# Infinite dimensional simple algebras

Are there examples of non-commutative finitely-presented infinite-dimensional simple algebras?

I am looking for examples and the only example I know is the Weyl algebra.

There are lots of examples among Leavitt path algebras. The easiest is the Leavitt algebra generated by x,x',y,y' and relations x'x=1=y'y and x'y=0=y'x and xx'+yy'=1. See https://arxiv.org/abs/math/0509494 for more details.

First, a couple of constructions. If $R$ is simple and finitely presented (f.p.) then the matrix ring $M_n(R)$ has the same properties. Moreover, $\operatorname{G.rk}(M_n(R))=n\operatorname{G.rk}(R)$, where $\operatorname{G.rk}$ is the Goldie rank. Thus if $R$ has finite Goldie rank (e.g. if $R$ is noncommutative domain, so that $\operatorname{G.rk}(R)=1$), all rings $M_n(R)$ are f.p., simple and mutually non-isomorphic. Likewise, if $S$ is a multiplicative Ore set and $R$ is simple, then so is the Ore localization $S^{-1}R$. If, moreover, $S$ is finitely generated and $R$ is f.p., then $S^{-1}R$ is simple and f.p. Frequently, one can show that $S^{-1}R$ is not isomorphic to $R$ by considering their units. Both constructions can be applied to the Weyl algebra $A_n$ (and iterated), yielding plenty of examples of non-isomorphic f.p. simple infinite-dimensional algebras. You should check standard sources on noncommutative ring theory, such as McConnell and Robson, for this theory and other standard examples, as well as for the filtered algebra argument mentioned below.

An important family of examples of simple (and more generally, primitive) f.p. rings dear to my heart arises from universal enveloping algebras $U(\frak{g})$ of semisimple Lie algebras $\frak{g}$: namely, the quotient of $U(\frak{g})$ by a maximal (resp. primitive) ideal is a simple algebra; if the ideal is completely prime, one even gets a noncommutative domain. These simple quotients are generically infinite-dimensional and cannot be isomorphic to Weyl algebras. Note that since $A=U(\frak{g})$ is a filtered algebra whose associated graded algebra is a free polynomial ring (namely, the symmetric algebra of $\frak{g}$), it follows by standard filtered techniques that every left (or right) ideal of $A$ is finitely generated — the only assumption here is that $\frak{g}$ is finite-dimensional. Thus, every quotient $A/I$ is finitely presented. The upshot of the theory of primitive ideals (assuming the ground field is the complex numbers) is that maximal ideals in $U(\frak{g})$ are parametrized by $W$-orbits on $\frak{h}^{*}$ and the quotient algebra is finite-dimensional if and only if the corresponding orbit $W\lambda$ is integral and non-singular (i.e. $\lambda$ has these properties). In particular, most of these simple quotients are f.p. simple infinite-dimensional algebras, as desired. The somewhat involved technical details, which require knowledge of semisimple Lie algebras and their representation theory, and more explicit examples are described below the fold. This theory is exposed in the books of Dixmier and Jantzen on the enveloping algebras (and in subsequent original papers), for those interested in the proofs.

Primitive ideals in $U(\frak{g})$ for a complex semisimple Lie algebra $\frak{g}$ have been completely classified in the work of Dixmier, Duflo, Borho, Jantzen, Joseph, Lusztig, Vogan, Barbasch, and others. Each primitive ideal $I$ has "infinitesimal character", determined by the maximal ideal $I\cap Z(\frak{g})$ and there is a unique maximal ideal in each infinitesimal character ($Z(\frak{g})$ denotes the center of $U(\frak{g})$). In more detail, maximal ideals of $U(\frak{g})$ have the form $I(\lambda)=\operatorname{Ann} L(\lambda-\rho)$, where $\lambda\in\frak{h}^*$ is maximal in its $W$-orbit (the orbit corresponds to the infinitesimal character), $L(\lambda)$ is the simple highest weight module with highest weight $\lambda$, and the linear dual of the Cartan subalgebra $\frak{h}^*$ is partially ordered by $\lambda\geq \mu \iff \lambda-\mu$ is a non-negative linear combination of simple roots $\{\alpha_i, 1\leq i\leq l\}$, where $l=\operatorname{rank}{\frak{g}}$ is the rank of $\frak{g}$. The ideal $I(\lambda)$ has finite codimension if and only if the corresponding simple module is finite-dimensional, i.e. the highest weight $\lambda-\rho$ is dominant integral (i.e., $(\lambda,\alpha_i)$ is a positive integer for all $1\leq i\leq l$); otherwise, $U({\frak{g}})/I(\lambda)$ is an infinite-dimensional primitive ring.

