Directed graphs and Lie superalgebras of matrices (Q1770462)

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scientific article; zbMATH DE number 2153330
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Directed graphs and Lie superalgebras of matrices
scientific article; zbMATH DE number 2153330

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    Directed graphs and Lie superalgebras of matrices (English)
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    7 April 2005
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    For a finite directed graph \(D=(V,E)\) without multiple edges but possibly with loops, with set of vertices \(V=\{1,\ldots,n\}\), one considers the vector subspace \(L\) of the \(n\times n\) matrix algebra over a field with basis \(\{e_{ij}\mid (i,j)\in E\}\). In the paper under review the author studies the following problem (related to the more general problem of characterizing gradings of matrix algebras): Let us define a mapping \(\alpha:E\to {\mathbb Z}_2=\{0,1\}\). It induces a \({\mathbb Z}_2\)-grading of \(L(D,\alpha)=L=L_0\oplus L_1\). Describe the graphs \(D\) and the mappings \(\alpha\) such that \(L(D,\alpha)\) is a Lie superalgebra. The main result of the paper gives a complete answer. Namely, \(L(D,\alpha)\) being a Lie superalgebra is equivalent to the condition that it is an associative superalgebra with respect to the matrix multiplication. This happens if and only if the set of edges \(E\) is transitive, i.e. \((i,j),(j,k)\in E\) implies \((i,k)\in E\), and \(\alpha\) is a homomorphism, i.e. \(\alpha(i,j)+\alpha(j,k)=\alpha(i,k)\). It turns out that if \(E\) is transitive, then there exists a subset \(B\) of \(E\) (called a superbasis) such that every mapping \(B\to {\mathbb Z}_2\) induces a superstructure on \(L\).
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    directed graphs
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    Lie superalgebras, blocked matrices
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    graded Lie superalgebras
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