Zoran Skoda bicategory of cleft extensions

Description

$H$ is a fixed Hopf algebra over a fixed ground ring $k$.

Objects

Pairs $(E,\gamma)$ of a right $H$-comodule algebra and a convolution invertible map of $H$-comodules $\gamma:H\to E$.

1-cells

A 1-cell from $(E,\gamma)$ to $(E',\gamma')$ is a $E$-$E'$-bimodule in the category of $H$-comodules.

2-cells

A 2-cell is a morphism $D$ in the category ${}_E\mathcal{M}^H_{E'}$.

Vertical composition

Composition in the category ${}_E\mathcal{M}^H_{E'}$.

Horizontal composition

Tensor product over the middle $H$-comodule algebra.

Base category (for coinvariants)

Objects

An algebra $U$ with $H$-measuring $\triangleright$ and (normalized) cocycle $\sigma:H\otimes H\to U$.

Cocycle means:

$$...$$

Morphisms

Bimodules with (left) bimodule measuring compatible with the cocycle.

2-cells

Morphisms of bimodules commuting with bimodule measuring.

$\phi: C\to C'$ of $U$-$U'$-bimodules.

Equivalence of bicategories

$(E,\gamma)\mapsto (E^{co H},\triangleright_\gamma)$, $h\triangleright u = \gamma(h_{(1)})u\gamma^{-1}(h_{(2)})$.

${}_U D_V\mapsto {}_U D^{co H}_V$

$h\triangleright c = \gamma_U(h_{(1)})u\gamma^{-1}_V (h_{(2)})$

2-cell to restriction.

$\sigma(h,k) = \gamma(h_{(1)})\gamma(k_{(1)})\gamma^{-1}(h_{(2)}k_{(2)})$.

Other direction:

Tensoring with $H$ for objects.

On 1-cells: $C\otimes H$ has $U\sharp H$-$U'\sharp H$-bimodule structure

$$(u\otimes h)(c\otimes g)(u'\otimes h') = u (h_{(1)}\triangleright c)(h_{(2)}g_{(1)}\triangleright u')\otimes h_{(3)}g_{(2)}h'$$

for the case of trivial cocycle for $U'$

Discuss the cocycle for the bimodule measuring. Then instead of expression like $(h_{(1)}\triangleright c)\otimes h_{(2)}g$ we need $(h_{(1)}\triangleright c)\sigma'(h_{(2)},g_{(1)})\otimes h_{(3)}g_{(2)}$

$$(u\otimes h)(c\otimes g)(u'\otimes h') = u ((h_{(1)}\triangleright c)(\sigma'(h_{(2)},g_{(1)})\cdot (h_{(3)}g_{(2)}\triangleright u'))\sigma'(h_{(4)}g_{(3)},h'_{(1)})\otimes h_{(5)}g_{(3)}h'_{(1)}$$

Need a compatibility condition here (to have a bimodule!).

Left action only:

$$(u\otimes h).(c\otimes g) = u.(h_{(1)}\triangleright c).\sigma'(h_{(2)},g_{(1)})\otimes h_{(3)}g_{(2)}$$

Right action only:

$$(c\otimes g).(v\otimes k) = c.((g_{(1)}\triangleright v)\sigma'(g_{(2)},k_{(1)}))\otimes g_{(3)} k_{(2)}$$

