# Braided monoidal category, example

Let $$H$$ be a cocommutative hopf algebra. Let $$M$$ be the category of $$H$$-bimodules. Does the category $$M$$ form a braided monoidal category with tensor product $$\otimes_{H}$$ ?

• Welcome to MO ! I think not, why should it ? The tensor product doesn't even involve the Hopf structure. Mar 25 at 19:29

The answer is no in general.

Here is a counter example. Let us work over a ground field $$k$$, and let $$H = \oplus_n k$$ be the direct sum of $$n$$ copies of $$k$$, with $$n \geq 2$$. This is a commutative, cocommutative Hopf algebra. Of course, as others have commented, the monoidal category $$({}_H Mod_H, \otimes_H)$$ only depends on the algebra structure of $$H$$.

The category of $$H$$-$$H$$-bimdoules with $$\otimes_H$$ monoidal structure has a nice interpretation. An $$H$$-$$H$$-bimodule $$M$$ can be thought of as an $$n \times n$$-matrix of vector spaces. The $$(i,j)$$th entry of this matrix is obtained by multiplying $$M$$ by the $$i$$th minimal idempotent on one side and the $$j$$th minimal idempotent on the other side.

Under this identification the monoidal structure $$\otimes_H$$ is easy to describe: it is "matrix multiplication", but where you replace the usual addition and multiplication operations with direct sum and tensor product of vector spaces.

However, now we see the problem emerge. For $$n \geq 2$$ matrix multiplication is not commutative, and it is impossible to equip $$({}_H Mod_H, \otimes_H)$$ with a braiding.

For a concrete example we can set

$$M = \begin{bmatrix}0 & V\\0 & 0\end{bmatrix}$$

$$N = \begin{bmatrix}0 & 0\\0 & W\end{bmatrix}$$

Then $$N \otimes_H M = 0$$, while

$$M \otimes_H N = \begin{bmatrix}0 & V \otimes_k W \\0 & 0\end{bmatrix}$$

There can be no isomorphism between $$N \otimes_H M$$ and $$M \otimes_H N$$.