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A not necessarily commutative algebra A (over C, say) is called formally smooth (or quasi-free) if, given any map $f:A \to B/I$, where $I \subset B$ is a nilpotent ideal, there is a lifting $F:A \to B$ that commutes with the projection. (The reason for the terminology is that if we restrict to the category of finitely generated commutative algebras, this condition is equivalent to Spec(A) being smooth. For more info see the paper "algebra extensions and nonsingularity" by Cuntz and Quillen, 1995.) It isn't hard to see that if A is formally smooth, then the representation varieties $Rep_\mathbb{C}(A,V)$ are smooth (V is finite dimensional). Does anyone know of an example of an algebra that is not formally smooth, but whose representation varieties are smooth?

One almost-answer is the Weyl algebra $A = \mathbb{C}\langle x,y\rangle/(xy - yx = 1)$. This isn't formally smooth, but its representation varieties are all empty. (To see this, take the trace of $xy - yx = 1$ to get $0 = n$.) This doesn't seem like it should count as answer, does anyone know a better one?

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Take any semi-simple lie algebra g and consider its enveloping algebra U(g). As all finite dimensional representations are semi-simple, every representation variety rep_n U(g) is a finite union of orbits, whence smooth. No such U(g) is formally smooth.

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Nice example! (A very slight nitpick, if the lie algebra is $\mathbb{C}$, then the enveloping algebra is $\mathbb{C}[x]$, which is formally smooth. But $\mathbb{C}$ is a silly lie algebra.) –  Peter Samuelson Dec 26 '09 at 4:03
    
Well, i dont consider C to be a semisimple Lie algebra as the Killing form in degenerate, whereas you'd say it has no proper Lie ideals. It's a bit like the eternal discussion whether 1 is a prime or not... Anyway, obviously the above example extends to all algebras having the property that Ext^1(M,M)=0 for all fin dml reps M. –  lieven lebruyn Dec 26 '09 at 9:44

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