Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $X$ be an algebraic variety (over any field). The definition of the Lie bracket of two vector fields on $X$ (i.e. sections of the tangent sheaf) which I know characterizes vector fields as derivations of the structure sheaf, and then one checks that the commutator of two derivations is a derivation. My first question is whether this definition works when $X$ is not smooth.

I am more concerned about my second question if the answer to the first is no, but I am interested regardless: is there an "intrinsic" definition of the Lie bracket which does not require us think of vector fields as operating on functions?

share|improve this question
    
For your second question: what do you want to think about? The usual definition is quite intrinsic! –  Mariano Suárez-Alvarez Apr 17 '12 at 0:15
    
Agreed: intrinsic is the wrong word. My point is that it is possible to think about vector fields without considering them as differential operators, so I wonder if one can define the Lie bracket as such. I am partially motivated by the case of smooth manifolds, where there is such an alternative definition of the Lie bracket, but this one uses limits in an essential way and does not have any obvious algebraic analogue. –  Justin Campbell Apr 17 '12 at 0:29
    
In qchu.wordpress.com/2011/02/26/… I give a definition of the Lie bracket of two derivations that doesn't require limits and doesn't require that you verify that the commutator of two derivations is a derivation; the whole thing naturally falls out of the careful use of nilpotents. Since the Lie bracket is local, for varieties it suffices to piece together the story of that post on open affines. Does that answer part of your question? –  Qiaochu Yuan Apr 17 '12 at 0:54
    
@Qiaochu: I'm not convinced there's a large conceptual difference between nilpotents and limits. @Justin: You can try to say that "a vector field is a section of the tangent bundle", but then what is the tangent bundle if not the collection of point-derivations? I guess you'll say "it's the vector bundle glued together with such-and-such transition functions", which is OK, but clearly less elegant than the other definition. –  Theo Johnson-Freyd Apr 17 '12 at 3:56
    
@Theo: okay, but Justin didn't seem to think that the limit definition extended to the algebraic setting, and recasting it in terms of nilpotents makes it clear that it does. –  Qiaochu Yuan Apr 17 '12 at 5:49
show 2 more comments

1 Answer

up vote 1 down vote accepted

In an old blog post I give a definition of the Lie bracket of two derivations (of a ring, not necessarily commutative, in which $2$ is invertible) that doesn't require limits and doesn't require that you verify that the commutator of two derivations is a derivation; the whole thing naturally falls out of the careful use of nilpotents. Since the Lie bracket is local, for varieties it suffices to piece together the story of that post on open affines.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.