# Hopf BiGalois Theory for Hopf Algebroids

Xiao Han, Peter Schauenburg

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Xiao Han, Peter Schauenburg. Hopf BiGalois Theory for Hopf Algebroids. 2025. hal-05320217

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Distributed under a Creative Commons Attribution 4.0 International License
HOPF BIGALOIS THEORY FOR HOPF ALGEBROIDS

XIAO HAN AND PETER SCHAUENBURG

ABSTRACT. We develop a theory of Hopf BiGalois extensions for Hopf algebroids. We understand these to be left bialgebroids (whose left module categories are monoidal categories) fulfilling a condition that is equivalent to being Hopf in the case of ordinary bialgebras, but does not entail the existence of an antipode map.

The immediate obstacle to developing a full biGalois theory for such Hopf algebroids is simple: The condition to be a left Hopf Galois extension can be defined in complete analogy to the Hopf case, but the Galois map for a right comodule algebra is not a well defined map. We find that this obstacle can be circumvented using bialgebroids fulfilling a condition that still does not entail the existence of an antipode, but is equivalent, for ordinary bialgebras, to being Hopf with bijective antipode.

The key technical tool is a result of Chemla [6] allowing to switch left and right
comodule structures under flatness conditions much like one would do using an an-
tipode. Using this, we arrive at a left-right symmetric theory of biGalois extensions,
including the construction of an Ehresmann Hopf algebroid making a one-sided Hopf-
Galois extension into a biGalois one.

Moreover, we apply a more general 2-cocycle twist theory[15] to Ehresmann Hopf
algebroids. As the 2-cocycle is only left $\bar{B}$-linear the base algebra is also twisted. We
also study the Ehresmann Hopf algebroids of quantum Hopf fibrations and quantum
homogeneous space as examples.

1. INTRODUCTION

Similarly to the relation between Hopf algebras and groups, Hopf algebroids can be
viewed as a quantization of groupoids. There are several different definitions of Hopf
algebroids, in this paper, we will consider the Hopf algebroids introduced in [23] and
[25]. In particular, our Hopf algebroids do not necessarily have antipodes, but we require
the bijectivity of certain maps that would be equivalent to the existence of a bijective
antipode in the case of bialgebras over a commutative ring. Generalizations of these
maps also define the notion of Hopf-Galois extension.

For a left bialgebroid $\mathcal{L}$ over a noncommutative ring $B$, a left $\mathcal{L}$-Galois extension can be defined in a natural way: The definition requires bijectivity of the "Galois map" which can be written down by the same formula as in the ordinary Hopf case, although now not only the source but also the target is a tensor product over a possibly noncommutative algebra rather than the base field. However, there is no obvious definition of a Hopf Galois extension on the right-hand side; the direct analog of the canonical map in the definition of "Galois" for the ordinary Hopf case is not well defined in the Hopf algebroid case. This makes it impossible to arrive at a naive direct generalization of the notion of Hopf biGalois extension as in [27] to the Hopf algebroid setting.

To deal with this problem we rather study anti-right $\mathcal{L}$-Galois extensions. A short definition of this notion is that $P$ is an anti-right $\mathcal{L}$-Galois extension iff it is a left $\mathcal{L}^{\text{cop}}$-Galois extension over the co-opposite bialgebroid. If $\mathcal{L}$ is an ordinary Hopf algebra with bijective antipode, it was already observed in [22] that the Galois and anti-Galois

<footnote n="2020">2020 Mathematics Subject Classification. 16T05,16T15,18M05,18M15.</footnote>

<footnote>Key words and phrases. Hopf algebroid, bialgebroid, quantum group, Hopf Galois extensions.</footnote>

<footnote>For the purpose of Open Access, a CC-BY public copyright licence has been applied by the author
to the present document and will be applied to all subsequent versions up to the Author Accepted
Manuscript arising from this submission.</footnote>
conditions are equivalent; in the Hopf algebroid case however, only one of them is defined for left comodule algebras, and only the other one is defined for right comodule algebras.

As a regular left comodule of itself, $\mathcal{L}$ is a left Hopf algebroid, if and only if $\bar{B} \subseteq \mathcal{L}$ is a left $\bar{B}$-Galois extension. In the case of ordinary bialgebras this condition is equivalent to the existence of an antipode. Similarly, as a regular right $\mathcal{L}$-comodule, $\mathcal{L}$ is an anti-left Hopf algebroid if and only if $B \subseteq \mathcal{L}$ is an anti-right $\mathcal{L}$-Galois extension. In the ordinary bialgebra case this is equivalent to the existence of a skew antipode. In particular the condition on a bialgebroid to be both a left Hopf algebroid and an anti-left Hopf algebroid is equivalent, in the ordinary bialgebra case, to having a bijective antipode. We will also see that under some faithful flatness conditions, a bialgebroid admitting a left Galois (resp. anti-right Galois) extension is necessarily (anti-)Hopf, similarly to the classical case.

For a classical Hopf algebra $H$ with bijective antipode and a left $H$-comodule $P$, one can always make $P$ into a right $H$ comodule, by defining the right coaction $\delta : p \to p_{(0)} \otimes S^{-1}(p_{(-1)})$. In fact this construction can be generalized to Hopf algebroids in absence of an antipode (and under some mild flatness conditions). This was shown in [7]; the general result best suited to our purposes is due to Chemla [6].

A consistent use of this side-switching trick will allow us to establish a full and left-right symmetric theory of Hopf biGalois extensions, always assuming our Hopf algebroids fulfill the analogous condition to having a bijective antipode (while not having an antipode), and suitable (faithful) flatness conditions over the base. For the rest of this introduction we will not mention these important conditions, nor the (faithful) flatness conditions needed frequently on our extensions and Hopf algebroids to make things work.

The facts on one-sided Hopf Galois extensions that we need will be established in section 3, including a study of the analogs of Schneider's structure theorems on Hopf modules, and an extensive discussion of the technical details of switching sides as alluded to above.

Section 4 contains the main general results on biGalois theory. In particular, following a result on the structure of Hopf bimodules based on the structure theorems obtained in section 3, we proceed to construct an Ehresmann Hopf algebroid for any one-sided Galois extension over a Hopf algebroid. The Ehresmann Hopf algebroid makes the one-sided Galois extension into a biGalois extension (which we define) in the essentially unique way. We prove that biGalois extensions form a groupoid under cotensor product, with biGalois extensions giving rise to equivalences between comodule categories over Hopf algebroids.

In section 5 we prove several structure theorems on Hopf bimodules based on the generalizations of Schneider's structure theorem. In particular, Hopf modules with two module structures over a Galois extension are classified (as a monoidal category) as modules over the Ehresmann Hopf algebroid, and Hopf modules with two module structures over a biGalois extension and two comodule structures over the left and right Hopf algebroids correspond to Yetter-Drinfeld modules. This generalizes the structure theorems for Hopf modules over a Hopf algebroid that appeared in [8] while the authors were finalizing the present paper. The paper [8] also discusses the braidings we discuss here.

In Section 6, we apply a new 2-cocycle twist theory[15] to the Ehresmann Hopf algebroid. More precisely, compared with the twist theory in [2] and [13], a 2-cocycle in the new theory is only left $\bar{B}$-linear (the 2-cocycle in [2] is left $B^e$-bilinear). This results in a deformed Hopf algebroid whose base is also deformed. Moreover, the coproduct and product are simultaneously deformed. Here we show Hopf Galois extensions can be twisted by a 2-cocycle with a deformed coaction and product. In particular, we show for any Hopf algebra $H$, if a $H$-Galois extension $B \subseteq P$ has an extra symmetry (left coaction which cocommutes with the right $H$-coaction) associated to a Hopf algebra $K$, then a 2-cocycle $\gamma$ on $K$ induces a 2-cocycle $\Gamma$ on the corresponding Ehresmann Hopf algebroid
$L(P, H)$. Moreover $L(P, H)^{\Gamma} \cong L(\gamma P, H)$, where $\gamma B \subseteq \gamma P$ is another H-Galois extension invented in [1] with a deformed base. Finally, we present two examples of this setup. The first one is a deformation of Hopf fibration $S^7 \to S^4$, the second one is quantum homogeneous spaces.

**Acknowledgements.** Xiao Han was supported by the European Union's Horizon 2020 research and innovation program under the Marie Skłodowska-Curie grant agreement No 101027463 and Leverhulme Trust project grant RPG-2024-177. This article is also based upon work from COST Action CaLISTA CA21109 supported by COST (European Cooperation in Science and Technology). Xiao Han is grateful to the Institut de Mathématiques de Bourgogne (Dijon) for hospitality.

## 2. BASIC ALGEBRAIC PRELIMINARIES

In this section, we will recall some basic definitions and notation. Let $B$ be an unital algebra over a field $k$. We denote the opposite algebra by $\overline{B}$ and let $B \to \overline{B}$, $b \mapsto \overline{b}$ for any $b \in B$ be the obvious $k$-algebra antiisomorphism. Define $B^e := B \otimes \overline{B}$, so $B$ and $\overline{B}$ are obvious subalgebras of $B^e$. Let $M, N$ be $B^e$-bimodules. We define

$$M \diamond_{B} N := \int_{b} \overline{b} M \otimes_{b} N := M \otimes N / \langle \overline{b}m \otimes n - m \otimes bn | b \in B, m \in M, n \in N \rangle$$

$$M \otimes_{B} N := \int_{b} M_{b} \otimes_{b} N := M \otimes N / \langle mb \otimes n - m \otimes bn | b \in B, m \in M, n \in N \rangle$$

$$M \otimes_{\overline{B}} N := \int_{b} M_{\overline{b}} \otimes_{\overline{b}} N := M \otimes N / \langle m\overline{b} \otimes n - m \otimes \overline{bn} | b \in B, m \in M, n \in N \rangle$$

For convenience, we also define $N \otimes^{B} M = \int_{b} bN \otimes M_{b}$ and $N \otimes^{\overline{B}} M = \int_{b} \overline{b}N \otimes M_{\overline{b}}$. Moreover, we define

$$\int^{b} M_{\overline{b}} \otimes N_{b} := \left\{ \sum_{i} m_{i} \otimes n_{i} \in M \otimes N \mid m_{i} \overline{b} \otimes n_{i} = m_{i} \otimes n_{ib}, \forall b \in B \right\}.$$

Although a subspace in a tensor product like the one above is usually not spanned by simple tensors (so that the element in question was written as a sum above), we will usually abuse notations and write such an element in the form $m \otimes n$ just the same, although it is not the simple tensor coming from single elements $m$ and $n$.

The symbols $\int^{b}$ and $\int^{c}$ commute, also $\int_{b}$ and $\int_{c}$ commute. However, in general, the symbol $\int^{b}$ and $\int_{c}$ doesn't commute. For any $\overline{B}$-bimodule $M$ and any $B$-bimodule $N$, we also define

$$M \times_{B} N := \int^{a} \int_{b} M \overline{a} \otimes_{b} N_{a}.$$

$M \times_{B} N$ is called Takeuchi product of $M$ and $N$. If $P$ is a $B^e$-bimodule, then $P \times_{B} N$ is a $B$-bimodule with $B$ acting on $P$. Similarly, $M \times_{B} P$ is a $\overline{B}$-bimodule with $\overline{B}$ acting on $P$. If both $M$ and $N$ are $B^e$-bimodule, then $M \times N$ is also a $B^e$-bimodule. However, the product $\times_{B}$ is neither associative nor unital on the category of $B^e$-bimodules. For any $M, N, P \in B^e M_{B^e}$, we can define

$$M \times_{B} P \times_{B} N := \int^{a,b} \int_{c,d} \overline{c} M \overline{a} \otimes_{c,d} P_{a,b} \otimes_{d} N_{b},$$

where $\int^{a,b} := \int^{a} \int^{b}$ and $\int_{c,d} := \int_{c} \int_{d}$. There are maps

$\alpha : (M \times_{B} P) \times_{B} N \to M \times_{B} P \times_{B} N, \quad m \otimes p \otimes n \mapsto m \otimes p \otimes n,$

$\alpha' : M \times_{B} (P \times_{B} N) \to M \times_{B} P \times_{B} N, \quad m \otimes p \otimes n \mapsto m \otimes p \otimes n.$
Notice that neither $\alpha$ nor $\alpha'$ are isomorphisms in general. We note that $\alpha$ is an isomorphism if $N$ is flat as a left $B$-module, and $\alpha'$ is an isomorphism if $M$ is flat as a left $\bar{B}$-module. In many instances in the paper, some of the spaces we use in Takeuchi products will be faithfully flat as left modules over $B$ and $\bar{B}$ (either by assumption or because we can prove that this is the case), which makes iterated Takeuchi products somewhat well behaved.

**2.1. Hopf algebroids.** Here we recall the basic definitions (cf. [5], [2]). Let $B$ be a unital algebra over a field $k$. A $B$-ring means a unital algebra in the monoidal category $B\mathcal{M}_B$ of $B$-bimodules. Likewise, a $B$-coring is a coalgebra in $B\mathcal{M}_B$. Morphisms of $B$-(co)rings are defined to be morphisms of (co)algebras, but in the category $B\mathcal{M}_B$.

Specifying a unital $B$-ring $\mathcal{L}$ is equivalent to specifying a unital algebra $\mathcal{L}$ (over $k$) and an algebra map $\eta: B \to \mathcal{L}$. Left and right multiplication in $\mathcal{L}$ pull back to left and right $B$-actions as a bimodule (so $bXc = \eta(b)X\eta(c)$ for all $b, c \in B$ and $X \in \mathcal{L}$) and the product descends to the product $\mu_B: \mathcal{L} \otimes_B \mathcal{L} \to \mathcal{L}$ with $\eta$ the unit map. Conversely, given $\mu_B$ we can pull back to an associative product on $\mathcal{L}$ with unit $\eta(1)$.

Now suppose that $s: B \to \mathcal{L}$ and $t: \bar{B} \to \mathcal{L}$ are algebra maps with images that commute. Then $\eta(b \otimes c) = s(b)t(c)$ is an algebra map $\eta: B^e \to \mathcal{L}$, where $B^e = B \otimes \bar{B}$, and is equivalent to making $\mathcal{L}$ a $B^e$-ring. The left $B^e$-action part of this is equivalent to a $B$-bimodule structure

$$b.X.c = b\bar{c}X := s(b)t(c)X \qquad (2.1)$$

for all $b, c \in B$ and $X \in \mathcal{L}$. Similarly, the right $B^e$-action part of this is equivalent to another $B$-bimodule structure

$$c.X.b = Xb\bar{c} := Xs(b)t(c). \qquad (2.2)$$

If $\mathcal{L}$ and $\mathcal{R}$ are two $B^e$-rings, then the Takeuchi product $\mathcal{L} \times_B \mathcal{R}$ is an algebra with multiplication given by $(\ell \otimes r)(\ell' \otimes r') = \ell\ell' \otimes rr'$ for $\ell \otimes r, \ell' \otimes r' \in \mathcal{L} \times_B \mathcal{R}$.

**Definition 2.1.** Let $B$ be a unital algebra. A left $B$-bialgebroid (or left bialgebroid over $B$) is an algebra $\mathcal{L}$ and ('source' and 'target') commuting algebra maps $s: B \to \mathcal{L}$ and $t: \bar{B} \to \mathcal{L}$ (thus making $\mathcal{L}$ a $B^e$-ring), such that

(i) There are two left $B^e$-module maps, the coproduct $\Delta: \mathcal{L} \to \mathcal{L} \times_B \mathcal{L}$ and counit $\varepsilon: \mathcal{L} \to B$ satisfying

$$\alpha \circ (\Delta \times_B \mathrm{id}) \circ \Delta = \alpha' \circ (\mathrm{id} \times_B \Delta) \circ \Delta : \mathcal{L} \to \mathcal{L} \times_B \mathcal{L} \times_B \mathcal{L} \\ \varepsilon(X_{(1)})X_{(2)} = \overline{\varepsilon(X_{(2)})}X_{(1)} = X,$$

for any $X \in \mathcal{L}$, where we use the sumless Sweedler index to denote the coproduct, namely, $\Delta(X) = X_{(1)} \otimes X_{(2)}$.

(ii) The coproduct is an algebra map. The counit $\varepsilon$ is a left character in the following sense:

$$\varepsilon(1_{\mathcal{L}}) = 1_B, \quad \varepsilon(X\varepsilon(Y)) = \varepsilon(XY) = \varepsilon(X\overline{\varepsilon(Y)})$$

for all $X, Y \in \mathcal{L}$ and $a \in B$.

**Remark 2.2.** Condition (i) implies $\mathcal{L}$ is a $B$-coring with the $B$-bimodule structure given by (2.1). By condition (ii), we can define an algebra map $\hat{\varepsilon}: \mathcal{L} \to \mathrm{End}(B)$ by $\hat{\varepsilon}(X)(b) := \varepsilon(Xb)$. Moreover, the image of $\alpha \circ (\Delta \times_B \mathrm{id}) \circ \Delta = \alpha' \circ (\mathrm{id} \times_B \Delta) \circ \Delta$ could be also denoted by the Sweedler index, namely, $\alpha \circ (\Delta \times_B \mathrm{id}) \circ \Delta(X) = \alpha' \circ (\mathrm{id} \times_B \Delta) \circ \Delta(X) = X_{(1)} \otimes X_{(2)} \otimes X_{(3)} \in \mathcal{L} \times_B \mathcal{L} \times_B \mathcal{L}$. Unlike the bialgebra case, the elements $X_{(1)(1)} \otimes X_{(1)(2)} \otimes X_{(2)}$, $X_{(1)} \otimes X_{(2)(1)} \otimes X_{(2)(2)}$ and $X_{(1)} \otimes X_{(2)} \otimes X_{(3)}$ are not the same, since they belong to three different vector spaces, even if in many interesting cases both $\alpha$ and $\alpha'$ are isomorphisms. Morphisms between left $B$-bialgebroids are $B$-coring maps which are also $B^e$-ring maps.
**Definition 2.3.** A left bialgebroid $\mathcal{L}$ over $B$ is a left Hopf algebroid ([23], *Thm and Def 3.5.) if

$$ \lambda : \mathcal{L} \otimes_{\bar{B}} \mathcal{L} \to \mathcal{L} \diamond_{\mathcal{B}} \mathcal{L}, \quad \lambda(X \otimes Y) = X_{(1)} \otimes X_{(2)} Y $$

is invertible.

Similarly, A left bialgebroid $\mathcal{L}$ over $B$ is a anti-left Hopf algebroid if

$$ \mu : \mathcal{L} \otimes^{\mathcal{B}} \mathcal{L} \to \mathcal{L} \diamond_{\mathcal{B}} \mathcal{L}, \quad \mu(X \otimes Y) = Y_{(1)} X \otimes Y_{(2)} $$

is invertible. A left B-bialgebroid is a B-Hopf algebroid if it is a left Hopf algebroid and anti-left Hopf algebroid. For convenience, we also define $\hat{\mu} : \mathcal{L} \otimes_{\mathcal{B}} \mathcal{L} \to \mathcal{L} \diamond_{\mathcal{B}} \mathcal{L}$ by $\hat{\mu}(X \otimes Y) = X_{(1)}Y \diamond_{\mathcal{B}} X_{(2)}$.

