MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $S$ be a finite set. Let $R$ be a complex vector space with basis indexed by subsets of $S$. Define a product on $R$ by defining it on the basis elements as $1_A\cdot 1_B=1_{A\Delta B}$, where $A\Delta B$ is the symmetric difference of $A$ and $B$. This gives $R$ the structure of a commutative and associative $C$-algebra.

Is this a well-understood algebra? Does it have a name?

share|cite|improve this question
up vote 5 down vote accepted

It is the complex group algebra over $((\mathbb{Z}/2)^S,+)$. This may be also described as the tensor product of $S$ copies of $\mathbb{C}[\mathbb{Z}/2] = \mathbb{C} \times \mathbb{C}$.

share|cite|improve this answer
Note that this description is also valid if $S$ is infinite. – Martin Brandenburg Sep 30 '10 at 12:46

This is the group algebra $\mathbb{C}G$ where $G$ is an elementary Abelian group of order $2^n$ where $n=|S|$ (the elements of $G$ are your $1_A$). Every group ring over $\mathbb{C}$ is isomorphic to the direct product of matrix algebras over $\mathbb{C}$. In this case as $G$ is Abelian, $\mathbb{C} G$ is isomorphic to the product of $2^n$ copies of $\mathbb{C}$.

share|cite|improve this answer

This is the group algebra of the additive group $(\mathbb Z/2\mathbb Z)^S$, hence it is the product of $2^{|S|}$ copies of $\mathbb C$.

share|cite|improve this answer
Synchronicity ! – Martin Brandenburg Sep 30 '10 at 12:22

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.