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I have a $\mathbb{C}$-algebra defined by two generators $a$ and $b$ with the relations that $a^2 = b^2 = 1$.

I would like to know of ways to think about this algebra (commutative or not).

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if you mean commutative algebra then you have just $\mathbb{C}\oplus\mathbb{C}\oplus\mathbb{C}\oplus\mathbb{C}$ –  user8268 Sep 12 '12 at 14:44
    
@user8268: Thanks. I actually didn't mean commutative. I had gotten that part wrong. –  Thomas Sep 12 '12 at 14:46
    
@rschwieb: it is a ring isomorphism - it is obtained by setting $a=\pm1$, $b=\pm1$ in all possible (four) ways –  user8268 Sep 12 '12 at 15:13
    
@user8268 I still find your description of an isomorphism incomprehensible, but yeah, it's obviously a commutative semisimple ring, hence a product of $\mathbb{C}$'s. I went ahead and deleted my mistaken comment. Thanks! –  rschwieb Sep 12 '12 at 16:10

2 Answers 2

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If the algebra is noncommutative, ie $ab\neq ba$ then you have the group algebra over $\mathbb{C}$ of $\mathbb{Z}/(2) *\mathbb{Z}/(2)$, the free product of $\mathbb{Z}/(2)$s. If the algebra is commutative, ie $ab=ba$ then you have the group algebra over $\mathbb{C}$ of $\mathbb{Z}/(2)\times \mathbb{Z}/(2)$, the Klein 4-group.

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If you require $a$ and $b$ to commute, then it looks like a graded algebra that one might see in algebraic geometry. It is a quotient of the symmetric algebra of a two-dimensional vector space.

If you require $a$ and $b$ to anticommute ($ab=-ba$) then you are looking at a complex Clifford algebra.

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Thanks for this. It gives me something to think about. –  Thomas Sep 12 '12 at 14:55

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