In the following, we will always use the balanced tensor product explained as above. If $B = k$ then this reduces to the map $\mathcal{L} \otimes \mathcal{L} \to \mathcal{L} \otimes \mathcal{L}$ given by $h \otimes g \mapsto h_{(1)} \otimes h_{(2)}g$ which for a usual bialgebra has an inverse, namely $h \otimes g \mapsto h_{(1)} \otimes (Sh_{(2)})g$ if and only if there is an antipode. We adopt the shorthand

$$ X_+ \otimes_{\bar{B}} X_- := \lambda^{-1}(X \diamond_{\mathcal{B}} 1) \qquad (2.3) $$

$$ X_{[-]} \otimes^{\mathcal{B}} X_{[+]} := \mu^{-1}(1 \diamond_{\mathcal{B}} X). \qquad (2.4) $$

So for a Hopf algebra $H$, $h_+ \otimes h_- = h_{(1)} \otimes S(h_{(2)})$ and $h_{[-]} \otimes h_{[+]} = S^{-1}(h_{(1)}) \otimes h_{(2)}$ for any $h \in H$. We recall from [23, Prop. 3.7] that for a left Hopf algebroid, and any $X, Y \in \mathcal{L}$ and $a, a', b, b' \in B$,

$$ X_{+(1)} \diamond_{\mathcal{B}} X_{+(2)} X_{-} = X \diamond_{\mathcal{B}} 1; \qquad (2.5) $$

$$ X_{(1)+} \otimes_{\bar{B}} X_{(1)-} - X_{(2)} = X \otimes_{\bar{B}} 1; \qquad (2.6) $$

$$ (XY)_+ \otimes_{\bar{B}} (XY)_- = X_+ Y_+ \otimes_{\bar{B}} Y_- X_-; \qquad (2.7) $$

$$ 1_+ \otimes_{\bar{B}} 1_- = 1 \otimes_{\bar{B}} 1; \qquad (2.8) $$

$$ X_{+(1)} \diamond_{\mathcal{B}} X_{+(2)} \otimes_{\bar{B}} X_{-} = X_{(1)} \diamond_{\mathcal{B}} X_{(2)} + \otimes_{\bar{B}} X_{(2)-}; \qquad (2.9) $$

$$ X_+ \otimes X_{-(1)} \otimes X_{-(2)} = X_{++} \otimes X_- \otimes X_{+-} \in \int_{c,d}^{a,b} \int_{\bar{a} \mathcal{L} \bar{c} \otimes \bar{d} \mathcal{L} \bar{b} \otimes \bar{c,d} \mathcal{L}_{b,\bar{a}}}; \qquad (2.10) $$

$$ X = X_+ \overline{\varepsilon(X_-)}; \qquad (2.11) $$

$$ X_+ X_- = \varepsilon(X); \qquad (2.12) $$

$$ (a\bar{a}'Xb\bar{b}')_+ \otimes_{\bar{B}} (a\bar{a}'Xb\bar{b}')_- = aX_+ b \otimes_{\bar{B}} b'X_- a'; \qquad (2.13) $$

$$ \bar{b}X_+ \otimes_{\bar{B}} X_- = X_+ \otimes_{\bar{B}} X_- \bar{b}. \qquad (2.14) $$

Similarly, for an anti-left Hopf algebroid, from [4] we have

$$ X_{[+](1)} X_{[-]\diamond_{\mathcal{B}} X_{[+](2)}} = 1 \diamond_{\mathcal{B}} X; \qquad (2.15) $$

$$ X_{(2)[-]} X_{(1)} \otimes^{\mathcal{B}} X_{(2)[+]} = 1 \otimes^{\mathcal{B}} X; \qquad (2.16) $$

$$ (XY)_{[-]} \otimes^{\mathcal{B}} (XY)_{[+]} = Y_{[-]} X_{[-]} \otimes^{\mathcal{B}} X_{[+]} Y_{[+]}; \qquad (2.17) $$

$$ 1_{[-]} \otimes^{\mathcal{B}} 1_{[+]} = 1 \otimes^{\mathcal{B}} 1; \qquad (2.18) $$

$$ X_{[-]} \otimes^{\mathcal{B}} X_{[+](1)} \diamond_{\mathcal{B}} X_{[+](2)} = X_{(1)[-]} \otimes^{\mathcal{B}} X_{(1)[+]} \diamond_{\mathcal{B}} X_{(2)}; \qquad (2.19) $$

$$ X_{[-](1)} \otimes X_{[-](2)} \otimes X_{[+]} = X_{[+)[-]} \otimes X_{[-]} \otimes X_{[+][+]} \in \int_{c,d}^{a,b} \int_{c,\bar{d}} \mathcal{L}_{a,\bar{b}} \otimes \mathcal{L}_{\bar{d}b} \otimes \mathcal{L}_{a\bar{c}}; \qquad (2.20) $$

$$ X = X_{[+]} \varepsilon(X_{[-]}); \qquad (2.21) $$

$$ X_{[+]} X_{[-]} = \overline{\varepsilon(X)}; \qquad (2.22) $$

$$ (a\bar{a}'Xb\bar{b}')_{[-]} \otimes^{\mathcal{B}} (a\bar{a}'Xb\bar{b}')_{[+]} = \bar{b}X_{[-]} \bar{a} \otimes^{\mathcal{B}} \bar{a}'X_{[+]} \bar{b}'; \qquad (2.23) $$
$$X_{[-]} b \otimes^{B} X_{-} = X_{[+]} \otimes^{B} b X_{-}. \qquad (2.24)$$

**Proposition 2.4.** If $\mathcal{L}$ is a left and anti-left Hopf algebroid, then we have

$$X_{[+]} \otimes X_{[-]+} \otimes X_{[-]-} = X_{(2)[+]} \otimes X_{(2)[-]} \otimes X_{(1)} \in \mathcal{L} \otimes_B \mathcal{L} \otimes_{\bar{B}} \mathcal{L};$$

$$X_{+} \otimes X_{-[+]} \otimes X_{[-]} = X_{(1)+} \otimes X_{(1)-} \otimes X_{(2)} \in \mathcal{L} \otimes_{\bar{B}} \mathcal{L} \otimes_B \mathcal{L};$$

$$X_{[+]} \otimes X_{[-]} \otimes X_{[+]} = X_{+[+]} \otimes X_{+[]} \otimes X_{-} \in \int^{c,d} \int_{a,b} c, \bar{d} \mathcal{L}_{\bar{a},b} \otimes b \mathcal{L}_c \otimes \bar{a} \mathcal{L}_{\bar{d}},$$

for any $X \in \mathcal{L}$.

*Proof.* The first equality can be proved by comparing the result of $\mathrm{id}_{\mathcal{L}} \otimes \lambda$ on both sides. The second equality can be proved by comparing the result of $\mathrm{id}_{\mathcal{L}} \otimes \mu$ on both sides. The third equality can be proved by comparing the result of $\mu \otimes \mathrm{id}_{\mathcal{L}}$ on both sides. $\square$

### 3. HOPF GALOIS EXTENSIONS OF HOPF ALGEBROIDS

In this section we will collect some results on (one-sided) Hopf-Galois extensions over Hopf algebroids, mostly in preparation for our work on biGalois theory. An important topic will be the question as to how one can switch between left and right Galois extensions. In the classical (Hopf algebra) theory this would simply be done by using the antipode. Here, this is not available, and also, the notion of a right Galois extension is not immediately well-defined.

We circumvent the latter problem by working with what we call "anti-right" Galois extensions. As to the problem of "switching sides" we lean heavily on an observation originating in [7]; the result we will use is in [6]. Namely, there is a way to switch between left and right comodule structures that are "regular" which is analogous to switching using a bijective antipode in the Hopf case. Moreover, under mild conditions, if our Hopf algebroid fulfills a condition analogous to having bijective antipode (namely, the bialgebroid and its coopposite need to be Hopf), then comodules are always regular.

**3.1. (Co)modules of left bialgebroids and regularity.** The following observation was made in [23]; with some technical modification it in fact characterizes bialgebroids:

*Remark 3.1.* Let $\mathcal{L}$ be a left bialgebroid over $B$. A left $\mathcal{L}$-module $M$ is a $B$-bimodule $M$ by

$$bm := s_L(b) \triangleright m, \quad mb := t_L(b) \triangleright m$$

for $b \in B$ and $m \in M$.

Thus, we have a functor $\mathcal{L}\mathcal{M} \to \mathcal{B}\mathcal{M}_B$. The category $\mathcal{L}\mathcal{M}$ is monoidal with tensor product of $M, N \in \mathcal{L}\mathcal{M}$ given by $M \otimes_B N$ with module structure

$$X \triangleright (mn) = (X_{(1)} \triangleright m) \otimes (X_{(2)} \triangleright n)$$

for any $m \in M, n \in N$.

Now the following definition of module algebra is natural.

**Definition 3.2.** Let $\mathcal{L}$ be a left bialgebroid over $B$. A left $\mathcal{L}$-module algebra is a left algebra in the monoidal category $\mathcal{L}\mathcal{M}$. Thus, it is a $B$-ring with a $\mathcal{L}$-module structure such that the $B$-bimodule structure of the $B$-ring coincides with that of the $\mathcal{L}$-module, and

$$X \triangleright b = \varepsilon(X b), \quad X \triangleright (p q) = (X_{(1)} \triangleright p)(X_{(2)} \triangleright q)$$

for any $p, q \in P$.

*One* (see remark below) suitable definition of comodules over a bialgebroid is given in [23].
**Definition 3.3.** Let $\mathcal{L}$ be a left bialgebroid over $B$. A left $\mathcal{L}$ comodule is a $B$-bimodule $\Gamma$, together with a $B$-bimodule map $L\delta : \Gamma \to \mathcal{L} \times_B \Gamma \subseteq \mathcal{L} \diamond_B \Gamma$, written $L\delta(p) = p_{(-1)} \otimes p_{(0)}$ ($L\delta$ is a $B$-bimodule map in the sense that $L\delta(bpb') = bp_{(-1)}b' \otimes p_{(0)}$), such that

$\alpha \circ (\Delta \times_B \mathrm{id}_\Gamma) \circ L\delta = \alpha' \circ (\mathrm{id}_\mathcal{L} \times_B L\delta) \circ L\delta : \Gamma \to \mathcal{L} \times_B \mathcal{L} \times_B \Gamma, p \mapsto p_{(-2)} \otimes p_{(-1)} \otimes p_{(0)}$ \\
$\varepsilon(p_{(-1)})p_{(0)} = p,$

The category $\mathcal{L}\mathcal{M}$ of left comodules is a monoidal category under the tensor product over $B$ with the codiagonal comodule structure

$$ \Gamma \otimes_B \Lambda \ni p \otimes q \mapsto p_{(-1)}q_{(-1)} \otimes p_{(0)} \otimes q_{(0)} \in \mathcal{L} \diamond_B (\Gamma \otimes_B \Lambda) $$

A right $\mathcal{L}$-comodule is a $\overline{B}$-bimodule $\Gamma$, together with a $\overline{B}$-bimodule map $\delta_L : \Gamma \to \Gamma \times_B \mathcal{L} \subseteq \Gamma \diamond_B \mathcal{L}$, written $\delta_L(p) = p_{(0)} \diamond_B p_{(1)}$ ($\delta_L$ is a $\overline{B}$-bimodule map in the sense that $\delta_L(\overline{b}p\overline{b}') = p_{(0)} \otimes \overline{b}p_{(1)}\overline{b}'$), such that

$$ \alpha \circ (\delta_L \times_B \mathrm{id}_\mathcal{L}) \circ \delta_L = \alpha' \circ (\mathrm{id}_\Gamma \times_B \Delta) \circ \delta_L : \Gamma \to \Gamma \times_B \mathcal{L} \times_B \mathcal{L}, p \mapsto p_{(0)} \otimes p_{(1)} \otimes p_{(2)} \\ \overline{\varepsilon(p_{(1)})p_{(0)}} = p, $$

for any $p \in \Gamma$.

**Remark 3.4.** Since a left bialgebroid $\mathcal{L}$ is in particular a coring over $B$, there is already a notion of comodule over $\mathcal{L}$. By definition, such a comodule over the coring $\mathcal{L}$ has just one $B$-module structure. As proved in [Section 1.4 of [21]], the two notions coincide: A left comodule over $\mathcal{L}$ in the sense of the definition above is quite obviously a left comodule over the coring $\mathcal{L}$, but also given a comodule over the coring $\mathcal{L}$, there is a unique second $B$-module structure which makes it a comodule in the bialgebroid sense.

**Remark 3.5.** For $\Gamma \in \mathcal{L}\mathcal{M}$ we have

$$ \mathrm{co}\mathcal{L}\Gamma := \{p \in \Gamma | \delta(p) = 1 \otimes p\} \subseteq \{p \in \Gamma | bp = pb \forall b \in B\} =: \Gamma^B $$

**Remark 3.6.** For a left $B$-bialgebroid $\mathcal{L}$, the coopposite $\mathcal{L}^{\mathrm{cop}}$ is a left $\overline{B}$-bialgebroid (while the opposite algebra of $\mathcal{L}$ does not have a left bialgebroid structure in general). The monoidal category of right (left) comodules over $\mathcal{L}$ is naturally isomorphic to the category of left (right) comodules over $\mathcal{L}^{\mathrm{cop}}$. In this way any notion or fact on left coactions of a left bialgebroid gives rise to a corresponding notion with equivalent properties for right coactions, and vice versa. For example the category $\mathcal{M}^\mathcal{L}$ of right comodules is monoidal with the tensor product of comodules taken over $\overline{B}$. Note also that the category of left $\mathcal{L}^{\mathrm{cop}}$-modules is monoidally equivalent to the category of $\mathcal{L}$-modules with opposite tensor product: $(\mathcal{L}^{\mathrm{cop}})\mathcal{M} \cong (\mathcal{L}\mathcal{M})^{\mathrm{sym}}$.

**Definition 3.7.** Let $\mathcal{L}$ be a left bialgebroid over $B$, A right $\mathcal{L}$-comodule $\Gamma$ is a regular right $\mathcal{L}$-comodule [[6, Definition 3.0.1]], if

$$ \psi : \Gamma \otimes_{\overline{B}} \mathcal{L} \to \Gamma \diamond_B \mathcal{L}, \quad \psi(p \otimes_{\overline{B}} X) = p_{(0)} \diamond_B p_{(1)} X $$

is invertible. We adopt the shorthand

$$ p_{[0]} \otimes_{\overline{B}} p_{[1]} = \psi^{-1}(p \diamond_B 1). \qquad (3.1) $$

so that $\psi^{-1}(p \diamond_B X) = p_{[0]} \otimes p_{[1]} X$.

We will call a left $\mathcal{L}$-comodule regular if it is regular as a right $\mathcal{L}^{\mathrm{cop}}$-comodule.

Thus a left comodule $\Gamma$ is regular if

$$ \phi : \mathcal{L} \otimes^B \Gamma \to \mathcal{L} \diamond_B \Gamma, \quad \phi(X \otimes p) = p_{(-1)}X \otimes p_{(0)} $$

is bijective. We note

$$ p_{[1]} \otimes^B p_{[0]} = \phi^{-1}(1 \diamond_B p). \qquad (3.2) $$
**Remark 3.8.** The definition of a regular right comodule is such that $\mathcal{L}$ is regular as a right comodule over itself if and only if it is a left Hopf algebroid. If $B = k$ and $\mathcal{L}$ is a Hopf algebra, then any right $\mathcal{L}$-comodule $\Gamma$ is regular and $\forall p \in \Gamma$ we have $p_{[0]} \otimes p_{[-1]} = p_{(0)} \otimes S(p_{(1)})$. By results of Chemla this can be maintained in absence of an antipode in the Hopf algebroid case under mild assumptions on the module structures of $\mathcal{L}$.

In the following proposition, we will see the properties of the inverse of the map $\phi$ generalize the properties of the inverse of the anti-Galois map.

**Proposition 3.9** ([6] Proposition 3.0.6). Let $\Gamma$ be a regular right comodule of a left $B$-bialgebroid $\mathcal{L}$, then we have

$$p_{[0] \diamond B} p_{[0] (1)} p_{[-1]} = 1 \diamond_B p, \qquad (3.3)$$

$$p_{(0)[0]} \otimes_{\bar{B}} p_{(0)[-1]} p_{(1)} = p \otimes_{\bar{B}} 1, \qquad (3.4)$$

$$(\bar{b}'p\bar{b})_{[0]} \otimes_{\bar{B}} (\bar{b}'p\bar{b})_{[-1]} = p_{[0]} \otimes_{\bar{B}} bp_{[-1]} b', \qquad (3.5)$$

$$\bar{b} p_{[0]} \otimes_{\bar{B}} p_{[-1]} = p_{[0]} \otimes_{\bar{B}} p_{[-1]} \bar{b}, \qquad (3.6)$$

Moreover, $\Gamma$ is a left comodule of $\mathcal{L}$, with $B$-bimodule structure $b'pb := \bar{b}p\bar{b}'$ and coaction $\mathcal{L}\delta(p) := p_{[-1]} \otimes p_{[0]}$. We will refer to this comodule structure as the reverse of the given one.

**Remark 3.10.** If we identify, as indicated, the $\bar{B}$-bimodule structure of a right $\mathcal{L}$-comodule with a $B$-bimodule structure, then (3.4) can be written

$$p_{(0)[-1]} p_{(1)} \diamond_B p_{(0)[0]} = 1 \diamond_B p \in \mathcal{L} \diamond_B \Gamma. \qquad (3.7)$$

As usual, there is a version of the above for regular left comodules. In the form (3.3). (3.7) of the condition that defines a regular right comodule and its reverse, it is obvious that the same condition describes regularity of the left comodule, and its reverse would be the structure we started from. Thus we have a bijection between right regular comodule structures and left regular comodule structures on the same space, given by taking the respective reverse.

We only note the rules for passing “from left to right” for later use:

**Proposition 3.11.** ([6] Proposition 3.0.6) Let $\Gamma$ be a regular left comodule of a left $B$-bialgebroid $\mathcal{L}$, then we have

$$p_{[0](-1)} p_{[1]} \diamond_B p_{[0](0)} = 1 \diamond_B p, \qquad (3.8)$$

$$p_{(0)[1]} p_{(-1)} \otimes^B p_{(0)[0]} = 1 \otimes^B p, \qquad (3.9)$$

$$(bpb')_{[1]} \otimes^B (bpb')_{[0]} = \bar{b}' p_{[1]} \bar{b} \otimes^B p_{[0]}, \qquad (3.10)$$

$$p_{[1]} b \otimes^B p_{[0]} = p_{[1]} \otimes^B bp_{[0]} \qquad (3.11)$$

Moreover, $\Gamma$ is a right comodule of $\mathcal{L}$, with coaction $\delta_L(p) := p_{[0]} \otimes p_{[1]}$, for any $p \in \Gamma$ and $b, b' \in B$; we will call this the reverse of the given coaction.

**Lemma 3.12.** ([6, Thm.3.0.4, Prop.3.0.6] ) If $\mathcal{L}$ is a left $B$-Hopf algebroid such that $\mathcal{L}$ is flat as a left $B$-module and left $\bar{B}$-module, then any right $\mathcal{L}$-comodule $\Gamma$ is regular. Moreover, we have

$$p_{[0](0)} \otimes p_{[0](1)} \otimes p_{[-1]} = p_{(0)} \otimes p_{(1)} + \otimes p_{(1)} - \in \int_{b,c}^{a,d} \int_{\bar{b}\Gamma\bar{a} \otimes b, \bar{d}\mathcal{L}_{a,\bar{c}} \otimes \bar{c}\mathcal{L}_{\bar{d}}} \qquad (3.12)$$

Similarly, if $\mathcal{L}$ is left anti-Hopf with the same flatness conditions, then every left $\mathcal{L}$-comodule is regular, and

$$p_{[1]} \otimes^B p_{[0](-1)} \diamond_B p_{[0](0)} = p_{(-1)[-]} \otimes^B p_{(-1)[+]} \diamond_B p_{(0)} \in \int_{b,c}^{a,d} \int_{b}^{b} \mathcal{L}_{a} \otimes_{a,\bar{c}} \mathcal{L}_{b,\bar{d}} \otimes_{c} \Gamma_{d}, \qquad (3.13)$$
for any $p \in \Gamma$.

We recall the definition of cotensor product: Let $P$ be a right comodule of a left $B$-bialgebroid $\mathcal{L}$ and $Q$ be a left comodule of $\mathcal{L}$, then the cotensor product $P\square^{\mathcal{L}}Q$ is defined to be the

$$P\square^{\mathcal{L}}Q := \{p \otimes q \in P\diamond_B Q \mid p_{(0)} \otimes p_{(1)} \otimes q = p \otimes q_{(-1)} \otimes q_{(0)}\} \subset P \times_B Q,$$

where the balanced tensor product $P\diamond_B Q$ above is induced by the $\bar{B}$-bimodule of $P$ and the $B$-bimodule of $Q$, i.e. $\bar{b}\diamond_B q = p\diamond_B bq$. One can check that $P\square^{\mathcal{L}}Q \subset P \times_B Q$. Indeed, if $p \otimes q \in P\square^{\mathcal{L}}Q$, then $\bar{p}\diamond_B q = p\diamond_B qb$, since

$$\begin{aligned} \delta_{L\diamond_B \mathrm{id}_Q}(\bar{p}\diamond_B q) &= p_{(0)}\diamond_B p_{(1)}t_L(b)\diamond_B q = p\diamond_B q_{(-1)}t_L(b)\diamond_B q_{(0)} \\ &= p\diamond_B q_{(-1)}\diamond_B q_{(0)}b = \delta_{L\diamond_B \mathrm{id}_Q}(p\diamond_B qb). \end{aligned}$$

*Remark 3.13.* If $U$ is a left $\mathcal{L}$-comodule then $U \cong \mathcal{L}\square^{\mathcal{L}}U$. Similarly, if $V$ is a right $\mathcal{L}$-comodule then $V \cong V\square^{\mathcal{L}}\mathcal{L}$.

