Euler's difference table and decomposition of tensor powers of adjoint representation of $A_n$ Lie algebra

TL;DR

This paper presents an explicit formula for decomposing tensor powers of the adjoint representation of the $A_n$ Lie algebra using Euler's difference table, applicable when $2k leq n+1$.

Contribution

It introduces a novel application of Euler's difference table to derive explicit decompositions of tensor powers of the adjoint representation of $A_n$ Lie algebra.

Findings

Explicit formula for tensor power decomposition derived

Applicable for $2k leq n+1$ cases

Simplifies understanding of representation structure

Abstract

By using of Euler's difference table, we obtain simple explicit formula for the decomposition of $k$-th tensor power of adjoint representation of $A_n$ Lie algebra at $2 k \le{n+1}$.

Equations

$$e_k^j=e_k^{j+1}-e_{k-1}^j,e_j^j=j!,$$

$$e_k^0=d_k.$$

$$d_k^j=\frac{e_k^j}{j!}.$$

$$d_k$$

$$\frac{e^{-x}}{1-x}=\sum _{k=0}^{\infty }\frac{d_k}{k!}{x}^k.$$

$$\begin{array}[]{ccccccccccccc}k&|&0&1&2&3&4&5&6&7&8&9&10\\ d_{k}&|&1&0&1&2&9&44&265&1854&14833&133496&1334961\end{array}$$

$$d_n^k=\frac{1}{k}(d_n^{k-1}+d_{n-1}^{k-1}).$$

$$d_n^k=\frac{1}{k!}\sum _{j=0}^k\left(\genfrac{}{}{0ex}{}{k}{j}\right)d_{n-j}.$$

$$\sum \frac{d_{n+k}^k}{n!}{x}^n=\frac{e^{-x}}{(1-x{)}^{k+1}}$$

$$\left[\begin{array}[]{llllllllll}1&0&0&0&0&0&0&0&0&0\\ 0&1&0&0&0&0&0&0&0&0\\ 1&1&2&0&0&0&0&0&0&0\\ 2&3&4&6&0&0&0&0&0&0\\ 9&11&14&18&24&0&0&0&0&0\\ 44&53&64&78&96&120&0&0&0&0\\ 265&309&362&426&504&600&720&0&0&0\\ 1854&2119&2428&2790&3216&3720&4320&5040&0&0\\ 14833&16687&18806&21234&24024&27240&30960&35280&40320&0\\ 133496&148329&165016&183822&205056&229080&256320&287280&322560&362880\end{array}\right]$$

$$\left[\begin{array}[]{llllllllll}1&0&0&0&0&0&0&0&0&0\\ 0&1&0&0&0&0&0&0&0&0\\ 1&1&1&0&0&0&0&0&0&0\\ 2&3&2&1&0&0&0&0&0&0\\ 9&11&7&3&1&0&0&0&0&0\\ 44&53&32&13&4&1&0&0&0&0\\ 265&309&181&71&21&5&1&0&0&0\\ 1854&2119&1214&465&134&31&6&1&0&0\\ 14833&16687&9403&3539&1001&227&43&7&1&0\\ 133496&148329&82508&30637&8544&1909&356&57&8&1\end{array}\right]$$

$$X_i^j,i,j=1,\dots ,n+1,$$

$$\sum _{i=1}^{n+1}{X}_i^i=0.$$

$$X_k=X_{i_1}^{j_1}{X}_{i_2}^{j_2}\cdots X_{i_k}^{j_k}$$

$$X_k=c_0^k{Y}_0+c_1^k{Y}_1+\cdots +c_k^k{Y}_k.$$

$$Y_p=R_p+...,R_p=X[p,0,...,0,p].$$

$$c_p^k=\left(\genfrac{}{}{0ex}{}{k}{p}\right)\frac{1}{p!}\sum _{l=0}^p\left(\genfrac{}{}{0ex}{}{p}{l}\right)d_{k-l}.$$

