A note on orthogonal decomposition of 𝔰𝔩n over commutative rings

Songpon Sriwongsa

doi:10.1515/math-2024-0092

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A note on orthogonal decomposition of 𝔰𝔩_n over commutative rings

Songpon Sriwongsa

Published/Copyright: November 16, 2024

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$Open Mathematics$

From the journal Open Mathematics Volume 22 Issue 1

Abstract

The orthogonal decomposition (OD) of special linear Lie algebras over the complex numbers was first explored in the early 1980s and has gained renewed interest in the past decade due to its applications in quantum information theory. In this study, we investigate the OD of special linear Lie algebras over commutative rings with identity. Additionally, we present examples of specific rings that facilitate the existence of such ODs for these Lie algebras.

Keywords: Cartan subalgebra; killing form; orthogonal decomposition

MSC 2010: 17B50; 13M05

1 Introduction

Over the field of complex numbers C , an orthogonal decomposition (OD) of a Lie algebra refers to a vector space decomposition into a direct sum of its Cartan subalgebras, which are pairwise orthogonal with respect to the Killing form. A notable instance of OD was studied by Thompson, who discovered an OD of the Lie algebra of type E 8 while constructing the sporadic simple group F 3 [1,2]. The theory of OD for simple Lie algebras over C was further developed by Korostikin et al. [3–6].

The problem of OD has gained significant attention due to its applications and connections to various fields. One prominent example is the study of mutually unbiased bases (MUBs) in C n , which is relevant to quantum information theory [7–9]. Specifically, a link between the existence problem of OD for sl n ( C ) and the construction of maximal collections of MUBs was established in [7]. A famous conjecture, known as the Winnie-the-Pooh conjecture, states that sl n ( C ) has an OD if and only if n is a prime power [6]. This conjecture, whimsically named after the translated work of Milne’s “Winnie-the-Pooh,” remains unsolved in its “only if” part, even for the case of n = 6 . Recent developments regarding sl 6 ( C ) can be found in [10,11], highlighting the conjecture’s impact on the MUB problem.

Furthermore, the Winnie-the-Pooh problem has been linked to an algebraic combinatorics problem [12]. Other notable works related to OD include a review of MUBs and OD applications in homological algebra and topology [13], the description of MUBs in dimension 7 [14,15], and the relationship between OD and symplectic geometry [16]. The p -adic model of quantum mechanics and its implications for MUBs and OD were explored in [17].

The question of OD for the Lie algebra sl n over fields of positive characteristic or other commutative rings has also been considered. Recent works on finite commutative rings include [18–20] that contain the work on the fields of positive characteristic, with this problem briefly mentioned in [6].

In this study, we explore the OD of the special linear Lie algebra over a commutative ring R with identity, focusing on the existence problem of such decompositions. We identify specific sufficient conditions that guarantee the occurrence of an OD and provide examples of rings that meet these conditions.

2 Existence of OD

Let R be a commutative ring with identity. Recall that the special linear Lie algebra of order n over R is defined as

sl n ( R ) = { traceless n × n matrices over R } .

A subalgebra H of sl n ( R ) is a Cartan subalgebra if it is a nilpotent subalgebra, which is its own normalizer. The orthogonality for the decomposition here is defined via the Killing form:

K ( A , B ) ≔ Tr ( ad A ⋅ ad B ) .

For the case of sl n , we have

K ( A , B ) = 2 n Tr ( A B ) ,

for all A , B ∈ sl n ( R ) where Tr is the trace of a matrix. Note that this simple formula can be proved with the same arguments as appeared in the case of sl n over a field due to the availability of the element 1. Therefore, an OD of sl n ( R ) is a decomposition

sl n ( R ) = H 0 ⊕ H 1 ⊕ … ⊕ H k ,

for some k ≥ 1 , where the H i ’s are pairwise orthogonal Cartan subalgebra of sl n ( R ) with respect to K .

The following theorem illustrates a construction of an OD of sl n ( R ) under certain assumptions. Note that the notation R × denotes the unit group of R .

Theorem 2.1

Let R be a commutative ring with 1 of non-even characteristic. For a prime power n = p m , if p ⋅ 1 ∈ R × and there exists a primitive pth root of unity u ∈ R × such that u − 1 ∈ R × , then sl n ( R ) has an OD.