For later use we note an analog of [31, Lemma 1.1]

**Lemma 3.14.** Let $P$ be a right $\mathcal{L}$-comodule and $V$ a regular left $\mathcal{L}$-comodule. Then an isomorphism

$$P\square^{\mathcal{L}}(V \otimes_B \mathcal{L}) \cong P\diamond_B V$$

is given by applying $\varepsilon$ to the right factor, i.e. $p \otimes v \otimes X \mapsto p \otimes v\varepsilon(X)$.

*Proof.* This is the isomorphism

$$P\cdot\square^{\mathcal{L}}(\cdot V \otimes_B \cdot \mathcal{L}) \cong P\cdot\square^{\mathcal{L}}(\cdot \mathcal{L}\diamond_B V) \cong (P\square^{\mathcal{L}}\mathcal{L})\diamond_B V \cong P\diamond_B V$$

in which the first stage is given by the map defining regularity, cotensored with $P$. $\square$

## 3.2. Hopf Galois extensions.

**Definition 3.15.** Given a left bialgebroid over $B$, a left $\mathcal{L}$-comodule algebra $P$ is a $B$-ring and a left $\mathcal{L}$-comodule, such that the coaction is a $B$-ring map. Let $N = \mathrm{co}\mathcal{L}P$ be the left invariant subalgebra of $P$, $N \subseteq P$ is called a left $\mathcal{L}$-Galois extension if the left canonical map $^L\mathrm{can} : P \otimes_N P \to \mathcal{L}\diamond_B P$ given by

$$^L\mathrm{can}(p \otimes_N q) = p_{(-1)}\diamond_B p_{(0)}q,$$

is bijective. If $N \subseteq P$ is a left $\mathcal{L}$-Galois extension, the inverse of $^L\mathrm{can}$ can be determined by the left translation map:

$$^L\tau := ^L\mathrm{can}^{-1}|_{\mathcal{L}\diamond_B 1} : \mathcal{L} \to P \otimes_N P, \quad X \mapsto X^{<1>} \otimes_N X^{<2>} \qquad (3.14)$$

If $P$ is faithfully flat as a right $N$-module we say it is a right faithfully flat extension, similarly for left faithful flatness. A faithfully flat extension will mean one that is both left and right faithfully flat.

We can see $\mathrm{can}^{-1}(X \otimes p) = X^{<1>} \otimes X^{<2>}p$. We note that $^L\mathrm{can}$ is a left $B$-module map in the sense that

$$^L\mathrm{can}(p \otimes bq) = p_{(-1)}\bar{b} \otimes p_{(0)}q \qquad (3.15)$$

because the image of the coaction map lies in the Takeuchi product. In addition, the map is obviously right $P$-linear, and also left $B$-linear in a suitable sense, because the coaction is.

In the theory of Hopf-Galois extensions over $k$-Hopf algebras, right comodule algebras with a Galois condition are more common. However, if one naively copies the definition of the canonical map for a right comodule algebra, namely $p \otimes q \mapsto pq_{(0)} \otimes p_{(1)}$, one ends up with a “map” that is not well-defined. Already in [22] a second canonical map is used; bijectivity of the two canonical maps is equivalent if the antipode of the Hopf algebra is
bijective. We will use an analog of this second canonical map as a Galois condition for right comodule algebras

**Definition 3.16.** Let $\mathcal{L}$ be a left $B$-bialgebroid. A right $\mathcal{L}$-comodule algebra $P$ is a $\overline{B}$-ring and a right $\mathcal{L}$-comodule, such that the coaction is a $\overline{B}$-ring map.

Thus, a right $\mathcal{L}$-comodule algebra is the same as a left $\mathcal{L}^{\text{cop}}$-comodule algebra.

Let $M := P^{\text{co}\mathcal{L}} = \text{co}\mathcal{L}^{\text{cop}}$ be the right invariant subalgebra of $P$. We say that $M \subseteq P$ is an anti-right $\mathcal{L}$-Galois extension if it is a left $\mathcal{L}^{\text{cop}}$-Galois extension.

Thus, $M \subset P$ is anti-right $\mathcal{L}$-Galois if the anti-right canonical map $\text{cân}^L : P \otimes_M P \to P \diamond_B \mathcal{L}$ given by

$$ \text{cân}^L(p \otimes_M q) = p_{(0)}q \diamond_B p_{(1)}, $$

is bijective. If $M \subseteq P$ is a anti-right $\mathcal{L}$-Galois extension, the inverse of $\text{cân}^L$ can be determined by the anti-right translation map:

$$ \hat{\tau}^L := (\text{cân}^L)^{-1}|_{1 \diamond_B \mathcal{L}} : \mathcal{L} \to P \otimes_M P, \quad X \mapsto X^{[1]} \otimes_M X^{[2]}. \quad (3.16) $$

The theory of right anti-Galois extensions is thus, in principle, entirely redundant. However, we will be looking at algebras that are naturally bi-comodule algebras over two Hopf algebroids, and it would be inconvenient to treat both structures as left comodule structures. In particular, a Hopf algebroid $\mathcal{L}$ that is also anti-Hopf is an example of a “biGalois object” with its obvious left and right comodule structures over itself and it would be inconvenient to treat the right comodule structure as a left $\mathcal{L}^{\text{cop}}$-comodule algebra structure.

In the sequel we will write out some notions and facts on (Galois) left comodule algebras in an extra version for right comodule algebras to have both explicitly. In other instances we might only mention a result on one side, but will freely use it on the other, if this presents no extra notational traps.

As a generalization of [24], we have

**Lemma 3.17.** If a left $B$-bialgebroid $\mathcal{L}$ admits left $\mathcal{L}$-Galois extension $N \subseteq P$ such that $P$ is faithfully flat as left $B$-module, then $\mathcal{L}$ is a left Hopf algebroid. More precisely,

$$ X_+ \otimes X_- \otimes 1 = X^{<1>}_{(-1)} \otimes X^{<2>}_{(-1)} \otimes X^{<1>}_{(0)} X^{<2>}_{(0)} \in \int_{a,b} \bar{\mathcal{L}}_{\bar{a}} \otimes \bar{\mathcal{L}}_{\bar{a}} \otimes_b P. $$

If in addition $P$ is (faithfully) flat as a left $N$-module, then $\mathcal{L}$ is (faithfully) flat as a left $B$-module.

*Proof.* It is sufficient to observe the following diagram commutes:

<figure><img src="image_2.png" /></figure>

where the map $\text{L}^{\text{can}}_{1,3} : P \otimes_N (\mathcal{L} \diamond_B P) \to \mathcal{L} \otimes_{\overline{B}} \mathcal{L} \diamond_B P$ is given by

$$ \text{L}^{\text{can}}_{1,3}(p \otimes X \otimes q) = p_{(-1)} \otimes X \otimes p_{(0)}q. $$

All maps in the diagram are well-defined as they can be viewed as tensor product of module maps with $P$ or $\mathcal{L}$. For the map $\text{L}^{\text{can}}_{1,3}$ in particular this is due to the linearity in (3.15), up to an identification. Now all the maps except for $\lambda \otimes P$ are isomorphisms by assumption, and so by faithful flatness of $P$ we can conclude that $\lambda$ is an isomorphism as well.
Now also assume that $_N P$ is (faithfully) flat. Since $(T \otimes_B \mathcal{L}) \diamond_B P \cong T \otimes_B (\mathcal{L} \diamond_B P) \cong T \otimes_B P \otimes_P P$ for $T \in \mathcal{M}_B$, we can conclude that $\mathcal{L}$ is (faithfully) flat over $B$ on the left if $P$ is (faithfully) flat over $N$ on the left. $\square$

**Lemma 3.18.** If a left $B$-bialgebroid $\mathcal{L}$ admits a left faithfully flat regular left $\mathcal{L}$-Galois extension $N \subseteq P$, then $\mathcal{L}$ is an anti-left Hopf algebroid. More precisely,

$$X_{[+]} \otimes 1 \otimes X_{[-]} = X^{<1>}_{[0](-1)} \otimes X^{<1>}_{[0](0)} X^{<2>} \otimes X^{<1>}_{[1]} \in \int_{a,b} \bar{a} \mathcal{L}_b \otimes_a P \otimes_b P.$$

*Proof.* It is sufficient to observe the following diagram commute:

$$\begin{tikzcd}
\mathcal{L} \otimes^B P \otimes_N P \arrow[r, "\phi \otimes P"] & \mathcal{L} \diamond_B P \otimes_N P \\
\mathcal{L} \otimes^B \mathcal{L} \diamond_B P \arrow[u, "\mathcal{L} \otimes^L \text{can}", "\mu \otimes P"'] \arrow[r, "\mathcal{L} \otimes^B \mathcal{L} \diamond_B P"] & \mathcal{L} \otimes^B \mathcal{L} \diamond_B P
\end{tikzcd}$$

it is not hard to see all the maps expect for $\mu \otimes P$ are well defined isomorphism, so by the faithful flatness of $P$, $\mu$ is bijective. $\square$

<figure><img src="image_3.png" /></figure>

There is a useful proposition for Hopf Galois extensions below:

**Proposition 3.19.** Let $N \subseteq P$ be a faithfully flat left $\mathcal{L}$-Galois extension, then we have

$$X^{<1>}_{(-1)} \diamond_B X^{<1>}_{(0)} \otimes_N X^{<2>} = X_{(1)} \diamond_B X_{(2)}^{<1>} \otimes_N X_{(2)}^{<2]}, \quad (3.17)$$

$$X^{<2>}_{(-1)} \otimes X^{<1>} \otimes X^{<2>}_{(0)} = X_{-} \otimes X_{+}^{<1>} \otimes X_{+}^{<2>} \in \int_{b} \bar{b} \mathcal{L} \otimes P \otimes_N bP, \quad (3.18)$$

$$X^{<1>}_{(-1)} \diamond_B X^{<1>}_{(0)} X^{<2>} = X \diamond_B 1, \quad (3.19)$$

$$p_{(-1)}^{<1>} \otimes_N p_{(-1)}^{<2>} p_{(0)} = p \otimes_N 1, \quad (3.20)$$

$$nX^{<1>} \otimes_N X^{<2>} = X^{<1>} \otimes_N X^{<2>} n, \quad (3.21)$$

$$(aXb)^{<1>} \otimes_N (aXb)^{<2>} = aX^{<1>} b \otimes_N X^{<2>}, \quad (3.22)$$

$$(\bar{a}X\bar{b})^{<1>} \otimes_N (\bar{a}X\bar{b})^{<2>} = X^{<1>} \otimes_N bX^{<2>} a, \quad (3.23)$$

$$X^{<1>} X^{<2>} = \varepsilon(X), \quad (3.24)$$

$$(XY)^{<1>} \otimes_N (XY)^{<2>} = X^{<1>} Y^{<1>} \otimes_N Y^{<2>} X^{<2>}, \quad (3.25)$$

$$X_{+}^{<1>} \otimes_N X_{-}^{<1>} \otimes_N X_{-}^{<2>} X_{+}^{<2>} = X_{+}^{<1>} \otimes_N X^{<2>} \otimes_N 1, \quad (3.26)$$

for any $X, Y \in \mathcal{L}$, $p \in P$, $n \in N$ and $a, b \in B$. If $P$ is a regular faithfully flat left $\mathcal{L}$-Galois extension, then we have

$$p_{[0]} p_{[1]}^{<1>} \otimes_N p_{[1]}^{<2>} = 1 \otimes_N p, \quad (3.27)$$

$$(pq)_{[1]} \otimes^B (pq)_{[0]} = q_{[1]} p_{[1]} \otimes^B p_{[0]} q_{[0]}, \quad (3.28)$$

$$X^{<1>} \otimes_N X^{<2>}_{[0]} \diamond_B X^{<2>}_{[1]} = X_{(1)}^{<1>} \otimes_N X_{(1)}^{<2>} \diamond_B X_{(2)}, \quad (3.29)$$

$$X^{<1>}_{[0]} \otimes_N X^{<2>} \otimes X^{<1>}_{[1]} = X_{[+]}^{<1>} \otimes_N X_{[+]}^{<2>} \otimes X_{[-]} \in \int_b P_b \otimes_N P \otimes_b \mathcal{L}, \quad (3.30)$$

$$X_{[+]}^{<1>} X_{[-]}^{<1>} \otimes_N X_{[-]}^{<2>} \otimes_N X_{[+]}^{<2>} = 1 \otimes_N X^{<1>} \otimes_N X^{<2>}. \quad (3.31)$$

*Proof.* We can see (3.19) and (3.20) can be given by the definition of translation map. (3.21), (3.22) and (3.23) can be shown as the image of the canonical map on both hand
sides of the equalities are the same. To show (3.18), we observe that $P \otimes_N P$ is a left $\mathcal{L}$-comodule with the coaction given by

$$ \tilde{\delta}(p \otimes_N q) = q_{(-1)} \diamond_B (p \otimes_N q_{(0)}), $$

where the underlying $B$-bimodule structure on $P \otimes_N P$ is given by $b.(p \otimes_N q).b' = p \otimes_N bqb'$, for any $b, b' \in B$ and $p, q \in P$, it is not hard to see the image of the coaction belongs to the Takeuchi product. We can also see $\mathcal{L} \diamond_B P$ is a $\mathcal{L}$-comodule with the coaction given by

$$ L\delta(X \diamond_B p) = X_{-p_{(-1)}} \diamond_B (X_{+\diamond_B p_{(0)}}), $$

where the underlying $B$-bimodule structure of $\mathcal{L} \diamond_B P$ is given by $b.(X \diamond_B p).b' = X \tilde{b} \diamond_B p b'$, for any $b, b' \in B$, $X \in \mathcal{L}$ and $p \in P$. It is not hard to see $L\delta$ is $B$-bimodule map with its image belonging to the Takeuchi product and $L\delta$ also factors through all the balanced tensor products. Moreover, by using

$$ X_+ \otimes X_{-(1)} \otimes X_{-(2)} = X_{++} \otimes X_- \otimes X_{+-}, $$

we can see the coaction is coassociative. It is not hard to see $L\mathcal{C}$ can is a $B$-bimodule map from $P \otimes_N P$ to $\mathcal{L} \diamond_B P$ and moreover, it is a left $\mathcal{L}$-comodule map. So the translation map is a $\mathcal{L}$-colinear map as well and this results (3.18). We can see (3.26) can be proved by (3.18) and (3.20). If $P$ is a skew regular left $\mathcal{L}$-comodule, then we can see $p_{[0]}p_{[1]}^{<1>} \otimes_N p_{[1]}^{<2>} \in N \otimes_N P$. Indeed,

$$ \begin{aligned} (L\delta \otimes_N \mathrm{id})(p_{[0]}p_{[1]}^{<1>} \otimes_N p_{[1]}^{<2>}) &= p_{[0]}_{(-1)}p_{[1]}^{<1>}_{(-1)} \otimes p_{[0]}_{(0)}p_{[1]}^{<1>}_{(0)} \otimes p_{[1]}^{<2>} \\ &= p_{[0]}_{(-1)}p_{[1]}_{(1)} \otimes p_{[0]}_{(0)}p_{[1]}_{(2)}^{<1>} \otimes p_{[1]}_{(2)}^{<2>} \\ &= p_{[0]}_{[0]}_{(-1)}p_{[0]}_{[1]} \otimes p_{[0]}_{[0]}_{(0)}p_{[1]}^{<1>}_{[1]} \otimes p_{[1]}^{<2>} \\ &= 1 \otimes p_{[0]}p_{[1]}^{<1>} \otimes p_{[1]}^{<2>}, \end{aligned} $$

where the 2nd step uses (3.17), the 3rd step uses coassociativity of the reverse comodule structure. So we can show (3.27), since $p_{[0]}p_{[1]}^{<1>} \otimes_N p_{[1]}^{<2>} = 1 \otimes_N p_{[0]}p_{[1]}^{<1>}p_{[1]}^{<2>} = 1 \otimes_N p$. We can show (3.29) by comparing the results of the map $\mathrm{id} \otimes \phi$ on both sides of (3.29). By Lemma 3.18, $\mathcal{L}$ is an anti-left Hopf algebroid, so we can show (3.30) by comparing the results of the map $L\mathcal{C}$ can $\otimes \mathrm{id}$ on both sides of (3.30). By applying $\mathrm{id} \otimes L\mathcal{C}$ can on the left hand side of (3.31), we get

$$ \begin{aligned} &X_{[+]}^{<1>} X_{[-]}^{<1>} \otimes X_{[-]}^{<2>}_{(-1)} \otimes X_{[-]}^{<2>}_{(0)} X_{[+]}^{<2>} \\ &= X_{[+]}^{<1>} X_{[-]}^{+} \otimes X_{[-]}^{+} \otimes X_{[-]}^{+} \otimes X_{[-]}^{<2>} X_{[+]}^{<2>} \\ &= X_{(2)}^{<1>} X_{(2)}^{<1>} \otimes X_{(1)}^{+} \otimes X_{(2)}^{<2>} X_{(2)}^{<2>} \\ &= 1 \otimes X \otimes 1, \end{aligned} $$

where the 1st step uses (3.18), the 2nd step uses Proposition 2.4. Since we have the same result by applying on the right hand side of (3.31), so we have (3.31). $\square$

The following is just the translation of the previous proposition for right anti-Galois extensions; we record it because we will use the notations.

**Proposition 3.20.** Let $M \subseteq P$ be a faithfully flat anti-right $\mathcal{L}$-Galois extension, then we have

$$ X^{[1]}_{(0)} \otimes_M X^{[2]} \otimes X^{[1]}_{(1)} = X^{[1]}_{(1)} \otimes_M X^{[2]}_{(1)} \otimes X^{[2]}_{(2)} \in \int_{\bar{b}} \bar{b} P \otimes_M P \otimes_b \mathcal{L}, \quad (3.32) $$

$$ X^{[1]} \otimes_M X^{[2]}_{(0)} \diamond_B X^{[2]}_{(1)} = X^{[1]}_{(+)} \otimes_M X^{[2]}_{(+)} \diamond_B X^{[2]}_{(-)}, \quad (3.33) $$

$$ X^{[1]}_{(0)} X^{[2]} \diamond_B X^{[1]}_{(1)} = 1 \diamond_B X, \quad (3.34) $$

$$ p_{(1)}^{[1]} \otimes_M p_{(1)}^{[2]} p_{(0)} = p \otimes_M 1, \quad (3.35) $$
$$mX^{[1]} \otimes_M X^{[2]} = X^{[1]} \otimes_M X^{[2]} m, \qquad (3.36)$$

$$(aXb)^{[1]} \otimes_M (aXb)^{[2]} = X^{[1]} \otimes_M \bar{b}X^{[2]}\bar{a}, \qquad (3.37)$$

$$(\bar{a}X\bar{b})^{[1]} \otimes_M (\bar{a}X\bar{b})^{[2]} = \bar{a}X^{[1]}\bar{b} \otimes_M X^{[2]}, \qquad (3.38)$$

$$X^{[1]} X^{[2]} = \overline{\varepsilon(X)}, \qquad (3.39)$$

$$(XY)^{[1]} \otimes_M (XY)^{[2]} = X^{[1]}Y^{[1]} \otimes_M Y^{[2]}X^{[2]}, \qquad (3.40)$$

$$X_{[+]}^{[1]} \otimes_M X_{[-]}^{[1]} \otimes_M X_{[-]}^{[2]} X_{[+]}^{[2]} = X^{[1]} \otimes_M X^{[2]} \otimes_M 1, \qquad (3.41)$$

for any $X, Y \in \mathcal{L}$, $p \in P$, $m \in M$ and $a, b \in B$. If $P$ is a regular right $\mathcal{L}$-comodule, then we have

$$p_{[0]}p_{[-1]}^{[1]} \otimes_M p_{[-1]}^{[2]} = 1 \otimes_M p, \qquad (3.42)$$

$$(pq)_{[0]} \otimes_{\bar{B}} (pq)_{[-1]} = p_{[0]}q_{[0]} \otimes_{\bar{B}} q_{[1]}p_{[1]} \qquad (3.43)$$

$$X^{[1]} \otimes_M X^{[2]}_{[0]} \diamond_B X^{[2]}_{[-1]} = X_{(2)}^{[1]} \otimes_M X_{(2)}^{[2]} \diamond_B X_{(1)}, \qquad (3.44)$$

$$X_{[0]}^{[1]} \otimes_M X^{[2]} \otimes X_{[-1]}^{[1]} = X_{+}^{[1]} \otimes_M X_{+}^{[2]} \otimes X_{-} \in \int_{\bar{b}} P_{\bar{b}} \otimes_M P \otimes_{\bar{b}} \mathcal{L}, \qquad (3.45)$$

$$X_{+}^{[1]} X_{-}^{[1]} \otimes X_{-}^{[2]} \otimes X_{+}^{[2]} = 1 \otimes X^{[1]} \otimes X^{[2]}. \qquad (3.46)$$

By 3.10 we have a bijection between regular left and right coactions. We now show that this restricts to Galois comodule algebra structures:

**Lemma 3.21.** Let $\mathcal{L}$ be a left $B$-bialgebroid and $P$ a regular left $\mathcal{L}$-comodule algebra. Then the right coaction $p \mapsto p_{[0]} \diamond_B p_{[1]}$ is a $\mathcal{L}$-comodule algebra structure on $P^{\text{op}}$.

