# Zoran Skoda formula for beta

Let $\hat{x}_1,\ldots,\hat{x}_n$ be a basis of a finite-dimensional Lie algebra $g$ over the field of characteristic $0$ and also the generators of the enveloping algebra $U(g)$. We denote the copy of the same elements but in symmetric algebra $Sym(g)$ by $x_1,\ldots,x_n$. The structure constants $C^k_{i j}$ of $g$ are defined by $[\hat{x}_i,\hat{x}_j] = C^k_{i j} \hat{x}_k$ with Einstein convention (summation over the repeated indices).

Let $e:Sym(g)\to U(g)$ be the symmetrization map

$e(x_{i_1}\cdots x_{i_k}) = \frac{1}{k!}\sum_{\sigma\in\Sigma(k)} \hat{x}_{i_{\sigma(1)}}\cdots\hat{x}_{i_{\sigma(k)}}.$

Denote furthermore by $\mathcal{C}$ the $n\times n$-matrix with entries $\mathcal{C}^i_j = C^i_{j k} \partial^k$ where $\partial^1,\ldots,\partial^n$ is the dual basis of $g^*$. For a multilabel $I= (i_1,\ldots, i_n)$ denote $x_I = x_1^{i_1}\cdots x_n^{i_n}$, $\hat{x}_I = \hat{x}_1^{i_1}\cdots \hat{x}_n^{i_n}$, $\partial^I = (\partial^1)^{i_1}\cdots(\partial^n)^{i_n}$. When we write $\sum_I x_I\otimes \partial^I$ the sum is going over all dinstinct ascendingly ordered multilabels, i.e. $\{x_I\}_I$ is the standard monomial basis of the space of polynomials in $n$-indeterminates.

Then for each $a\in\{1,\ldots,n\}$ one has the identity (MAIN FORMULA)

$\sum_{I, J} \frac{(-1)^{|J|}}{I! J!} e(x_I) \hat{x}_a e(x_J) \otimes \partial^{J+I} = \sum_{b =1}^n \hat{x}_b \otimes (e^{-\mathcal{C}})^b_a$

(Warning: The factoriels $I!= i_1!\cdots i_n!$ and $J!$ are not the ones from the expression for $e(x_I)$, $e(x_J)$, but rather extra ones, coming from a normalization of the pairing between $x$-s and $\partial$-s (in certain interpretation of the formula). One can replace $I!$ by $|I|!$ if one allows the permutations in $x_I$ (before applying $e$), i.e. overcounting basis of commutative polynomials (of course, this means that we work with a different version of multiindices (with sorts, and of arbitrary length)). Indeed then $x_1 x_1$ appears once, but $x_1 x_2$ appears along with $x_2 x_1$, hence the additional counting cancels with the difference between $I!$ and $|I|!$. It is of course simpler to compute not caring when the labels are indeed different, hence using $|I|!$ version.). This also shows that $\sum_I \frac{1}{I!} x_I\otimes\partial^I = exp(\sum_{i=1}^n x_i\otimes\partial^i)$.

The right hand side in the identity (main formula) is the series $\sum_{N = 0}^\infty \sum_{b=1}^n \frac{(-1)^N}{N!} \hat{x}_b \otimes (\mathcal{C}^N)^b_a$. The tensor product is over the ground field.

One filters the identity by the total degree of $\partial$-s (this appears the same as the degree in structure constants, as readily seen from the right hand side, but it is entirely non-obvious from the right hand side). Thus one can consider the above identity as a series of identities homogeneous in $\partial$-s for $N = 0,1,2,\ldots$:

$\sum_{|I|+|J| = N} \frac{(-1)^{|I|}}{I! J!} e(x_I) \hat{x}_a e(x_J) \otimes \partial^{J+I} = \frac{1}{N!}\sum_{b=1}^n \hat{x}_b \otimes (\mathcal{C}^N)^b_a$

We used here also $(-1)^{|I|} = (-1)^N (-1)^{|J|}$.

Up to the second degree (with the tensor product symbol skipped) the left hand side of the main formula is

$\hat{x}_a +\hat{x}_c \hat{x}_a \partial^c - \hat{x}_a\hat{x}_c \partial^c +\left(-\hat{x}_c \hat{x}_a \hat{x}_d +\frac{1}{4}(\hat{x}_{c}\hat{x}_{d}\hat{x}_{a}+\hat{x}_{d}\hat{x}_{c}\hat{x}_{a}+\hat{x}_{a}\hat{x}_{c}\hat{x}_{d}+\hat{x}_{a}\hat{x}_{d}\hat{x}_{c})\right)\partial^c\partial^d$

what can be easily calculated to

$\hat{x}_b(\delta^b_a - C^b_{a c}\partial^c+ \frac{1}{2} C^r_{a c} C^b_{r d}\partial^c\partial^d)$

Namely the coefficient in front of $\partial^c\partial^d$ can be rewritten as

$\frac{1}{4}([\hat{x}_a\hat{x}_d,\hat{x}_c]+[\hat{x}_d,\hat{x}_c\hat{x}_a] +\hat{x}_c[\hat{x}_d,\hat{x}_a]+[\hat{x}_a,\hat{x}_c]\hat{x}_d)$

what amounts to

$1/4 (2 C^b_{a c} \hat{x}_b \hat{x}_d + \hat{x}_a C^b_{d c} \hat{x}_b + C^b_{d c} \hat{x}_b \hat{x}_a + 2 C^b_{d a} \hat{x}_d\hat{x}_b)$

but the middle two summands involving $C^b_{d c}$ drop out as antisymmetric in $c\leftrightarrow d$, after contracting with the symmetric $\partial^d\partial^c$. The other two summands, after renaming, give another commutator what gives another structure constant factor and the result quadratic in $\mathcal{C}$.

The proof of the main formula is very simple. Indeed, $\partial^i$-s are in a different tensor factor so they can be viewed as parameters. If $e$ is the symmetrization map, then $e\hat\otimes id_{\hat{S}(\mathfrak{g}^*)}$ is an operator from $S(\mathfrak{g})\widehat{\otimes}\hat{S}(\mathfrak{g}^*)$ to $U(\mathfrak{g})\hat\otimes \hat{S}(\mathfrak{g}^*)$.

Then

$\sum_J \frac{1}{J!}\hat{x}^J \otimes\partial^J = (e\otimes 1)\left(\sum_J \frac{1}{J!}x^J \otimes\partial^J\right) = (e\otimes 1)exp(\sum_{i=1}^n x_i\otimes \partial^i) = exp(\sum_{j=1}^n\hat{x}_j\otimes \partial^j).$

Similarly,

$\sum_I (-1)^I\frac{1}{I!}\hat{x}^I \otimes\partial^I = exp(-\sum_{i=1}^n\hat{x}_i\otimes \partial^i).$

Thus we need to compute

$exp(-\sum_{i=1}^n\hat{x}_i\otimes \partial^i)(\hat{x}_a\otimes 1)exp(\sum_{j=1}^n\hat{x}_j\otimes \partial^j)$

what can be done by the Hadamard’s formula $exp(ad(A))B = exp(A)B exp(-A)$.

Thus we need to compute

$exp(-\sum_{i=1}^n ad(\hat{x}_i\otimes \partial^i)) (\hat{x}_a\otimes 1)$

what we do term by term and sum up to get $\sum_{b =1}^n \hat{x}_b \otimes (e^{-\mathcal{C}})^b_a$.

Last revised on October 30, 2015 at 14:55:58. See the history of this page for a list of all contributions to it.