Generalized Jacobi identities and Jacobi elements of the group ring of the symmetric group

TL;DR

This paper introduces the concept of Jacobi elements in the group ring of the symmetric group and provides a combinatorial criterion for identifying Jacobi subsets, extending classical Lie algebra identities.

Contribution

It defines Jacobi elements in the group ring of symmetric groups and characterizes Jacobi subsets through combinatorial conditions, generalizing Lie algebra identities.

Findings

Introduced Jacobi elements in the group ring of $S_n$.

Provided a combinatorial necessary and sufficient condition for Jacobi subsets.

Extended classical Lie algebra identities to broader algebraic structures.

Abstract

By definition the identities $[x_1, x_2] + [x_2, x_1] = 0$ and $[x_1, x_2, x_3] + [x_2, x_3, x_1] + [x_3, x_1, x_2] = 0$ hold in any Lie algebra. It is easy to check that the identity $[x_1, x_2, x_3, x_4] + [x_2, x_1, x_4, x_3] + [x_3, x_4, x_1, x_2] + [x_4, x_3, x_2, x_1] = 0$ holds in any Lie algebra as well. I. Alekseev in his recent work introduced the notion of Jacobi subset of the symmetric group $S_n$. It is a subset of $S_n$ that gives an identity of this kind. We introduce a notion of Jacobi element of the group ring $\mathbb{Z}[S_n]$ and describe them on the language of equations on coefficients. Using this description we obtain a purely combinatorial necessary and sufficient condition for a subset to be Jacobi.

Equations

$$\sum _{\sigma \in S_n}[x_{\sigma (1)},\dots ,x_{\sigma (n)}]=0.$$

$$\sum _{\sigma \in S_n}a(\sigma )[x_{\sigma (1)},\dots ,x_{\sigma (n)}]=0$$

$$\sum _{i=0}^{n-1}\sum _{(\alpha ,\beta )}(-1{)}^i\lambda \left(\tau \cdot {\left(\begin{array}{cccccc}1& \dots & i& i+1& \dots & n\\ \beta (i)& \dots & \beta (1)& \alpha (1)& \dots & \alpha (n-i)\end{array}\right)}^{-1}\right)=0,$$

$${\mathrm{I}}_n^{+}=\left\{{\left(\begin{array}{cccccc}1& \dots & i& i+1& \dots & n\\ \beta (i)& \dots & \beta (1)& \alpha (1)& \dots & \alpha (n-i)\end{array}\right)}^{-1}\mid (\alpha ,\beta )\in S{h}^1(n-i,i),i\text{mbox}iseven,0\le i\le n-1\right\},$$

$${\mathrm{I}}_n^{-}=\left\{{\left(\begin{array}{cccccc}1& \dots & i& i+1& \dots & n\\ \beta (i)& \dots & \beta (1)& \alpha (1)& \dots & \alpha (n-i)\end{array}\right)}^{-1}\mid (\alpha ,\beta )\in S{h}^1(n-i,i),i\text{mbox}isodd,0\le i\le n-1\right\}.$$

$$\mid T\cap \tau {\mathrm{I}}_n^{+}\mid =\mid T\cap \tau {\mathrm{I}}_n^{-}\mid .$$

$$[x_1,x_2,x_3,x_4]+[x_3,x_4,x_1,x_2]+[x_2,x_1,x_4,x_3]+[x_4,x_3,x_2,x_1]=0$$

$$[x_1,x_2,x_3,x_4]+[x_3,x_1,x_2,x_4]+[x_4,x_1,x_2,x_3]+[x_1,x_4,x_3,x_2]+[x_2,x_3,x_4,x_1]=0$$

$$[x_1,x_2,x_3,x_4]+[x_3,x_1,x_2,x_4]+[x_2,x_1,x_4,x_3]+[x_4,x_2,x_1,x_3]+[x_1,x_4,x_3,x_2]+[x_2,x_3,x_4,x_1]=0$$

$$[x_1,x_2,x_3,x_4]+[x_3,x_1,x_2,x_4]+[x_2,x_1,x_4,x_3]+[x_4,x_2,x_1,x_3]+[x_1,x_3,x_4,x_2]+[x_3,x_4,x_1,x_2]+[x_2,x_3,x_4,x_1]=0$$

$$\sum _{\sigma \in S_n}a(\sigma )[x_{\sigma (1)},\dots ,x_{\sigma (n)}]=0$$

$$⟨a,b⟩=\sum _{g\in G}a(g)\cdot b(g).$$

$$\mathsf{s}:\mathbb{Z}[G]\to \mathbb{Z}[G],\text{mbox}suchthat\mathsf{s}(a)=\sum _{g\in G}a(g)g^{-1},$$