If, moreover, $N \subseteq P$ is a left $\mathcal{L}$-Galois extension, then $N^{\text{op}} \subseteq P^{\text{op}}$ is a regular anti-right $\mathcal{L}$-Galois extension. The anti-right translation map is

$$\hat{\tau}^{\mathcal{L}} : X \mapsto X^{<2>} \otimes_{N^{\text{op}}} X^{<1>}$$

for any $p \in P$ and $X \in \mathcal{L}$.

Similarly, if $P$ is a regular right comodule algebra, then $P^{\text{op}}$ is a left $\mathcal{L}$-comodule algebra, and if $M \subset P$ is a regular anti-right $\mathcal{L}$-Galois extension, then $M^{\text{op}} \subseteq P^{\text{op}}$ is a skew regular left Hopf Galois extension of $\mathcal{L}$. The left translation map is

$$\mathcal{L}_{\tau} : X \to X^{[2]} \otimes_{M^{\text{op}}} X, \qquad [1]$$

for any $p \in P$ and $X \in \mathcal{L}$.

*Proof.* We only show the first half of the lemma. By (3.28), we can see

$$(qp)_{[1]} \otimes (qp)_{[0]} = p_{[1]}q_{[1]} \otimes q_{[0]}p_{[0]},$$

for any $p, q \in P$. Indeed, by applying $\phi$ on the right hand side, we have

$$\begin{aligned} \phi(p_{[1]}q_{[1]} \otimes q_{[0]}p_{[0]}) &= q_{[0]}(-1)p_{[0]}(-1)p_{[1]}q_{[1]} \otimes q_{[0]}(0)p_{[0]}(0) \\ &= q_{[0]}(-1)q_{[1]} \otimes q_{[0]}(0)p = 1 \otimes qp \\ &= \phi((qp)_{[1]} \otimes (qp)_{[0]}). \end{aligned}$$

Moreover, we can see the right coinvariant subalgebra is $N^{\text{op}}$. More precisely, if $n \in N^{\text{op}}$, then

$$n_{[0]} \diamond_B n_{[1]} = n_{(0)[0]} \diamond_B n_{(0)[1]} n_{(-1)} = n \diamond_B 1.$$

Conversely, if $m \in (P^{\text{op}})^{\text{co}\mathcal{L}}$, then

$$m_{(-1)} \diamond_B m_{(0)} = m_{[0]}(-1) m_{[1]} \diamond_B m_{[0]}(0) = 1 \diamond_B m.$$

By (3.27) and (3.29), we can check $\hat{\tau}^{\mathcal{L}}$ is the anti-right translation map. The regular right comodule structure is given by the original left $\mathcal{L}$-coaction. $\square$
$$M \mapsto M \otimes_N P$$
$$^{\text{co}\mathcal{L}}\Gamma \leftrightarrow \Gamma$$

is a category equivalence. We note the isomorphism

$$\Gamma \to {}^{\text{co}\mathcal{L}}\Gamma \otimes_N P; \quad p \mapsto p_{[0]} p_{[1]}^{<1>} \otimes_N p_{[1]}^{<2>}$$

(3) If $N \subset P$ is a right faithfully flat right $\mathcal{L}$-anti-Galois extension and a regular comodule, then

$$\mathcal{M}_N \to \mathcal{M}_P^\mathcal{L}$$
$$M \mapsto M \otimes_N P$$
$$\Gamma^{\text{co}\mathcal{L}} \leftrightarrow \Gamma$$

is a category equivalence. We note the isomorphism

$$\Gamma \to \Gamma^{\text{co}\mathcal{L}} \otimes_N P; \quad p \mapsto p_{[0]} p_{[-1]}^{[1]} \otimes_N p_{[-1]}^{[2]}$$

**Corollary 3.24.** Let $P$ be a right faithfully flat left $\mathcal{L}$-Galois extension of $N$, and consider $\Gamma \in \mathcal{L} \mathcal{M}_P$ endowed with a right $S$-module structure such that the action of each $s \in S$ is a Hopf module map. Then the obvious map $^{\text{co}\mathcal{L}}\Gamma \otimes_S T \to {}^{\text{co}\mathcal{L}}(\Gamma \otimes_S T)$ is an isomorphism for every $T \in {}_S\mathcal{M}$.

*Proof.* The equivalence from theorem 3.22 sends the map to the identity on $\Gamma \otimes_S T$. $\square$

The following theorem generalizes results of Ulbrich [31], [32] for the Hopf algebra case.

To prepare the proof we give an analog of [31, Lemma 1.3]

**Lemma 3.25.** Let $N \subset P$ be a left faithfully flat left $\mathcal{L}$-Galois extension. Then for any right $\mathcal{L}$-comodule $V$ we have

$$f: (V^{\square\mathcal{L}} P) \otimes_N P \cong V^{\diamond_B P}; \quad v \otimes p \otimes q \mapsto v \otimes pq. \qquad (3.47)$$

and $f(v \otimes p \otimes bq) = f(v \otimes p \otimes q)\bar{b}$, where the module structure on the target is on the left tensor factor.

*Proof.* We have

$$(V^{\square\mathcal{L}} P) \otimes_N P \cong V^{\square\mathcal{L}}(P \otimes_N P) \cong V^{\square\mathcal{L}}(\mathcal{L}_{\diamond_B} P) \cong V_{\diamond_B} P.$$

The first isomorphism stems from faithful flatness of $P$, the second uses the Galois map, and the last is the canonical isomorphism applying $\varepsilon$ to the middle factor. It is straightforward to check that the composition has the stated form. The additional $B$-linearity holds because the cotensor product is contained in the Takeuchi product. $\square$

**Theorem 3.26.** Let $\mathcal{L}$ be a left $B$-bialgebroid left flat over $B$ and $\bar{B}$, which is Hopf and anti-Hopf. If $N \subseteq P$ is a right faithfully flat anti-right $\mathcal{L}$-Galois extension then the functor

$$^{\mathcal{L}}\mathcal{M} \to {}_N\mathcal{M}_N$$
$$V \mapsto P^{\square\mathcal{L}}V$$

is monoidal with monoidal functor structure

$$\xi: (P^{\square\mathcal{L}}V) \otimes_N (P^{\square\mathcal{L}}W) \to P^{\square\mathcal{L}}(V \otimes_B W)$$
$$\xi: (p \diamond_B v) \otimes (q \diamond_B w) \mapsto pq \diamond_B (v \otimes_B w).$$

Similarly, if $P$ a left faithfully flat left $\mathcal{L}$-Galois extension of $N$ then the functor

$$\mathcal{M}^{\mathcal{L}} \to {}_N\mathcal{M}_N$$
$$V \mapsto V \square^\mathcal{L} P$$

is monoidal with monoidal functor structure

$$\zeta : (V \square^\mathcal{L} P) \otimes_N (W \square^\mathcal{L} P) \to (V \otimes_{\bar{B}} W) \square^\mathcal{L} P$$

$$\zeta : (v \diamond_B p) \otimes (w \diamond_B q) \mapsto (v \otimes_{\bar{B}} w) \diamond_B pq.$$

*Proof.* We can see $\xi$ is well defined as $P \square^\mathcal{L} V$ is contained in the Takeuchi product.

For bijectivity, we first note that applying Lemma 3.25 to $P$ as a $\mathcal{L}^{\text{cop}}$-comodule algebra gives the isomorphism

$$f = f_V : (P \square^\mathcal{L} V) \otimes_N P \to P \diamond_B V; p \otimes v \otimes q \mapsto pq \otimes v$$

We have

$$f(p \otimes v \otimes \bar{b}q) = p\bar{b}q \otimes v = pq \otimes vb = f(p \otimes v \otimes q)b$$

because the cotensor product is contained in the Takeuchi product. It is not hard to see $f_V^{-1}(p \otimes v) = v_{(-1)}^{[1]} \otimes v_{(0)} \otimes v_{(-1)}^{[2]} p$. Indeed, $f_V \circ f_V^{-1}(p \otimes v) = v_{(-1)}^{[1]} v_{(-1)}^{[2]} p \otimes v_{(0)} = p \otimes v$. Also,

$$\begin{align*} f_V^{-1} \circ f_V(p \otimes v \otimes q) &= v_{(-1)}^{[1]} \otimes v_{(0)} \otimes v_{(-1)}^{[2]} pq = p_{(1)}^{[1]} \otimes v \otimes p_{(1)}^{[2]} p_{(0)} q \\ &= p \otimes v \otimes q. \end{align*}$$

The diagram

$$\begin{tikzcd}[column sep=2.8em, row sep=2.8em]
(P \square^\mathcal{L} V) \otimes_N (P \square^\mathcal{L} W) \otimes_N P \arrow[onnee]{cc}{\xi_N \otimes P}{\arrow{->}} \arrow{dl}{P \square^\mathcal{L}(V \otimes_B W)} \arrow{dr}{P \square^\mathcal{L} W}
(P \square^\mathcal{L} V) \otimes_N f_W \arrow{dl}{f_V \otimes_W} \arrow{dr}{P \diamond_B W}
(P \square^\mathcal{L} V) \otimes_N P \diamond_B W \arrow{dl}{P \otimes_W} \arrow{dr}{P \diamond_B V \otimes_B W}
\end{tikzcd}$$

commutes, proving that $\xi$ is an isomorphism. We can also compute the inverse of $\xi$ which is given by

$$\xi^{-1}(p \diamond_B (v \otimes_B w)) = (v_{(-1)}^{[1]} \diamond_B v_{(0)}^{[1]} \otimes_N v_{(-1)}^{[2]} p \diamond_B w),$$

we can see $\xi^{-1}$ factors through all the balanced tensor products by applying (3.37) and (3.38). Indeed,

$$\begin{align*} (\mathrm{id}_{P \square^\mathcal{L} V} \otimes f_W^{-1}) \circ (f_V^{-1} \otimes \mathrm{id}_W) \circ f_{V \otimes_B W} (p \otimes v \otimes w \otimes 1) \\ &= (\mathrm{id}_{P \square^\mathcal{L} V} \otimes f_W^{-1}) \circ (f_V^{-1} \otimes \mathrm{id}_W) (p \otimes v \otimes w) \\ &= (\mathrm{id}_{P \square^\mathcal{L} V} \otimes f_W^{-1}) (v_{(-1)}^{[1]} \otimes v_{(0)} \otimes v_{(-1)}^{[2]} p \otimes w) \\ &= v_{(-1)}^{[1]} \otimes v_{(0)} \otimes w_{(-1)}^{[1]} \otimes w_{(0)} \otimes_N w_{(-1)}^{[2]} v_{(-1)}^{[2]} p \in (P \square^\mathcal{L} V) \otimes_N (P \square^\mathcal{L} V) \otimes_N N \\ &= v_{(-1)}^{[1]} \otimes v_{(0)} \otimes w_{(-1)}^{[1]} w_{(-1)}^{[2]} v_{(-1)}^{[2]} p \otimes w_{(0)} \otimes_N 1 \\ &= v_{(-1)}^{[1]} \otimes v_{(0)} \otimes v_{(-1)}^{[2]} p \otimes w \otimes_N 1. \end{align*}$$

Similarly, we could see $\zeta$ is well defined since $V \square^\mathcal{L} P$ is a subalgebra of $V \times_B P$. The inverse of $\zeta$ is given by

$$\zeta^{-1}((v \otimes_{\bar{B}} w) \diamond_B p) = (v_{(0)} \diamond_B v_{(1)}^{<1>}) \otimes_N (w \diamond_B v_{(1)}^{<2>}) p.$$

□

In the theory of Galois and biGalois objects, monoidal "fiber" functors are always given by Galois extensions [31, 32]. At this time we cannot prove a full analog of this statement, but we have the following partial result:

**Proposition 3.27.** Let $P$ be a right $\mathcal{L}$-comodule algebra such that we have a monoidal functor as in Theorem 3.26. Then $P$ is an anti-right $\mathcal{L}$-Galois extension.
*Proof.* We have a commutative diagram

<figure><img src="image_4.png" /></figure>

where $f$ is the isomorphism from Lemma 3.14.

<figure><img src="image_5.png" /></figure>

## 4. HOPF BIGALOIS EXTENSIONS

### 4.1. Some technical preparations.

*Remark 4.1.* Let $P, Q$ be left $\mathcal{L}$-comodule algebras and $M \in \mathcal{L}_P M_Q$. Then

$$P \otimes_B Q \otimes^{\mathcal{C}\mathcal{L}} M \ni p \otimes q \otimes m \mapsto pmq \in M$$

is well defined and a left $\mathcal{L}$-comodule map. In particular $^{\mathcal{C}\mathcal{L}}(P \otimes_B Q)$ is an algebra under $(p \otimes q)(p' \otimes q') = pp' \otimes q'q$ and there is a functor

$$^{\mathcal{L}}P M_Q \to ^{\mathcal{C}\mathcal{L}}(P \otimes_B Q) M, M \mapsto ^{\mathcal{C}\mathcal{L}}M.$$

Also note that $^{\mathcal{C}\mathcal{L}}P$ and $(^{\mathcal{C}\mathcal{L}}Q)^{\text{op}}$ are obvious subalgebras of $^{\mathcal{C}\mathcal{L}}(P \otimes_B Q)$.

Indeed the first map is well-defined by the previous remark, colinearity is obvious and implies that the map restricts to a map

$$^{\mathcal{C}\mathcal{L}}(P \otimes_B Q) \otimes^{\mathcal{C}\mathcal{L}} M \to^{\mathcal{C}\mathcal{L}} M$$

**Lemma 4.2.** Let $\mathcal{L}$ be a bialgebroid and $\Gamma, \Lambda \in \mathcal{L}M$. If $\Gamma$ is regular, then

$$^{\mathcal{C}\mathcal{L}}(\Gamma \otimes_B \Lambda) = \Gamma \Box^{\mathcal{L}} \Lambda$$

if we endow $\Gamma$ with the reverse right $\mathcal{L}$-comodule structure as in proposition 3.11

*Proof.* Let $p \otimes q \in ^{\mathcal{C}\mathcal{L}}(\Gamma \otimes_B \Lambda)$. We can see

$$(\phi \otimes \text{id})(p \otimes_B q_{(-1)} \diamond_B q_{(0)}) = p_{(0)} \otimes p_{(-1)} q_{(-1)} \otimes q_{(0)} = (\phi \otimes \text{id})(p_{[0]} \otimes_B p_{[1]} \diamond_B q).$$

So we have

$$p_{[0]} \otimes_B p_{[1]} \diamond_B q = p \otimes_B q_{(-1)} \diamond_B q_{(0)},$$

therefore, $p \otimes q \in \Gamma \Box^{\mathcal{L}} \Lambda$. Conversely, if $p \otimes q \in \Gamma \Box^{\mathcal{L}} \Lambda$, we have

$$\delta(p \otimes q) = p_{(-1)} q_{(-1)} \otimes p_{(0)} \otimes q_{(0)} = p_{[0]} p_{(-1)} \otimes p_{[1]} \otimes p_{[0]} \otimes q = 1 \otimes p \otimes q.$$

<figure><img src="image_6.png" /></figure>

**Proposition 4.3.** Let $\mathcal{L}$ be a left Hopf algebroid over $B$, and $N \subset P$ a right faithfully flat left $\mathcal{L}$-Galois extension. Then the functor

$$N \mathcal{M} \ni M \mapsto P \otimes_N M \in \mathcal{L} \mathcal{M}$$

is right adjoint. The left adjoint is

$$^{\mathcal{L}}\mathcal{M} \ni \Gamma \mapsto ^{\mathcal{C}\mathcal{L}}(P \otimes_B \Gamma) \in N\mathcal{M}.$$

The unit of the adjunction is given by

$$\alpha: \Gamma \ni p \mapsto p_{(-1)}^{<1>} \otimes p_{(-1)}^{<2>} \otimes p_{(0)} \in P \otimes_N^{\mathcal{C}\mathcal{L}}(P \otimes_B \Gamma).$$

The same formulas describe a pair of adjoint functors between $N\mathcal{M}_N$ and $^{\mathcal{C}\mathcal{L}}\mathcal{M}_N$.
*Proof.* This is just the composition between the adjoint equivalence of theorem 3.22 and the standard fact that the forgetful functor from the category $\mathcal{L}_P\mathcal{M}$ of left $P$ modules in the monoidal category $\mathcal{L}\mathcal{M}$ to the underlying category $\mathcal{L}\mathcal{M}$ is right adjoint to tensoring with $P$. $\square$

We stated proposition 4.3 for left comodules and left modules because it involves as little as possible rearrangements of tensor product, and uses fewer hypotheses. To make the Ehresmann Hopf algebroid construction work below, however, we need the twisted version using part (2) of corollary 3.23 instead. We will formulate this adjunction in terms of a universal property.

**Corollary 4.4.** Let $\mathcal{L}$ be a left Hopf algebroid over $B$, $N \subset P$ a left faithfully flat left $\mathcal{L}$-Galois extension, and assume that $P$ is a regular comodule. Write $\overline{N} = N^{\text{op}}$. Then for every $\Gamma \in \mathcal{L}\mathcal{M}$ we have a map

$$ \alpha: \Gamma \ni p \mapsto p_{[1]}^{<2>} \otimes p_{[0]} \otimes p_{[1]}^{<1>} \in P \diamond_{\overline{N}}^{\text{co}\mathcal{L}}(\Gamma \otimes_B P) $$

of left $\mathcal{L}$-comodules with the universal property that for all $M \in \overline{N}\mathcal{M}$ and every map $\beta: \Gamma \to P \otimes_M M$ of left $\mathcal{L}$-comodules there is a unique map $f: \text{co}\mathcal{L}(\Gamma \otimes_R P) \to M$ of left $\overline{N}$-modules making

<figure><img src="image_7.png" /></figure>

commute. If $A$ is a $k$-algebra and $\Gamma \in \mathcal{L}\mathcal{M}$ (meaning $\Gamma$ is in addition an $A$-module such that multiplication with an element of $A$ is a comodule map) $\alpha$ is an $A$-module map, and if $\beta$ is an $A$-module map then so is $f$.

