Talk:*-algebra

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 Field: Analysis

August 2003 discussion


Don't you think it's just a bit too redundant to list the same definitions in * algebra, B* algebra and C* algebra? Phys

I agree, but how do you suggest removing the redundancy? I created the page *-algebra because I needed it for the Stone-Weierstrass theorem and character (mathematics). At least the page on C* algebras should contain the whole definition because this is the concept with the best-developed theory. Then, it would be just wrong to say "a *-algebra is just like a C*-algebra, but without a norm". Now that you mention it, we also need a page about W*-algebras, but luckily those are defined in terms of C*-algebras, so there will be no redundancy there. If any page should be removed, I think it's B*-algebras, which by all reckoning are not studied much any more. I say let's leave it as it is. Redundancy is one of the strengths of natural language, not a weakness. -- Miguel


Post2009 discussion

[edit]

*-subalgebras[edit]

I'm having a hard time finding material on *-subalgebras, which are (typographically at least) hard to search for. Wouldn't this be a good place to put at least a definition? Can anyone step in? Episanty (talk) 19:34, 19 June 2011 (UTC)


Is 1* = 1 necessary?[edit]

Am I just confused, or does this follow from the other 3 conditions?

. is multiplication

1* = 1.(1*) = (1**).(1*) = (1.(1*))* = 1** = 1

169.231.27.96 (talk) 23:36, 10 September 2009 (UTC) Ricky

It seems to me that the second condition is enough:

\begin{array}{rll}
  x^* & = (x \cdot 1)^* & \mbox{(by ring axioms)} \\
     & = x^* \cdot 1^*  & \mbox{(by second condition)}
\end{array}

but x* = x* · 1* establishes 1* as a unit.

Cacadril (talk) 22:02, 19 December 2009 (UTC)

No, x* = x* · 1* does not establish 1* as a unit.
Consider zero function on a nontrivial ring.
You need x** = x to get useful results.
JumpDiscont (talk) 05:55, 12 January 2010 (UTC)

It might be helpful to point out why you need x**=x to "get results." You can show that 1*=1, not just any old unit. x**=x for all x means that (_)* is a bijection. In particular, (_)* is injective. Then from (a1*)* = (1*)*a*=1a*=a*, we can conclude that a1*=a. Since the identity in a ring is unique, 1*=1. Also, since the map preserves addition, 0*=0. —Preceding unsigned comment added by Jdgallag84 (talkcontribs) 18:45, 2 May 2011 (UTC)

operator[edit]

Is the "* operation" referred to here? ⇔ ChristTrekker 23:23, 22 December 2009 (UTC)

"Associative ring" seems redundant[edit]

This Terminology section currently begins: "In mathematics, a *-ring is an associative ring... " Associative ring redirects to Ring, and according to that article, a ring is associative by definition.

Therefore it seems clear that the phrase "associative ring" is redundant, something like writing "the discrete integers." Dratman (talk) 22:52, 21 October 2013 (UTC)