Journal of Nonlinear Mathematical Physics

Volume 27, Issue 1, October 2019, Pages 1 - 6

Euler’s triangle and the decomposition of tensor powers of the adjoint 𝔰𝔩(2)-module

Authors
Askold M. Perelomov
Institute for Theoretical and Experimental Physics, 117259, Moscow, Russia, aperelomo@gmail.com
Received 6 June 2019, Accepted 2 July 2019, Available Online 25 October 2019.
DOI
https://doi.org/10.1080/14029251.2020.1684001How to use a DOI?
Abstract

By considering a relation between Euler’s trinomial problem and the problem of decomposing tensor powers of the adjoint 𝔰𝔩(2)-module I derive some new results for both problems, as announced in arXiv:1902.08065.

Copyright
Β© 2020 The Authors. Published by Atlantis and Taylor & Francis
Open Access
This is an open access article distributed under the CC BY-NC 4.0 license (http://creativecommons.org/licenses/by-nc/4.0/).

1. Introduction

In 1765, Euler [1] investigated the coefficients of trinomial

(1+x+x2)n=βˆ‘k=βˆ’nnan(k)xn+k.(1.1)

For central trinomial coefficients an(0) he found the generating function and a two-term recurrence relation. For a discussion of properties of the an(k), see [3].

Let us change variable x by exp(iΞΈ) and rewrite the left-hand side of (1.1) as

(1+x+x2)n=xnXn,   where   X=1+2cosΞΈ.

Note that X is the character Ο‡1 of the adjoint 𝔰𝔩(2)-module. In what follows, Xn denotes both the representation with character Xn, and the corresponding module.

So, Euler’s problem is equivalent to the problem of multiplicities of weights in the representation with character Xn. I also consider, related to the above, the problem of decomposing Xn into irreducible 𝔰𝔩(2)-modules.

2. Euler’s triangle

It is evident that an(βˆ’k)=an(k). So, it suffices to consider only quantities an(k) for k β‰₯ 0. It is convenient to arrange these coefficients in a triangle. I give here the table of these numbers till n = 10:

(2.1)

Eq. (1.1) immediately implies the three-term recurrence relation

an+1(k)=an(kβˆ’1)+an(k)+an(k+1).(2.2)

Introduce the generating function F(t) for the central trinomial coefficients:

F(t)=βˆ‘n=0∞antn,   where   an=an(0).

Theorem 2.1 (Euler 1765).

The following statements hold.

  1. 1)

    The generating function F(t) has the form

    F(t)=1(1βˆ’2tβˆ’3t2).(2.3)

  2. 2)

    For the an, the following two-term recurrence relation takes place

    nan=(2nβˆ’1)anβˆ’1+3(nβˆ’1)anβˆ’2.(2.4)

    We give here a very short proof of item 1); it is different from Euler’s.

Proof.

Note that

an=1Ο€βˆ«0Ο€XndΞΈ,   where   X=1+2cosΞΈ.

So,

F(t)=1Ο€βˆ«0Ο€dΞΈ1βˆ’tβˆ’2tcosΞΈ.

Evaluating this integral we obtain formula (2.3).

Item 2) is a special subcase of the following more general statement.

Theorem 2.2.

For the an(k), there is the following two-term recurrence relation

(n2βˆ’k2)an(k)=n(2nβˆ’1)anβˆ’1(k)+3n(nβˆ’1)anβˆ’2(k).(2.5)

Proof.

We have

an(k)=1Ο€βˆ«0Ο€XncoskΞΈdΞΈ,
and
∫0Ο€Xn[(d2dΞΈ2+k2)coskΞΈ]dΞΈ=0=∫0Ο€coskΞΈ[(d2dΞΈ2+k2)Xn]dΞΈ.

But,

d2XndΞΈ2=βˆ’n2Xn+n(2nβˆ’1)Xnβˆ’1+3n(nβˆ’1)Xnβˆ’2.

This implies formula (2.5).

Theorem 2.3.

