User rodrigo vargas - MathOverflow

Generalizing groups via the Hall-Witt identity

2012-05-25T22:12:21Z

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?

UPDATE:

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 http://lamington.wordpress.com/2011/11/20/the-hall-witt-identity/), 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.

Is $G$ a group? I'm guessing the answer is no, so:
Have things like $G$ been considered in the literature?
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?

Answer by Rodrigo Vargas for monomials in the universal enveloping of a Lie algebra in terms of the symmetric basis

2011-12-13T12:39:18Z

An explicit formula is given in this paper by L. Solomon. I copy the abstract here:

Let g be a Lie algebra over a field of characteristic zero. Let T be the tensor algebra of g, let S be the subspace of symmetric tensors and let J be the two-sided ideal of T generated by tensors x⊗y−y⊗x−[x, y]. One formulation of the P-B-W theorem states that T=S⊕J, direct sum. In this paper we give an explicit formula for the projection of T on S defined by this direct decomposition.

Characterizing non-singularity of varieties through properties of their derivations

2011-11-18T16:35:30Z

I am interested in knowing about the possible implications between the following properties of a commutative, complex algebra:

Its spectrum is non-singular.
Its derivation module is projective and finitely generated.
All pointwise derivations (as explained below) admit an extension to an open set of the spectrum.

Let me note, in case it is not already clear, that I have a weak to non-existent background in commutative algebra and algebraic geometry.

Notational preliminaries

Let $A$ be a commutative algebra over $\mathbb C$. I want to think of it as an algebra of functions on its character space $$ M = \left\{ \chi\in A^*\ \vert\ (\forall a,b\in A)\ \chi(ab) = \chi(a)\chi(b) \right\}, $$ so I will suppose that the Gelfand transform $$ a\in A\mapsto \hat a\in \{\text{functions } M\rightarrow \mathbb C\},\quad \hat a(\chi)=\chi(a) $$ is injective (or equivalently, that the Jacobson radical of $A$ is trivial). A derivation is a linear application $\delta:A\rightarrow A$ such that $$ \delta(ab) = \delta(a)b + a\delta(b),\quad a,b\in A, $$ and a derivation over a character $\chi$ is a linear application $\delta_\chi:A\rightarrow \mathbb C$ such that $$ \delta_\chi(ab) = \delta_\chi(a)\chi(b) + \chi(a)\delta_\chi(b),\quad a,b\in A. $$ The $A$-modules of derivations and derivations over $\chi$ will be denoted by $\text{der}(A)$ and $\text{der}_\chi(A)$, respectively.

I will suppose that $M$ is endowed with the trace of the Zariski topology (I hope that there are no problems in considering only maximal ideals), and I will write $\mathcal O_A$ for the structure sheaf of $M$.

Questions

The Serre-Swan theorem establishes an equivalence between the categories of vector bundles and projective, finitely generated modules. So, here is my first question:

Is the non-singularity of $M$ equivalent to $\text{der}(A)$ being projective and finitely generated?

Given $\chi\in U\subseteq M$, observe that there is a natural map $\text{der}(\mathcal O_A(U)) \rightarrow \text{der}_\chi(A)$ given by $$ \delta \mapsto \delta_\chi = \chi\circ \delta. $$ It seems to me geometrically plausible that there is an affirmative answer to my second question:

Is the non-singularity of $M$ equivalent to the surjectivity of the map above, for every $\chi\in M$ and a suitable neighborhood $U\ni\chi$?

Now, if both questions above have affirmative answers, my third one is superfluous, but it must be stated anyways. It is actually the problem that interests me the most, independently of singularity considerations:

Under what conditions are the following two properties equivalent? Does at least one imply the other?

$\text{der}(A)$ is projective and finitely generated.

For every $\chi\in M$, there exists a neighborhood $U\ni\chi$ such that $\text{der}(\mathcal O_A(U))\rightarrow \text{der}_\chi(A)$ is surjective.

Answer by Rodrigo Vargas for Mappings between states on *-algebras

2011-10-13T22:51:13Z

Let me first stick to more conventional notation and, given a positive operator $f:C\rightarrow A$, denote by $f^\ast$ the map that you call $f^{-1}$. If $B=A\otimes C$ and $f$ is defined by $f(a\otimes c) = a\phi(c)$ with $\phi$ a state on $C$, then you get a special case of what is called a conditional expectation, and $f^\ast$ coincides with your $t_\phi$ above. If you are willing to enlarge $\mathrm{Mor}(s(A),s(B))$ to contain such maps too, then Stinespring dilation theorem tells you that any completely positive map, being the composition of an algebra morphism followed by a conditional expectation, will induce an admissible morphism on the state spaces.

Sheafification of Arens-Michael algebra-valued presheaves

2011-09-09T15:11:00Z

Let $\mathcal A$ be the category of Arens-Michael algebras, that is, projective limits of Banach algebras. Since $\mathcal A$ is a concrete category, an $\mathcal A$-valued presheaf $A$ admits a set-valued sheafification $A_{S.}$ I would like to know if there is a good way to associate an $\mathcal A$-valued sheaf to $A_{S.}$

Now, as I gather from nLab's article on sheafification and Kashiwara and Schapiro's Categories and Sheafs, if the category $\mathcal C$ is such that:

Small projective and small inductive limits exist,
Small filtrant limits are exact,
The IPC property holds,

then $\mathcal C$-valued presheaves admit a $\mathcal C$-valued sheafification. Thus, my question is: does the category $\mathcal A$ of Arens-Michael algebras have these properties? And, in the negative case, is there a category of topological algebras, containing $\mathcal A$ as a subcategory, having these properties?

Comment by Rodrigo Vargas

2012-06-16T21:14:59Z

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...

Comment by Rodrigo Vargas

2012-06-16T20:30:59Z

Marty, sorry for the delay in answering. I have edited the question with your commentary in mind. I think now it is much clearer.

Comment by Rodrigo Vargas

2011-11-18T19:43:07Z

Thanks for your comments, knowing that my first question is not completely settled is already quite helpful. I will check the references you gave me. Any idea regarding the third question will also be very appreciated.