$$⟨a\cdot b,c⟩=⟨a,c\cdot \mathsf{s}(b)⟩.$$

$$⟨a\cdot b,c⟩=\sum _{g\in G}(a\cdot b)(g)\cdot c(g)=\sum _{g\in G}\sum _{\begin{array}{c}f,h\in G\\ f\cdot h=g\end{array}}a(f)\cdot b(h)\cdot c(g)$$

$$⟨a,c\cdot \mathsf{s}(b)⟩=\sum _{g\in G}a(g)\cdot (c\cdot \mathsf{s}(b))(g)=\sum _{g\in G}\sum _{\begin{array}{c}f,h\in G\\ f\cdot h=g\end{array}}a(g)\cdot b(h^{-1})\cdot c(f).$$

$$X^{\perp }=⟨X{⟩}^{\perp },$$

$$Ker(\cdot \lambda )={\left(G\cdot \mathsf{s}(\lambda )\right)}^{\perp },$$

$$\forall x\in \mathbb{Z}[G]$$

$$⟨a\cdot \lambda ,x⟩=0$$

$$⟨a,x\cdot \mathsf{s}(\lambda )⟩=0$$

$$⟨b,x\cdot \mathsf{s}(\lambda )⟩=0$$

$$⟨b\cdot \lambda ,x⟩=0$$

$$Ker(\cdot \lambda )={\left(\mathbb{Z}[G]\cdot \mathsf{s}(\lambda )\right)}^{\perp }={\left(G\cdot \mathsf{s}(\lambda )\right)}^{\perp }.$$

$$\omega_{n}=\sum_{i=0}^{n-1}\sum_{(\alpha,\beta)}(-1)^{i}\left(\begin{array}[]{cccccc}1&\dots&i&i+1&\dots&n\\ \beta(i)&\dots&\beta(1)&\alpha(1)&\dots&\alpha(n-i)\end{array}\right),$$

$$\phi (\sum _{\sigma \in S_n}\alpha (\sigma )\cdot x_{\sigma (1)}\dots x_{\sigma (n)})=\sum _{\sigma \in S_n}\alpha (\sigma )\sigma .$$

$$\phi (g)=\sum _{\sigma \in S_n}g(\sigma )\sigma$$

$${\tilde{\beta }}_n(\phi (a))=\sum _{\sigma \in S_n}g(\sigma )[x_{\sigma (1)},\dots ,$$

$$\phi ({\tilde{\beta }}_n(a))=\sum _{\sigma \in S_n}a(\sigma )(\sigma \cdot {\omega }_n)={\Omega }_n(a).$$

$${\mathsf{J}}_n=Ker{\tilde{\beta }}_n=Ker{\Omega }_n$$

$${\Omega }_n(a)=\phi ({\tilde{\beta }}_n(a))=\phi (0)=0⟹a\in Ker{\Omega }_n$$

$${\mathsf{J}}_n=Ker{\tilde{\beta }}_n=Ker{\Omega }_n={\left(S_n\cdot s({\omega }_n)\right)}^{\perp }.$$

Full text

Introduction

By a Lie ring we mean a Lie algebra over $\mathbb{Z}$. Any Lie algebra can be considered as a Lie ring. By definition the identities $[x_{1},x_{2}]+[x_{2},x_{1}]=0$ and $[x_{1},x_{2},x_{3}]+[x_{2},x_{3},x_{1}]+[x_{3},x_{1},x_{2}]=0$ hold in any Lie ring where $[x_{1},...,x_{n}]$ denotes the left-normed bracket: $[x_{1},...,x_{n}]=[[x_{1},...,x_{n-1}],x_{n}]$. Moreover, it is easy to check that there is one more identity that holds in any Lie ring: $[x_{1},x_{2},x_{3},x_{4}]+[x_{2},x_{1},x_{4},x_{3}]+[x_{4},x_{3},x_{2},x_{1}]+[x_{3},x_{4},x_{1},x_{2}]=0$. Following I. Alekseev we call a subset $T\subseteq S_{n}$ Jacobi if the following identity holds in any Lie ring:

[TABLE]

In this paper we study a more general notion. An element of the group ring $a\in\mathbb{Z}[S_{n}]$ is called Jacobi element if the following identity is satisfied in any Lie ring.

[TABLE]

By $\mathsf{J}_{n}$ we denote the set formed by all Jacobi elements of $\mathbb{Z}[S_{n}]$.

