Definition talk:B-Algebra

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Hi LF. I can't find that either. Looking at: It may be Boolean Algebra. Other sources abbreviate Banach Algebra to B-Algebra (but I don't think that's anything to do with them).

There's not that much research on B-Algebras in general. ProofWiki can lead the way with them then. --Jshflynn 08:52, 20 July 2012 (UTC)

PM not LF. Boolean algebras are something different, and Banach algebras are something different again. I found a few papers on them, but so far as you say there's little out there. Go for it. --prime mover 08:58, 20 July 2012 (UTC)
Sorry. LF was doing a lot of work last night cleaning up this new section (which I am thankful for) and I just assumed it was him. They seem like a very arbitrary structure but they have a connection to group theory which I will get to eventually. --Jshflynn 10:58, 20 July 2012 (UTC)
B-algebras appear to be generalisations of Boolean algebras, having the $x \circ x = 0$ axiom. That's where I'd put my money on. --Lord_Farin 09:03, 20 July 2012 (UTC)
To add to the confusion there are also Borel algebras which are a different thing yet again. --prime mover 21:48, 20 July 2012 (UTC)

Fundamental theorem

By Group/Examples/Self-Inverse and Cancellable Elements, I was brought to think about what kept a B-algebra from being induced by a group, i.e. fitting also the second condition of said page (the first is established by a theorem already). Follow my reasoning please (not up to house standards):

$(x\circ y) \circ(z \circ y) = x \circ((z \circ y) \circ(0 \circ y))$, Axiom 3
$(z\circ y)\circ(0\circ y) = z \circ((0\circ y)\circ(0 \circ y))$, applying above
$ = z\circ 0 = z$, Axioms 1,2

Hence $(x\circ y)\circ(z\circ y) = x \circ z$ and the B-algebra defines a group through the Group/Examples/Self-Inverse and Cancellable Elements. --Lord_Farin 15:31, 21 July 2012 (UTC)

Thus, it appears that every B-algebra is induced by a group (the group it so induces). Thought I'd mention this. --Lord_Farin 15:31, 21 July 2012 (UTC)

Excellent. I suspected this since I first looked for "Sub-$B$-Algebras" of the example $B$-Algebra on this page and found it had as many and of the same order as the Dihedral group of order 6. This is a powerful result of yours. I will clean up the rest of this category as best as I can before moving further. --Jshflynn 21:34, 21 July 2012 (UTC)
Quick verification yielded the two induced operations are mutually inverse. Thus we established $B$-algebras are simply a different perspective to groups. This can provide interesting lines of thought in the future. --Lord_Farin 21:40, 21 July 2012 (UTC)
Note this is mentioned in the source under T4.3 and P4.4 (for what it's worth, I thought of it myself). --Lord_Farin 21:43, 21 July 2012 (UTC)
The way I understand it, a B-algebra is a generalisation of the concept of subtraction in the general group. --prime mover 22:06, 21 July 2012 (UTC)
The theorem in fact proves that it is in every possible respect subtraction in a group. Every $B$-algebra so arises. --Lord_Farin 04:45, 22 July 2012 (UTC)

The final step consists of proving that the two induced structures are mutually compatible, i.e., composing them gives the identity operation. --Lord_Farin 22:40, 22 July 2012 (UTC)

Actually, the final step consists of proving that the B-algebra induced by the group induced by the B-Algebra is that B-algebra, and vice versa. That is, every B-algebra and every group exist in pairs. And then that reduces the entire field of investigation into an exercise in group theory, and can be considered as being complete. --prime mover 05:07, 23 July 2012 (UTC)
I tried to say that, but apparently failed. --Lord_Farin 06:18, 23 July 2012 (UTC)

Cleaning up

Prime.mover. Your result Group Induces $B$-Algebra can be used for the example. It is induced from the dihedral group of order 6. Alternatively a different more workable example could be used.

I e-mailed Nora O. Al-Shehrie who wrote a paper about $B$-Algebras asking why they're named so. But no answer yet. Who knows. --Jshflynn 22:32, 22 July 2012 (UTC)