Proof

Begin with m = 1 . For p = 2 , let u be a primitive square root of unity of R such that u − 1 is a unit. We can decompose sl 2 ( R ) as follows:

sl 2 ( R ) = 1 0 0 u R ⊕ 0 1 u 0 R ⊕ 0 1 1 0 R ,

where ⟨ ⋅ ⟩ R denotes a free R -module with the given elements as a basis. Note that in this case, u is actually − 1 . Verifying this decomposition is straightforward.

Assume now that p > 2 and R contains a primitive p th root of unity u such that u − 1 is a unit in R . Let

D = diag ( 1 , u , … , u p − 1 ) and P = 0 0 ⋯ 0 1 1 0 ⋯ 0 0 0 1 ⋯ 0 0 ⋮ ⋮ ⋱ ⋮ ⋮ 0 0 ⋯ 1 0 .

The trace of D is equal to Tr ( D ) = 1 + u + u 2 + … + u p − 1 = 0 because u p − 1 = 0 and u − 1 ∈ R × . Thus, D and P are matrices in sl p ( R ) and p is the smallest positive integer such that D p = P p = I p . For any a , b ∈ Z p , let J ( a , b ) = D a P b . We have

(2.1) Tr ( J ( a , b ) ) = 0 ⇔ ( a , b ) ≠ ( 0 , 0 )

and

(2.2) P b D a = u − a b D a P b .

This implies

(2.3) J ( a , b ) J ( c , d ) = u − b c J ( a + c , b + d ) and

(2.4) [ J ( a , b ) , J ( c , d ) ] = ( u − b c − u − a d ) J ( a + c , b + d ) ,

for a , b , c , d ∈ Z p . For a , k ∈ Z p with a ≠ 0 , J ( a , k a ) and J ( 0 , a ) are the elements of sl p ( R ) by (2.1). For a fixed k ∈ Z p , it follows immediately from the definitions of D and P that J ( 1 , k ) , J ( 2 , 2 k ) , … , J ( p − 1 , k ( p − 1 ) ) are linearly independent. Construct the following subalgebras:

H k = ⟨ J ( a , k a ) ∣ a ∈ Z p × ⟩ R , k ∈ Z p and H ∞ = ⟨ J ( 0 , a ) ∣ a ∈ Z p × ⟩ R = ⟨ P , P 2 , … , P p − 1 ⟩ R .

Let

X = 1 u p ( p − 1 ) 2 u ( p − 1 ) ( p − 2 ) 2 ⋯ u 3 u u 1 u p ( p − 1 ) 2 ⋯ u 6 u 3 u 3 u 1 ⋯ u 10 u 6 ⋮ ⋮ ⋮ ⋱ ⋮ ⋮ u ( p − 1 ) ( p − 2 ) 2 u ( p − 2 ) ( p − 3 ) 2 u ( p − 3 ) ( p − 4 ) 2 ⋯ 1 u p ( p − 1 ) 2 u p ( p − 1 ) 2 u ( p − 1 ) ( p − 2 ) 2 u ( p − 2 ) ( p − 3 ) 2 ⋯ u 1 .

Note that X is a p × p circulant matrix. Since p > 2 and u − 1 is a unit in R , X is invertible. It is straightforward to verify that D P X = X D and P X = X P , so X − 1 D P X = D and X − 1 P X = P . Thus, by (2.2), conjugation by the matrix X shifts H 0 , H 1 , … , H p − 1 cyclically and fixes H ∞ . This implies that all H j , j = 0 , 1 , … , p − 1 are Cartan subalgebras.

Next, we show that

(2.5) sl p ( R ) = H ∞ ⊕ H 0 ⊕ H 1 ⊕ … ⊕ H p − 1 ,

as an R -module. It is clear from the construction that H 0 ∩ ∑ j ≠ 0 H j = { 0 } . In particular, the sum is direct for H 0 and H ∞ . Thus, the sums for all H i s are also direct and H ∞ ⊕ H 0 ⊕ H 1 ⊕ … ⊕ H p − 1 is a free R -module of sl p ( R ) . Note that any element in sl p ( R ) can be expressed as a linear combination of matrices in the right-hand side of (2.5). Indeed, all traceless diagonal matrices are in H 0 and all matrices E i j , i ≠ j whose i j th entry is one and zero elsewhere can be written as a sum of matrices in H 1 , H 2 , … , H p − 1 by the fact that 1 + u + u 2 + … + u p − 1 = 0 and the assumption that p ⋅ 1 is a unit in R . Hence, we have equation (2.5).