$$c_j^k=\left(\genfrac{}{}{0ex}{}{k}{j}\right)d_k^j,$$

$$c_{j+1}^k=\frac{1}{(j+1{)}^2}\left((k-j)c_j^k+k{c}_j^{k-1}\right),c_0^k=d_k,$$

$$X_1$$

$$X_2$$

$$X_3$$

$$X_4$$

$$X_5$$

$$X_6$$

$$+30Y_5+Y_6$$

$$X_7$$

$$+651Y_5+42Y_6+Y_7$$

$$X_8$$

$$+12712Y_5+1204Y_6+56Y_7+Y_8$$

$$X_9$$

$$+240534Y_5+29904Y_6+2052Y_7+72Y_8+Y_9$$

Full text

1. Introduction

Decomposition of tensor products of representations of simple complex Lie algebras is complicate problem.111For the basic notations of Lie algebras see for example [Ov]. For the Lie algebras of small rank $A_{2},C_{2},G_{2}$ and $A_{3}$ some results were obtained in recent papers [FGP1] - [FGP4] (see also references there). For the general case see, for example, papers [Ha],[BD] and references there.

In present note we consider the problem of decomposition of $k$-th tensor power of adjoint representation of $A_{n}$ Lie algebra at stability domain $2k\leq{n+1}$ , where the results dont depend on $n$ [Ma]. Using Euler’s difference table [Eu1]-[Eu2], (see also [Ri]) we obtain simple explicit formulas (19),  (20) for such decomposition, which on best author knowlidge are new one.

2. Euler’s difference table

In 1753 Euler with relation to the card game ”Jeu de Recontre” investigated in details the permutations of $k$ numbers [1,2,…,k] without fixed points [Eu1], [Eu2] (see also [Ri]). Such permutation has the name derangement and we denote the number of derangements as $d_{k}$.

In order to find derangement numbers $d_{k}$ Euler constructed first the difference table from numbers $e_{k}^{j}$,  $j\leq k$,  that determined by basic recurrence relation

[TABLE]

and proved that

[TABLE]

Note that $e_{k}^{j}$ is divided to $j!$ and it is useful to introduce higher derangement numbers

[TABLE]

For derangement numbers we have two basic recurrence formulae [9], [11]

[TABLE]

and the generating function

[TABLE]

The first ten derangement numbers are

[TABLE]

From (1) follows recurrence relation for numbers $d_{n}^{k}$

[TABLE]

Iterating it we obtain

[TABLE]

Let us give also the generating function

[TABLE]

and the tables of numbers $e_{n}^{k}$ and $d_{n}^{k}$

[TABLE]

[TABLE]

3. Decomposition of tensor powers

It is convenient to consider adjoint representation $ad$ of Lie algebra $A_{n}$ as tensor

[TABLE]

which satisfies the condition

[TABLE]

The $k$th tensor power of $ad$

[TABLE]

has the decomposition

[TABLE]

The quantities $Y_{p}\,,p=0,1,...,k$ decompose into irreducible representations of Lie algebra $A_{n}$

[TABLE]

The coefficients of terms denoted as … are the Littlewood - Richardson coefficients , see [Ma].

The quantity $Y_{p}$ obtained by contractions in (15) $k-p$ upper indeces with $k-p$ lower indeces. So we choose ordered subset of $p$ quantities $X_{i}^{j}$ from $k$ such quantities. The number of such subsets is $\frac{1}{p!}{k\choose p}$. Then we choose from $p$ quantities $X_{i}^{j}$ $l$ quantities of type $X_{i}^{i}$. This gives factor${p\choose l}$. The contraction on rest $k-l$ indeces gives the quantity $d_{k-l}$. As result we have the formula

[TABLE]

Taking into account (9) we obtain the main formula of present note

[TABLE]

where ${k\choose j}$ are binomial coefficients, $d_{k}^{j}$ are higher derangement numbers (3).

Note that coefficients $c_{j}^{k}$ satisfied the recurrence relation

[TABLE]

which is consequence of recurrence relation (8).

4. Conclusion

In conclusion we give the decomposition of first ten powers of adjoint representation of Lie algebra $A_{n}$ at $2k\leq{n+1}$.

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]