Left action axiom

$$\begin{array}{l} (u\sharp h).((\tilde{u}\sharp\tilde{h}).(m\otimes g)) = (u\sharp h).(\tilde{u}.(\tilde{h}_{(1)}\triangleright_M m).\tau(\tilde{h}_{(2)},g_{(1)})\otimes \tilde{h}_{(3)}g_{(2)})\\ = u.(h_{(1)}\triangleright_M (\tilde{u}.(\tilde{h}_{(1)}\triangleright_M m).\tau(\tilde{h}_{(2)},g_{(1)})).\tau(h_{(2)},\tilde{h}_{(3)}g_{(2)}))\otimes h_{(3)}\tilde{h}_{(4)}g_{(3)}\\ = u(h_{(1)}\triangleright_U\tilde{u}).(h_{(2)}\triangleright_M(\tilde{h}_1\triangleright_M m)).(h_{(3)}\triangleright_V\tau(\tilde{h}_{(2)},g_{(1)}))\tau(h_{(4)},\tilde{h}_{(3)}g_{(2)})\otimes h_{(5)}\tilde{h}_{(4)}g_{(3)}\\ = u(h_{(1)}\triangleright_U\tilde{u})\sigma(h_{(2)},\tilde{h}_{(1)}).(h_{(3)},\tilde{h}_{(2)}\triangleright_M m).\tau^{-1}(h_{(4)},\tilde{h}_{(3)})\tau(h_{(5)},\tilde{h}_{(4)})\tau(h_{(6)}\tilde{h}_{(5)},g_{(1)})\otimes h_{(7)}\tilde{h}_{(6)}g_{(2)}\\ = u(h_{(1)}\triangleright_U\tilde{u})\sigma(h_{(2)},\tilde{h}_{(1)}).(h_{(3)},\tilde{h}_{(2)}\triangleright_M m).\tau(h_{(4)}\tilde{h}_{(3)},g_{(1)})\otimes h_{(5)}\tilde{h}_{(4)}g_{(2)} \\ = (u(h_{(1)}\triangleright_U\tilde{u})\sigma(h_{(2)},\tilde{h}_{(1)})\otimes h_{(3)}\tilde{h}_{(2)}).(m\otimes g)\\ = ((u\sharp h)(\tilde{u}\sharp\tilde{h})).(m\otimes g) \end{array}$$

To check the right action axiom we proceed as

$$\begin{array}{l} (m\otimes g).((v\sharp h))(w\sharp k)) = (m\otimes g).(v(h_{(1)}\triangleright_V w)\tau(h_{(2)},k_{(1)})\sharp h_{(3)}k_{(2)})\\ = m.(g_{(1)}\triangleright_V v)(g_{(2)}\triangleright_V(h_{(1)}\triangleright_V w))(g_{(3)}\triangleright_V \tau(h_{(2)},k_{(1)}))\tau(g_{(4)},h_{(3)}k_{(2)})\sharp g_{(5)}h_{(4)}k_{(3)}\\ = m.(g_{(1)}\triangleright_V v)\tau(g_{(2)},h_{(1)})(g_{(3)}h_{(2)}\triangleright_V w)\tau^{-1}(g_{(4)},h_{(3)})\tau(g_{(5)},h_{(4)})\tau(g_{(6)}h_{(5)},k_{(1)})\sharp g_{(7)}h_{(6)}k_{(2)}\\ = m.(g_{(1)}\triangleright_V v)\tau(g_{(2)},h_{(1)})(g_{(3)}h_{(2)}\triangleright_V w)\tau(g_{(4)}h_{(3)},k_{(1)})\sharp g_{(5)}h_{(4)}k_{(2)}\\ = (m.(g_{(1)}\triangleright_V v)\tau(g_{(2)},h_{(1)})\otimes g_{(3)}h_{(2)}).(w\sharp k)\\ = ((m\otimes g).(v\sharp h)).(w\sharp k). \end{array}$$

The unitality of both the left action and the right action is straightforward.

Finally, we check that the left and right actions commute,

$$\begin{array}{l} (u\sharp h).((m\otimes g).(v\sharp k)) = (u\sharp h)(m.((g_{(1)}\triangleright_V v)\tau(g_{(2)},k_{(1)}))\otimes g_{(3)}k_{(2)}) \\ = u.(h_{(1)}\triangleright_M m).(h_{(2)}\triangleright(g_{(1)}\triangleright_V v))(h_{(3)}\triangleright_V\tau(g_{(2)},k_{(1)}))\tau(h_{(4)},g_{(3)}k_{(2)})\otimes h_{(5)}g_{(4)}k_{(3)} \\ = u.(h_{(1)}\triangleright_M m).\tau(h_{(2)},g_{(1)})(h_{(3)}g_{(2)}\triangleright_V v)\tau^{-1}(h_{(4)},g_{(3)})\tau(h_{(5)},g_{(4)}))\tau(h_{(6)}g_{(5)},k_{(1)})\otimes h_{(7)}g_{(6)}k_{(2)} \\ = u.(h_{(1)}\triangleright_M m).\tau(h_{(2)},g_{(1)})(h_{(3)}g_{(2)}\triangleright_V v)\tau(h_{(4)}g_{(3)},k_{(1)})\otimes h_{(5)}g_{(4)}k_{(2)} \\ =(u.(h_{(1)}\triangleright_M m).\tau(h_{(2)},g_{(1)})\otimes h_{(3)}g_{(2)}).(v\sharp k)\\ = ((u\sharp h).(m\otimes g)).(v\sharp k). \end{array}$$