2
$\begingroup$

Proposition 8.2 of Majid's primer on quantum groups says that if $H$ is a finite dimensional Hopf algebra with quantum double $D(H)$, then this is a factorizable Hopf algebra with quasi-triangular structure $$ R=\sum_a (f^a\otimes 1)\otimes (1\otimes e_a) $$ where $\{e_a\}$ is a basis of $H$, and $\{f^a\}$ the dual basis.

The axiom $(\Delta\otimes\operatorname{id})R=R_{13}R_{23}$ translates to $$ (f^a_{(1)}\otimes 1)\otimes (f^a_{(2)}\otimes 1)\otimes (1\otimes e_a)=(f^a\otimes 1)\otimes (f^b\otimes 1)\otimes (1\otimes e_ae_b). \quad(\ast) $$

It says evaluating against $\phi\in H^\ast$ in the third factor gives both sides of this identity as $\phi_{(1)}\otimes 1\otimes\phi_{(2)}\otimes 1\otimes 1$.

What does it mean to "evaluate against $\phi\in H^\ast$ in the third factor? Does that mean compute $$ \langle (f^a_{(1)}\otimes 1)\otimes (f^a_{(2)}\otimes 1)\otimes (1\otimes e_a),(1\otimes 1)\otimes (1\otimes 1)\otimes (\phi\otimes 1)\rangle$$

If so, I don't know what to make of this expression, I know $H$ and $H^\ast$ are dually paired by the evaluation $\langle \phi,h\rangle=\phi(h)$, but I don't know what this means for $D(H)\otimes D(H)\otimes D(H)$, nor how it would lead to $\phi_{(1)}\otimes 1\otimes\phi_{(2)}\otimes 1\otimes 1$.

So I think evaluation pairing of $H^\ast$ and $H$ induces a pairing of $D(H)=H^\ast\otimes H$ and $D(H)^\ast=H^{\ast\ast}\otimes H^{\ast}=H\otimes H^\ast$ by $$ \langle \phi\otimes g,h\otimes \psi\rangle=\langle \phi,h\rangle\langle g,\psi\rangle=\phi(h)\psi(g) $$ which induces of pairing of $D(H)^{\otimes 3}$ and $D(H)^{\ast\ \otimes 3}$ by $$ \langle (\phi\otimes g)\otimes (\psi\otimes h)\otimes (\chi\otimes k),(a\otimes \alpha)\otimes(b\otimes\beta)\otimes(c\otimes \gamma)\rangle=\langle \phi\otimes g,a\otimes\alpha\rangle\langle \psi\otimes h,b\otimes\beta\rangle\langle \chi\otimes k,c\otimes\gamma\rangle = \phi(a)\alpha(g)\psi(b)\beta(h)\chi(c)\gamma(k) $$

Edit: So wouldn't evaluating both sides of $(\ast)$ above against $(1\otimes 1)\otimes (1\otimes 1)\otimes (1\otimes \phi)$ give $$ \epsilon(f^a_{(1)})\epsilon(f^a_{(2)})\phi(e_a) $$ and $$ \epsilon(f^a)\epsilon(f^b)\phi(e_ae_b)? $$ How does this lead to $\phi_{(1)}\otimes 1\otimes \phi_{(2)}\otimes 1\otimes 1$?

$\endgroup$
2
$\begingroup$

The pairing of $ H^* $ and $ H $ gives a pairing of $ D(H) \otimes D(H)^* $, which induces a pairing of $ D(H) \otimes D(H) \otimes D(H) $ and $ D(H)^* \otimes D(H)^* \otimes D(H)^* $. You are evaluating the partial pairing with $ 1 \otimes \phi $ in the third tensor product.

More precisely, in the equation: $$ (f^a_{(1)}\otimes 1)\otimes (f^a_{(2)}\otimes 1)\otimes (1\otimes e_a)=(f^a\otimes 1)\otimes (f^b\otimes 1)\otimes (1\otimes e_ae_b). $$ each side is inside $ D(H) \otimes D(H) \otimes D(H) $, or dually, it as a map $ D(H)^* \rightarrow D(H) \otimes D(H) $. 'Evaluating against' $ 1 \otimes \phi \in H \otimes H^* \approx D(H)^* $ is just checking the image of $ \phi $ on each side.

EDIT: Generally, one can write $ T: V \rightarrow V $ as $ T = \sum_i v_i \otimes \alpha^i $ for $ v_i \in V $ and $ \alpha^i \in V^* $. Then, for $ x \in V $, $ T(x) = \sum v_i \langle \alpha_i x \rangle $.

Now, returning to our case, by definition of dual basis, a vector $ v \in H $ can be written $ v = \sum_a e_a \langle f^a, v \rangle $, which means the identity matrix is $ \sum_a e_a \otimes f^a $. Similarly the identity matrix on $ H \otimes H $ is $ \sum_{a,b} e_a \otimes e_b \; f^a \otimes f^b $.

Next, since $ \Delta f^a = f^a_{(1)} \otimes f^a_{(2)} $, one has $ \Delta \psi = \sum_a f^a_{(1)} \otimes f^a_{(2)} \langle e_a, \psi \rangle $.

Finally, we have $$ f^a \otimes f^b \langle e_a e_b, \psi \rangle = f^a \otimes f^b \langle e_a \otimes e_b, \Delta \psi \rangle \\ = f^a \otimes f^b \langle e_a \otimes e_b, \psi_{(1)} \otimes \psi_{(2)} \rangle \\ = \psi_{(1)} \otimes \psi_{(2)} $$

Now let us compute the image of $(1 \otimes \phi ) $: $$ (f^a_{(1)}\otimes 1)\otimes (f^a_{(2)}\otimes 1)\otimes (1\otimes e_a) \cdot (1 \otimes \phi ) = f^a_{(1)}\otimes 1 \otimes f^a_{(2)}\otimes 1 \langle e_a, \psi \rangle \\ = \psi_{(1)} \otimes 1 \otimes \psi_{(2)} \otimes 1 $$

The other side, I hope you will try.

$\endgroup$
  • $\begingroup$ Thanks, I've tried to flesh this out as an edit, but I don't get what it means computationally to check the image of $\phi$ on each side. (Like what's the explicit formula and how it leads to $\phi_{(1)}\otimes 1\otimes\phi_{(2)}\otimes 1\otimes 1$.) $\endgroup$ – Kristina Thai Aug 3 '15 at 17:38
  • $\begingroup$ Thank you, seeing an explicit computation was very helpful. $\endgroup$ – Kristina Thai Aug 3 '15 at 20:21
  • $\begingroup$ That last expression can't possibly be right. You can't have an $f_{(2)}$ without an $f_{(1)}$ somewhere. Similar issue with a $\psi_{(1)}$ all by itself. $\endgroup$ – zibadawa timmy Aug 4 '15 at 1:34
  • $\begingroup$ @zibadawa timmy, thank you. it was a typo. $\endgroup$ – user226970 Aug 4 '15 at 3:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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