Timeline for Bounded operators $T: B(H)\to H$ whose Kernel is a Lie algebra
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Jul 4, 2016 at 14:10 | vote | accept | Ali Taghavi | ||
| Jul 2, 2016 at 18:45 | answer | added | jjcale | timeline score: 1 | |
| Jul 2, 2016 at 11:11 | comment | added | YCor | A variant of the previous one: pick now an isometric isomorphism $v$ from the $\ell^2$-direct sum of infinitely many copies of $H$ onto $H$. Let's assume that $H$ is separable and let $(e_n)_{n\ge 1}$ be an orthonormal basis. Then $T\mapsto v(2^{-1}Te_1,2^{-2}Te_2,\dots)$ is a well-defined bounded operator and is injective. That is, the kernel is $\{0\}$, which is not the case for any $\phi_h$. | |
| Jul 2, 2016 at 11:07 | comment | added | YCor | My previous comments provides example where the kernel is the same as that of some $\phi_h$. Here's a way to get other kernels. Let $u:H\times H\to H$ be a bounded isomorphism. Let $h_1,h_2$ be non-collinear vectors. Define $s:T\mapsto u(\phi_{h_1}(t),\phi_{h_2}(t))=u(Th_1,Th_2)$. Then $s$ is a bounded linear operator and its kernel is $\mathrm{ker}\phi_{h_1}\cap \mathrm{ker}\phi_{h_2}$, which is distinct from any $\mathrm{ker}\phi_{h}$ and still is a Lie subalgebra. | |
| Jul 2, 2016 at 9:09 | comment | added | YCor | Yes. Write $\phi_h(T)=Th$. Then $\mathrm{ker}\phi_h$ determines the line generated by $h$ (namely, this line is the intersection of all kernels of all $T\in\mathrm{ker}\phi_h$). So if $\phi$ has the same kernel as $\phi_h$ but is not of the form $\phi_{th}$ for any scalar $t$ then $\phi$ is not of the form $\phi_{h'}$ for any $h'$. With the help of the Hahn-Banach theorem, it's then easy to cook up examples . | |
| Jul 2, 2016 at 8:28 | history | edited | Ali Taghavi | CC BY-SA 3.0 |
added 10 characters in body
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| Jul 2, 2016 at 8:22 | history | asked | Ali Taghavi | CC BY-SA 3.0 |