Abstract

0.1. If R is any algebra over a field F and xi E R, i= l,..., k, let IIX 1 ,..., .Y~] 1 E R denote the alternating sum of products Z: sg 0 x,,(,) *** x,,(~) over all permutations 0 of l,..., k. One says that R satisfies the k-fold standard identity if [IX, ,..., xk]] = 0 for any choice xi E R. Of course if PC_ R is any subspace of R one can restrict one’s attention to P and say that P satisfies the k-fold standard identity if 1 [x1 ,..,,x~]] = 0 for any choice xi f P. Now the Amitsur-Levitski theorem asserts that the n X n matrix algebra M(n, F) satisfies the 2n-fold standard identity. (For a survey and history of this theorem see [ 9 1. See also 17 I.) It is of course enough to consider the case where F = C. Some time ago (see [3]) we came upon a proof of the Amitsur-Levitski theorem from the point of Lie algebra cohomology. The cohomology of M(n, C) as a Lie algebra is an exterior algebra with primitive generators having degrees, 1, 2,..., 2n 1. The failure of M(n, Cc) to have a primitive class in degree 2n + 1 leads to the trace relation tr[ [x, ,...,x~~+,]] = 0 for any xi E M(n, C;). for i = I ,..., 2n + 1. But this quickly implies the Amitsur-Levitski theorem. In fact the argument can be pushed further. If d, G M(n. Cc) denotes the Lie algebra of all n x n skew-symmetric matrices one knows that d,, has no primitive class in degree 2n 1 in case n is even. This led to a new identity in [3 1. For iz even

Talk to us

Join us for a 30 min session where you can share your feedback and ask us any queries you have

Schedule a call

Disclaimer: All third-party content on this website/platform is and will remain the property of their respective owners and is provided on "as is" basis without any warranties, express or implied. Use of third-party content does not indicate any affiliation, sponsorship with or endorsement by them. Any references to third-party content is to identify the corresponding services and shall be considered fair use under The CopyrightLaw.