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arxiv: math/0312467 · v1 · pith:KWH4Q2NXnew · submitted 2003-12-26 · 🧮 math.CO

Nonintersecting Subspaces Based on Finite Alphabets

classification 🧮 math.CO
keywords nonintersectingsubspacescasefieldonlycomplexfinitenumber
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Two subspaces of a vector space are here called ``nonintersecting'' if they meet only in the zero vector. The following problem arises in the design of noncoherent multiple-antenna communications systems. How many pairwise nonintersecting M_t-dimensional subspaces of an m-dimensional vector space V over a field F can be found, if the generator matrices for the subspaces may contain only symbols from a given finite alphabet A subseteq F? The most important case is when F is the field of complex numbers C; then M_t is the number of antennas. If A = F = GF(q) it is shown that the number of nonintersecting subspaces is at most (q^m-1)/(q^{M_t}-1), and that this bound can be attained if and only if m is divisible by M_t. Furthermore these subspaces remain nonintersecting when ``lifted'' to the complex field. Thus the finite field case is essentially completely solved. In the case when F = C only the case M_t=2 is considered. It is shown that if A is a PSK-configuration, consisting of the 2^r complex roots of unity, the number of nonintersecting planes is at least 2^{r(m-2)} and at most 2^{r(m-1)-1} (the lower bound may in fact be the best that can be achieved).

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