Here are some more specific examples. At one extreme (generic case), if $(\lambda,\alpha_i)\notin\Bbb{Z}\setminus 0$ for all $1\leq i\leq l$, the Verma module $M(\lambda-\rho)$ is a simple infinite-dimensional module whose annihilator is a maximal ideal. A theorem of Dixmier asserts that this annihilator is generated by a maximal ideal of $Z(\frak{g})$, and $Z(\frak{g})$ is known to be a polynomial algebra in $l=\operatorname{rank}{\frak{g}}$ generators, making the quotient manifestly finitely presented. The ideals just described have maximal Gelfand-Kirillov (GK) dimension $\dim{\frak{g}}-l$. At the other extreme are the Joseph ideal for types other than $A$, which is a unique maximal completely prime ideal of minimal positive GK dimension $2l$ (and its analogues in type $A$ that form a one-parameter family). Rather than going through the rather involved construction, let me give the most elementary example: for ${\frak{g}}={\frak{sp}}_{2n}$, the quotient is the even Weyl algebra $A_{n}^{\Bbb{Z}_2}$ (the nontrivial element of cyclic group of order 2 acts by $-1$ on the Weyl algebra; it is easy to see that this action preserves the algebra structure). Thus, the even Weyl algebra is an infinite-dimensional simple ring of GK dimension $2n$, in fact, a non-commutative domain, that is not isomorphic to a Weyl algebra.

The preceding theory can be made fairly explicit in type $A$, including the description of generators for $I(\lambda)$ in several important cases (see e.g. https://arxiv.org/abs/0802.1952, where some examples are worked out, but without spelling out the maximality condition).

• Why do maximal ideals of infinite codimension exist, and why is the quotient finitely presented? – YCor Apr 4 '17 at 22:16
• I was wondering exactly the same thing. I think that mentioning a concrete example of a maximal ideal of infinite codimension, inside the UEA of some semisimple Lie algebra would clarify things. Also, mentioning the quotient and demonstrating that it is simple and fp, would add value to both the answer and the question as well. – Konstantinos Kanakoglou Apr 4 '17 at 23:34
• This a well-developed theory, and it is not clear to me how much to say before it becomes too much (primarily, because the question itself is overly broad). As the first example, the annihilator of the Verma module with maximally non-integral infinitesimal character (generated by the corresponding maximal ideal of the center) works. I will add more refs when I have time. The fact that ideals of universal enveloping algebras are finitely generated is standard and follows from the fact that the associated graded algebra is Noetherian. In a way, the best clarification is "read the literature". – Victor Protsak Apr 5 '17 at 2:57
• I have added some technical details. I am still unconvinced that this adds clarity for anyone unfamiliar with representation theory of semisimple Lie algebras (and primitive ideals). – Victor Protsak Apr 5 '17 at 7:28

You can look at Kac-Moody algebras.

• I don't think those are usually algebras in the sense the OP is asking; I think the OP wants multiplication algebras (i.e. having multiplication, not just a Lie bracket). The corresponding universal enveloping algebra wouldn't be simple - in fact, it has a two-sided ideal coming from the filtration derived from the grading on $T^*(\mathfrak{g})$. – user44191 Apr 4 '17 at 8:34
• There are plenty of simple quotients, though (see my answer). – Victor Protsak Apr 4 '17 at 16:24
• @user44191 I don't see how you get a 2-sided ideal in the enveloping algebra from the grading, could you explain? which filtration do you mean? – YCor Apr 4 '17 at 16:26
• @YCor I was referring to the augmentation ideal, although I had forgotten the term for it. I apologize though; while my description can be technically accurate, it is misleading. The grading on $T^*(\mathfrak{g})$ gives a decreasing filtration $F^i = \cup_{j \geq i} T^{j}(\mathfrak{g})$; this descends to a filtration on $U(\mathfrak{g})$ with $F^0 = U(\mathfrak{g})$, and $F^{i} = F^1$ for $i \geq 1$. This is an ideal $U(\mathfrak{g})_+$ which consists of all elements with no constant term, also known as the augmentation ideal. Note that for any $\mathfrak{g}$, $U/U+ \simeq k$, the base field. – user44191 Apr 4 '17 at 21:45
• @user44191 thanks. Actually this can also be seen without reference to filtration by considering, for any Lie algebra $\mathfrak{g}$ the zero Lie algebra homomorphism $\mathfrak{g}\to k$ and use the universal property to get an (unital associative) algebra homomorphism $U(\mathfrak{g})\to k$. – YCor Apr 6 '17 at 5:38