# Journal of Nonlinear Mathematical Physics

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

# Eulerβs triangle and the decomposition of tensor powers of the adjoint π°π©(2)-module

Authors
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
10.1080/14029251.2020.1684001How to use a DOI?
Abstract

By considering a relation between Eulerβs trinomial problem and the problem of decomposing tensor powers of the adjoint π°π©(2)-module I derive some new results for both problems, as announced in arXiv:1902.08065.

Open Access

## 1. Introduction

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

(1+x+x2)n=βk=β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,βββwhereβββX=1+2cosΞΈ.

Note that X is the character Ο1 of the adjoint π°π©(2)-module. In what follows, Xn denotes both the representation with character Xn, and the corresponding module.

So, Eulerβ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 π°π©(2)-modules.

## 2. Eulerβs triangle

It is evident that an(β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(kβ1)+an(k)+an(k+1).(2.2)

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

F(t)=βn=0βantn,βββwhereβββ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(1β2tβ3t2).(2.3)

2. 2)

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

nan=(2nβ1)anβ1+3(nβ1)anβ2.(2.4)

We give here a very short proof of item 1); it is different from Eulerβs.

### Proof.

Note that

an=1Οβ«0ΟXndΞΈ,βββwhereβββX=1+2cosΞΈ.

So,

F(t)=1Οβ«0ΟdΞΈ1βtβ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

(n2βk2)an(k)=n(2nβ1)anβ1(k)+3n(nβ1)anβ2(k).(2.5)

### Proof.

We have

an(k)=1Οβ«0ΟXncoskΞΈdΞΈ,
and
β«0ΟXn[(d2dΞΈ2+k2)coskΞΈ]dΞΈ=0=β«0ΟcoskΞΈ[(d2dΞΈ2+k2)Xn]dΞΈ.

But,

d2XndΞΈ2=βn2Xn+n(2nβ1)Xnβ1+3n(nβ1)Xnβ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(kβ1)βan(k+1)),(2.6)
(nβk+1)an(kβ1)=kan(k)+(n+k+1)an(k+1),(2.7)
(nβ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(kβ1)).(2.9)

### Proof.

From the identity

β«0Ο[ddΞΈ(XnsinkΞΈ)]dΞΈ=0,
we obtain relation (2.6). Combining this relation with (2.2), we obtain relations (2.7)β(2.9).

Note that eq. (2.2) implies

an(1)=12(an+1βan),an(2)=12(an+2β2an+1βan),an(3)=12(an+3β3an+2+2an),an(4)=12(an+4β4an+3+2an+2+4an+1βan).

### Corollary 2.1.

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

an(nβ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)=(nβk)Qk(n)+k(2nβk+1)Qkβ1(n).

Here are the explicit expressions for the first ten polynomials.

Q0=1;βββQ1=n;βββQ2=n(n+1);βββQ3=(nβ1)n(n+4);Q4=(nβ1)n(n2+7nβ6);Q5=(nβ2)(nβ1)n(n+1)(n+12);Q6=(nβ2)(nβ1)n(n3+18n2+17nβ120);Q7=(nβ3)(nβ2)(nβ1)n(n3+27n2+116nβ120);Q8=(nβ3)(nβ2)(nβ1)n(n+1)(n+10)(n2+23nβ84);Q9=n(nβ1)(nβ2)(nβ3)(nβ4)(n4+46n3+467n3+86nβ3360);Q10=n(nβ1)(nβ2)(nβ3)(nβ4)(n5+55n4+665n3β895n2β16626n+15120).

## 3. Decomposition of Xn into irreducible representations

This problem is equivalent to expanding Xn in terms of characters of π°π©(2)-modules:

Xn=β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β«0ΟΟk(ΞΈ)Οl(ΞΈ)(1βcos(ΞΈ))dΞΈ=Ξ΄k,l,
and we have
bn(k)=1Οβ«0ΟXnfk(ΞΈ)dΞΈ,βββwhereβββfk(ΞΈ)=cos(kΞΈ)βcos((k+1)ΞΈ).

This implies the basic relation

bn(k)=an(k)βan(k+1),
and a three-term recurrence relation similar to relation (2.2)
bn+1(k)=bn(kβ1)+bn(k)+bn(k+1)βββforββnβ₯2,ββkβ₯1,
as well as the following relations
bn=bn(0)=12(3anβan+1),bn(1)=bn+1,βββbn(2)=bn+2βbn+1βbn,bn(3)=bn+3β2bn+2βbn+1+bn,bn(4)=bn+4β3bn+3+3bnβ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)=βn=0βbntn is of the form

G(t)=12t(1β1β3t(1+t)).

### Proof.

Taking into account the identity

1βcos(ΞΈ)1βtβ2tcos(ΞΈ)=12t(1β1β3t1βtβ2tcos(ΞΈ))
we reduce the proof to the proof for F(t). We also have the recurrence relation
(n+1)bn=(nβ1)(2bnβ1+3bnβ2)
which follows from eq. (2.4) and the equality bn=anβan(1).

### Theorem 3.2.

There is a four-term recurrence relation

An,kbn(k)+Bn,kbnβ1(k)+Cn,kbnβ2(k)+Dn,kbnβ3(k)+En,kbnβ4(k)=0,
where
An,k=(n2β(k+1)2)(n2βk2);Bn,k=β2n(2nβ1)(n+k)(nβkβ1);Cn,k=β2n(nβ1)(n2β2n+3β3k(k+1));Dn,k=6n(nβ1)(nβ2)(2nβ3);En,k=9n(nβ1)(nβ2)(nβ3).(3.2)

### Proof.

We have

bn(k)=1Οβ«0ΟXnfk(ΞΈ)dΞΈ,(3.3)
where
X=1+2cos(ΞΈ),βββfk(ΞΈ)=cos(kΞΈ)βcos((k+1)ΞΈ),
and
Akfk(ΞΈ)=0,βββwhereβββAk=(d2dΞΈ2+k2)(d2dΞΈ2+(k+1)2).

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

1Οβ«0Οfk(ΞΈ)(AkXn)dΞΈ=0.

### Theorem 3.3.

There is the following three-term recurrence relation

(k+1)(n+1βk)bn(kβ1)=(k(k+1)βnβ1)bn(k)+k(n+k+2)bn(k+1).

### Proof.

This follows from eq. (2.7) and the relation bn(k)=an(k)β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
[2]
[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
2019/10/25
ISSN (Online)
1776-0852
ISSN (Print)
1402-9251
DOI
10.1080/14029251.2020.1684001How to use a DOI?
Open Access

TY  - JOUR
PY  - 2019
DA  - 2019/10/25
TI  - Eulerβs triangle and the decomposition of tensor powers of the adjoint π°π©(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  - 10.1080/14029251.2020.1684001
ID  - Perelomov2019
ER  -