Throughout the paper we use the following definition of $(s,t)$–shuffle. It is a pair $(\alpha,\beta)$ such that $\alpha:\{1,\dots,s\}\to\{1,\dots,s+t\}$ and $\beta:\{1,\dots,t\}\to\{1,\dots,s+t\}$ are strictly monotonic functions with non intersecting images. The set of all $(s,t)$–shuffles is denoted by ${\sf Sh}(s,t)$. By ${\sf Sh}^{1}(s,t)$ we denote the set of all $(s,t)$–shuffles with $\alpha(1)=1$. One of the main results is the following description of all Jacobi elements of $\mathbb{Z}[S_{n}]$.

Theorem 1**.**

Let $\lambda=\sum\limits_{\sigma\in S_{n}}\lambda(\sigma)\sigma$ be an element of $\mathbb{Z}[S_{n}].$ Then $\lambda$ is a Jacobi element if and only if for any $\tau\in S_{n}$

[TABLE]

*where $(\alpha,\beta)$ runs over ${\sf Sh}^{1}(n-i,i).$ *

The main result of this paper is a necessary and sufficient condition for a subset to be Jacobi. Consider two subsets of $S_{n}$:

[TABLE]

[TABLE]

Theorem 2**.**

Let $T$ be a subset of $S_{n}$. Then $T$ is Jacobi if and only if for any $\tau\in S_{n}$

[TABLE]

Using this theorem we obtain the following interesting identities that hold in any Lie ring.

[TABLE]

[TABLE]

[TABLE]

[TABLE]

The paper is organised as follows. In the first section we describe Jacobi elements and prove Theorem 1 with some additional statements. In the second section we consider Jacobi subsets using the language of Jacobi elements and prove Theorem 2.

1. Jacobi elements

Definition 1.1**.**

An element $a=\sum_{\sigma\in S_{n}}a(\sigma)\sigma$ of $\mathbb{Z}[S_{n}]$ is called a Jacobi element if the following identity is satisfied in any Lie ring.

[TABLE]

The set of all Jacobi elements of $\mathbb{Z}[S_{n}]$ is denoted by $\mathsf{J}_{n}$.

Definition 1.2**.**

Let $G$ be a finite group, $\mathbb{Z}[G]$ be the group ring and $a=\sum_{g\in G}a(g)g,b=\sum_{g\in G}b(g)g$ be elements of $\mathbb{Z}[G].$ The scalar product of $a$ and $b$ is defined as follows

[TABLE]

(By the scalar product in the group ring some authors mean $\sum_{g\in G}a(g)\cdot b(g^{-1})$. We do not use this definition.)

The antipode is the map

[TABLE]

where $a=\sum\limits_{g\in G}a(g)g.$

Lemma 1.3**.**

Let $a,b,c\in\mathbb{Z}[G].$ Then

[TABLE]

Proof.

Computations show that

[TABLE]

[TABLE]

Then both scalar products are the same sums of elements $a(g_{1})\cdot b(g_{2})\cdot c(g_{3})$, where $g_{1},g_{2},g_{3}\in G$ and $g_{1}\cdot g_{2}=g_{3}$. ∎

If $X\subseteq\mathbb{Z}[G]$ we set $X^{\perp}=\{a\in\mathbb{Z}[G]\mid\forall x\in X\ \langle a,x\rangle=0\}$ and call it orthogonal complement.

Remark 1.4**.**

For any $X\subseteq\mathbb{Z}[G]$ the following is satisfied:

[TABLE]

where $\langle X\rangle$ is the abelian subgroup generated by $X$.

Lemma 1.5**.**

Let $G$ be a finite group and $\lambda\in\mathbb{Z}[G]$. Then the following is satisfied:

[TABLE]

where $\cdot\,\lambda$ is a map such that $(\cdot\,\lambda)(a)=a\cdot\lambda$ for every $a\in\mathbb{Z}[G]$.

Proof.

First, we need to prove that $Ker\,(\cdot\,\lambda)=\left(\mathbb{Z}[G]\cdot{\sf s}(\lambda)\right)^{\perp}$. Consider $a\in Ker\,(\cdot\,\lambda)$ then $a\cdot\lambda=0$. According to basic properties of scalar product and lemma 1.3 the following is satisfied:

[TABLE]

Hence $a\in\left(\mathbb{Z}[G]\cdot{\sf s}(\lambda)\right)^{\perp}$. Consider $b\in\left(\mathbb{Z}[G]\cdot{\sf s}(\lambda)\right)^{\perp}$ then $\forall c\in\mathbb{Z}[G]\cdot\mathsf{s}(\lambda)\>\>\exists x\in\mathbb{Z}[G]\>\>c=x\cdot\mathsf{s}(\lambda)$

