Let $L$ be a finite-dimensional Lie algebra over a field $k$ of characteristic zero and $e_1,\ldots, e_n$ some basis of $L$. The formula $[e_i,e_j] = \sum_k C_{ij}^k e_k$ determines the structure coefficients $C_{ij}^k$. Given any ordered $k$-tuple $I = (i_1,\ldots,i_k)\in \lbrace 1,\ldots,n \rbrace^k$, define $e_I = e_{i_1}\cdots e_{i_k}\subset U(L)$ and $$ e^S_I = \frac{1}{n!}\sum_{\sigma\in\Sigma(k)} e_{\sigma(i_1)}\cdots e_{\sigma(i_k)}\in U(L). $$ As it is well known, from various forms of a PBW theorem, $e_I$, for all $I$ with $i_1\lt i_2\lt \ldots \lt i_k$, $k \geq 0$, form a basis and also $e^S_I$ (for the same set of $I$-s) form a basis. I need explicit formulas for $e_I$ in the linear basis of $e^S_J$-s where the coefficients are expressed in terms of the structure constants $C_{ij}^k$ (and combinatorial factors). In fact, for my present purposes, it would be enough to know explicitly the deepest, lowest order, linear term (but it is of course the hardest summand in the expansion). For example, for the easiest nontrivial case $k = 2$, $$e_{(i,j)} = e_i e_j = \frac{1}{2}(e_i e_j + e_j e_i) + \frac{1}{2} \sum_k C_{ij}^k e_k = e_{(i,j)}^S + \frac{1}{2} \sum_k C_{ij}^k e_k,$$ hence the linear term is $\frac{1}{2} \sum_k C_{ij}^k e_k$.

  • $\begingroup$ It would be extremely helpful if you could specify what you mean by "explicit formulas". For instance, some formulas can be extracted from the Kontsevich's universal quantisation formula applied to the Kostant-Kirillov Poisson bracket, but I presume that's not explicit enough for most needs... $\endgroup$ – Vladimir Dotsenko Dec 12 '11 at 13:25
  • $\begingroup$ As I said I am happy with knowing the linear term which is of the form $\sum_r f^r(\{C^l_{ij}\})e_r$ where $f^r$ is a polynomial in $C^l_{ij}$-s with some combinatorial coefficients. $\endgroup$ – Zoran Skoda Dec 12 '11 at 17:35
  • $\begingroup$ So you are looking for the (explicit formula for the) inverse of the symmetrization map $\mathrm{Sym}\mathfrak g\to U\left(\mathfrak g\right)$. This inverse is called the "symbol map $\sigma$" in Emanuela Petracci's thesis iecn.u-nancy.fr/~petracci/tesi.pdf , and is given an algebraic definition (not involving inverses, but involving lots of things like "define a map on the generators of an algebra and extend as an algebra morphism") on page 27. I am far from being sure of whether this is even close to what you want... $\endgroup$ – darij grinberg Dec 12 '11 at 17:57
  • $\begingroup$ I would not expect answer to be known. It is might not be very direct, but I think from answer to your question it would be possible to answer to mine mathoverflow.net/questions/77550 which as far as I known is not known... $\endgroup$ – Alexander Chervov Dec 12 '11 at 18:41
  • $\begingroup$ Darij: It is trivial to say the inverse of symmetrization map in terms of generators by the method of realizations via differential operators. But this reuires lots of commutation to get an explicit formula for a given $k$ and $n$. I need a single polynomial in structure constants for given $n$ and $k$, not an abstract theory if these maps which I know very well. $\endgroup$ – Zoran Skoda Dec 12 '11 at 19:02

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.

  • $\begingroup$ I need to digest this. It looks interesting and very complicated. The right hand side is a priori a symmetric tensor, but not yet in terms of the distinguished basis of the sapce of symmetric tensor (though maybe the remaining step is not far). It also maybe simplifies for the linear term. $\endgroup$ – Zoran Skoda Dec 13 '11 at 15:13
  • $\begingroup$ Zoran: As far as understand, the degree-$1$ part (and that's what you want, isn't it?) corresponds to the $m=1$ addend on the right hand side of (1.3), and that's exactly $\theta\left(a_1\otimes a_2\otimes ...\otimes a_n\right)$ - and that is very beautifully computed in (1.2). You'll have to translate it into your structure constants if you wish, but I don't see any symmetrization that needs to be done here. $\endgroup$ – darij grinberg Dec 13 '11 at 16:03
  • $\begingroup$ Yes, that's the simplification for the linear term I alluded to (was not sure at the time). Somehow when one computes the things by my method the deeper you go from trivial $m=n$ down, $n-1$ $n-2$ etc. the things are more complicated. But for Solomon, one gets at $1$ with simpler expression with the hardest about the middle, say around $m=n/2$. $\endgroup$ – Zoran Skoda Dec 13 '11 at 17:05
  • $\begingroup$ Can someone share this article with me? $\endgroup$ – Alexander Chervov Dec 13 '11 at 20:06
  • $\begingroup$ I will send it. $\endgroup$ – Zoran Skoda Dec 14 '11 at 7:39

This will not be an honest answer but just a long remark: it reminds me a bit on the relation between Weyl quantization and standard ordered quantization. Here one has the polynomials in $q$ and $p$ with their canonical Poisson bracket $\lbrace q, p\rbrace = 1$ which should be quantized into operators as usual. For higher polynomials one has to choose an ordering, e.g. Weyl or standard or many more... Pulling back the operator product gives then a star product, depending on the choice of the ordering. All of them are isomorphic by explicit isomorphism. In case of Weyl/standard the isomorphism is given by the exponential of the indefinite Laplacian $\frac{\partial^2}{\partial q \partial p}$. This is very explicit and allows for many nice formulas and computations.

Now in your situation it seems to me that you would like to have some similar isomorphism between the two quantizations of the Polynomials on the dual $\mathfrak{g}^*$ which you obtain by total symmetrization (= Weyl) and a standard ordering with respect to the choice of a basis of $\mathfrak{g}$. The point is that the standard ordering you are considering is much less canonical as in the case of canonical quantization: you really have to specify a basis.

I don't think that this has been worked out in detail, but maybe the old work of Simone Gutt from 1983 in Lett. Math. Phys. might be inspiring.

EDIT: Maybe I have misunderstood the question in the first place. So if your intention is to find an explicit formula for the Weyl ordered case alone, then one indeed has an "explicit" formula: I'm not sure where this showed up first, but the formula goes as follows: given $\xi, \eta \in \mathfrak{g}$ one considers the (formal) BCH series $BCH(\hbar \xi, \hbar\eta)$ with $\hbar$ being a formal parameter. Then the Weyl product of the formal exponentials $\exp(\hbar\xi)$ and $\exp(\hbar\eta)$ is given by $$\exp(\hbar\xi) \star \exp(\hbar\eta) =\exp (\hbar^{-1} BCH(\hbar \xi, \hbar\eta)) $$

Well, from this one get's the formula for monomials by differentiating and polarization. But you see, you won't get what you want: this is not at all explicit for two reasons. First, the differentation and polarization has to be done, very ugly. But second, and this is the more severe point, you have to know the BCH series.

Now on the other hand, this formula shows that you probabaly can not expect to get a simpler formula without using BCH. I fear, one can not go beyond this... :(

  • $\begingroup$ I know, I can write the solution in many symbolic forms, using Fock space etc. but I do not see at the moment how to traslate it into an explicit combinatorial formula. $\endgroup$ – Zoran Skoda Dec 12 '11 at 17:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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