For the an(k), there are the following two-term recurrence relations:

kan+1(k)=(n+1)(an(kβˆ’1)βˆ’an(k+1)),(2.6)
(nβˆ’k+1)an(kβˆ’1)=kan(k)+(n+k+1)an(k+1),(2.7)
(nβˆ’k+1)an+1(k)=(n+1)(an(k)+2an(k+1)),(2.8)
(n+k+1)an+1(k)=(n+1)(an(k)+2an(kβˆ’1)).(2.9)

Proof.

From the identity

∫0Ο€[ddΞΈ(XnsinkΞΈ)]dΞΈ=0,
we obtain relation (2.6). Combining this relation with (2.2), we obtain relations (2.7)–(2.9).

Note that eq. (2.2) implies

an(1)=12(an+1βˆ’an),an(2)=12(an+2βˆ’2an+1βˆ’an),an(3)=12(an+3βˆ’3an+2+2an),an(4)=12(an+4βˆ’4an+3+2an+2+4an+1βˆ’an).

Corollary 2.1.

Explicit expressions for quantities an(nβˆ’k) for k small can be obtained from eqs. (2.5) and (2.7) and we have

an(nβˆ’k)=1k!Qk(n),
where Qk(n) is a degree k polynomial in n.

The recurrence relation for these polynomials follows from eq. (2.7):

Qk+1(n)=(nβˆ’k)Qk(n)+k(2nβˆ’k+1)Qkβˆ’1(n).

Here are the explicit expressions for the first ten polynomials.

Q0=1;   Q1=n;   Q2=n(n+1);   Q3=(nβˆ’1)n(n+4);Q4=(nβˆ’1)n(n2+7nβˆ’6);Q5=(nβˆ’2)(nβˆ’1)n(n+1)(n+12);Q6=(nβˆ’2)(nβˆ’1)n(n3+18n2+17nβˆ’120);Q7=(nβˆ’3)(nβˆ’2)(nβˆ’1)n(n3+27n2+116nβˆ’120);Q8=(nβˆ’3)(nβˆ’2)(nβˆ’1)n(n+1)(n+10)(n2+23nβˆ’84);Q9=n(nβˆ’1)(nβˆ’2)(nβˆ’3)(nβˆ’4)(n4+46n3+467n3+86nβˆ’3360);Q10=n(nβˆ’1)(nβˆ’2)(nβˆ’3)(nβˆ’4)(n5+55n4+665n3βˆ’895n2βˆ’16626n+15120).

3. Decomposition of Xn into irreducible representations

This problem is equivalent to expanding Xn in terms of characters of 𝔰𝔩(2)-modules:

Xn=βˆ‘k=0nbn(k)Ο‡k(ΞΈ).

These characters are well known (see, for example, [4]):

Ο‡k=1+2cos(ΞΈ)+2cos(2ΞΈ)+β‹―+2cos(kΞΈ).

They are orthogonal

1Ο€βˆ«0πχk(ΞΈ)Ο‡l(ΞΈ)(1βˆ’cos(ΞΈ))dΞΈ=Ξ΄k,l,
and we have
bn(k)=1Ο€βˆ«0Ο€Xnfk(ΞΈ)dΞΈ,   where   fk(ΞΈ)=cos(kΞΈ)βˆ’cos((k+1)ΞΈ).

This implies the basic relation

bn(k)=an(k)βˆ’an(k+1),
and a three-term recurrence relation similar to relation (2.2)
bn+1(k)=bn(kβˆ’1)+bn(k)+bn(k+1)   for  nβ‰₯2,  kβ‰₯1,
as well as the following relations
bn=bn(0)=12(3anβˆ’an+1),bn(1)=bn+1,   bn(2)=bn+2βˆ’bn+1βˆ’bn,bn(3)=bn+3βˆ’2bn+2βˆ’bn+1+bn,bn(4)=bn+4βˆ’3bn+3+3bnβˆ’1.

The triangle for the numbers bn(k) analogous to the triangle (2.1) is as follows.

(3.1)

Theorem 3.1.

