Timeline for Unitary operators in $\mathcal{U}(L^2(X,\mu))$ not coming from $\operatorname{Aut}(X,\mu)$
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 8 at 13:33 | comment | added | David Gao | @KajalDas This really should just be a MSE question instead of being asked here. Bilateral shifts on the Fourier series, basically all Fourier multipliers with values in the unit circle, and multiplication by any function in $L^\infty([0, 1], \mathbb{T})$ (which is not a.s. $1$) are all examples. Are these “well-studied” enough for you? | |
| Jun 7 at 9:48 | comment | added | ors | @KajalDas The Fourier series gives you an isometry from $L^2([0,1])$ to $l^2$ (the space of square summable sequences. What I was suggesting was to construct a unitary on $L^2([0,1])$ by mapping to $l^2$, then applying some simple unitary on $l^2$ (for example a diagonal one, which just multiplies each component by a complex number of absolute value $a$) then applying the inverse Fourier transform to this new $l^2$ sequence which gives you a new $L^2([0,1])$ function. The Riesz–Fischer theorem ensures this process gives you a unitary on $L^2([0,1])$. | |
| Jun 7 at 6:46 | comment | added | Kajal Das | @ors : I am looking for some motivating examples of unitary operators in $\mathcal{U}(L^2(X,\mu))$ which are well studied and which are not image of Aut$(X,\mu)$ as explained above. Could you explain rigorously what do you mean by 'stick some phases on some of the components'? | |
| Jun 6 at 18:19 | comment | added | Jochen Glueck | I think I don't really understand the questiion. Map an orthonormal basis to another one and you get a unitary operator. Make sure that it doesn't map the constant 1-function to itself (or that it maps some positive function to a non-positive function) and it doesn't stem from a measure preserving transformation. | |
| Jun 6 at 15:39 | comment | added | Ben Johnsrude | Generalizing your elementary construction to the case $X\simeq[0,1]$ (or anything else, really), you can consider multiplication operators $M_\psi: f\mapsto f\psi$, where $\psi:X\to\mathbb{C}$ has $|\psi|=1$ almost everywhere. | |
| Jun 6 at 14:21 | comment | added | ors | Is it possible to be more specific about what you're looking for? There are a lot of examples you can cook up. One easy approach if your space is $L^2([0,1])$ is to mess with the Fourier series, e.g. Fourier transform your vector, change the Fourier series somehow (apply a permutation or stick some phases on some of the components or something) and Fourier transform back. | |
| Jun 6 at 13:30 | history | edited | LSpice | CC BY-SA 4.0 |
`\DeclareMathOperator`
|
| Jun 6 at 7:27 | history | asked | Kajal Das | CC BY-SA 4.0 |