Zoran Skoda bicategory of cleft extensions

Description

HH is a fixed Hopf algebra over a fixed ground ring kk.

Objects

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

1-cells

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

2-cells

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

Vertical composition

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

Horizontal composition

Tensor product over the middle HH-comodule algebra.

Base category (for coinvariants)

Objects

An algebra UU with HH-measuring \triangleright and (normalized) cocycle σ:HHU\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.

ϕ:CC\phi: C\to C' of UU-UU'-bimodules.

Equivalence of bicategories

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

UD V UD V coH{}_U D_V\mapsto {}_U D^{co H}_V

hc=γ U(h (1))uγ V 1(h (2))h\triangleright c = \gamma_U(h_{(1)})u\gamma^{-1}_V (h_{(2)})

2-cell to restriction.

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

Other direction:

Tensoring with HH for objects.

On 1-cells: CHC\otimes H has UHU\sharp H-UHU'\sharp H-bimodule structure

(uh)(cg)(uh)=u(h (1)c)(h (2)g (1)u)h (3)g (2)h(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 UU'

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

(uh)(cg)(uh)=u((h (1)c)(σ(h (2),g (1))(h (3)g (2)u))σ(h (4)g (3),h (1))h (5)g (3)h (1)(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:

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

Right action only:

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

Left action axiom

(uh).((u˜h˜).(mg))=(uh).(u˜.(h˜ (1) Mm).τ(h˜ (2),g (1))h˜ (3)g (2)) =u.(h (1) M(u˜.(h˜ (1) Mm).τ(h˜ (2),g (1))).τ(h (2),h˜ (3)g (2)))h (3)h˜ (4)g (3) =u(h (1) Uu˜).(h (2) M(h˜ 1 Mm)).(h (3) Vτ(h˜ (2),g (1)))τ(h (4),h˜ (3)g (2))h (5)h˜ (4)g (3) =u(h (1) Uu˜)σ(h (2),h˜ (1)).(h (3),h˜ (2) Mm).τ 1(h (4),h˜ (3))τ(h (5),h˜ (4))τ(h (6)h˜ (5),g (1))h (7)h˜ (6)g (2) =u(h (1) Uu˜)σ(h (2),h˜ (1)).(h (3),h˜ (2) Mm).τ(h (4)h˜ (3),g (1))h (5)h˜ (4)g (2) =(u(h (1) Uu˜)σ(h (2),h˜ (1))h (3)h˜ (2)).(mg) =((uh)(u˜h˜)).(mg)\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

(mg).((vh))(wk))=(mg).(v(h (1) Vw)τ(h (2),k (1))h (3)k (2)) =m.(g (1) Vv)(g (2) V(h (1) Vw))(g (3) Vτ(h (2),k (1)))τ(g (4),h (3)k (2))g (5)h (4)k (3) =m.(g (1) Vv)τ(g (2),h (1))(g (3)h (2) Vw)τ 1(g (4),h (3))τ(g (5),h (4))τ(g (6)h (5),k (1))g (7)h (6)k (2) =m.(g (1) Vv)τ(g (2),h (1))(g (3)h (2) Vw)τ(g (4)h (3),k (1))g (5)h (4)k (2) =(m.(g (1) Vv)τ(g (2),h (1))g (3)h (2)).(wk) =((mg).(vh)).(wk).\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,

(uh).((mg).(vk))=(uh)(m.((g (1) Vv)τ(g (2),k (1)))g (3)k (2)) =u.(h (1) Mm).(h (2)(g (1) Vv))(h (3) Vτ(g (2),k (1)))τ(h (4),g (3)k (2))h (5)g (4)k (3) =u.(h (1) Mm).τ(h (2),g (1))(h (3)g (2) Vv)τ 1(h (4),g (3))τ(h (5),g (4)))τ(h (6)g (5),k (1))h (7)g (6)k (2) =u.(h (1) Mm).τ(h (2),g (1))(h (3)g (2) Vv)τ(h (4)g (3),k (1))h (5)g (4)k (2) =(u.(h (1) Mm).τ(h (2),g (1))h (3)g (2)).(vk) =((uh).(mg)).(vk).\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}

Last revised on May 9, 2024 at 15:53:15. See the history of this page for a list of all contributions to it.