Definition talk:C*-Algebra

FWIW this is somewhat antiquated terminology, the modern terminology is "C*-algebra". Caliburn (talk) 13:57, 6 January 2023 (UTC)


 * Any sources for this are welcome. --prime mover (talk) 14:46, 6 January 2023 (UTC)


 * I will try to track one down. But I think we should definitely work with "C*-algebra" since Wikipedia suggests the B* term hasn't been used actively since the 40s. C* algebras are definitely on my to do list so this should get done in due time. Caliburn (talk) 21:41, 6 January 2023 (UTC)


 * I added a reference that is AFAIK a modern one. In the old literature, this definition was called $B^*$-algebra, while $C^*$-algebra was defined replacing $(B^*5)$ by
 * $(C^* 5):\quad \norm {x x^*} = \norm {x^*} \norm x$
 * But after the equivalence of $B^*$-algebra and $C^*$-algebra was shown, the old $B^*$-algebra's definition is preferably taken to define $C^*$-algebra. --Usagiop (talk) 21:50, 6 January 2023 (UTC)


 * Yep okay, this is something I know little about. You've seen how this sort of thing is handled, feel free to tidy this up. My only source is my beloved Borwein and Borowsky. Brain no longer sharp but heart still red. --prime mover (talk) 23:18, 6 January 2023 (UTC)


 * Okay job done, we now have this as $C$-algebra, and a redirect and an also known as. Someone want to craft a historical note? Let's capture this stuff. --prime mover (talk) 23:24, 6 January 2023 (UTC)


 * Yes, in Einsiedler's book (and for me personally) this is the definition of the $C^*$-algebra. However, there are also books where the $C^*$-algebra is defined using $(C^* 5)$ condition. Someone should add more notes. Caliburn, what is your definition? --Usagiop (talk) 01:21, 7 January 2023 (UTC)


 * If you've got the book and it gives a specific definition that's different from what is here, then go ahead and sort out what needs to be sorted out.
 * I have gone back to the source book I used (the Borowski and Borwein dictionary) and it gives the definition as given here, calling it a B* algebra. Then it goes on to say: "The prototype is the of a matrix or an operator on Hilbert space. A B*-algebra of operators with this involution is called a C*-algebra."
 * We need to do one of the following:


 * a) Return this page to B*-algebra and set up another page as C*-algebra, then write whatever equivalence proof needed to show etc.
 * b) Set up a double definition plus equivalence proof in the manner it is done on this site. Everyone in this discussion has sufficient rights as to be able to rename pages, and everyone by now ought to have managed to familiarise themselves with how this is done here.


 * My money is on a) as there is already a source work which defines it as a B*-algebra.


 * What we really need is someone who knows about this, to a greater extent than getting it off Wikipedia. --prime mover (talk) 09:01, 7 January 2023 (UTC)


 * All $C^\ast$ algebras are (isometrically $\ast$-)isomorphic to a norm and adjoint closed subalgebra of $\map B H$ for a Hilbert space $H$, (this is a famous theorem of Gelfand & Neimark) so these formulations are equivalent. My vision for this page would be: have $\norm {x^\ast x} = \norm x \norm {x^\ast}$ and $\norm {x^\ast x} = \norm x^2$ as two definitions, indicate that the latter was once referred to as the $B^\ast$ identity and include a bit of history, and then show equivalence. Maybe definition 3 can then be a historical one, a special type of subalgebra of $\map B H$. Also worth mentioning that the first 4 conditions will eventually be absorbed away by defining $\ast$-algebras. Caliburn (talk) 12:58, 7 January 2023 (UTC)


 * I prefer your plan (as well as $(a)$ of prime mover). Please do it. I have also found that Peter Lax uses in his 2002's book the terminology $B^*$-algebra, so it is probably not so ancient. I would like to have a page for $H^*$-algebra,too. --Usagiop (talk) 23:10, 7 January 2023 (UTC)


 * If you know about (and can source) such things, please go ahead. I have various projects ongoing, not least of which is attacking the backlog of redlinks. --prime mover (talk) 00:41, 8 January 2023 (UTC)

Anti-homomorphism means $\map f {x y} = \map f y \map f x$, so the same as a homomorphism in commutative structures. So anti-automorphism here would just be a bijective anti-homomorphism (ie. "anti-isomorphism") from the algebra to itself. Surjectivity is obvious since $\paren {x^\ast}^\ast = x$ and injectivity also follows from this. These definitions should not be hard to source. Caliburn (talk) 13:40, 10 January 2023 (UTC)