Note that $\alpha$ is defined so that the composition

$$ \Gamma \xrightarrow{\alpha} P \diamond_{\overline{N}}^{\text{co}\mathcal{L}}(\Gamma \otimes_B P) \cong \Gamma \otimes_B P $$

maps $p$ to $p \otimes 1$. Also note that the map $f$ corresponding to $\beta$ is the coinvariant part of

$$ \Gamma \otimes_B P \ni p \otimes q \mapsto \beta(p)q \quad (4.1) $$

**4.2. Ehresmann Hopf algebroids.** As a generalization of [26], given a Hopf Galois extension we have the following definition

**Lemma 4.5.** Let $\mathcal{L}$ be a left $B$-Hopf algebroid and $N \subseteq P$ be a left faithfully flat left $\mathcal{L}$-Hopf Galois extension with regular comodule structure. We set $\overline{N} = N^{\text{op}}$. Define $R(P, \mathcal{L}) := \text{co}\mathcal{L}(P \otimes_B P)$. Then $R(P, \mathcal{L})$ is a left $\overline{N}$-bialgebroid with

$$ (p \otimes q)(p' \otimes q') = pp' \otimes q'q $$

$$ s: \overline{N} \ni \bar{n} \mapsto 1 \otimes n \in R(P, \mathcal{L}) $$

$$ t: N \ni n \mapsto n \otimes 1 \in R(P, \mathcal{L}) $$

$$ \begin{aligned} \Delta(p \otimes q) &= p_{[1]}^{<2>} \otimes q \otimes p_{[0]} \otimes p_{[1]}^{<1>} \\ &= q_{(-1)}^{<2>} \otimes_B q_{(0)} \diamond_{\overline{N}} p \otimes_B q_{(-1)}^{<1>} \end{aligned} $$

$$ \varepsilon(p \otimes q) = pq. $$

$P$ is a right $R(P, \mathcal{L})$-comodule algebra by

$$ \delta: P \to P \diamond_{\overline{N}} R(P, \mathcal{L}); p \mapsto p_{[1]}^{<2>} \otimes p_{[0]} \otimes p_{[1]}^{<1>} $$
In this way $B \subset P$ is an anti-right $R(P, \mathcal{L})$-Galois extension. If $P$ is also right faithfully flat over $N$ then it is a regular right $R(P, \mathcal{L})$-comodule, with reverse comodule structure

$$R \delta(p) = p_{[-1]} \diamond_{\bar{N}} p_{[0]} = p_{(-1)}^{<2>} \otimes_B p_{(0)} \diamond_{\bar{N}} p_{(-1)}^{<1>} \quad (4.2)$$

and the anti-translation map is

$$(p \otimes_B q)^{[1]} \otimes_B (p \otimes_B q)^{[2]} = p \otimes_B q. \quad (4.3)$$

*Proof.* We already remarked that the algebra structure is well defined. The proposed comodule structure on $P$ is an instance of the universal map in Corollary 4.4. The image of $\delta$ is in $P \times_{\bar{N}} R(P, \mathcal{L})$ by eq. (3.21). It is an algebra map by Lemma 3.21 and eq. (3.25). It follows from the universal property (with $A = N$) that $R(P, \mathcal{L})$ has a unique coring structure $R(P, \mathcal{L}) \to R(P, \mathcal{L}) \diamond_{\bar{N}} R(P, \mathcal{L})$ for which $P$ is a comodule, namely the unique $\Delta$ making

<figure><img src="image_8.png" /></figure>

commute. It is a standard argument how to derive coassociativity of $\Delta$ from this definition.

By (4.1), we can see the second formula for comultiplication is given by the formula above. Indeed, we first observe that

$$\begin{align*}
\beta(p) &= p_{[1]}^{<2>} p_{[1]}^{<2>} \otimes p_{[1]}^{<2>} p_{[0]}^{<2>} \otimes p_{[1]}^{<2>} p_{[1]}^{<1>} \otimes p_{[0]} p_{[1]}^{<1>} \\
&= p_{[1]}^{<2>} p_{[1]}^{<2>} \otimes p_{[1]}^{<2>} p_{[1]}^{<1>} \otimes p_{[0]} p_{[1]}^{<1>}.
\end{align*}$$

Thus, for any $p \otimes q \in R(P, \mathcal{L})$ we have

$$\begin{align*}
\Delta(p \otimes q) &= \beta(p)q \\
&= p_{[1](2)}^{<2>} q \otimes p_{[1](1)}^{<2>} \otimes p_{[1](2)}^{<1>} \otimes p_{[0]} \otimes p_{[1](1)}^{<1>} \\
&= q_{(-1)}^{<2>} q_{(0)} \otimes q_{(-2)}^{<2>} \otimes q_{(-1)}^{<1>} \otimes p \otimes q_{(-2)}^{<1>} \\
&= 1 \otimes q_{(-1)}^{<2>} \otimes q_{(0)} \otimes p \otimes q_{(-1)}^{<1>}.
\end{align*}$$

where we use Lemma 4.2 in the second step. The first formula for comultiplication is a consequence of Lemma 4.2. It is an algebra map by the same argument as for the coaction, and its image is in the Takeuchi product by the same argument as well. The composition

$$P \xrightarrow{\delta} P \diamond_{\bar{N}} R(P, \mathcal{L}) \cong P \otimes_B P$$

is given by $p \mapsto p \otimes 1$, so that $P^{\text{coR}(P, \mathcal{L})} \subset P$ is the equalizer of $\iota \otimes_B P, P \otimes_B \iota: P \to P \otimes_B P$ by the diagram

<figure><img src="image_9.png" /></figure>

and thus $P^{\text{coR}(P, \mathcal{L})} = B$ by faithfully flat descent.

The anti-right Galois map for the right $\mathcal{L}$-extension $B \subset P$ is the isomorphism (coming from Theorem 3.22)

$$P \otimes_B P \cong P \diamond_{\bar{N}} R(P, \mathcal{L})$$

and thus $P$ is anti-right Galois; its inverse maps $1 \otimes p \otimes q$ to $p \otimes q$ proving that the translation map has the claimed form.
By definition of $\delta$, the diagram

<figure><img src="image_10.png" /></figure>

commutes, where $\psi$ is from the definition of regular comodule which is given by $\psi(p' \otimes (p \otimes q)) = p'_{[1]} <^{2>} \otimes p'_{[0]} p \otimes q p'_{[1]} <^{1}>$; $f$ is the counit of the adjoint equivalence from Theorem 3.22, which is given by $f(p' \otimes p \otimes q) = p'p \otimes q$ with inverse $f^{-1}(p \otimes q) = p q_{(-1)} <^{1>} \otimes q_{(-1)} <^{2>} \otimes q_{(0)}$; and $g$, given by $g(p' \otimes p \otimes q) = p \otimes qp'$, is, up to the obvious identification, the counit of the variant (2) from Corollary 3.23, thus also is an isomorphism under the additional flatness hypothesis. Thus $P$ is regular as claimed. To find the reverse comodule structure explicitly, we simply calculate

$$ 
\begin{align*}
\psi^{-1}(p \otimes 1) &= f^{-1}g(p \otimes 1) \\
&= f^{-1}(1 \otimes p) \\
&= p_{(-1)} <^{1>} \otimes p_{(-1)} <^{2>} \otimes p_{(0)}.
\end{align*}
$$

□

**Proposition 4.6.** Let $\mathcal{L}$ be a left $B$-bialgebroid and $N \subseteq P$ be a regular faithfully flat left $\mathcal{L}$-Hopf Galois extension. Then

$$ R(P, \mathcal{L}) = \{p \otimes_B q \in P \otimes_B P \mid pq_{(-1)} <^{1>} \otimes_N q_{(-1)} <^{2>} \otimes_B q_{(0)} = 1 \otimes_N p \otimes_B q\}, \quad (4.4) $$

$$ R(P, \mathcal{L}) = \{p \otimes_B q \in P \otimes_B P \mid p_{[0]} \otimes_B p_{[1]} <^{1>} \otimes_N p_{[1]} <^{2>} q = p \diamond_B q \otimes_N 1\}. \quad (4.5) $$

*Proof.* For (4.4), by definition, $p \otimes q \in P \otimes_P P$ belongs to $R(P, \mathcal{L})$ if and only if

$$ 1 \otimes p \otimes q = p_{(-1)} q_{(-1)} \otimes p_{(0)} \otimes q_{(0)} \in \mathcal{L} \diamond_B P \otimes_P P $$

where, again, we'll have to plug the proper tensor products. Now applying the translation map to the first two factors maps the left hand side to $1 \otimes p \otimes q \in P \otimes_N P \otimes_B P$, and the right hand side to

$$ 
\begin{align*}
(p_{(-1)} q_{(-1)})^{<1>} \otimes (p_{(-1)} q_{(-1)})^{<2>} p_{(0)} \otimes q_{(0)} \\
&= p_{(-1)}^{<1>} q_{(-1)}^{<1>} \otimes q_{(-1)}^{<2>} p_{(-1)}^{<2>} p_{(0)} \otimes q_{(0)} \\
&= pq_{(-1)}^{<1>} \otimes q_{(-1)}^{<2>} \otimes q_{(0)},
\end{align*}
$$

using successively (3.25) and (3.20).

The proof of (4.5) is similar.

□

### 4.3. BiGalois Extensions.

**Definition 4.7.** Let $B, N$ be two algebras. Let $\mathcal{L}$ be a $B$-bialgebroid which is Hopf and anti-Hopf and faithfully flat as left module over $B$ and $\bar{B}$, and let $\mathcal{H}$ an $\bar{N}$-bialgebroid which is Hopf and anti-Hopf and left faithfully flat over $N$ and $\bar{N}$.

A $\mathcal{L}$-$\mathcal{H}$-Hopf biGalois extension is an algebra $P$, such that $N \subseteq P$ is a faithfully flat left $\mathcal{L}$-Galois extension and $B \subseteq P$ is a faithfully flat anti-right $\mathcal{H}$-Galois extension, in such a way that the two comodule structures make $P$ a $\mathcal{L}$-$\mathcal{H}$-bicomodule. We denote the collection of $\mathcal{L}$-$\mathcal{H}$-Hopf biGalois extension by $\text{BiGal}(\mathcal{L}, \mathcal{H})$.

**Proposition 4.8.** Let $P \in \text{BiGal}(\mathcal{L}, \mathcal{H})$, then we have

$$ \alpha^{<1>}_{[0]} \otimes \alpha^{<1>}_{[-1]} \otimes \alpha^{<2>} = \alpha^{<1>} \otimes \alpha^{<2>}_{(1)} \otimes \alpha^{<2>}_{(0)}; \quad (4.6) $$

$$ X^{[1]}_{[0]} \otimes X^{[1]}_{[1]} \otimes X^{[2]} = X^{[1]} \otimes X^{[2]}_{(-1)} \otimes X^{[2]}_{(0)}, \quad (4.7) $$

for any $\alpha \in \mathcal{L}$ and $X \in \mathcal{H}$.
Proof. First, we can see the both side of (4.6) belongs to $\int^{a,b} \int_{c,d} \bar{b} P_{\bar{c}} \otimes_{\bar{c},\bar{d}} \mathcal{L}_{a,\bar{b}} \otimes \bar{d} P_{\bar{a}}$. Indeed, the map

$$ \phi_{(10.1)} : \int_{b}^{a} \int_{\bar{a}} P_{\bar{b}} \otimes_{b} P_{a} \to \int_{c,d}^{a,b} \int_{\bar{b}} P_{\bar{c}} \otimes_{\bar{c},\bar{d}} \mathcal{L}_{a,\bar{b}} \otimes \bar{d} P_{\bar{a}} $$

given by

$$ \phi_{(10.1)}(p \otimes q) = p_{[0]} \otimes p_{[-1]} \otimes q $$

is well defined. Similarly, the map

$$ \phi_{(10.2)} : \int_{b}^{a} \int_{\bar{a}} P_{\bar{b}} \otimes_{b} P_{a} \to \int_{c,d}^{a,b} \int_{\bar{b}} P_{\bar{c}} \otimes_{\bar{c},\bar{d}} \mathcal{L}_{a,\bar{b}} \otimes \bar{d} P_{\bar{a}} $$

given by

$$ \phi_{(10.2)}(p \otimes q) = p \otimes q_{(1)} \otimes q_{(0)} $$

is well defined. Apply $^{L}$can$_{1,3} \circ (\psi \otimes \mathrm{id}_P)$ on both sides of (4.6). On the one hand,

$$ \begin{aligned} ^L\mathrm{can}_{1,3} \circ (\psi \otimes \mathrm{id}_P)(\alpha^{<1>}_{[0]} \otimes \alpha^{<1>}_{[-1]} \otimes \alpha^{<2>}_{(0)}) &= ^L\mathrm{can}_{1,3}(\alpha^{<1>} \otimes 1 \otimes \alpha^{<2>}_{(0)}) \\ &= \alpha \otimes 1 \otimes 1. \end{aligned} $$

On the other hand,

$$ \begin{aligned} ^L\mathrm{can}_{1,3} \circ (\psi \otimes \mathrm{id}_P)(\alpha^{<1>} \otimes \alpha^{<2>}_{(1)} \otimes \alpha^{<2>}_{(0)}) &= ^L\mathrm{can}_{1,3}(\alpha^{<1>}_{(0)} \otimes \alpha^{<1>}_{(1)} \alpha^{<2>}_{(1)} \otimes \alpha^{<2>}_{(0)}) \\ &= \alpha^{<1>}_{(-1)} \otimes \alpha^{<1>}_{(1)} \alpha^{<2>}_{(1)} \otimes \alpha^{<1>}_{(0)} \alpha^{<2>}_{(0)} \\ &= \alpha^{<1>}_{(-1)} \otimes (\alpha^{<1>}_{(0)} \alpha^{<2>}_{(0)})_{(1)} \otimes (\alpha^{<1>}_{(0)} \alpha^{<2>}_{(0)})_{(0)} \\ &= \alpha \otimes 1 \otimes 1. \end{aligned} $$

Since $^{L}$can$_{1,3} \circ (\psi \otimes \mathrm{id}_P)$ is isomorphism, we have (4.6). The equation (4.7) can be similarly proved. $\square$

<figure><img src="image_11.png" /></figure>

As a result of Lemma 4.5, we have

**Theorem 4.9.** Let $\mathcal{L}$ be a left $\mathcal{B}$-bialgebra that is Hopf and anti-Hopf, and faithfully flat as left $\mathcal{B}$ and $\overline{\mathcal{B}}$-module. Let $N \subseteq P$ be a faithfully flat left $\mathcal{L}$-Hopf Galois extension such that $P$ is also faithfully flat over $R$ on both sides. Then $P$ is a $\mathcal{L}$-$R(P, \mathcal{L})$-biGalois extension, and this structure is unique in the sense that for every $\mathcal{L}$-$\mathcal{H}$-biGalois structure extending the given $\mathcal{L}$-structure there is a unique isomorphism $R(P, \mathcal{L}) \cong \mathcal{H}$ identifying the structures in the obvious sense.

Proof. By Lemma 3.12 comodule structures over $\mathcal{L}$ are always regular. We first show that $P$ is $\mathcal{L}$-$R(P, \mathcal{L})$-biGalois, which means showing that $R(P, \mathcal{L})$ is Hopf, anti-Hopf and fulfills the faithful flatness conditions.

By the coopposite version of Lemma 3.17 $R(P, \mathcal{L})$ is left faithfully flat over $\overline{N}$ and anti-Hopf. Using the reverse of the right comodule structure on $P^{\mathrm{op}}$, which is also a Galois coaction by Lemma 3.21, so we can use Lemma 3.17 again to conclude that $R(P, \mathcal{L})$ is also left flat over $\overline{N}$ and Hopf.

Now assume $P$ is a $\mathcal{L}$-$\mathcal{H}$-bicomodule algebra for some $\overline{N}$-bialgebroid $\mathcal{H}$. Denote the comodule structure by $\beta: p \mapsto p_{(0)} \otimes p_{(1)}$. The corresponding map $f: R(P, \mathcal{L}) \to \mathcal{H}$ as in Corollary 4.4 is given by $f(p \otimes q) = \beta(p)q$ as an element of $^{\mathrm{co}\mathcal{L}}(P \diamond_{\overline{N}} \mathcal{H}) \cong \mathcal{H}$, so

$$ f(p \otimes q) = p_{(0)}q \otimes p_{(1)} = \overline{p_{(0)}} \overline{q} p_{(1)} \in \mathcal{H} \subset P \diamond_{\overline{N}} \mathcal{H}. $$

In other words, if we restrict the anti-Galois map

$$ P \otimes_P P \to P \diamond_{\overline{N}} \mathcal{H} $$
for the right $\mathcal{H}$-comodule algebra $P$, which is a map of Hopf modules in $\mathcal{L}_P\mathcal{M}$, to the left $\mathcal{L}$-coinvariants, we obtain

$$f: R(P, \mathcal{L}) = \mathcal{L} \otimes_P P \to \mathcal{L} \otimes_P (P \diamond \overline{\mathcal{N}} \mathcal{H}) \cong \mathcal{H}.$$

In particular, if $P$ is anti-Galois, then $f$ is an isomorphism.

Furthermore, $f$ is multiplicative because the image of $\beta$ lies in the Takeuchi product:

$$f(pp' \otimes q'q) = \overline{p_{(0)}p'_{(0)}q'qp_{(1)}} p'_{(1)} = \overline{p_{(0)}} \overline{q} p_{(1)} \overline{p'_{(0)}} q' p'_{(1)} = f(p \otimes q)f(p' \otimes q').$$

From the definition of both $f$ and the comultiplication of $R(P, \mathcal{L})$ through the universal property in Corollary 4.4 it is clear that $f$ is also a coring map, and thus it is a morphism of $\overline{\mathcal{N}}$-bialgebroids. $\square$

By the general correspondence between left comodule structures and right comodule structures over the coopposite, we get a corresponding procedure to uniquely complete a right anti-Galois extension to give a biGalois extension.

**Corollary 4.10.** Let $\mathcal{H}$ be a left $\overline{B}$-bialgebra that is Hopf and anti-Hopf, and faithfully flat as left $B$ and $\overline{B}$-module. Let $N \subseteq P$ be a faithfully flat right $\mathcal{H}$-anti-Hopf Galois extension such that $P$ is also faithfully flat over $B$ on both sides. Then $P$ is a biGalois extension in a unique way. Namely, $P$ is a $L(P, \mathcal{H})$-$\mathcal{H}$-biGalois extension, where $L(P, \mathcal{H}) := (P \otimes_{\overline{B}} P)^{\text{coH}}$ is a $N$-Hopf algebroid by

$$s_L(n) = n \otimes_{\overline{B}} 1, \quad t_L(n) = 1 \otimes_{\overline{B}} n, \quad (p \otimes_{\overline{B}} q)(p' \otimes_{\overline{B}} q') = pp' \otimes_{\overline{B}} q'q,$$

for any $p \otimes_{\overline{B}} q, p' \otimes_{\overline{B}} q' \in L(P, \mathcal{H})$. The left coaction of $L(P, \mathcal{H})$ and its reverse are given by

$$L\delta(p) = p_{(-1)} \otimes_N p_{(0)} = p_{[0]} \otimes_{\overline{B}} p_{(-1)}^{[1]} \otimes_N p_{(-1)}^{[2]}, \quad (4.8)$$

$$\delta_L(p) = p_{[0]} \otimes_N p_{[1]} = p_{(1)}^{[1]} \otimes_N p_{(1)}^{[2]} \otimes_{\overline{B}} p_{(0)}, \quad (4.9)$$

and the translation map:

$$(p \otimes_{\overline{B}} q)^{<1>} \otimes_{\overline{B}} (p \otimes_{\overline{B}} q)^{<2>} = p \otimes_{\overline{B}} q. \quad (4.10)$$

**Proposition 4.11.** Let $\mathcal{H}$ be a left $\overline{B}$-bialgebroid and $N \subseteq P$ be a regular faithfully flat anti-right $\mathcal{H}$-Hopf Galois extension, then for any $p \otimes_{\overline{B}} q \in L(P, \mathcal{H})$, $\overline{bp} \otimes_{\overline{B}} q = p \otimes_{\overline{B}} q\overline{b}$ and

$$L(P, \mathcal{H}) = \{p \otimes_{\overline{B}} q \in P \otimes_{\overline{B}} P \mid pq_{(1)}^{[1]} \otimes_N q_{(1)}^{[2]} \otimes_{\overline{B}} q_{(0)} = 1 \otimes_N p \otimes_{\overline{B}} q\} \quad (4.11)$$

$$L(P, \mathcal{H}) = \{p \otimes_{\overline{B}} q \in P \otimes_{\overline{B}} P \mid p_{[0]} \otimes p_{[-1]} \otimes q = p \otimes q_{(1)} \otimes q_{(0)}\} \quad (4.12)$$

$$L(P, \mathcal{H}) = \{p \otimes_{\overline{B}} q \in P \otimes_{\overline{B}} P \mid p_{[0]} \otimes_{\overline{B}} p_{[-1]}^{[1]} \otimes_N p_{[-1]}^{[2]} q = p \otimes_{\overline{B}} q \otimes_N 1\} \quad (4.13)$$

**Remark 4.12.** Recall that for any Hopf algebra $H$ and a right $\mathcal{H}$-comodule algebra $P$, $N = P^{\text{coH}} \subseteq P$ is a $\mathcal{H}$-Galois extension if the map

$$\text{can}^H : P \otimes_N P \to P \otimes H, \quad p \otimes q \mapsto pq_{(0)} \otimes q_{(1)},$$

is bijective. If $H$ has a bijective antipode, then $N = P^{\text{coH}} \subseteq P$ is an $\mathcal{H}$-Galois extension if and only if $N = P^{\text{coH}} \subseteq P$ is an anti-right $\mathcal{H}$-Galois extension. More precisely, if we denote the translation map for the classical Hopf Galois extension by

$$\text{can}^{H-1}(1 \otimes h) = h^{(1)} \otimes_N h^{(2)}, \quad \forall h \in H.$$

Then we have the anti-right translation map

$$h^{[1]} \otimes_N h^{[2]} = S^{-1}(h)^{(1)} \otimes_N S^{-1}(h)^{(2)}.$$

Recall that in [26], $L(P, H)$ has the coproduct

$$\Delta(p \otimes q) = p_{(0)} \otimes p_{(1)}^{(1)} \otimes_N p^{(2)} \otimes q, \quad \forall p \otimes q \in L(P, H).$$

Therefore, $L(P, H)$ in Lemma 4.10 recovers the Ehresmann Hopf algebroid in [26].
**Lemma 4.13.** Let $\mathcal{H}$ be a left $B$-bialgebroid and $N \subseteq P$ be a regular faithfully flat anti-right $\mathcal{H}$-Hopf Galois extension. Then $L(P, \mathcal{H})$ is a Hopf algebroid. More precisely, for any $X = p \otimes_{\overline{B}} q \in L(P, \mathcal{H})$

$$X_+ \otimes_{\overline{N}} X_- = p \otimes_{\overline{B}} q_{[-1]}^{[2]} \otimes_{\overline{N}} q_{[0]} \otimes_{\overline{B}} q_{[-1]}^{[1]} \quad (4.14)$$

$$X_{[-]} \otimes^{N} X_{[+]} = p_{(1)}^{[2]} \otimes_{\overline{B}} p_{(0)} \otimes^{N} p_{(1)}^{[1]} \otimes_{\overline{B}} q. \quad (4.15)$$

**Corollary 4.14.** If $N \subseteq P$ is a regular left Hopf Galois extension of $\mathcal{L}$, then $L(P, R(P, \mathcal{L})) \simeq \mathcal{L}$. Similarly, if $M \subseteq P$ is a regular anti-Hopf Galois extension of $\mathcal{L}$, then $R(P, L(P, \mathcal{L})) \simeq \mathcal{L}$. More precisely, the isomorphism $L(P, R(P, \mathcal{L})) \simeq \mathcal{L}$ can be given by left translation map

$$L_{\tau}: \mathcal{L} \to L(P, R(P, \mathcal{L})), \quad X \mapsto X^{<1>} \otimes_{N} X^{<2>}.$$

Similarly, the isomorphism $R(P, L(P, \mathcal{L})) \simeq \mathcal{L}$ can be given by anti-right translation map

$$\hat{\tau}^L: \mathcal{L} \to R(P, L(P, \mathcal{L})), \quad X \mapsto X^{[1]} \otimes_M X^{[2]},$$

for any $X \in \mathcal{L}$.

