The aim of this paper is to generalize the classical formula
$e^xye^{-x}=\sum\limits_{k\ge 0} \frac{1}{k!} (ad~x)^k(y)$ by replacing $e^x$
with any formal power series $\displaystyle {f(x)=1+\sum_{k\ge 1} a_kx^k}$. We
also obtain combinatorial applications to $q$-exponentials, $q$-binomials, and
Hall-Littlewood polynomials.Comment: 5 pages, LaTeX, typos corrected, to appear in Journal of Lie Theor

Berenstein, Arkady

Retakh, Vladimir

English

arXiv

The aim of this paper is to generalize the classical formula $e^xye^{-x}=\sum\limits_{k\ge 0} \frac{1}{k!} (ad~x)^k(y)$ by replacing $e^x$ with any formal power series $\displaystyle {f(x)=1+\sum_{k\ge 1} a_kx^k}$. We also obtain combinatorial applications to $q$-exponentials, $q$- binomials, and Hall-Littlewood polynomials

Berenstein, A.

Retakh, V.

MPG.PuRe

arXiv:1506.07071v4  [math.QA]  22 Jul 2015GENERALIZED ADJOINT ACTIONSARKADY BERENSTEIN AND VLADIMIR RETAKHAbstract. The aim of this paper is to generalize the classical formula exye−x =∑k≥01k!(ad x)k(y) by re-placing ex with any formal power series f(x) = 1 +∑k≥1akxk. We also obtain combinatorial applications toq-exponentials, q-binomials, and Hall-Littlewood polynomials.1. Notation and main resultsOne of the most fundamental tools in Lie theory, the adjoint action of Lie groups on their Lie algebras, isbased on the following formula:(1.1) exye−x = ead x(y) =∑k≥01k!(ad x)k(y) ,where (ad x)k(y) = [x, [x, . . . , [x, y], . . .]] and [a, b] = ab− ba.The aim of this paper is to generalize (1.1) by replacing et with any formal power series(1.2) f = f(t) = 1 +∑k≥1aktkover a field k.For any formal power series (1.2) over k define polynomialsPk(t) = Pf,k(t) = (−1)k det1 a1t a2t2 . . . aktk1 a1 a2 . . . ak0 1 a1 . . . ak−1....... . .. . .. . .0 0 . . . 1 a1for k = 0, 1, 2, . . .. (with the convention that P0(t) = 1). Clearly, Pk(1) = 0 for k ≥ 1. Using Cramer’s rulewith respect to the last column, one obtains a recursion Pk(t) = aktk −k∑i=1aiPk−i(t).The following result is, probably, well-known (for readers’ convenience, we prove it in Section 2).Theorem 1.1. For any power series f(t) as in (1.2), one has(1.3)f(tx)f(x)=∑k≥0Pf,k(t) · xkand(1.4) Pf,k(st) =k∑i=0Pf,i(s)Pf,k−i(t)tifor all k ≥ 0.This work was partially supported by the NSF grant DMS-1403527 (A. B.) and by the NSA grant H98230-14-1-0148 (V. R.).12 ARKADY BERENSTEIN AND VLADIMIR RETAKHFurthermore, for any algebra A over k, a subset q = {q1, . . . , qk} ⊂ k, x, y ∈ A, and k ≥ 1 define(ad x)q(y) = [x, [x, . . . , [x, y]q1 , . . .]qk−1 ]qkwhere [a, b]q := ab− qba. It is easy to see that(1.5) (ad x)q(y) =k∑j=0(−1)jej(q1, . . . , qk) · xk−jyxj ,where ej(q1, . . . , qk) is the j-th elementary symmetric function.Theorem 1.2. Let A be k-algebra and suppose that f is any power series (1.2) with ak 6= 0 for k ≥ 1. Then(1.6) f(x)yf(x)−1 = y +∑k≥1ak(ad x)qk(y) ,for any x, y ∈ A, where qk = {q1k, . . . , qkk} is the set of roots of Pf,k(t).Remark 1.3. A formula for f(x)yf(x)−1 without assumption that all ak 6= 0 is given in Proposition 2.3.Remark 1.4. Strictly speaking, the formula (1.6), similarly to (1.1) requires a completion of A. One canbypass this by replacing x with τ · x where τ is a purely transcendental element of k so that the right handside of (1.6) becomes