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


I am looking for an example of a commutative algebra object in a braided monoidal category C which it can also be turned into a commutative Frobenius algebra. If you have any examples could you also tell me what the multiplication and unit are?

Thank you


share|cite|improve this question
I have a feeling that your question is strongly related to this one:… –  Scott Carter Oct 25 '11 at 12:44
Thanks but it didn't help much. I need a concrete example of a commutative algebra object. I have something in mind but I am not sure: The algebra of polynomials C[x], divided by the ideal <x^d>. If correct, what is its dimension, multiplication and unit in this case? –  user18768 Oct 25 '11 at 13:12

3 Answers 3

up vote 4 down vote accepted

The standard example here is where the braided tensor category is the Drinfeld center Z(C) and the algebra object is the induction of the trivial object from C to Z(C). If C is semsimple over an algebraically closed field then this can be written explicitly as $\sum_x x \otimes x^*$ with half braiding given by Theorem 2.3 of Kirillov-Balsam.

There are plenty of trivial examples when the category is allowed to be symmetric (which presumably you don't want), for example any ordinary commutative algebra is an algebra object in the symmetric (and hence braided) tensor category Vec.

share|cite|improve this answer

The group ring of any group $G$ yields a special case of Noah's answer, where $C$ is the monoidal category of $G$-graded vector spaces. I wrote this up in a blog post a few years ago.

share|cite|improve this answer

As an example of a commutative algebra object which can also be turned into a Frobenius algebra, consider the algebra of polynomials $\mathbb{C}[X]$ in one indeterminate $X$ over the field $\mathbb{C}$, divided by the ideal $\langle X^d \rangle$, i.e. $A=\mathbb{C}[X]/\langle X^d \rangle$. This is a commutative algebra object in the category of finite dimensional complex vector spaces $\mathbf{Vect}_{\mathbb{C}}$. The reason one divides by the ideal $\langle X^d \rangle$ is to make the algebra object (viewed as a vector space over $\mathbb{C}$) finite dimensional, in order to be able to turn it into a Frobenius algebra. Thus in the case at hand $\dim A=d$. The tensor product bifunctor $\otimes_{\mathbb{C}} \colon \mathbf{Vect}_{\mathbb{C}} \times \mathbf{Vect}_{\mathbb{C}} \to \mathbf{Vect}_{\mathbb{C}}$ is given by $\cdot \colon A \times A \to A$. More explicitly the multiplication is $$ \left( \sum_{i=0}^{d-1} a_i X^i \right)\cdot \left( \sum_{j=0}^{d-1} b_j X^j \right) = \sum_{k=0}^{d-1}\left( \sum_{i+j=k} a_ib_j \right)X^k ~, $$ where $a_i,b_j\in \mathbb{C}$ and $X^i\in A$. The monoidal unit is given by the underlying field $\mathbb{C}$.

share|cite|improve this answer

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.