Twisted tensor products of $K^n$ with $K^m$

Abstract

We find three families of twisting maps of K m with K n . One of them is related to truncated quiver algebras, the second one consists of deformations of the first and the third one requires m = n and yields algebras isomorphic to Mn(K). Using these families and some exceptional cases we construct all twisting maps of K 3 with K 3 .

On the classification of graded twisted planes

“…This proves item (1). In order to prove item (2), note that M i,j = 0 if L k < i < L k + n and j = 0, 1. Hence the only terms that survive in the sum of (7.3) for j = 3, .…”

“…, k−1. Multiplying Y n−1 M = M n−1 M by M 2 = M Y +Y M +Y 2 from the left we obtainM n+1 M = M Y n M + Y M Y n−1 M + Y n+1 M and combined with M Y n M + Y M Y n−1 M = aY n+1 M which holds by(2), this yieldsM n+1 M = (a + 1)Y n+1 M . Using (a + 1)d = 1 this gives Y n+1 M = dM n+1 M , which is (A.2) for k = n + 1.…”

“…We assume that K is a field, and consider the basic problem of classifying all twisted tensor products of A with B for a given pair of algebras A and B. In general this problem is out of reach, although some results have been obtained, mainly for finite dimensional algebras (see [1], [2], [4], [7], [9] and [10]). In particular, in [7] some families of twisted tensor products of K[x] with K[y] were found.…”

“…For example the commutation relation yx = qxy gives the quantum plane, and in Example 2.1 we explore the case yx = bxy + cy 2 . In general, the relation yx = ax 2 + bxy + cy 2 yields a quadratic algebra, provided that ac = 1 and that (b, ac) is not a root of any member of a certain family of polynomials.…”

“…Note that in item i) we have (a, d) = (0, 1), since M n * = M 0 * . On one hand i), ii) and iii) imply condition (1) and on the other hand iv) implies condition (2). Since items i)-iv) cover all possible cases, it suffices to prove these items in order to show that one of the conditions (1) or (2) necessarily holds.…”