We prove that the decomposition (2.5) is pairwise orthogonal with respect to the Killing form K ( A , B ) = 2 p Tr ( A B ) . It is obvious that H ∞ is orthogonal to all the others H i ’s. Let a , b ∈ Z p × and k 1 , k 2 ∈ Z p with k 1 ≠ k 2 . Then, ( a + b , k 1 a + k 2 b ) ≠ ( 0 , 0 ) , and so by (2.3),

K ( J ( a , k 1 a ) , J ( b , k 2 b ) ) = 2 p Tr ( J ( a , k 1 a ) J ( b , k 2 b ) ) = 2 p u − k 1 a b Tr ( J ( a + b , k 1 a + k 2 b ) ) = 0 .

Thus, H i and H j are orthogonal for all i , j ∈ Z p and i ≠ j .

Note that H ∞ is abelian due to (2.4). It remains to verify that H ∞ is self-normalizing. Recall that for all k ∈ Z p and a , b ∈ Z p × , [ J ( a , k a ) , J ( 0 , b ) ] = ( 1 − u − a b ) J ( a , k a + b ) is in H c for some c ∈ Z p . Now, let A ∈ N sl p ( R ) ( H ∞ ) . Then, by (2.5), we can write

A = ∑ c = 1 p − 1 ∑ j = 0 p − 1 ( α ( c , j ) J ( c , j c ) ) + β c J ( 0 , c ) ,

where α ( c , j ) , β c ∈ R . For any basis element J ( 0 , a ) of H ∞ , we have

[ A , J ( 0 , a ) ] = ∑ c = 1 p ∑ j = 0 p ( α ( c , j ) [ J ( c , j c ) , J ( 0 , a ) ] ) + β c [ J ( 0 , c ) , J ( 0 , a ) ] ∈ H ∞ .

This implies

∑ c = 1 p − 1 ∑ j = 0 p − 1 ( α ( c , j ) ( 1 − u − a c ) J ( c , j c + a ) ) = ∑ c = 1 p − 1 ∑ j = 0 p − 1 ( α ( c , j ) [ J ( c , j c ) , J ( 0 , a ) ] ) ∈ H ∞ .

This summation is also in ⊕ i = 0 p − 1 H i . Then, by (2.5), it must be zero. For any c ∈ Z p × , j ∈ Z p , we can choose a = − c − 1 so the scalar 1 − u − a c = 1 − u is a unit in R . So, α ( c , j ) = 0 . Hence, H ∞ = N sl p ( R ) ( H ∞ ) . This completes the proof for the case m = 1 .

Next, suppose that m ≥ 2 . Let F p m be the finite field of p m elements and W = F p m ⊕ F p m a 2 m -dimensional vector space over Z p equipped with a symplectic form ⟨ ⋅ , ⋅ ⟩ : W × W → Z p defined by the field trace^[1] as follows: for any elements w → = ( α ; β ) , w → ′ = ( α ′ ; β ′ ) ∈ W ,

⟨ w → , w → ′ ⟩ ≔ Tr F p m ⁄ Z p ( α β ′ − α ′ β ) .

Then, by Corollary 3.3 of [21], W possesses a symplectic basis ℬ = { e → 1 , … , e → m , f → 1 , … , f → m } , where { e → 1 , … , e → m } and { f → 1 , … , f → m } span the first and the second factors, respectively, such that

⟨ w → , w → ′ ⟩ = ∑ i = 1 m ( a i b ′ i − a ′ i b i ) ,

where w → = ∑ i = 1 m ( a i e → i + b i f → i ) and w → ′ = ∑ i = 1 m ( a ′ i e → i + b ′ i f → i ) . With the basis ℬ , write each vector w → ∈ W as

w → = ( a 1 , … , a m ; b 1 , … , b m ) ,

and associate it with a matrix

J w → = J ( a 1 , b 1 ) ⊗ J ( a 2 , b 2 ) ⊗ … ⊗ J ( a m , b m ) ,

where ⊗ denotes the Kronecker product of matrices,^[2] and J ( a i , b i ) is given as in the case m = 1 with a given primitive p th root of unity u ∈ R × such that u − 1 ∈ R × for all i = 1 , 2 , … , m . Then, the set { J w → ∣ 0 ≠ w → ∈ W } forms a basis of sl p m ( R ) . By properties of the Kronecker product, we have the following identities:

(2.6) J w → J w → ′ = u − B ( w → , w → ′ ) J w → + w → ′ and

(2.7) [ J w → , J w → ′ ] = ( u − B ( w → , w → ′ ) − u − B ( w → ′ , w → ) ) J w → + w → ′ = u − B ( w → ′ , w → ) ( u ⟨ w → , w → ′ ⟩ − 1 ) J w → + w → ′ ,

where

B ( w → , w → ′ ) = ∑ i = 1 m a ′ i b i ,

for all w → = ( a 1 , … , a m ; b 1 , … , b m ) , w → ′ = ( a ′ 1 , … , a ′ m ; b ′ 1 , … , b ′ m ) ∈ W .