[TABLE]

Hence $b\cdot\lambda=0$ and $b\in Ker\,(\cdot\,\lambda)$. It is obvious that $\mathbb{Z}[G]\cdot{\sf s}(\lambda)=\langle G\cdot{\sf s}(\lambda)\rangle$ then according to remark 1.4 and statement above

[TABLE]

∎

By $\omega_{n}$ we denote an element of $\mathbb{Z}[S_{n}]$ which is defined as follows:

[TABLE]

where $(\alpha,\beta)\in Sh^{1}(n-i,i)$.

In the work we used the following notation.

$\gamma_{n}$ is the abelian subgroup of the free associative ring $\mathbb{Z}\langle x_{1},\dots,x_{n}\rangle$ generated by monomials $x_{\sigma(1)}\dots x_{\sigma(n)},$ where $\sigma\in S_{n}.$

$\beta_{n}:\gamma_{n}\longrightarrow\gamma_{n}$ is the homomorphism that is defined on the basis as $\beta_{n}(x_{\sigma(1)}\dots x_{\sigma(n)})=[x_{\sigma(1)},\dots,x_{\sigma(n)}].$

$\widetilde{\beta}_{n}:\mathbb{Z}[S_{n}]\longrightarrow\gamma_{n}$ is the homomorphism such that $\widetilde{\beta_{n}}(\sum_{\sigma\in S_{n}}\alpha(\sigma)\sigma)=\sum_{\sigma\in S_{n}}\alpha(\sigma)[x_{\sigma(1)},\dots,x_{\sigma(n)}].$

$\varphi:\gamma_{n}\longrightarrow\mathbb{Z}[S_{n}]$ is the isomorphism given by:

[TABLE]

It is important to mention that $\varphi([x_{1},\dots,x_{n}])=\omega_{n}$.

$\Omega_{n}:\mathbb{Z}[S_{n}]\rightarrow\mathbb{Z}[S_{n}]$ is a homomorphism such that for all $a\in\mathbb{Z}[S_{n}]\quad\Omega_{n}(a)=a\cdot\omega_{n}$.

There is a connection between all these maps.

Lemma 1.6**.**

The following diagram is commutative:

{\gamma_{n}}$${\gamma_{n}}$${\mathbb{Z}[S_{n}]}$${\mathbb{Z}[S_{n}].}$$\scriptstyle{\beta_{n}}$$\scriptstyle{\varphi}$$\scriptstyle{\varphi}$$\scriptstyle{\widetilde{\beta}_{n}}$$\scriptstyle{\Omega_{n}}

Proof.

To prove the fact that this diagram is commutative we need to show that $\widetilde{\beta}_{n}\circ\varphi=\beta_{n}$ and $\varphi\circ\widetilde{\beta}_{n}=\Omega_{n}$.

The first equality. Consider an element $g\in\gamma_{n}$ then $g=\sum\limits_{\sigma\in S_{n}}g(\sigma)x_{\sigma(1)}{\dots}x_{\sigma(n)}$. It is obvious that

[TABLE]

The second equality. Consider an element $a\in\mathbb{Z}[S_{n}]$ then $a=\sum_{\sigma\in S_{n}}a(\sigma)\sigma$. We can assume that $\widetilde{\beta}_{n}(a)=\sum_{\sigma\in S_{n}}a(\sigma)[x_{\sigma(1)},\dots,x_{\sigma(n)}]$ and $\varphi([x_{\sigma(1)},\dots,x_{\sigma(n)}])=\sigma\cdot\omega_{n}$ (see reference References, identity (2.1)). It is correct because $\sigma$ permutes corresponding letters (indices) in obtained formula for bracket $[x_{1},\dots,x_{n}]$. Hence the following is satisfied:

[TABLE]

∎

Lemma 1.7**.**

The following is satisfied:

[TABLE]

Proof.

