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The following result is proved using diagrams in Lemma 14.2 on page 85 in the book a quantum groups primer by S. Majid.

Let $B, C$ be two algebras in a braided monoidal category. Then the braided tensor product algebra $B \underline{\otimes} C$ has the structure of an algebra in the category.

My question is: how to prove this lemma directly? My solution: let $b,b',b'' \in B$ and $c, c', c'' \in C$. We want to prove that \begin{align} ((b \otimes c)(b' \otimes c'))(b'' \otimes c'') = (b \otimes c)((b' \otimes c')(b'' \otimes c'')). \end{align}

We have \begin{align} (b \otimes c)(b' \otimes c') \\ = b \Psi(c \otimes b') c', \end{align} where $\Psi: C \otimes B \to B \otimes C$ is a braiding. $\Psi$ satisfies Hexagon identities: \begin{align} \Psi_{V,W \otimes Z} = \Psi_{V, Z} \circ \Psi_{V, W}, \\ \Psi_{V \otimes W, Z} = \Psi_{V, Z} \circ \Psi_{W, Z}. \end{align}

We have \begin{align} ((b \otimes c)(b' \otimes c'))(b'' \otimes c'') \\ = ( b \Psi(c \otimes b') c' )(b'' \otimes c''). \end{align} But how to multiply $( b \Psi(c \otimes b') c' )$ and $(b'' \otimes c'')$? How to prove \begin{align} ((b \otimes c)(b' \otimes c'))(b'' \otimes c'') = (b \otimes c)((b' \otimes c')(b'' \otimes c''))? \end{align} Thank you very much.

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In this prove we don't use Hexagon identities. The only thing we need is the functoriality of braiding - in our case it means that $\Psi$ commutes with multiplication : $$\Psi(m\otimes\mathrm{id})=(\mathrm{id}\otimes m)(\Psi\otimes \mathrm{id})(\mathrm{id}\otimes \Psi)$$ We use it twice - in the first and the last step.

Remember that by definition $\Psi$ gives us functorial isomorphisms $\Psi_{W,V} :W\otimes V \rightarrow V\otimes W$, which, in particular, in case of algebra means commutation with multiplication and unit morphisms. For details see for example S.Majid Foundations of Quantum Group Theory, Def. 9.2.1. and discussion below them.

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