**Definition 4.15** ([14]). Let $\mathcal{L}$ be a Hopf algebroid over $B$, a vertical bisection is a $B$-bilinear map $\sigma: \mathcal{L} \to B$, namely $\sigma(b\bar{b}'X) = b\sigma(X)b'$ such that $\sigma(XY) = \sigma(X\sigma(Y)) = \sigma(X\overline{\sigma(Y)})$ for any $X, Y \in \mathcal{L}$ and $b, b' \in B$.

We know the collection of vertical bisections of a Hopf algebroid is a group with convolution product as the product and counit as the unit. We denote the group of bisections of $\mathcal{L}$ by $Z^1(\mathcal{L}, B)$. We can also generalize the main theorem of [11].

**Lemma 4.16.** Let $\mathcal{L}$ be a Hopf algebroid over $B$, and $N \subseteq P$ be an anti-right $\mathcal{L}$-Galois extension, then the automorphism $\text{Aut}(P)$ (the group of morphisms of right $\mathcal{L}$-comodule algebra map from $P$ to $P$ which preserve the coinvariant subalgebra $N$) is isomorphic to $Z^1(L(P, \mathcal{L}), N)$.

Similarly, let $\mathcal{L}$ be a Hopf algebroid over $B$, and $N \subseteq P$ be a left $\mathcal{L}$-Galois extension, then $\text{Aut}(P)$ (the group of morphisms of left $\mathcal{L}$-comodule algebra map from $P$ to $P$ which preserve the coinvariant subalgebra $N$) is isomorphic to $Z^1(R(P, \mathcal{L}), N^{op})$.

*Proof.* For the first half of the statement, we will present the one-to-one correspondence between these two groups. Given a vertexie bisection $\sigma \in Z^1(L(P, \mathcal{L}), N)$, we can construct $F_{\sigma} \in \text{Aut}(P)$ by $F_{\sigma}(p) := \sigma(p_{[0]} \otimes_{\overline{B}} p_{[-1]}^{[1]})p_{[-1]}^{[2]}$ for any $p \in P$, which is well defined since $\sigma$ is $N$-bilinear. We can see $F_{\sigma}$ is an algebra map. Indeed,

$$\begin{align*}
F_{\sigma}(pq) &= \sigma(p_{[0]}q_{[0]} \otimes_{\overline{B}} q_{[-1]}^{[1]}p_{[-1]}^{[1]})p_{[-1]}^{[2]}q_{[-1]}^{[2]} \\
&= \sigma(p_{[0]} \otimes_{\overline{B}} \sigma(q_{[0]} \otimes_{\overline{B}} q_{[-1]}^{[1]})p_{[-1]}^{[1]})p_{[-1]}^{[2]}q_{[-1]}^{[2]} \\
&= \sigma(p_{[0]} \otimes_{\overline{B}} p_{[-1]}^{[1]})p_{[-1]}^{[2]}\sigma(q_{[0]} \otimes_{\overline{B}} q_{[-1]}^{[1]})q_{[-1]}^{[2]} \\
&= F_{\sigma}(p)F_{\sigma}(q),
\end{align*}$$

It is also not hard to see $F_{\sigma}$ is $N$-bilinear. Moreover, we can check $F_{\sigma}$ is right $\mathcal{L}$-colinear. Indeed,

$$\begin{align*}
F_{\sigma}(p)_{(0)} \diamond_{B} F_{\sigma}(p)_{(1)} &= \sigma(p_{[0]} \otimes_{\overline{B}} p_{[-1]}^{[1]}) \otimes_{N} p_{[-1]}^{[2]} \delta_{(0)} \diamond_{B} p_{[-1]}^{[2]} \delta_{(1)} \\
&= \sigma(p_{[0]} \otimes_{\overline{B}} p_{[-1]}^{[+1]}) \otimes_{N} p_{[-1]}^{[+1]} \delta_{B} p_{[-1]}^{-[1]} \\
&= \sigma(p_{(0)[0]} \otimes_{\overline{B}} p_{(0)[-1]}^{[1]}) \otimes_{N} p_{(0)[-1]}^{[2]} \delta_{B} p_{(1)} \\
&= F_{\sigma}(p_{(0)}) \diamond_{B} p_{(1)},
\end{align*}$$

where the 2nd step uses (3.33) and the 3rd step uses Lemma 3.12.

Conversely, given $F \in \text{Aut}(P)$, we can construct $\sigma_F(p \otimes_{\overline{B}} q) := F(p)q$, which is also well defined since $F$ is a $\mathcal{L}$-comodule map. Clearly, $\sigma_F$ is $N$-bilinear. We can also see

$$\sigma_F((p \otimes_{\overline{B}} q)(p' \otimes_{\overline{B}} q')) = \sigma_F(pp' \otimes_{\overline{B}} q'q) = F(pp')q'q = F(p)F(p')q'q$$
$$ \begin{aligned} &=\sigma_F(p \otimes_{\overline{B}} \sigma_F(p' \otimes_{\overline{B}} q')q) \\ &=\sigma_F((p \otimes_{\overline{B}} q)\sigma_F(p' \otimes_{\overline{B}} q')), \end{aligned} $$

and,

$$ \sigma_F((p \otimes_{\overline{B}} q)\sigma_F(p' \otimes_{\overline{B}} q')) = F(pF(p')q')q = F(p)F(p')q'q = \sigma_F((p \otimes_{\overline{B}} q)\overline{\sigma_F(p' \otimes_{\overline{B}} q')}), $$

For the 'round trip', we have

$$ F_{\sigma_F}(p) = \sigma_F(p_{[0]} \otimes_{\overline{B}} p_{[-1]}^{[1]}) p_{[-1]}^{[2]} = F(p_{[0]}) p_{[-1]}^{[1]} p_{[-1]}^{[2]} = F(p), $$

for any $p \in P$. Also we have

$$ \sigma_{F_{\sigma}}(p \otimes_{\overline{B}} q) = F_{\sigma}(p)q = F(p_{[0]} \otimes_{\overline{B}} p_{[-1]}^{[1]}) p_{[-1]}^{[2]} q = \sigma(p \otimes_{\overline{B}} q), $$

for any $p \otimes_{\overline{B}} q \in L(P, \mathcal{L})$. The second half of the statement is similar. $\square$

Moreover, by Lemma 3.3 of [14], let Aut($\mathcal{L}$) be the group of automorphism of Hopf algebroid $\mathcal{L}$, we can construct a group morphism $\text{Ad} : Z^1(\mathcal{L}, B) \to \text{Aut}(\mathcal{L})$ by

$$ \text{Ad}_{\sigma}(X) := \sigma(X_{(1)}) \overline{\sigma(X_{(3)})} X_{(2)}, \qquad (4.16) $$

for any $X \in \mathcal{L}$. We define

$$ \text{CoInn}(\mathcal{L}) := \{\text{Ad}_{\sigma} | \sigma \in Z^1(\mathcal{L}, B)\} $$

to be the set of coinner automorphism of the Hopf algebroid $\mathcal{L}$. By Theorem of 3.4 of [14], we know CoInn($\mathcal{L}$) is a normal subgroup of Aut($\mathcal{L}$). More precisely, for any $\Phi \in \text{Aut}(\mathcal{L})$ and $\sigma \in Z^1(\mathcal{L}, B)$, we have

$$ \Phi \circ \text{Ad}_{\sigma} \circ \Phi^{-1} = \text{Ad}_{\Phi \triangleright \sigma}, $$

where $\Phi \triangleright \sigma$ is a vertical bisection given by $\Phi \triangleright \sigma := \sigma \circ \Phi^{-1}$. So we denote $\text{CoOut}(\mathcal{L}) = \text{Aut}(\mathcal{L})/\text{CoInn}(\mathcal{L})$ the group of co-outer automorphisms of $\mathcal{L}$. As a generalization of Lemma 3.11 of [27], we have

**Lemma 4.17.** Let $\mathcal{L}$ be a Hopf algebroid over $B$ and $\mathcal{H}$ be a Hopf algebroid over $N$ and $\text{BiGal}(\mathcal{L}, \mathcal{H})$ be the isomorphism class of $\mathcal{L} - \mathcal{H}$-biGalois extension, then $\text{CoOut}(\mathcal{L})$ acts freely on $\text{BiGal}(\mathcal{L}, \mathcal{H})$.

*Proof.* Let $\Phi \in \text{Aut}(\mathcal{L})$ and $P \in \text{BiGal}(\mathcal{L}, \mathcal{H})$. We define $^{\Phi}P := \Phi \triangleright P$ be the $\mathcal{L} - \mathcal{H}$-biGalois extension with the modified left $\mathcal{L}$-comodule structure given by

$$ \mathcal{L}^{\Phi} \delta(p) = \Phi(p_{(-1)}) \diamond_{B} p_{(0)}, $$

for all $p \in {}^{\Phi}P$. The rest structures of $^{\Phi}P$ is the same with $P$. If $N \subseteq P$ is a left $\mathcal{L}$-Galois extension, and $\overline{B} \subseteq \mathcal{H}$ is an anti-right $\mathcal{H}$-Galois extension then it is not hard to see $N \subseteq {}^{\Phi}P$ is a left $\mathcal{L}$-Galois extension with the deformed coaction. Indeed, the coaction is well defined since $\Phi$ is $B$-bilinear. The coaction is also an algebra map since $\Phi$ is an algebra map. And $N \subseteq {}^{\Phi}P$ is a left $\mathcal{L}$-Galois extension with the deformed translation map given by

$$ X^{<1>} \otimes_N X^{<2>} = \Phi^{-1}(X)^{<1>} \otimes_N \Phi^{-1}(X)^{<2>}. $$

$\square$

**4.4. The groupoid of BiGalois extensions.** As we know $P \square^{\mathcal{L}} Q \in P \times_B Q$, we have

**Proposition 4.18.** If $P$ is a right comodule algebra of $\mathcal{L}$ and $Q$ is a left comodule algebra of $\mathcal{L}$, then $P \square^{\mathcal{L}} Q$ is a subalgebra of the Takeuchi product $P \times_B Q$ algebra.

**Theorem 4.19.** The category whose objects are bialgebroids that are Hopf, anti-Hopf, and left faithfully flat over source and target algebra, and whose morphisms are isomorphism classes of Hopf biGalois extensions, is a groupoid, with composition of morphisms given by cotensor product.
Proof. Let $\mathcal{G}$, $\mathcal{H}$ and $\mathcal{L}$ be three Hopf and anti-Hopf algebroids over $A$, $B$ and $C$ respectively, left faithfully flat over their source and target subalgebras. Let $P \in \mathrm{BiGal}(\mathcal{G}, \mathcal{H})$, $Q \in \mathrm{BiGal}(\mathcal{H}, \mathcal{L})$. $A \subseteq P^{\square^H}Q$ is a $\mathcal{L}$-anti-Hopf Galois extension of $A$ with $\mathrm{id}^{\square^H}\delta_{\mathcal{L}}$ as right $\mathcal{L}$-coaction. Namely, the composition

$$ (P^{\square^H}Q) \underset{A}{\otimes} (P^{\square^H}Q) \xrightarrow{\xi} P^{\square^H}(Q \underset{B}{\otimes} Q) \xrightarrow{P^{\square \text{c}\text{\'a}\text{n}}} P^{\square^H}(Q \diamond_C \mathcal{L}) \cong (P^{\square^H}Q) \diamond_C \mathcal{L} $$

is the anti-right Galois map for $P^{\square^H}Q$. Similarly $P^{\square^H}Q$ is a left Hopf-Galois extension of $C$.

Cotensor product $-\square^H Q$ preserves and reflects exact sequences because $\bar{C}Q$ is faithfully flat and thus

$$ (M^{\square^H}Q) \underset{\bar{C}}{\otimes} Q \cong M^{\square^H}(Q \underset{\bar{C}}{\otimes} Q) \cong M^{\square^H}(\mathcal{H} \diamond_B Q) \cong (M^{\square^H}\mathcal{H}) \diamond_B Q \cong M \diamond_B Q $$

for every right $\mathcal{H}$-comodule $M$, and $BQ$ is faithfully flat. Similarly (although writing out the proof requires some awkward changes in the order of factors in tensor products) $P^{\square^H}$ preserves and reflects exact sequences. This fact (for all biGalois extensions) implies that the cotensor product of biGalois extensions is associative. Also $P \underset{A}{\otimes} (P^{\square^H}Q) \cong P \diamond_B Q$ as $\bar{C}$-bimodules implies that $P^{\square^H}Q$ is a faithfully flat left and right $\bar{C}$-module. Similarly, $P^{\square^H}Q$ is a faithfully flat left and right $A$-module.

It remains to show that every biGalois extension has an inverse under cotensor product. We claim the inverse can be given by $P^{-1} = P^{\mathrm{op}}$ with the reverse comodule structures of those of $P$. In fact by Lemma 4.2

$$ P^{-1} \square^G P = \mathrm{coG}(P \underset{A}{\otimes} P) \cong R(P, \mathcal{G}) \cong \mathcal{H} $$

and it is easy to see that this is an isomorphism of $\mathcal{H}$-bicomodule algebras. The proof that $P^{-1}$ is also the right inverse of $P$ is similar.

<figure><img src="image_12.png" /></figure>

**Proposition 4.20.** There is a morphism from the groupoid of Hopf biGalois extensions to the groupoid of monoidal equivalences between comodule categories of Hopf bialgebroids. More precisely, $P \in \mathrm{BiGal}(\mathcal{L}, \mathcal{H})$ gives rise to a monoical category equivalence

$$ {}^{\mathcal{H}}\mathcal{M} \ni V \mapsto P^{\square^H}V \in \mathcal{L}\mathcal{M}. $$

Proof. The inverse of the category equivalence is given by the inverse of $P$ in the groupoid above; the fact that the equivalences are monoidal was essentially checked in Theorem 3.26.

<figure><img src="image_13.png" /></figure>

# 5. STRUCTURE THEOREMS FOR HOPF MODULES

**Theorem 5.1.** Let $\mathcal{L}$ be a left $B$-bialgebroid and $N \subseteq P$ be a regular left faithfully flat anti-right $\mathcal{L}$-Hopf Galois extension, then we have the monoidal category equivalences

$$ (P\mathcal{M}_P^L, \otimes_P) \cong (L(P, \mathcal{L})\mathcal{M}, \diamond_N). $$

Proof. The equivalence can be given via

$$ P\mathcal{M}_P^L \to L(P, \mathcal{L})\mathcal{M}, \quad M \mapsto M^{\mathrm{coL}} \qquad (5.1) $$

$$ L(P, \mathcal{L})\mathcal{M} \to P\mathcal{M}_P^L, \quad \Lambda \mapsto \Lambda \otimes_N P. \qquad (5.2) $$

More precisely, for any $p \otimes q \in L := L(P, \mathcal{L})$ and $\eta \in M^{\mathrm{coL}}$, the left $L$-module structure is given by

$$ (p \otimes_{\bar{B}} q) \bullet \eta = p \eta q. $$

For any $\Lambda \in L(P, \mathcal{L})\mathcal{M}$, $\Lambda \otimes_N P$ has the $P$-bimodule structure

$$ p(\eta \otimes_N q)p' = p_{[0]}\eta p_{[-1]}^{[1]} \otimes_N p_{[-1]}^{[2]}qp' = p_{(-1)} \bullet \eta \otimes p_{(0)}qp', $$
and right $\mathcal{L}$-comodule structure on the right factor. The isomorphism $\Lambda \simeq (\Lambda \otimes_N P)^{\mathcal{C}\mathcal{O}\mathcal{L}}$ for any $\Lambda \in L(P, \mathcal{L})\mathcal{M}$ can be given by $\eta \mapsto \eta \otimes 1$. The isomorphism $M \simeq M^{\mathcal{C}\mathcal{O}\mathcal{L}} \otimes_N P$ for any $M \in P\mathcal{M}_P^\mathcal{L}$ can be given by $m \mapsto m_{[0]}m_{[-1]}^{[1]} \otimes_N m_{[-1]}^{[2]}$ with inverse $\eta \otimes_N p \mapsto \eta p$. The monoidal functor structure of (5.2) is given by the obvious isomorphism

$$ (\Lambda \otimes_N P) \otimes_P (\Lambda' \otimes_N P) \to (\Lambda \otimes_N \Lambda') \otimes_N P. \qquad (5.3) $$

It is immediate to check that it is a morphism in the appropriate category, and it is obviously coherent. $\square$

**Theorem 5.2.** Let $\mathcal{L}$ be a left Bialgebroid and $N \subseteq P$ be a regular left faithfully flat left $\mathcal{L}$-Hopf Galois extension, then we have the monoidal category equivalences

$$ (\mathcal{L}_P \mathcal{M}_P, \otimes_P) \simeq (R(P, \mathcal{L})^{\text{cop}} \mathcal{M}, \diamond_N) \simeq (R(P, \mathcal{L}) \mathcal{M}, \diamond_{\bar{N}})^{\text{sym}}. $$

*Proof.* The equivalence can be given via

$$ \mathcal{L}_P \mathcal{M}_P \to R(P, \mathcal{L})^{\text{cop}} \mathcal{M}, \quad M \mapsto \mathcal{C}\mathcal{O}\mathcal{L} M \qquad (5.4) $$

$$ R(P, \mathcal{L})^{\text{cop}} \mathcal{M} \to \mathcal{L}_P \mathcal{M}_P, \quad \Lambda \mapsto \Lambda \otimes_N P. \qquad (5.5) $$

Given $M \in \mathcal{L}_P \mathcal{M}_P$, $^{\mathcal{C}\mathcal{O}\mathcal{L}}M$ is a left $R(P, \mathcal{L})$-module with the left action given by

$$ (p \otimes_B q) \bullet m = pmq, $$

for any $p \otimes_B q \in R(P, \mathcal{L})$ and $m \in \mathcal{C}\mathcal{O}\mathcal{L}M$. We can see $bm = mb$ for any $b \in B$ and $m \in \mathcal{C}\mathcal{O}\mathcal{L}M$ as both sides have the same images of the left coaction. Moreover, for any $p \otimes_B q \in R(P, \mathcal{L})$ and $m \in \mathcal{C}\mathcal{O}\mathcal{L}M$, $pmq \in \mathcal{C}\mathcal{O}\mathcal{L}M$. So the left action is well defined, which is clearly associative and unital. Also, $^{\mathcal{C}\mathcal{O}\mathcal{L}}M$ is a $N$-bimodule with the structure given by $n \triangleright m \triangleleft n' = nmn'$ which is also a $\bar{N}$-bimodule with $\bar{n}' \triangleright m \triangleleft \bar{n} = nmn'$. Conversely, given $\Lambda \in R(P, \mathcal{L})^{\text{cop}} \mathcal{M}$, $\Lambda \otimes_N P \in \mathcal{L}_P \mathcal{M}_P$, with the left $\mathcal{L}$-coaction given by the right factor and the $P$-bimodule structure given by

$$ p \triangleright (\eta \otimes_N q) \triangleleft p' = p_{[0]} \eta p_{[1]}^{<1>} \otimes_N p_{[1]}^{<2>} q p' = p_{(1)} \triangleright \eta \otimes p_{(0)} q p', $$

for any $\eta \otimes_N q \in \Lambda \otimes_N P$. $\square$

**Definition 5.3** ([23][Def.4.2].) Let $\mathcal{L}$ be a left bialgebroid over $B$, a left-left Yetter-Drinfeld module of $\mathcal{L}$ is a left $\mathcal{L}$-comodule and a left $\mathcal{L}$-module $\Lambda$, such that

* $s_L(b) \bullet \rho = b\rho$ and $t_L(b) \bullet \rho = \rho b$, $\forall b \in B, \rho \in \Lambda$.

