All Questions
57 questions
5
votes
1
answer
516
views
Stinespring's dilation without $C^{\ast}$-algebras
Does Stinespring's dilation theorem hold if the algebra of interest is a topological $\ast$-algebra instead of the usual $C^{\ast}$-algebra?
I will now state the version of Stinespring's dilation ...
- 8,098
4
votes
1
answer
386
views
Invertible unbounded linear maps defined on a Hilbert space
It is well-known that, assuming the axiom of choice, there are unbounded linear maps defined not only on a dense subset but on all of Hilbert space. Is it possible that such a map is invertible?
- 8,086
4
votes
1
answer
384
views
A Hilbert-space completion of a Hilbert $ C^{*} $-module over a separable $ C^{*} $-algebra
Let $ B $ be a separable $ C^{*} $-algebra and $ \mathcal{E} $ a Hilbert $ B $-module. We know that $ B $ has a faithful state $ \phi $. Using $ \phi $, we can construct a $ \mathbb{C} $-valued pre-...
- 66
2
votes
1
answer
926
views
Eigenvalues and Compact Resolvent
For $A$ an unbounded (densely defined) operator on a separable Hilbert space, what conditions on its eigenvalues will show that, for $\lambda \notin $spec$(A)$, we have that $(A-\lambda)^{-1}$ is a ...
- 52.3k
3
votes
1
answer
1k
views
Self-adjoint bounded operator, resolution of the identity, def. of the diagonal
Let $A$ be a self adjoint bounded linear operator with a continuous spectrum
$\sigma(A)=[a,b]$ which acts on a separable Hilbert space. Let
$E_\lambda$ be its resolution of the identity.
For ...
0
votes
2
answers
377
views
"Frobenius-finite" linear operators on a Hilbert Space
Let $H = L_2(S)$ be the complex Hilbert space over $S$ with the counting measure. (There might be another term for this concept, but) I define a continuous linear operator $L$ on $H$ with matrix ...
- 15.3k
9
votes
5
answers
870
views
Abelianization of GL(H)
This is related to Theo's question about the abelianizations of finite dimensionsal Lie groups.
I am interested in a specific (infinite-dimensional) case of the above question. Let H be an infinite-...
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