What is so amazing about Wigner's semicircle law?

[@Jaysharings]

This question may not relevant to this topic, but I just wanted to ask, in a von neumann algebra, suppose we have a dense sub algebra, whose centre is trivial can we say the closure of the dense sub algebra which is the whole von neumann algebra is factor?

[@tofu-munchingCoalition.ofChaos]

If I made no mistake it's in general not a factor. Let H be an infinity dimensional Hilbert space. Let the von Neumann algebra be the direct sum B(H)+B(H). It's obviously not a factor. The dense sub algebra is given by the finite rank operators in B(H)+B(H). The center of a dense sub algebra is a subset of the center (density argument). But the only elements in the center contain operators of infinite rank and zero. Therefore the center of the sub algebra is zero. If you want a unital sub algebra you can choose the operators in B(H)+B(H) that are of the form r*Id+(finite rank). Again an element of the center of this algebra must be an element the center of B(H)+B(H). But the only element in the center of B(H)+B(H) that is of the form are multiples of the identity. Therefore the center of this sub algebra is trivial (multiples of the unit).