a power series in τ (and, maybe extending k if it lack such an element).Remark 1.5. The subsets qk may belong to an extension of k, however, the operators (ad x)qk are definedover k due to (1.5) because all symmetric functions in qk belong to k.It is easy to see that if ak =1k! for all k, then Pk(t) =(t−1)kk! which immediately recovers (1.1). Supposenow that ak =1[k]q !for all k, where kq! = [1]q · · · [k]q is the q-factorial and [ℓ]q = 1 + q + · · ·+ qℓ−1. We willshow (Proposition 2.5) that Pf,k(t) =(t− 1)(t− q) · · · (t− qk−1)[k]q!for f(t) = etq =∑k≥0tk[k]q!, therefore, recoverthe following famous result (see e.g., [3]).Theorem 1.6. Let exq =∑k≥0xk[k]q!be the q-exponential. Then exq · y · (exq )−1 =∑k≥01[k]q!(ad x){1,q,...,qk−1}(y).On the other hand, combining Theorem 1.1 and Proposition 2.5, we recover the following well-knownproperties of q-exponentials and q-binomials:eqnxq = exq(1 +n∑k=1(qn − 1)(qn − q) · · · (qn − qk−1)[k]q!xk)for n ≥ 0, in particular,eqxq = exq · (1 + (q − 1)x)and1 +n∑k=1(qn − 1)(qn − q) · · · (qn − qk−1)[k]q!xk =n∏i=1(1 + (q − 1)qi−1x) .We conclude with a curious observation that the polynomials Pf,k(t) are related to the Hall-Littlewoodsymmetric polynomials.Proposition 1.7. Suppose that f(t) =∏k≥1(1− xkt). ThenPf,k(t) = Q(k)(x; t)GENERALIZED ADJOINT ACTIONS 3for all k ≥ 0, where x = {xk, k ≥ 0} is viewed as an infinite set of variables, Qλ(x; t) is Hall-Littlewoodpolynomial ([2, Section 3.2]), and (k) is a one-row Young diagram with k cells. In particular,Q(k)(x; t) = (−1)k det1 −e1t e2t2 . . . (−1)kektk1 −e1 e2 . . . (−1)kek0 1 −e1 . . . (−1)k−1ek−1....... . .. . .. . .0 0 . . . 1 −e1for all k ≥ 0, where ek = ek(x) is the k-th elementary symmetric function.Acknowledgments. We gratefully acknowledge the support of Centre de Recerca Matema`tica, Barcelona,where this work was accomplished. We are grateful to Eric Rains for pointing out the connection withHall-Littlewood polynomials.2. ProofsProof of Theorem 1.1. We need the following well-known fact (attributed to Wronski, see e.g., [1]).Lemma 2.1. Let f be any formal power series (1.2). Then1f(t)= 1 +∑k≥1Dk(f)tk, whereDk(f) = (−1)k deta1 a2 a3 . . . ak1 a1 a2 . . . ak−10 1 a1 . . . ak−2....... . .. . .. . .0 0 . . . 1 a1(with the convention D0(f) = 1).The following generalization of Lemma 2.1 is, apparently, well-known (for readers’ convenience we proveit here).Lemma 2.2. Let f(t) = 1 +∑k≥1aktk, g(t) = 1 +∑k≥1bktk be formal power series. Theng(t)f(t)=∑k≥0Dk(g, f)tk ,where Dk(g, f) = (−1)k det1 b1 . . . bk−1 bk1 a1 . . . ak−1 ak0 1 . . . ak−2 ak−1....... . .. . .. . .0 0 . . . 1 a1 (with the convention D0(g, f) = 1).Proof. Indeed, using Lemma 2.1, we obtain (with the convention b0 = 1):g(t)f(t)= g(t) ·1f(t)=∑i≥0biti∑j≥0Dj(f)tj =∑k≥0dktkwheredk =k∑i=0biDk−i(f) = (−1)k detb0 b1 . . . bk−1 bk1 a1 . . . ak−1 ak0 1 . . . ak−2 ak−1....... . .. . .. . .0 0 . . . 1 a14 ARKADY BERENSTEIN AND VLADIMIR RETAKHby Cramer rule because biDk−i(f) = (−1)k det0 0 . . . bi . . . 01 a1 . . . ai . . . ak0 1 . . . ai+1 . . . ak−1....... . .. . .. . .. . .0 0 . . . 1 a1 a20 0 . . . . . . 