Write w → = ( α ; β ) ∈ W , where α = ( a 1 , a 2 , … , a m ) and β = ( b 1 , b 2 , … , b m ) . Define

H ∞ = ⟨ J ( 0 ; λ ) ∣ λ ∈ F p m × ⟩ R and H α = ⟨ J ( λ ; α λ ) ∣ λ ∈ F p m × ⟩ R ,

where α ∈ F p m . It is similar to the case of m = 1 that all J w → ’s are the basis elements, and we have

(2.8) sl p m ( R ) = H ∞ ⊕ ( ⊕ α ∈ F p m H α ) .

It is clear that ⟨ ( λ ; α λ ) , ( λ ′ ; α λ ′ ) ⟩ = ⟨ ( 0 ; λ ) , ( 0 ; λ ′ ) ⟩ = 0 , so by (2.7), all H α and H ∞ are abelian. To show that all H α s and H ∞ are their own normalizers, we first show that for α ≠ α ′ ∈ F p m and λ ′ ∈ F p m × ,

there is an λ ∈ F p m × such that ⟨ ( λ ; α λ ) , ( λ ′ ; α ′ λ ′ ) ⟩ = 1 and
there is an λ ∈ F p m × such that ⟨ ( λ ; α λ ) , ( 0 ; λ ′ ) ⟩ = 1 .

Since the field trace is surjective, there exists γ ∈ F p m such that Tr F p m ⁄ F p ( γ ) = 1 . Thus, we can choose λ = γ ( λ ′ ( α ′ − α ) ) − 1 for 1 and choose λ = ( λ ′ ) − 1 for 2. Now, for any α ∈ F p m and A ∈ N sl p m ( R ) ( H α ) ,

A = ∑ λ ′ ∈ F p m × ∑ α ′ ∈ F p m a ( λ ′ , α ′ ) J ( λ ′ , α ′ λ ′ ) + b λ ′ J ( 0 , λ ′ ) .

For any basis element J ( λ , α λ ) ∈ H α , we have

(2.9) ∑ λ ′ ∈ F p m × ∑ α ′ ∈ F p m α ′ ≠ α a ( λ ′ , α ′ ) [ J ( λ ′ , α ′ λ ′ ) , J ( λ , α λ ) ] + b λ ′ [ J ( 0 , λ ′ ) , J ( λ , α λ ) ] ∈ H α .

Note that

[ J ( λ ′ , α ′ λ ′ ) , J ( λ , α λ ) ] = u − B ( ( λ , α λ ) , ( λ ′ , α ′ λ ′ ) ) ( u ⟨ ( λ ′ , α ′ λ ′ ) , ( λ , α λ ) ⟩ − 1 ) J ( λ ′ + λ , α ′ λ ′ + α λ ) , [ J ( 0 , λ ′ ) , J ( λ , α λ ) ] = u − B ( ( λ , α λ ) , ( 0 , λ ′ ) ) ( u ⟨ ( 0 , λ ′ ) , ( λ , α λ ) ⟩ − 1 ) J ( λ , λ ′ + α λ ) .

The summation in (2.9) is also in ∑ i ≠ α H i . For any ( λ ′ , α ′ ) , by 1, we can choose a suitable λ for which u ⟨ ( λ ′ ; α ′ λ ′ ) , ( λ ; α λ ) ⟩ − 1 = u − 1 is a unit in R . This implies a ( λ ′ , α ′ ) is zero because the sums in (2.8) are direct. By 2, we can show that any b λ ′ is also zero. Thus, A ∈ H α , and so, N sl p m ( R ) ( H k ) = H k . Using similar arguments, we can show N sl p m ( R ) ( H ∞ ) = H ∞ .

