The C*-algebra of a locally compact group
... However, if A is not abelian, then the topology of Ab is in general not Hausdorff. A net in Ab can have many limit points and simultaneously many cluster points (see [3, 1, 5, 6] for details). On the other hand, for most C*-algebras, either its dual space is not known or if it is known, the topology ...
... However, if A is not abelian, then the topology of Ab is in general not Hausdorff. A net in Ab can have many limit points and simultaneously many cluster points (see [3, 1, 5, 6] for details). On the other hand, for most C*-algebras, either its dual space is not known or if it is known, the topology ...
Class Field Theory - Purdue Math
... To show that 1 + pi is closed under inverses, one need only observe that if 1 + xπ i is a member of this set, then its inverse is the infinite sum 1 − xπ i + x2 π 2i − · · · , with −xπ i + x2 π 2i − · · · ∈ pi . This series converges because |xπ i |p goes to 0. We also state, but do not prove, a gen ...
... To show that 1 + pi is closed under inverses, one need only observe that if 1 + xπ i is a member of this set, then its inverse is the infinite sum 1 − xπ i + x2 π 2i − · · · , with −xπ i + x2 π 2i − · · · ∈ pi . This series converges because |xπ i |p goes to 0. We also state, but do not prove, a gen ...
DRINFELD ASSOCIATORS, BRAID GROUPS AND EXPLICIT
... for sl2 ([R]), and Vergne ([V]) and Alekseev–Meinrenken ([AM1]) proved it for quadratic Lie algebras; it turns out ([AT1]) that in the latter case all solutions of equation (KV1) solve equation (KV3). All these constructions lead to explicit formulas for solutions of the KV conjecture, which are bot ...
... for sl2 ([R]), and Vergne ([V]) and Alekseev–Meinrenken ([AM1]) proved it for quadratic Lie algebras; it turns out ([AT1]) that in the latter case all solutions of equation (KV1) solve equation (KV3). All these constructions lead to explicit formulas for solutions of the KV conjecture, which are bot ...
School of Mathematics and Statistics The University of Sydney
... But the corresponding subset of Z, namely, {z ∈ Z : 2z = 0} = {0} has only one element. Hence these two rings cannot be isomorphic. The same argument shows that Z ⊕ Z2 is not isomorphic to Z ⊕ Z2 ⊕ Z2 because the same subset of Z ⊕ Z2 ⊕ Z2 is {(0, 0, 0), (0, 1, 0), (0, 0, 1), (0, 1, 1)} and has four ...
... But the corresponding subset of Z, namely, {z ∈ Z : 2z = 0} = {0} has only one element. Hence these two rings cannot be isomorphic. The same argument shows that Z ⊕ Z2 is not isomorphic to Z ⊕ Z2 ⊕ Z2 because the same subset of Z ⊕ Z2 ⊕ Z2 is {(0, 0, 0), (0, 1, 0), (0, 0, 1), (0, 1, 1)} and has four ...
