MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In studying the integrability problem for Lie algebra representations, I have been led to wonder whether generalizing the notion of group by dropping associativity, while keeping the Hall-Witt identity, might be useful. Of course, I am assuming that associativity is stronger than Hall-Witt... So, let me ask first: is this the case? If yes, has such a notion of generalized group, or a similar one, been considered in the literature? Is there a reason I am missing why this might not be a good idea, either in itself or to study non-integrable Lie algebra representations?


Since the question is somewhat vague, let me try to explain myself better. Let us think of a representation of a Lie algebra by means of vector fields over a manifold. This representation can fail to integrate, i.e. give rise to an action of the corresponding Lie group, even if the vector fields are individually integrable. Nelson, in the paper where he introduces analytic vectors, gives a nice example with two commuting vector fields $X$, $Y$ on a manifold $M$, with the following property: let us start at a point $x_0\in M$, and move along the vector field $X$ for an amount of time, say $t=1$, to end up at $x_1$ (i.e. solve the equation $\dot x_t = X(x_t)$). Then, from $x_1$ let us move along the vector field $Y$ for a time $t=1$ to end up at $x_2$. Well, it is possible that, if one reverses the order of the displacements, i.e. if one moves first along $Y$ and then along $X$, one ends up at a point which is different from $x_2$. Thus, the representation of the Lie algebra $\mathbb R^2$ given by $X$ and $Y$ acting on, say, $C_0^\infty(M)$, cannot exponentiate to give an action of the Lie group $\mathbb R^2$ on $M$. Now, the idea is that these vector fields might still give rise to a an action of some other kind of structure---perhaps a nonassociative one. Since Hall-Witt has a geometric meaning (see, it seems plausible to me that it might still make sense for the looked-for nonassociative object. Given that we are talking about actions on manifolds, one should perhaps define Hall-Witt in such a context associating from the left, i.e. defining $g_1g_2\cdots g_n = g_1(g_2(\cdots g_n))$. If one does so, then my questions can be rephrased as follows: let $G$ be the not necessarily associative algebraic object generated by taking products of a finite set of symbols together with their inverses, and quotienting away Hall-Witt.

  1. Is $G$ a group? I'm guessing the answer is no, so:
  2. Have things like $G$ been considered in the literature?
  3. Is there any reason I'm failing to see why $G$ might not be an interesting object, either in itself or to study the integrability of Lie algebra representations?
share|cite|improve this question
Asking if associativity is stronger than Hall-Witt amounts to asking if Hall-Witt impliee associativity. But what would that mean? To state Hall-Witt, you need commutators and conjugation. How would you define those in a nonassociative context? – Marty Isaacs Jun 9 '12 at 22:31
Marty, sorry for the delay in answering. I have edited the question with your commentary in mind. I think now it is much clearer. – Rodrigo Vargas Jun 16 '12 at 20:30
This seems problematic. It is far from clear that your parenthesization is the one that gives the "right" Hall-Witt identity. – Qiaochu Yuan Jun 16 '12 at 20:33
Could you be more precise regarding this? Indeed, what is the right parenthesization should be understood as part of the question, so the reasons you have to believe just associating from the right may be wrong could actually evolve into a complete answer... – Rodrigo Vargas Jun 16 '12 at 21:14
I just mean that there's no good reason to prefer this particular parenthesization over any of the others. One might prefer a parenthesization which computes the conjugations first (somehow), then the commutators (somehow), then the actual products, for example. – Qiaochu Yuan Jun 16 '12 at 22:22

Your Answer


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

Browse other questions tagged or ask your own question.