It remains to show that they are pairwise orthogonal. Note that if ( γ ; δ ) ≠ ( − α ; − β ) , then Tr ( J ( α ; β ) J ( γ ; δ ) ) = 0 . Indeed, if λ = ( a 1 , … , a m ) , β = ( b 1 , … , b m ) , γ = ( a ′ 1 , … , a ′ m ) , δ = ( b ′ 1 , … , b ′ m ) , and a i ≠ − a ′ i for some i ∈ { 1 , … , m } , then a i + a ′ i ≠ 0 and Tr ( J ( a i + a ′ i , b i + b ′ i ) ) = 0 (as in the case m = 1 ). By (2.6) and the trace property of the Kronecker product,

Tr ( J ( α ; β ) J ( γ ; δ ) ) = u − B ( ( α ; β ) , ( γ ; δ ) ) Tr ( J ( a 1 + a 1 ′ , … , a m + a m ′ ; b 1 + b 1 ′ , … , b m + b m ′ ) ) = u − B ( ( α ; β ) , ( γ ; δ ) ) Tr ( ⊗ j = 1 m J ( a j + a j ′ , b j + b j ′ ) ) = u − B ( ( α ; β ) , ( γ ; δ ) ) ∏ j = 1 m Tr ( J ( a j + a j ′ , b j + b j ′ ) ) = 0 .

This completes the proof.□

We note that the proof of Theorem 2.1 is inspired by the main result in [19], where the ring is assumed to be finite. However, the result we present here is more general. Additionally, we would like to acknowledge Kostrikin et al., whose foundational ideas for proving results over C influenced our approach. Certain aspects of their proof can be adapted to the case of rings.

3 Examples

From the main result (Theorem 2.1), we identify three sufficient conditions to ensure the existence of an OD of sl n over a commutative ring R with identity and non-even characteristic:

n = p m is a prime power.
p ⋅ 1 is a unit in R .
R possesses a primitive p th root of unity u such that u − 1 is a unit in R .

Example 1

For the case n = 2 , it is obvious that any commutative ring R with 1 that has char ( R ) is either zero or odd satisfies the desired conditions. We can choose such a primitive p th root of unity u = − 1 . Then, an OD of sl 2 ( R ) is as follows:

sl 2 ( R ) = 1 0 0 − 1 R ⊕ 0 1 − 1 0 R ⊕ 0 1 1 0 R .

Example 2

A classic example that satisfies these conditions for any prime power n is the field of complex numbers C . In fact, any field containing a primitive p th root of unity meets the required criteria. For a less trivial example, let ζ p be a primitive p th root of unity in C and R = Z [ 1 ⁄ p , ζ p ] . Then, ζ p − 1 is a unit in R . Thus, the ring R has the desired properties. We present one example of OD that can be constructed by Theorem 2.1 for sl 3 ( R ) . Let i = − 1 , ω = e 2 π i 3 and R = Z [ 1 ⁄ 3 , ω ] . We have the following OD:

sl 3 ( R ) = H 0 ⊕ H 1 ⊕ H 2 ⊕ H 3 ,

where

H 1 = 0 1 0 0 0 1 1 0 0 , 0 0 1 1 0 0 0 1 0 R , H 2 = 0 1 0 0 0 ω ω 2 0 0 , 0 0 1 ω 2 0 0 0 ω 0 R , H 3 = 0 1 0 0 0 ω 2 ω 0 0 , 0 0 1 ω 0 0 0 ω 2 0 R .

4 Concluding remarks

There are two sufficient conditions for Theorem 2.1. A natural question arises: what happens if one of these conditions fails? To the best of the author’s knowledge, the case when n is not a prime power remains an open problem. Furthermore, the outcome when a primitive p th root of unity, as required in the theorem, does not exist in R is also unknown. This presents an interesting direction for future investigation, which could begin by studying small rings and utilizing computer algebra systems to explore potential results.

As mentioned in Section 1, the problem of ODs over complex numbers is related to the MUBs problem. Since the discovery of ODs in Lie algebras, it has taken several decades to uncover significant applications. In this work, we have investigated aspects of the OD problem over commutative rings with identity, and we are optimistic that potential applications of our results will be explored in future research.

Acknowledgement

The author would like to thank Yotsanan Meemark for his earlier suggestions of this manuscript.

Funding information: This work was supported by Office of the Permanent Secretary, Ministry of Higher Education, Science, Research and Innovation (OPS MHESI), Thailand Science Research and Innovation (TSRI), and King Mongkut’s University of Technology Thonburi (Grant No. RGNS 64-096).
Author contributions: The author confirms the sole responsibility for the conception of the study, presented results, and prepared the manuscript.
Conflict of interest: The author states no conflict of interest.
Data availability statement: Data sharing is not applicable to this article as no datasets were generated or analysed during the current study.

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Received: 2024-07-31

Revised: 2024-10-14

Accepted: 2024-10-17

Published Online: 2024-11-16

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https://doi.org/10.1515/math-2024-0092

Keywords for this article

Cartan subalgebra; killing form; orthogonal decomposition

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