Zoran Skoda
formula for beta

Let x^ 1,,x^ n\hat{x}_1,\ldots,\hat{x}_n be a basis of a finite-dimensional Lie algebra gg over the field of characteristic 00 and also the generators of the enveloping algebra U(g)U(g). We denote the copy of the same elements but in symmetric algebra Sym(g)Sym(g) by x 1,,x nx_1,\ldots,x_n. The structure constants C ij kC^k_{i j} of gg are defined by [x^ i,x^ j]=C ij kx^ k[\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)U(g)e:Sym(g)\to U(g) be the symmetrization map

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

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

I,J(1) |J|I!J!e(x I)x^ ae(x J) J+I= b=1 nx^ b(e 𝒞) a b \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!i n!I!= i_1!\cdots i_n! and J!J! are not the ones from the expression for e(x I)e(x_I), e(x J)e(x_J), but rather extra ones, coming from a normalization of the pairing between xx-s and \partial-s (in certain interpretation of the formula). One can replace I!I! by |I|!|I|! if one allows the permutations in x Ix_I (before applying ee), 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 1x 1x_1 x_1 appears once, but x 1x 2x_1 x_2 appears along with x 2x 1x_2 x_1, hence the additional counting cancels with the difference between I!I! and |I|!|I|!. It is of course simpler to compute not caring when the labels are indeed different, hence using |I|!|I|! version.). This also shows that I1I!x I I=exp( i=1 nx i i)\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 N=0 b=1 n(1) NN!x^ b(𝒞 N) a b\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,N = 0,1,2,\ldots:

|I|+|J|=N(1) |I|I!J!e(x I)x^ ae(x J) J+I=1N! b=1 nx^ b(𝒞 N) a b \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|(-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

x^ a+x^ cx^ a cx^ ax^ c c+(x^ cx^ ax^ d+14(x^ cx^ dx^ a+x^ dx^ cx^ a+x^ ax^ cx^ d+x^ ax^ dx^ c)) c d \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

x^ b(δ a bC ac b c+12C ac rC rd b c d) \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 c d\partial^c\partial^d can be rewritten as

14([x^ ax^ d,x^ c]+[x^ d,x^ cx^ a]+x^ c[x^ d,x^ a]+[x^ a,x^ c]x^ d) \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(2C ac bx^ bx^ d+x^ aC dc bx^ b+C dc bx^ bx^ a+2C da bx^ dx^ b) 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 dc bC^b_{d c} drop out as antisymmetric in cdc\leftrightarrow d, after contracting with the symmetric d c\partial^d\partial^c. The other two summands, after renaming, give another commutator what gives another structure constant factor and the result quadratic in 𝒞\mathcal{C}.

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


J1J!x^ J J=(e1)( J1J!x J J)=(e1)exp( i=1 nx i i)=exp( j=1 nx^ j j).\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).


I(1) I1I!x^ I I=exp( i=1 nx^ i i).\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( i=1 nx^ i i)(x^ a1)exp( j=1 nx^ j j) 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)Bexp(A)exp(ad(A))B = exp(A)B exp(-A).

Thus we need to compute

exp( i=1 nad(x^ i i))(x^ a1) 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 b=1 nx^ b(e 𝒞) a b\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.