppose you had a finite dimensional representation, and take the trace of both sides of the above.

It implies that the identity acts as 0 so the whole representation does.

  • $\begingroup$ I was going to say this, but I wasn't sure if it still works when the underlying ring is noncommutative. $\endgroup$ Nov 9, 2009 at 15:01
  • $\begingroup$ Which ring are you calling underlying? The base field? This is part of the standard proof that W_n is simple. In fact not only can it not have finite dimensional representations, but it's smallest representations have so-called Gelfand Kirillov dimension n, meaning that they are infinite-dimensional, and graded, and the dimension of the kth piece is on the order of k^{n-1}. These are called holonomic modules. The prototype example is C[x_1,...x_n] with its natural action. The dimension of the kth graded part is \choose{k+n-1}{n-1}, which is approximately k^{n-1} as k-->\infty. $\endgroup$ Nov 9, 2009 at 16:05
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    $\begingroup$ The question is about matrix rings over division rings, not over fields. Unless you're just stating a weaker result? $\endgroup$ Nov 9, 2009 at 23:13
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    $\begingroup$ ...and, moreover, the division rings in questions need not be finite dimensional algebras over the basefield of $A_n$. $\endgroup$ Jun 26, 2011 at 17:03

This is not an answer to the original question. However, it is related and I think that it is worth mentioning.

Assuming that ring morphisms take identity elements to identity elements, we can show the following for the $n$:th Weyl algebra, with very basic methods.

Choose an arbitrary positive integer $n$ and put $A_n := \mathbb{C}\langle x_1,\ldots,x_n,y_1,\ldots,y_n \rangle / I$ where $I$ is the ideal generated by the elements $y_1x_1-x_1y_1-1,\ldots,y_nx_n-x_ny_n-1$ and $x_i x_j-x_jx_i, y_i y_j - y_j y_i$ for $i,j \in 1,\ldots,n$.

There does not exist a positive integer $m$ and an associative, commutative and unital ring $R$ such that there is a ring morphism

$$ \phi : A_n \to M_m(R).$$

Seeking for a contradiction, suppose that there is some $m$ and some associative, commutative and unital ring $R$ such that $\phi$ exists. Denote the images in $M_m(R)$ of $x_1$ respectively $y_1$, under $\phi$, by $A:=\phi(x_1)$ respectively $B:=\phi(y_1)$. The image of $1$ will be the identity matrix $I$.

Consider the element $y_1x_1-x_1y_1=1$, the image of which, under $\phi$, is equal to

$$ BA-AB=\phi(y_1x_1-x_1y_1)=\phi(1)=I. $$

Hence the matrices $A$ and $B$ have to satisfy $BA-AB=I$. Taking the trace of the left hand side of this equality yields

$$ tr(BA-AB)=tr(BA)-tr(AB)=tr(AB)-tr(AB)=0 $$

whereas the trace of the right hand side is equal to $tr(I)=m$. This is a contradiction.

Corollary of the above proof:
The same claim holds if we replace $M_m(R)$ by any unital Banach algebra.

This is easily seen by using the following well-known fact:
The identity element of a unital Banach algebra can not be a commutator, i.e. $ab-ba\neq 1$ for any elements $a,b$ of the Banach algebra.

This applies to the case $M_m(R)$ with $R=\mathbb{C}$, because $M_m(\mathbb{C})$ is a unital C*-algebra and in particular a unital Banach algebra. So in this particular case, this gives us an alternative way of making the desired conclusion without using the trace.

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    $\begingroup$ I think by "unital ring" you mean "unital commutative (associative) ring". Otherwise, there are obviously ring homomorphisms. Similarly tr(AB) = tr(BA) requires commutativity (it has counterexamples in every non-commutative ring). Unfortunately, the original question specifically allows matrix rings over non-commutative rings, so this doesn't answer the question. $\endgroup$ Jun 21, 2011 at 20:27
  • $\begingroup$ Jack, you are of course right. Thanks for pointing this out. I have tried to make my "answer" a bit more clear now. $\endgroup$ Jun 26, 2011 at 16:16

A matrix ring contains zero-divisors.


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