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Nov
7 |
awarded | Nice Question |
Nov
7 |
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Unital $C^{*}$ algebras which all elements have path connected spectrum
Example of a connected C∗ algebra : See B.E. Blackadar. A simple unital projectionless C∗-algebra. J. Operator Theory, 5:63–71, 1981. By the comments of Hannes Thiel and Sam Evington this algebra is an example of a connected C* algebra. But is it also path connected (remark by Ali Taghavi) ? |
Sep
12 |
awarded | Nice Answer |
Aug
19 |
awarded | Yearling |
Jul
8 |
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Quantum Fields and Infinite Tensor Products
'tiny separable subset' = GNS construction ? |
Jun
11 |
accepted | Do non-normal states exist in the Solovay model? |
May
28 |
awarded | Enthusiast |
May
26 |
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Do non-normal states exist in the Solovay model?
This should be a nonnormal state on a nonseparable Hilbert space in the Solovay model : Let $H = l^2([0,1])$ and $\lbrace e_x : x \in [0,1] \rbrace$ an orthonormal basis of H . Then $f$ defined by $f(A) = \int_0^1 <e_x, A e_x> dx$ for $A \in B(H)$ is a nonnormal state . |
May
26 |
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Do non-normal states exist in the Solovay model?
@Ashutosh : to lo.logic + math.physics : See the comments to Arnold Neumaiers answer in physicsoverflow.org/31006/why-isnt-the-path-integral-rigorous |
May
25 |
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Do non-normal states exist in the Solovay model?
@Ashutosh : Yes, at least for separable Hilbert spaces. |
May
25 |
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Do non-normal states exist in the Solovay model?
@Ashutosh : According to Gleasons theorem for separable Hilbert spaces of dimension other than 2 states correspond to finitely additive measures on pairwise orthogonal projections but normal states correspond to sigma-finite measures, see ncatlab.org/nlab/show/Gleason%27s+theorem . |
May
25 |
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Do non-normal states exist in the Solovay model?
@Asaf Karagila : All states have norm 1, but non-normal states are discontinous in some other topologies (e.g. ultraweak), see ncatlab.org/nlab/show/state+on+an+operator+algebra . |
May
25 |
asked | Do non-normal states exist in the Solovay model? |
May
9 |
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How to prove this determinant is positive?
@TerryTao : $T$ could also be of the form $\begin{pmatrix} A & A B \\ 0 & A^{*-1}\end{pmatrix}$, where $B^{*} = - B$ . |
May
3 |
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How to prove this determinant is positive?
@GH from MO : ok, my mistake was that the Baker–Campbell–Hausdorff formula is not always convergent. |
May
3 |
revised |
How to prove this determinant is positive?
added 67 characters in body |
May
3 |
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How to prove this determinant is positive?
@Christian Remling choose $M = \left[\begin{matrix}0 & \pi & 0 & 0\\- \pi & 0 & 0 & 0\\0 & 0 & 0 & \pi\\0 & 0 & - \pi & 0\end{matrix}\right]$ |
May
3 |
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How to prove this determinant is positive?
@user23765 : This formular doesn't evaluate log(-1), it evaluates the log auf the product of 2 exponentials, which is simply $X + Y$ in the commuting case but more complicated in the non-commuting case. But the result is still real, if X and Y are real. |
May
3 |
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How to prove this determinant is positive?
@user23765 : See en.wikipedia.org/wiki/… : From the Baker–Campbell–Hausdorff formula it follows, that a real M exists. |
May
3 |
revised |
How to prove this determinant is positive?
added 1 character in body |