${\sf J}_{n}$ consists of elements $\sum a(\sigma)\sigma\in\mathbb{Z}[S_{n}]$ such that $\sum a(\sigma)[l_{\sigma(1)},\dots,l_{\sigma(n)}]=0$ for any elements $l_{1},\dots,l_{n}$ of any Lie ring $L.$ Denote by $\mathcal{L}(x_{1},\dots,x_{n})$ the Lie subalgebra of $\mathbb{Z}\langle x_{1},\dots,x_{n}\rangle$ generated by $x_{1},\dots,x_{n}.$ It is the free Lie ring generated by $x_{1},\dots,x_{n}$ (see [2]). Then there is a homomorphism of Lie rings $\Phi:\mathcal{L}(x_{1},\dots,x_{n})\to L$ such that $\Phi(x_{i})=l_{i}.$ It follows that an element $a$ of $\mathbb{Z}[S_{n}]$ lies in ${\sf J}_{n}$ if and only if $\sum a(\sigma)[x_{\sigma(1)},\dots,x_{\sigma(n)}]=0$ in $\mathcal{L}(x_{1},\dots,x_{n}).$ Then ${\sf J}_{n}=Ker\>\tilde{\beta}_{n}.$

To prove the second equality we need to use Lemma 1.6. According to it $\Omega_{n}=\varphi\circ\widetilde{\beta}_{n}$ and $\mathsf{J}_{n}=Ker\,\widetilde{\beta}_{n}$. Since $\varphi$ is an isomorphism then $\varphi(0)=0$ hence for all $a\in\mathsf{J}_{n}=Ker\,\widetilde{\beta}_{n}$

[TABLE]

Consider an element $b\in Ker\;\Omega_{n}$ then $\Omega_{n}(b)=\varphi(\widetilde{\beta}_{n}(b))=0$. Hence $\widetilde{\beta}_{n}(b)=0$ and $b\in\mathsf{J}_{n}$. ∎

Corollary 1.8**.**

[TABLE]

Proof.

$S_{n}$ is a finite group and $\omega_{n}\in\mathbb{Z}[S_{n}]$ then conditions of Lemma 1.5 are fulfilled and according to lemma 1.7 we get the assertion. ∎

Theorem 1.9**.**

Let $\lambda=\sum\limits_{\sigma\in S_{n}}\lambda(\sigma)\sigma$ be an element of $\mathbb{Z}[S_{n}].$ Then $\lambda$ is a Jacobi element if and only if for any $\tau\in S_{n}$

[TABLE]

where $(\alpha,\beta)$ runs over ${\sf Sh}^{1}(n-i,i).$

Proof.

Consider $\lambda=\sum\limits_{\sigma\in S_{n}}\lambda(\sigma)\sigma\in\mathbb{Z}[S_{n}]$. Then $\lambda\in{\sf J}_{n}$ is equivalent to $\lambda\in(S_{n}\cdot\mathsf{s}(\omega_{n}))^{\perp}$ according to corollary 1.8. Hence for all $a=\tau\cdot\mathsf{s}(\omega_{n})\in(S_{n}\cdot\mathsf{s}(\omega_{n}))$ it is true that $\langle\lambda,\tau\cdot\mathsf{s}(\omega_{n})\rangle=0$. Expanding $\omega_{n}$ by definition we get:

[TABLE]

Then it is equivalent to the following statement

[TABLE]

∎

2. Jacobi subsets

Definition 2.1**.**

The subset $T\subseteq S_{n}$ is called a Jacobi subset if the following identity holds in any Lie ring:

[TABLE]

So, we can consider Jacobi subset $T$ as a Jacobi element $a$ in $\mathbb{Z}[S_{n}]$ by defining it as $a=\sum_{\sigma\in S_{n}}\alpha_{T}(\sigma)\sigma$, where coefficient is defined as $\alpha_{T}(\sigma)=\{\begin{smallmatrix}1,\,\sigma\in T\\ 0,\,\sigma\not\in T\end{smallmatrix}$.

Sets $\mathrm{I}_{n}^{+}$ and $\mathrm{I}_{n}^{-}$ are defined as follows:

[TABLE]

[TABLE]

Theorem 2.2**.**

Let $T$ be a subset of $S_{n}$. Then $T$ is Jacobi if and only if for any $\tau\in S_{n}$

[TABLE]

Proof.

As it was said in the beginning any Jacobi subset $T$ can be represented as a Jacobi element $a=\sum_{\sigma\in S_{n}}\alpha_{T}(\sigma)\sigma$ and theorem 1.9. can be applied to it. So, the subset $T$ is Jacobi if and only if

[TABLE]

where $(\alpha,\beta)\in{\sf Sh}^{1}(n-i,i).$ We can move terms with odd $i$ to the right side.

[TABLE]

Notice that coefficient $\alpha_{T}$ shows that some permutation belongs to the subset $T$ or not. Hence sum of these coefficients is equal to cardinality of the intersection of $T$ and the set of considered permutations. Then the following is satisfied:

[TABLE]

[TABLE]

So, the previous equality can be represented as

[TABLE]

As a consequence $T$ is Jacobi if and only if

[TABLE]

∎