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This is taken from Timmermann's Invitation to Quantum Groups and Duality.

Hi folks I am struggling a little with a small calculation in the above text.

I will just get right into it.

Lemma 3.2.5 Let $(A,\Delta)$ be a Hopf $*$-algebra with a normalised integral $h$, let $\chi_V$ and $\chi_W$ be corepresentations on finite-dimensional vector spaces $V$ and $W$, respectively, and let $Q\in \hom(V,W)$. Denote by $X$ and $Y$ the corepresentation operators corresponding to $\chi_V$ and $\chi_W$, respectively, and define $R,T\in\hom(V,W)$ by $$R:=(I\otimes h)(Y^{-1}(Q\otimes 1)X)\,\,\,\textit{ and }\,\,\, T:=(I\otimes h)(Y(Q\otimes 1)X^{-1}).$$ Then $R,T\in\hom(\chi_V,\chi_W)$. If $Q\in\hom(\chi_V,\chi_W)$, then $R=T=Q$.

Proof Since $Y$ and $X$ are corepresentations, the composition

$$\begin{align} Y^{-1}(R\otimes 1)X&=Y^{-1}(I\otimes h)(Y^{-1}(Q\otimes 1)X)X \\&=(I\otimes h\otimes I)\left(Y_{[13]}^{-1}Y_{[12]}^{-1}(Q\otimes 1\otimes 1)X_{[12]}X_{[13]}\right) \end{align}$$

Question: Where does the second equality come from?

The Rest of the Proof

This second expression can be rewritten in the form $$(I\otimes h\otimes I)((I\otimes \Delta)\left(Y^{-1}\right)(Q\otimes 1\otimes 1)(I\otimes\Delta)(X)).$$

By right-invariance of $h$, this composition is equal to

$$(I\otimes h)\left(Y^{-1}(Q\otimes 1)X\right)\otimes 1=R\otimes 1.$$

Thus $(R\otimes 1)X=Y(R\otimes 1)$. A similar calculation shows that $(T\otimes 1)X=Y(T\otimes 1)$ (using left-invariance of $h$). A previous result shows that $R$ and $T$ are then elements of $\hom(\chi_V,\chi_W)$.

If $Q\in\hom(\chi_V,\chi_W)$, then $Y^{-1}(Q\otimes 1)X=Q\otimes 1=Y(Q\otimes 1)X^{-1}$ by that previous result and so $R=Q=T$.

Further Details

Normalised Integral: A linear functional $h:A\rightarrow \mathbb{C}$ is a normalised integral if it is both left-and right-invariant:

$$(I\otimes h)\Delta(a)=h(a)1_{\mathcal{M}(A)}\,\,\,\text{ and }\,\,\,(h\otimes I)\Delta(a)=h(a)1_{\mathcal{M}(A)},$$

(where $\mathcal{M}(A)$ is the multiplier algebra of the quantum group $A$), as well as normalised

$$h(1_A)=1.$$

Corepresentation Operator: Where $\chi:V\rightarrow V\otimes A$ is a corepresentation of $A$ on a vector space $V$, the corepresentation operator associated with $\chi$ is an $X\in\hom(V)\otimes A$ such that $$X=\sum_{i,j}e_{ij}\otimes a_{ij},$$

where the $a_{ij}$ are elements of $A$ given by

$$\chi(e_j)=\sum_i e_i\otimes a_{ij},$$

so that $\chi(v)=X(v\otimes 1_A)$.

The inverse of $X$ is given by $X^{-1}=(I\otimes S)(X)$ where $S$ is the antipode.

Leg Notation: The leg notation is given by $X_{[12]}=X_0\otimes 1_A$ and $X_{[13]}=(I\otimes\tau)(X\otimes 1_A)$ where $\tau:A\otimes B\rightarrow B\otimes A$ is the flip map and we can show that $$(I\otimes\Delta )X=X_{[12]}X_{[13]}.$$

Intertwiners: An intertwiner is a map $\varphi:V\rightarrow W$ such that $$\chi_W\varphi=(\varphi\otimes 1_A)\chi_V.$$ The space of intertwiners from a corepresentation $\chi_V:V\rightarrow V\otimes A$ to a corepresentation $\chi_W:W\rightarrow W\otimes A$ is denoted by $\hom(\chi_V,\chi_W)$.

My Thoughts

...are few and far between. If anything I think it should be $$Y^{-1}(R\otimes 1)X=Y^{-1}((I\otimes h)(Y^{-1}(Q\otimes 1)X)\otimes 1)X$$

I had previously tried to unravel this lemma but Timmermann called $R$ by $S$ which I confused with the antipode eventually to experience "death by a million linear maps".

Further information available of course on request.

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The formula should be

$$\begin{align} Y^{-1}(R\otimes 1)X&=Y^{-1}(I\otimes h)(Y^{-1}(Q\otimes 1)X)X \\&=(I\otimes h\otimes I)\left(Y_{[13]}^{-1}Y_{[12]}^{-1}(Q\otimes 1\otimes 1)X_{[12]}X_{[13]}\right) \end{align}$$

i.e. the $Q$ in the first term should be an $R$.

The first equality is just the substitution of the definition of $R$ into the expression $Y^{-1}(R\otimes 1)X$. In the second equality the Haar state $h$ gets pulled to the outside of the expression and the resulting triple tensor product expression is re-written in leg notation (since the $Y^{-1}$ and the $X$ on the outside of the previous expression now act on the first and third factor in the tensor product).

I hope this helps.

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  • $\begingroup$ I fixed the typo and thank you for your help. $\endgroup$ – JP McCarthy Nov 6 '13 at 12:20

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