The four subspace problem: an elementary solution.
Let \(K\) be an arbitrary field. We recall that a quadruple of \(K\)-vector spaces is a system \(U=(U_0,U_1,U_2,U_3,U_4)\), where \(U_0\) is a finite dimensional \(K\)-vector space and \(U_1,U_2,U_3,U_4\) are subspaces of \(U_0\). A morphism \(f\colon U\to U'\) between two quadruples \(U\) and \(U'\) is a linear map \(f\colon U_0\to U'_0\) such that \(f(U_j)\subseteq U'_j\), for \(j=1,2,3,4\). The morphism \(f\colon U\to U'\) is an isomorphism if the linear map \(f\colon U_0\to U'_0\) is bijective. We set \(U\oplus U'=(U_0\oplus U'_0,U_1\oplus U'_1,U_2\oplus U'_2,U_3\oplus U'_3,U_4\oplus U'_4)\), and we call \(V\) indecomposable if \(V\) is not isomorphic to a direct sum \(U\oplus U'\) of two non-zero quadruples. The Four Subspace Problem is the problem of classifying all indecomposable quadruples, up to isomorphism. It was solved over 30 years ago by Gelfand and Ponomarev (over an algebraically closed field), and independently by Nazarova (over an arbitrary field). The paper contains a new elementary and short proof of the problem. The endomorphism algebra \(\text{End}(U)\) of any indecomposable quadruple is described. One of the classification tools used in the paper is a bipartition matrix interpretation of quadruples, and the following useful lemma. If \(U\) is a quadruple without ``trivial'' direct summands and \(U'\) is the quadruple defined by setting \(U'_0=(U_1+U_2)\cap(U_3+U_4)\) and \(U'_j=U_j\cap U_0\), for \(j= 1,2,3,4\), then \(U\) is indecomposable if and only if \(U'\) is indecomposable.
scientific article; zbMATH DE number 2119124