* $(X_{(1)} \bullet \rho)_{(-1)} X_{(2)} \diamond_B (X_{(1)} \bullet \rho)_{(0)} = X_{(1)} \rho_{(-1)} \diamond_B X_{(2)} \bullet \rho_{(0)}, \forall X \in \mathcal{L}, \rho \in \Lambda$.

We denote the category of left-left Yetter-Drinfeld modules of $\mathcal{L}$ by $\mathcal{L}\mathcal{Y}\mathcal{D}^\mathcal{L}$.

We define right-left Yetter-Drinfeld modules so that they are left-left ones over the coopposite bialgebroid.

**Definition 5.4.** Let $\mathcal{L}$ be a left bialgebroid over $B$, a right-left Yetter-Drinfeld module of $\mathcal{L}$ is a right $\mathcal{L}$-comodule and a left $\mathcal{L}$-module $\Lambda$, such that

* $s_L(b) \bullet \rho = \rho \bar{b}$ and $t_L(b) \bullet \rho = \bar{b} \rho$, $\forall b \in B, \rho \in \Lambda$.

* $(X_{(2)} \bullet \rho)_{(0)} \diamond_B (X_{(2)} \bullet \rho)_{(1)} X_{(1)} = X_{(1)} \bullet \rho_{(0)} \diamond_B X_{(2)} \rho_{(1)}$.

The category of right-left Yetter-Drinfeld modules of $\mathcal{L}$ is denoted by $\mathcal{L}\mathcal{Y}\mathcal{D}^\mathcal{L}$.

Recall that if $\mathcal{L}$ is an anti-left Hopf algebroid, then every left comodule of $\mathcal{L}$ is skew regular, so we can generalize the results in [10] (where a full Hopf algebroid is assumed) and [23][Prop.4.4] (where the center is replaced by the weak center):

**Lemma 5.5.** If $\mathcal{L}$ is an anti-left Hopf algebroid and left flat over $B$ and $\bar{B}$, then $(\mathcal{L}\mathcal{Y}\mathcal{D}, \diamond_B)$ is the Drinfeld center of $\mathcal{L}\mathcal{M}$, and in particular a is a braided monoidal category. The half-braiding and its inverse are given by

$$ \sigma(v \diamond_B w) = v_{(-1)} \bullet w \diamond_B v_{(0)} \qquad (5.6) $$
$$ \sigma^{-1}(w \diamond_B v) = v_{[0]} \diamond_B v_{[1]} \bullet w, \qquad (5.7) $$

for any a Yetter-Drinfeld module $V \in \mathcal{L}\mathcal{Y}\mathcal{D}$, module $W \in \mathcal{L}\mathcal{M}$, and $v \in V, w \in W$.

The only novelty is that the inverse of the half-braiding exists and is given by the reverse comodule structure. Similarly, we have

**Lemma 5.6.** If $\mathcal{L}$ is a left Hopf algebroid left flat over $B$ and $\bar{B}$, then $(\mathcal{L}\mathcal{Y}\mathcal{D}^{\mathcal{L}}, \diamond_B)$ is the Drinfeld center of $\mathcal{L}\mathcal{M}$; the half-braiding and its inverse are given by

$$ \tilde{\sigma}(v \diamond_B w) = w_{(0)} \diamond_B w_{(1)} \bullet v \qquad (5.8) $$

$$ \tilde{\sigma}^{-1}(w \diamond_B v) = w_{[-1]} \bullet v \diamond_B w_{[0]}, \qquad (5.9) $$

for $w \in W \in \mathcal{L}\mathcal{Y}\mathcal{D}^{\mathcal{L}}$ and $v \in V \in \mathcal{L}\mathcal{M}$.

It may be worth noting that in this case the underlying functor from Yetter-Drinfeld modules to comodules reverses tensor products. We also note that from the two previous results it is obvious that in case that $\mathcal{L}$ is Hopf and anti-Hopf and suitably flat, there is a bijection between left-left and right-left Yetter-Drinfeld module structures given by taking reverse comodule structures. We will omit the proof, as the proof is similar to [10] and [26].

**Theorem 5.7.** Let $L$ be a $N$-Hopf algebroid and $\mathcal{L}$ be a $B$-Hopf algebroid, if $P \in \mathrm{BiGal}(L, \mathcal{L})$, then we have the monoidal category equivalences

$$ (L\mathcal{Y}\mathcal{D}, \otimes_N) \simeq (L_P\mathcal{M}_P^{\mathcal{L}}, \otimes_P) \simeq (L\mathcal{Y}\mathcal{D}^{\mathcal{L}}, \diamond_B)^{\mathrm{sym}}. $$

*Proof.* The first equivalence can be given via

$$ L_P\mathcal{M}_P^{\mathcal{L}} \to L\mathcal{Y}\mathcal{D}, \quad M \mapsto M^{\mathrm{co}\mathcal{L}} \qquad (5.10) $$

$$ L\mathcal{Y}\mathcal{D} \to L_P\mathcal{M}_P^{\mathcal{L}}, \quad \Lambda \mapsto \Lambda \otimes_N P. \qquad (5.11) $$

We already know these assignments define an equivalence $L\mathcal{M} \to P\mathcal{M}_P^{\mathcal{L}}$ from Theorem 5.1. By the identification of $L$ with $L(P, \mathcal{L})$ the action of $L$ on $M^{\mathrm{co}\mathcal{L}}$ is given by

$$ \alpha \bullet \eta = \alpha^{<1>} \eta \alpha^{<2>}, $$

for any $\alpha \in L$ and $\eta \in M^{\mathrm{co}\mathcal{L}}$. If $M \in P\mathcal{M}_P^{\mathcal{L}}$ then the additional left $L$-comodule structure restricts to a left $L$-comodule structure on $M^{\mathrm{co}\mathcal{L}}$. Now, we can check the Yetter-Drinfeld conditions on $M^{\mathrm{co}\mathcal{L}}$. First, we can see $s_L(n) \bullet \eta = n\eta$ and $t_L(n) \bullet \eta = \eta n$ for any $n \in N$ and $\eta \in M^{\mathrm{co}\mathcal{L}}$. Second, for any $\alpha \in L$.

$$ \begin{aligned} (\alpha_{(1)} \bullet \eta)_{(-1)} \alpha_{(2)} \otimes_N (\alpha_{(1)} \bullet \eta)_{(0)} &= \alpha_{(1)}^{<1>} \eta_{(-1)} \alpha_{(1)}^{<2>} \alpha_{(2)} \otimes \alpha_{(1)}^{<1>} \eta_{(0)} \alpha_{(1)}^{<2>} \alpha_{(0)}^{<2>} \\ &= \alpha_{(1)}^{<1>} + \alpha_{(-1)}^{<1>} \eta_{(-1)} \alpha_{(1)} - \alpha_{(2)} \otimes \alpha_{(1)}^{<1>} + \alpha_{(0)}^{<1>} \eta_{(0)} \alpha_{(1)}^{<2>} \\ &= \alpha^{<1>} \eta_{(-1)} \otimes \alpha^{<1>} \eta_{(0)} \alpha_{(2)}^{<2>} \\ &= \alpha_{(1)} \eta_{(-1)} \otimes \alpha_{(2)}^{<1>} \eta_{(0)} \alpha_{(2)}^{<2>} \\ &= \alpha_{(1)} \eta_{(-1)} \otimes \alpha_{(2)} \bullet \eta_{(0)}, \end{aligned} $$

where the 2nd step uses (3.18), the 4th step uses (3.17).

For inverse equivalence 5.11, we already know from Theorem 5.1 that the $P$-bimodule structure on $\Lambda \otimes_N P$ is given by

$$ p(\eta \otimes_N q)p' = p_{(-1)} \bullet \eta \otimes_N p_{(0)}qp'. $$

We define the left $L$-comodule structure as the codiagonal coaction on $\Lambda \otimes_N P$. We know that $\Lambda \otimes_N P \in P\mathcal{M}_P^{\mathcal{L}}$, and it is obviously a bicomodule. So we only need to check $\Lambda \otimes_N P \in L_P\mathcal{M}_P$,

$$ \begin{aligned} L\delta(p(\eta \otimes q)p') &= (p_{(-2)} \bullet \eta)_{(-1)} p_{(-1)} q_{(-1)} p'_{(-1)} \otimes (p_{(-2)} \bullet \eta)_{(0)} \otimes p_{(0)} q_{(0)} p'_{(0)} \\ &= p_{(-2)} \eta_{(-1)} q_{(-1)} p'_{(-1)} \otimes p_{(-1)} \bullet \eta_{(0)} \otimes p_{(0)} q_{(0)} p'_{(0)}. \end{aligned} $$
It is easy to check that the adjunction morphisms in the category equivalence from Theorem 5.1 respect the additional L-structures. As for the monoidal functor structure, we only have to add that the isomorphism (5.3) is also an L-comodule map in our present situation.

The second equivalence can be given via the coopposite version of the first

$$ L_P M_P^L \to \mathcal{L} \mathcal{Y} \mathcal{D}^L, \quad M \mapsto \mathcal{c}^{\mathcal{L}} M \qquad (5.12) $$

$$ \mathcal{L} \mathcal{Y} \mathcal{D}^L \to L_P M_P^L, \quad \Lambda \mapsto \Lambda \otimes_{\bar{B}} P. \qquad (5.13) $$

More precisely, for (5.12), the left L-module action is given by

$$ X \bullet \rho = X^{[1]} \rho X^{[2]}, $$

for any $\rho \in \mathcal{c}^{\mathcal{L}} M$ and $X \in \mathcal{L}$. For (5.13), the balanced tensor product of $\Lambda \otimes_{\bar{B}} P$ is given by $s_L(b) \bullet \eta \otimes_{\bar{B}} p = \eta \otimes_{\bar{B}} \bar{b}p$, the left L-comodule structure is given by

$$ L \delta(\eta \otimes_{\bar{B}} p) = p_{(-1)} \otimes_N \eta \otimes_{\bar{B}} p_{(0)}. $$

the right $\mathcal{L}$-comodule structure is given by

$$ \delta_{\mathcal{L}}(\eta \otimes_{\bar{B}} p) = \eta_{(0)} \otimes_{\bar{B}} p_{(0)} \diamond_{B} \eta_{(1)} p_{(1)}, $$

The P-bimodule structure is given by

$$ p(\eta \otimes_{\bar{B}} q)p' = p_{(1)} \bullet \eta \otimes_{\bar{B}} p_{(0)}qp', $$

□

As a result, we have

**Lemma 5.8.** Let $L$ be a $N$-Hopf algebroid and $\mathcal{L}$ be a $B$-Hopf algebroid, if $P \in \text{BiGal}(L, \mathcal{L})$, then the Hopf bimodule category $(L_P M_P^L, \otimes_P)$ is a braided monoidal category, namely the center of both monoidal categories $(P M_P^L, \otimes_P)$ and $(L_P M_P, \otimes_P)$. The half-braidings are given by

$$ \sigma(m \otimes_P n) = m_{(-1)}^{<1>} n_{[0]} n_{[-1]}^{[1]} \otimes_P m_{(-1)}^{<2>} m_{(0)} n_{[-1]}^{[2]} \qquad (5.14) $$

$$ \sigma^{-1}(n \otimes_P m) = n_{(1)}^{[1]} m_{[0]} m_{[1]}^{<1>} \otimes_P n_{(1)}^{[2]} m_{(0)} m_{[1]}^{<2>}, \qquad (5.15) $$

defined for $m \in M \in \mathcal{L} M_P$ and $n \in M' \in P M_P^L$.

We omit the proof as it is similar to [26].

## 6. APPLICATIONS AND EXAMPLES

In this section, we will study the 2-cocycle twist theory of Hopf Galois extensions with the base algebra deformed by a 2-cocycle. Recall that [15],

**Definition 6.1.** Let $\mathcal{L}$ be a left $B$-bialgebroid. An left 2-cocycle on $\mathcal{L}$ is an element $\Gamma \in \text{Hom}_{\bar{B}}(\mathcal{L} \otimes_{\bar{B}} \mathcal{L}, B)$, such that

$$ (1) \quad \Gamma(X, \Gamma(Y_{(1)}, Z_{(1)})Y_{(2)}Z_{(2)}) = \Gamma(\Gamma(X_{(1)}, Y_{(1)})X_{(2)}Y_{(2)}, Z), $$

$$ (2) \quad \Gamma(1_{\mathcal{L}}, X) = \varepsilon(X) = \Gamma(X, 1_{\mathcal{L}}), $$

for all $X, Y, Z \in \mathcal{L}$.

Given a 2-cocycle $\Gamma$ on a left $B$-bialgebroid, we can deform the base algebra $B$ with a new product

$$ a \cdot_{\Gamma} b = \Gamma(a, b), $$

for any $a, b \in B$. We denote the new algebra by $B^{\Gamma}$. Moreover, for any left $\mathcal{L}$-comodule $M$, $M$ has a $B^{\Gamma}$-bimodule structure given by

$$ a \cdot_{\Gamma} m = \Gamma(a, m_{(-1)}) m_{(0)}, \quad m \cdot_{\Gamma} a = \Gamma(m_{(-1)}, a) m_{(0)}, \qquad (6.1) $$
for any $a \in B$ and $m \in M$. For any $N, M \in \mathcal{L}\mathcal{M}$, we can define $\Gamma^{\#} : M \otimes_{B^\Gamma} N \to M \otimes_B N$
given by

$$ \Gamma^{\#}(m \otimes_{B^{\Gamma}} n) = \Gamma(m_{(-1)}, n_{(-1)}) m_{(0)} \otimes_B n_{(0)}, $$

for any $m \in M, n \in N$. We call $\Gamma$ invertible, if $\Gamma^{\#}$ is invertible for any left $\mathcal{L}$-comodule $M, N$.

**Theorem 6.2.** [15] If $\Gamma$ is an invertible left 2-cocycle on a left $B$-Hopf algebroid then $\mathcal{L}\Gamma$ is a Hopf algebroid over $B^\Gamma$ with the $B^\Gamma$-ring structure

$$ X \cdot_{\Gamma} Y := \Gamma(X_{(1)}, Y_{(1)}) X_{(2)} + Y_{(2)} + \overline{\Gamma(Y_{(2)}, X_{(2)})}, $$

and $B^\Gamma$-coring structure

$$ \Delta^{\Gamma}(X) = \Gamma^{\#^{-1}}(X_{(1)} \diamond_{B} X_{(2)}), \quad \varepsilon^{\Gamma}(X) = \Gamma(X_{+}, X_{-}), $$

where $\Gamma^{\#} : \mathcal{L} \diamond_{B^\Gamma} \mathcal{L} \to \mathcal{L} \diamond_{B} \mathcal{L}$ given by

$$ \Gamma^{\#}(X \diamond_{B^{\Gamma}} Y) = X_{+} \overline{\Gamma(X_{-}, Y_{(1)})} \diamond_{B} Y_{(2)}, $$

is invertible as we consider the left term of $\mathcal{L} \diamond_{B} \mathcal{L}$ has the regular comodule structure induced by its coproduct. We denote the image of the twisted coproduct by

$$ \Delta^{\Gamma}(X) = X_{\{1\}} \diamond_{B^{\Gamma}} X_{\{2\}}, $$

for any $X \in \mathcal{L}$.

Let $M$ be a left $\mathcal{L}$-comodule, we can see $M$ is also a left $\mathcal{L}^{\Gamma}$ comodule with the coproduct given by

$$ \mathcal{L}^{\Gamma} \delta := \Gamma^{\#^{-1}} \circ \mathcal{L} \delta, $$

where $\Gamma^{\#} : \mathcal{L} \diamond_{B^\Gamma} M \to \mathcal{L} \diamond_{B} M$ given by

$$ \Gamma^{\#}(X \diamond_{B^{\Gamma}} m) = X_{+} \overline{\Gamma(X_{-}, m_{(-1)})} \diamond_{B} m_{(0)}, $$

is invertible as $\Gamma$ is invertible. We denote the image of the twisted coaction by

$$ \mathcal{L}^{\Gamma} \delta(m) = m_{\{-1\}} \diamond_{B^{\Gamma}} m_{\{0\}}, $$

**Proposition 6.3.** If $P$ is a left $\mathcal{L}$-comodule algebra then $P$ is a left $\mathcal{L}^{\Gamma}$-comodule algebra with a deformed product

$$ p \cdot_{\Gamma} q = \Gamma(p_{(-1)}, q_{(-1)}) p_{(0)} q_{(0)}, $$

for any $p, q \in P$. We denote the new algebra by $\Gamma P$.