1 a1.The lemma is proved. Then taking bk = aktk for k ≥ 1 in Lemma 2.2, we obtain (1.3).To prove (1.4), compute f(stx)f(x) in two ways, using the first assertion:f(stx)f(x)=∑k≥0Pf,k(st) · xkandf(stx)f(x)=f(stx)f(tx)·f(tx)f(x)=∑i≥0Pf,i(s) · (tx)i∑j≥0Pf,j(t) · xj. Comparing the coefficients of xk inboth series, we obtain (1.4).Theorem 1.1 is proved. Proof of Theorem 1.2. We need the following result.Proposition 2.3. For any power series f as in (1.2) one has:(a) Pf,k(t) =k∑j=0ak−jDj(f) · tj for all k ≥ 0.(b) f(x)yf(x)−1 = y + z1 + z2 + · · · , where zk =k∑i=0aiDk−i(f) · xiyxk−i for all k ≥ 1.Proof. Prove (a). Indeed, using Lemma 2.1, we obtain:f(tx)f(x)=∑i,j≥0(aitixi) (Dj(f)xj)=∑k≥0(k∑i=0aiDk−i(f) · ti).This together with Theorem 1.1 proves (a).Prove (b) now. Indeed,f(x)yf(x)−1 =∑i,j≥0(aixi)y(Dj(f)xj)=∑k≥0 k∑j=0aiDk−i(f) · xiyxk−iThis proves (b).Now we can finish the proof of Theorem 1.2. Indeed, suppose that Pf,k(t) is factored asPf,k(t) = ak(t− q1k) · · · (t− qkk) .Then, by Proposition 2.3(a), aiDk−i = ak(−1)k−iek−i(q1k, . . . , qkk) for i = 0, . . . , k. Therefore, in the notationof Proposition 2.3(b),zk =k∑i=0ak(−1)k−iek−i(q1k, . . . , qkk) · xiyxk−i = ak(ad x)qk(y)for all k ≥ 1, which together with Proposition 2.3(b) verifies (1.6).Theorem 1.2 is proved. Proposition 2.4. Petq ,k(t) =(t− 1)(t− q) · · · (t− qk−1)[k]q!for all k ≥ 1.GENERALIZED ADJOINT ACTIONS 5Proof. It suffices to show that Petq ,k(qa) = 0 for all 0 ≤ a < k. We proceed by induction in such pairs (a, k).If a = 0, then we have nothing to prove since Pf,k(1) = 0 for all f .Using Theorem 1.1, we obtain:Pf,k(qa) =k∑i=0Pf,i(qb)Pf,k−i(qa−b)q(a−b)i .Taking f = etq, 1 ≤ b ≤ a < k, and using the inductive hypothesis, this gives Pf,k(qa) = 0 for any 1 ≤ a < k.The proposition is proved. Corollary 2.5. For all k ≥ 1 one has: det1 t[1]qt2[2]q. . . tk[k]q1 1[1]q1[2]q. . . 1[k]q0 1 1[1]q . . .1[k−1]q. . . . . . . . . . . . . . .0 0 . . . 1 1[1]q =(1− t)(q − t) · · · (qk−1 − t)[k]q!.Proof of Proposition 1.7. Indeed, if f(t) is as in Proposition 1.7, thenf(tu)f(u)=∏k≥11− xktu1− xku=∑k≥0Q(k)(x; t)ukby [2, Equations (2.10) and (2.13)]. This and Theorem 1.1 imply that Pf,k = Q(k)(x; t) for all k ≥ 0, whichproves the first assertion of Proposition 1.7.To prove the second assertion, note that ak = (−1)kek(x) for all k ≥ 0 because of the well-known formula(see e.g., [2, Section 1.2]): ∏k≥1(1 − xkt) =∑k≥0(−1)kek(x)tk .This and the first assertion of Proposition 1.7 imply the second assertion of the proposition. References[1] A. Inselberg, On determinants of Toeplitz-Hessenberg matrices arising in power series, J. Math. Anal. Appl. 63 (1978),no. 2, 347–353.[2] I. Macdonald, Symmetric functions and Hall polynomials Second edition. With contributions by A. Zelevinsky. OxfordMathematical Monographs. Oxford Science Publications. The Clarendon Press, Oxford University Press, New York, 1995.[3] A. Volkov, Beyond the “pentagon identity,” Lett. Math. Phys. 39 (1997), no. 4, 393–397.Department of Mathematics, University of Oregon, Eugene, OR 97403, USAE-mail address: arkadiy@math.uoregon.eduDepartment of Mathematics, Rutgers University, Piscataway, NJ 08854, USAE-mail address: vretakh@math.rutgers.edu

Generalized adjoint actions

The aim of this paper is to generalize the classical formula $ \displaystyle {e^xye^{-x}\!= }$ \linebreak $ \displaystyle{\sum_{k\ge 0} \frac{1}{k!} (ad~x)^k(y)}$ 

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