If the statistical version of an uncertainty principle (i.e. in terms of standard deviation) is assumed to hold in an algebra of observables, can we ever deduce from this that the ambient operator algebra is necessarily noncommutative? (Say we give the notion of standard deviation as in the Robertson uncertainty principle.)
EDIT: In light of the first answer, I should clarify: If there is a universal lower bound for the measurement of the standard deviations in question, must there be noncommutativity?
Let me rephrase the question:
Question: Why should the presence of a statistical uncertainty relation imply that we need a noncommutative quantum theory?
I guess the point is that Heisenberg's "matrix" explanation for Rydberg-Ritz shockingly implies the uncertainty relation. Particularly, imagining that the classical phase space is a circle, the set of real-valued functions on the space are the observables, and we assume that these functions can be expanded in Fourier series...and then by the Fourier transform the algebra of observables is identified with the convolution algebra of the integers. In the quantum case, Rydberg-Ritz gives that we should replace the above abelian convolution algebra with the convolution algebra of a groupoid...hence a noncommutative matrix algebra. From this noncommutativity, the uncertainty relation can be deduced. Is it possible that a fundamental uncertainty relation exist between commuting observables? No reason exists experimentally to think so, except for the fact that the noncommutative theory predicts it for noncommuting observables...I should be a bit ashamed of this question, as there is no physical reason to think it would be true as of now. Yet I ask anyhow...
I think that the first answer does the trick, and that the answer is no, but I'd be interested in any further thoughts on the degree to which uncertainty should force a noncommutative quantum theory.