# How do you evaluate elements of the Drinfeld double $D(H)$ against elements of $H^\ast$ or $H$?

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 $\phi_{(1)}\otimes 1\otimes \phi_{(2)}\otimes 1\otimes 1$?

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.

• 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 $\phi_{(1)}\otimes 1\otimes\phi_{(2)}\otimes 1\otimes 1$.) – Kristina Thai Aug 3 '15 at 17:38
• Thank you, seeing an explicit computation was very helpful. – Kristina Thai Aug 3 '15 at 20:21
• 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. – zibadawa timmy Aug 4 '15 at 1:34
• @zibadawa timmy, thank you. it was a typo. – user226970 Aug 4 '15 at 3:09