The generating function G(t)=βˆ‘n=0∞bntn is of the form

G(t)=12t(1βˆ’1βˆ’3t(1+t)).

Proof.

Taking into account the identity

1βˆ’cos(ΞΈ)1βˆ’tβˆ’2tcos(ΞΈ)=12t(1βˆ’1βˆ’3t1βˆ’tβˆ’2tcos(ΞΈ))
we reduce the proof to the proof for F(t). We also have the recurrence relation
(n+1)bn=(nβˆ’1)(2bnβˆ’1+3bnβˆ’2)
which follows from eq. (2.4) and the equality bn=anβˆ’an(1).

Theorem 3.2.

There is a four-term recurrence relation

An,kbn(k)+Bn,kbnβˆ’1(k)+Cn,kbnβˆ’2(k)+Dn,kbnβˆ’3(k)+En,kbnβˆ’4(k)=0,
where
An,k=(n2βˆ’(k+1)2)(n2βˆ’k2);Bn,k=βˆ’2n(2nβˆ’1)(n+k)(nβˆ’kβˆ’1);Cn,k=βˆ’2n(nβˆ’1)(n2βˆ’2n+3βˆ’3k(k+1));Dn,k=6n(nβˆ’1)(nβˆ’2)(2nβˆ’3);En,k=9n(nβˆ’1)(nβˆ’2)(nβˆ’3).(3.2)

Proof.

We have

bn(k)=1Ο€βˆ«0Ο€Xnfk(ΞΈ)dΞΈ,(3.3)
where
X=1+2cos(ΞΈ),   fk(ΞΈ)=cos(kΞΈ)βˆ’cos((k+1)ΞΈ),
and
Akfk(ΞΈ)=0,   where   Ak=(d2dΞΈ2+k2)(d2dΞΈ2+(k+1)2).

Integrating by parts in (3.3) we get (3.2) and

1Ο€βˆ«0Ο€fk(ΞΈ)(AkXn)dΞΈ=0.

Theorem 3.3.

There is the following three-term recurrence relation

(k+1)(n+1βˆ’k)bn(kβˆ’1)=(k(k+1)βˆ’nβˆ’1)bn(k)+k(n+k+2)bn(k+1).

Proof.

This follows from eq. (2.7) and the relation bn(k)=an(k)βˆ’an(k+1).

Acknowledgments

I am thankful to D. Leites who improved my English in this letter.

References

[1]L. Euler, Observationes analyticae Novi Comm. Acad. Sci. Petropolitanae, Vol. 11, 1765/1767, pp. 124-143. reprinted in: Opera Omnia. Teubner, Leipzig, Series (1), Vol. 15, (1911) p. 54; see also http://eulerarchive.maa.org/docs/originals/E326.pdf
[3]J. Riordan, An Introduction to Combinatorial Analysis, Reidel, 1974.
[4]H. Weyl, The Classical Groups. Their Invariants and Representations, Princeton, 1939.
Journal
Journal of Nonlinear Mathematical Physics
Volume-Issue
27 - 1
Pages
1 - 6
Publication Date
2019/10
ISSN (Online)
1776-0852
ISSN (Print)
1402-9251
DOI
https://doi.org/10.1080/14029251.2020.1684001How to use a DOI?
Copyright
Β© 2020 The Authors. Published by Atlantis and Taylor & Francis
Open Access
This is an open access article distributed under the CC BY-NC 4.0 license (http://creativecommons.org/licenses/by-nc/4.0/).

Cite this article

TY  - JOUR
AU  - Askold M. Perelomov
PY  - 2019
DA  - 2019/10
TI  - Euler’s triangle and the decomposition of tensor powers of the adjoint 𝔰𝔩(2)-module
JO  - Journal of Nonlinear Mathematical Physics
SP  - 1
EP  - 6
VL  - 27
IS  - 1
SN  - 1776-0852
UR  - https://doi.org/10.1080/14029251.2020.1684001
DO  - https://doi.org/10.1080/14029251.2020.1684001
ID  - Perelomov2019
ER  -