Timeline for adjoint of multiplication operator in a commutative algebra
Current License: CC BY-SA 3.0
4 events
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| Aug 27, 2011 at 2:43 | comment | added | Dima Shlyakhtenko | @Chris: what I meant was that you are embedding $A$ into linear operators $End(A)$ on the vector space $A$ and that you have (by the choice of your inner product on $A$) chosen a $*$-operation on $End(A)$; together with the operator norm $\Vert T\Vert = \sup_{\langle xi,xi\rangle=1}\langle A\xi ,A\xi\rangle$ and this $*$-operation, $End(V)$ is a von Neumann algebra. To require that $L_a^* \in L_A$ you end up requiring that the image of $A$ (under $a\mapsto L_a$) is closed under the $*$-operation on $End(V)$, i.e. be a von Neumann subalgebra of $End(V)$. | |
| Aug 26, 2011 at 10:14 | comment | added | Chris Heunen | Nice example! The last paragraph confuses me, though. All finite-dimensional von Neumann algebras are direct sums of full matrix algebras, and hence must be H*-algebras. So that cannot be the characterizing condition for each $L_a^*$ to be a multiplication operator again. But the norm induced by the trace inner product does not satisfy the C*-condition $\|X\|^2=\|X^*X\|$, so at least there is no contradiction. | |
| Aug 26, 2011 at 7:58 | comment | added | Tom De Medts | Thanks for this explicit counterexample; it was indeed my intuition that this condition would not be automatically fulfilled. That brings me back to the original question... | |
| Aug 25, 2011 at 15:58 | history | answered | Dima Shlyakhtenko | CC BY-SA 3.0 |