*Proof.* We can check that the coaction is an algebra map. Indeed,

$$
\begin{align*}
\Gamma^{\#}(p_{\{-1\}}, \Gamma q_{\{-1\}} \diamond_{B^{\Gamma}} p_{\{0\}} \Gamma p_{\{0\}}) \\
&= \Gamma^{\#}(\Gamma(p_{\{-1\}}(1), q_{\{-1\}}(1)) p_{\{-1\}}(2) + q_{\{-1\}}(2) + \overline{\Gamma(q_{\{-1\}}(2), p_{\{-1\}}(2))}) \diamond_{B^{\Gamma}} \Gamma(p_{\{0\}}(-1), q_{\{0\}}(-1)) p_{\{0\}}(0) q_{\{0\}}(0)) \\
&= \Gamma(p_{\{-1\}}(1), q_{\{-1\}}(1)) p_{\{-1\}}(2) + \overline{\Gamma(q_{\{-1\}}(2), p_{\{-1\}}(2))} q_{\{-1\}}(2) + \overline{\Gamma(p_{\{0\}}(-1), q_{\{0\}}(-1))} q_{\{0\}}(1) \\
&= \Gamma(p_{\{-1\}}(1), q_{\{-1\}}(1)) p_{\{-1\}}(2) + \overline{\Gamma(p_{\{0\}}(-1), q_{\{0\}}(2))} q_{\{0\}}(1) + \overline{\Gamma(p_{\{0\}}(-2), q_{\{0\}}(2))} p_{\{0\}}(1) q_{\{0\}}(1) \\
&\qquad \diamond_{B} p_{\{0\}}(0) q_{\{0\}}(0) \\
&= \Gamma(p_{\{-1\}}(1), q_{\{-1\}}(1)) p_{\{-1\}}(2) + \overline{\Gamma(p_{\{0\}}(-1), q_{\{0\}}(2))} q_{\{0\}}(1) + \overline{\Gamma(p_{\{0\}}(-2), q_{\{0\}}(2))} p_{\{0\}}(1) q_{\{0\}}(1) \\
&\qquad \diamond_{B} p_{\{0\}}(0) q_{\{0\}}(0) \\
&= \Gamma(p_{\{-1\}}(1), q_{\{-1\}}(1)) p_{\{-1\}}(2) + \overline{\Gamma(p_{\{0\}}(-1), q_{\{0\}}(2))} p_{\{0\}}(1) q_{\{0\}}(1) + \overline{\Gamma(p_{\{0\}}(-2), q_{\{0\}}(2))} p_{\{0\}}(1) q_{\{0\}}(1) \\
&\qquad \diamond_{B} p_{\{0\}}(0) q_{\{0\}}(0) \\
&= \Gamma(p_{\{-1\}}(1), q_{\{-1\}}(1)) p_{\{-1\}}(2) + \overline{\Gamma(p_{\{0\}}(-1), q_{\{0\}}(2))} p_{\{0\}}(1) q_{\{0\}}(1) + \overline{\Gamma(p_{\{0\}}(-2), q_{\{0\}}(2))} p_{\{0\}}(1) q_{\{0\}}(1)
\end{align*}
$$
$$
\begin{align*}
& \overline{\Gamma(q_{\{-1\} (2)} - \Gamma(\Gamma(p_{\{-1\} (2)} - (1), p_{\{0\} (-3)}) p_{\{-1\} (2)} - (2) p_{\{0\} (-2)}, q_{\{0\} (-2)}) p_{\{-1\} (2)} - (3) p_{\{0\} (-1)} q_{\{0\} (-1)})} \\
& \diamond BP_{\{0\} (0)} q_{\{0\} (0)} \\
& = \Gamma(p_{\{-1\} (1)}, q_{\{-1\} (1)}) p_{\{-1\} (2)} + q_{\{-1\} (2)} + \\
& \overline{\Gamma(q_{\{-1\} (2)} - \Gamma(\Gamma(p_{\{-1\} (2)} - p_{\{0\} (-3)}) p_{\{-1\} (2)} + (1) p_{\{0\} (-2)}, q_{\{0\} (-2)}) p_{\{-1\} (2)} + (2) p_{\{0\} (-1)} q_{\{0\} (-1)})} \\
& \diamond BP_{\{0\} (0)} q_{\{0\} (0)} \\
& = \Gamma(p_{\{-3\}}, q_{\{-1\} (1)}) p_{\{-2\}} + q_{\{-1\} (2)} + \\
& \overline{\Gamma(q_{\{-1\} (2)} - \Gamma(p_{\{-2\}} - (1) p_{\{-1\} (1)}, q_{\{0\} (-2)}) p_{\{-2\}} - (2) p_{\{-1\} (2)} q_{\{0\} (-1)})} \diamond BP_{\{0\} (0)} q_{\{0\} (0)} \\
& = \Gamma(p_{\{-2\}}, q_{\{-1\} (1)}) p_{\{-1\}} q_{\{-1\} (2)} + \overline{\Gamma(q_{\{-1\} (2)} - q_{\{0\} (-1)})} \diamond BP_{\{0\} (0)} q_{\{0\} (0)} \\
& = \Gamma(p_{\{-2\}}, q_{\{-2\}}) p_{\{-1\}} q_{\{-1\}} \diamond BP_{\{0\}} q_{\{0\}} \\
& = \mathcal{L} \delta(p \cdot \Gamma q) = \Gamma^{\#} (\mathcal{L}^{\Gamma} \delta(p \cdot \Gamma q))
\end{align*}
$$

where the 7th and 9th steps use the fact that

$$
p_{\{-1\}} + \overline{\Gamma(p_{\{-1\}} - p_{\{0\}(-1)})} \diamond BP_{\{0\}(0)} = p_{\{-1\}} \diamond BP_{\{0\}}.
$$

□

**Lemma 6.4.** If $N \subseteq P$ is a left $\mathcal{L}$- Galois extension and $\Gamma$ is an invertible left 2-cocycle on $\mathcal{L}$ then $N \subseteq \Gamma P$ is a left $\mathcal{L}^{\Gamma}$- Galois extension.

*Proof.* First, we can see that the coinvariant subalgebra $N$ is not changed. It is sufficient to show the following diagram commute:

<figure><img src="image_14.png" /></figure>

On the one hand,

$$
\begin{align*}
\Gamma^{\#} \circ \mathcal{L}^{\Gamma} \operatorname{can}(p \otimes q) &= \Gamma^{\#} (p_{\{-1\}} \diamond_{B^{\Gamma}} p_{\{0\}} \cdot \Gamma q) \\
&= \Gamma^{\#} (p_{\{-1\}} \otimes \Gamma(p_{\{0\}(-1)}, q_{(-1)}) p_{\{0\}(0)} q_{(0)}) \\
&= p_{\{-1\}} + \overline{\Gamma(p_{\{-1\}} - \Gamma(p_{\{0\}(-2)}, q_{(-2)}) p_{\{0\}(-1)} q_{(-1)})} \diamond BP_{\{0\}(0)} q_{(0)} \\
&= p_{\{-1\}} + \overline{\Gamma(\Gamma(p_{\{-1\}} - (1), p_{\{0\}(-2)}) p_{\{-1\}} - (2) p_{\{0\}(-1)}, q_{(-1)})} \diamond BP_{\{0\}(0)} q_{(0)} \\
&= p_{\{-1\}} + \overline{\Gamma(p_{\{-1\}} - p_{\{0\}(-2)}) p_{\{-1\}} + (-p_{\{0\}(-1)}, q_{(-1)})} \diamond BP_{\{0\}(0)} q_{(0)} \\
&= p_{\{-2\}} + \overline{\Gamma(p_{\{-2\}} - p_{\{-1\}}, q_{(-1)})} \diamond BP_{\{0\}} q_{(0)} \\
&= p_{\{-1\}} \diamond BP_{\{0\}} q \\
&= \mathcal{L} \operatorname{can}(p \otimes q),
\end{align*}
$$

where the 6th step use the fact that

$$
p_{\{-1\}} + \overline{\Gamma(p_{\{-1\}} - p_{\{0\}(-1)})} \diamond BP_{\{0\}(0)} = p_{\{-1\}} \diamond BP_{\{0\}},
$$

for any $p \in P$.

□

**Lemma 6.5.** Let $N \subseteq P$ be a left $\mathcal{L}$-Galois extension and $\Gamma$ is a 2-cocycle on $\mathcal{L}$. Then $R(\Gamma P, \mathcal{L}^{\Gamma}) \cong R(P, \mathcal{L})$.

*Proof.* By Theorem 4.9, it is sufficient to show $\Gamma P \in \operatorname{BiGal}(\mathcal{L}^{\Gamma}, R(P, \mathcal{L}))$. First, we show that the $\Gamma P$ is a right $R(P, \mathcal{L})$-comodule algebra with the original coaction $\delta_R$. Indeed,

$$
\delta_R(p \cdot \Gamma q) = \delta_R(\Gamma(p_{(-1)}, q_{(-1)}) p_{(0)} q_{(0)}) = \Gamma(p_{(-1)}, q_{(-1)}) p_{(0)} q_{(0)} \otimes p_{(1)} q_{(1)} = \delta_R(p) \cdot \Gamma \delta_R(q).
$$
Second, we show $\Gamma B \subseteq \Gamma P$ is an anti-right $R(P, \mathcal{L})$-Galois extension. It is sufficient to show the following diagram commute:

<figure><img src="image_15.png" /></figure>

Third, we can show the left and right coaction cocommute. In other words, we need to show

$$ (\mathcal{L}^\Gamma \delta \diamond \bar{\mathcal{N}} \text{ id}) \circ \delta_R = (\text{id} \diamond_{B^\Gamma} \delta_R) \circ \mathcal{L}^\Gamma \delta. $$

So it is equivalent to show

$$ (\Gamma^{\#} \otimes \text{id}) \circ (\mathcal{L}^{\Gamma} \delta \diamond \bar{\mathcal{N}} \text{ id}) \circ \delta_R = (\Gamma^{\#} \otimes \text{id}) \circ (\text{id} \diamond_{B^{\Gamma}} \delta_R) \circ \mathcal{L}^{\Gamma} \delta. $$

On the one hand,

$$ (\Gamma^{\#} \otimes \text{id}) \circ (\mathcal{L}^{\Gamma} \delta \otimes \text{id}) \circ \delta_R(p) = p_{(-1)} \otimes p_{(0)} \otimes p_{(1)}. $$

On the other hand,

$$ \begin{align*} (\Gamma^{\#} \otimes \text{id}) \circ (\text{id} \diamond_{B^{\Gamma}} \delta_R) \circ \mathcal{L}^{\Gamma} \delta(p) & \\ &= p_{\{-1\}} + \overline{\Gamma(p_{\{-1\}} - p_{(0)\{-1\}})} \otimes p_{(0)\{0\}} \otimes p_{\{0\}(1)} \\ &= p_{\{-1\}} \otimes p_{(0)} \otimes p_{(1)}. \end{align*} $$

□

**6.1. Twisted Ehresmann Hopf algebroids.** Let $H$ be a Hopf algebra and $P$ be a right $H$-comodule algebra such that $B \subseteq P$ is a right $H$-Galois extension. Assume $K$ is another Hopf algebra, and $P$ is a $K$ comodule algebra such that $P$ is a $K$-$H$-bicomodule. If $\gamma: K \otimes K \to k$ is an invertible 2-cocycle then $\gamma P$ is a left $K^\gamma$ comodule algebra in the sense of Proposition 6.3. By [1], define $\gamma B := \gamma(P^{\text{coH}}) = (\gamma P)^{\text{coH}}$ being the deformed algebra of $B$ with product

$$ a \cdot_{\gamma} b = \gamma(a_{(-1)}, b_{(-1)})a_{(0)}b_{(0)}, \quad \forall a, b \in B, $$

we know $\gamma B \subseteq \gamma P$ is a $H$-Galois extension with the twisted translation map given by

$$ h^{(1)'} \otimes_{\gamma N} h^{(2)'} = \gamma^{-1}(h^{(1)}_{(-1)}, h^{(2)}_{(-1)}) h^{(1)}_{(0)} \otimes_{\gamma N} h^{(2)}_{(0)}, \quad (6.2) $$

where $h^{(1)} \otimes_{\gamma N} h^{(2)}$ is the image of the original translation map.

**Lemma 6.6.** For $(H, K, P, B)$ be as above and $\gamma$ be an invertible 2-cocycle on $K$, we can construct an invertible left 2-cocycle $\Gamma$ on $L(P, H)$ by

$$ \Gamma(p \otimes q, p' \otimes q') = \gamma(p_{(-1)}, p'_{(-1)}) p_{(0)} p'_{(0)} q', $$

for any $p \otimes q, p' \otimes q' \in L(P, H)$.

*Proof.* It is not hard to see $\Gamma$ factor through $\otimes_{\bar{B}}$ and left $\bar{B}$-linear. We can also see the image of $\Gamma$ belongs to $B$. Moreover, $\Gamma$ is unital as $\gamma$ is. Now, let's check $\Gamma$ satisfies the 2-cocycle condition. Let $X = p \otimes q, Y = p' \otimes q', Z = p'' \otimes q'' \in L(P, H)$. We have on the one hand,

$$ \begin{align*} \Gamma(\Gamma(X_{(1)}, Y_{(1)})X_{(2)}Y_{(2)}, Z) &= \Gamma((\gamma(p_{(-1)}, p'_{(-1)}))p_{(0)}p'_{(0)}p'_{(1)} p_{(1)}^{(1)} p_{(1)}^{(2)} \otimes q)(p'_{(1)}^{(2)} \otimes q'), p'' \otimes q'') \\ &= \Gamma(\gamma(p_{(-1)}, p'_{(-1)}))p_{(0)}p'_{(0)} \otimes q'q, p'' \otimes q'') \\ &= \gamma(\gamma(p_{(-2)}, p'_{(-2)}))p_{(-1)}p'_{(-1)}, p''_{(-1)} p_{(0)} p'_{(0)} p''_{(0)} q''q'q. \end{align*} $$

On the other hand,

$$ \Gamma(X, \Gamma(Y_{(1)}, Z_{(1)})Y_{(2)}Z_{(2)}) $$
$$
\begin{align*}
&= \Gamma(p \otimes q, \gamma(p'_{(-1)}, p''_{(-1)}) p'_{(0)} p''_{(0)} \otimes q''q') \\
&= \gamma(p_{(-1)}, \gamma(p'_{(-2)}, p''_{(-2)}) p'_{(-1)} p''_{(-1)}) p_{(0)} p'_{(0)} p''_{(0)} q''q'q.
\end{align*}
$$

They are equal as $\gamma$ is a 2-cocycle. Next, we will show $\Gamma^{\#} : M \otimes_B N \to M \otimes_B N$ is invertible for any left $L(P, H)$-comodules $M$ and $N$. We can see the map $\kappa\delta : L(P, H) \to K \otimes L(P, H)$, given by

$$
\kappa \delta(p \otimes q) = p_{(-1)} \otimes (p_{(0)} \otimes q), \quad \forall p \otimes q \in L(P, H),
$$

is a left $K$-coaction on $L(P, H)$. By applying the fact that $M \cong L(P, H) \square M$ as left $L(P, H)$-comodule, it is not hard to see $\Gamma^\#$ can be written explicitly by

$$
\Gamma^{\#}(m \otimes n) = \gamma(m_{(-1)(-1)}, n_{(-1)(-1)}) \varepsilon(m_{(-1)(0)} n_{(-1)(0)}) m_{(0)} \otimes n_{(0)}, \quad \forall m \in M, n \in M.
$$

Therefore, we can define $\Gamma^{\#^{-1}} : M \otimes_B N \to M \otimes_{B^\Gamma} N$ by

$$
\Gamma^{\#^{-1}}(m \otimes n) = \gamma^{-1}(m_{(-1)(-1)}, n_{(-1)(-1)}) \varepsilon(m_{(-1)(0)} n_{(-1)(0)}) m_{(0)} \otimes n_{(0)}, \quad \forall m \in M, n \in M.
$$

As $\gamma$ is convolution invertible, $\Gamma^\#$ is invertible. Indeed, let $p \otimes q \otimes m \in L(P, H) \square M$ and $p' \otimes q' \otimes n \in L(P, H) \square N$. We can see

$$
\Gamma^{\#}(p \otimes q \otimes m, p' \otimes q' \otimes n) = \gamma(p_{(-1)}, p'_{(-1)}) p_{(0)} p'_{(0)} q' q \otimes m \otimes n.
$$

As a result,

$$
\begin{align*}
\Gamma^{\#} \circ \Gamma^{\#^{-1}} (p \otimes q \otimes m, p' \otimes q' \otimes n)
&= \gamma^{-1}(p_{(-1)}, p'_{(-1)}) p_{(0)} p'_{(0)} p'_{(1)} p'_{(1)} \gamma(p_{(1)}^{(2)}(-1), p'_{(1)}^{(2)}(-1)) p_{(1)}^{(2)}(0) p'_{(1)}^{(2)}(0) q' q \otimes m \otimes n \\
&= \gamma^{-1}(p_{(-1)}, p'_{(-1)}) \gamma(p_{(0)}(-1), p'_{(0)}(-1)) p_{(0)}(0) p'_{(0)}(0) q' q \otimes m \otimes n \\
&= pp'q'q \otimes m \otimes n \\
&= 1 \otimes m \otimes n.
\end{align*}
$$

<figure><img src="image_16.png" /></figure>

**Theorem 6.7.** For $(H, K, P, B)$ as above and $\gamma$ be an invertible 2-cocycle on $K$, we have $L(\gamma P, H) = L(P, H)^\Gamma$, where $\Gamma$ is an invertible left 2-cocycle given by Lemma 6.6.

*Proof.* We first observe that $B^\Gamma = \gamma B$. Indeed,

$$
a \cdot \Gamma b = \Gamma(a, b) = \gamma(a_{(-1)}, b_{(-1)}) a_{(0)} b_{(0)} = a \cdot \gamma b,
$$

for any $a, b \in B$. We can see the $B^{\Gamma^e}$-ring structure on $L(P, H)^{\Gamma}$ is given by

$$
b \cdot \Gamma (p \otimes q) = \gamma(b, p_{(-1)}) b_{(0)} p_{(0)} \otimes q; \quad (p \otimes q) \cdot \Gamma b = \gamma(p_{(-1)}, b) p_{(0)} b_{(0)} \otimes q,
$$

and

$$
\begin{align*}
\bar{b} \cdot \Gamma (p \otimes q) &= (p \otimes q_{(1)}^{(2)}) \overline{\Gamma((q_{(0)} \otimes q_{(1)}^{(1)}), b)} \\
&= p \otimes \gamma(q_{(-1)}, b_{(-1)}) q_{(0)} b_{(0)} q_{(1)}^{(1)} q_{(1)}^{(2)} \\
&= p \otimes \gamma(q_{(-1)}, b_{(-1)}) q_{(0)} b_{(0)},
\end{align*}
$$

where we use

$$
(p \otimes q)_+ \otimes_{\bar{B}} (p \otimes q)_- = p \otimes q_{(1)}^{(2)} \otimes_{\bar{B}} q_{(0)} \otimes q_{(1)}^{(1)}
$$

in the 2nd step which is shown in [13]. Similarly,

$$
(p \otimes q) \cdot \Gamma \bar{b} = p \otimes \gamma(b_{(-1)}, q_{(-1)}) b_{(0)} q_{(0)}.
$$

We can also see,

$$
\begin{align*}
(p \otimes q) \cdot \Gamma(p' \otimes q')
&= \Gamma((p_{(0)} \otimes p_{(1)}^{(1)}), (p'_{(0)} \otimes p'_{(1)}^{(1)})) (p_{(1)}^{(2)} p'_{(1)}^{(2)} \otimes q'_{(1)}^{(2)} q_{(1)}^{(2)}) \overline{\Gamma((q'_{(0)} \otimes q'_{(1)}^{(1)}), (q_{(0)} \otimes q_{(1)}^{(1)}))} \\
&= \gamma(p_{(-1)}, p'_{(-1)}) p_{(0)} p'_{(0)} p'_{(1)}^{(1)} p_{(1)}^{(1)} p_{(1)}^{(2)} p'_{(1)}^{(2)} \otimes \gamma(q'_{(-1)}, q_{(-1)}) q'_{(0)} q_{(0)} q_{(1)}^{(1)} q'_{(1)}^{(1)} q'_{(1)}^{(2)} q_{(1)}^{(2)} \\
&= \gamma(p_{(-1)}, p'_{(-1)}) p_{(0)} p'_{(0)} \otimes \gamma(q'_{(-1)}, q_{(-1)}) q'_{(0)} q_{(0)}.
\end{align*}
$$
So $L(\gamma P, H) = L(P, H)^\Gamma$ as a $\gamma B^e$-ring. Now, let's check that they have the same coring structure. We can see

$$
\begin{align*}
\varepsilon^{\Gamma}(p \otimes q) &= \Gamma((p \otimes q_{(1)}^{(2)}), (q_{(0)} \otimes q_{(1)}^{(1)})) = \gamma(p_{(-1)}, q_{(-1)}) p_{(0)} q_{(0)} q_{(1)}^{(1)} q_{(1)}^{(2)} \\
&= \gamma(p_{(-1)}, q_{(-1)}) p_{(0)} q_{(0)}.
\end{align*}
$$

We can also see

$$
\Delta^{\Gamma}(p \otimes q) = p_{(0)} \otimes p_{(1)}^{(1)}_{(0)} \gamma^{-1}(p_{(1)}^{(1)}_{(-1)}, p_{(1)}^{(2)}_{(-1)}) \diamond_{B^{\Gamma}} p_{(1)}^{(2)}_{(0)} \otimes q.
$$

Indeed, by applying $\Gamma^\#$ on the right hand side, we get

$$
\begin{align*}
& (p_{(0)} \otimes p_{(1)}^{(1)}_{(1)}{}^{(2)}) \gamma^{-1} (p_{(1)}^{(1)}_{(-1)}, p_{(1)}^{(2)}_{(-1)}) \overline{\Gamma((p_{(1)}^{(1)}_{(0)} \otimes p_{(1)}^{(1)}_{(1)}{}^{(1)}), (p_{(1)}^{(2)}_{(0)} \otimes p_{(1)}^{(2)}_{(1)}{}^{(1)}))} \\
& \qquad \diamond_B p_{(1)}^{(2)}{}_{(1)}^{(2)} \otimes q \\
&= (p_{(0)} \otimes p_{(1)}^{(1)}{}_{(0)} p_{(1)}^{(2)}{}_{(0)} p_{(1)}^{(2)}{}_{(1)}{}^{(1)} p_{(1)}^{(1)}{}_{(1)}{}^{(1)} p_{(1)}^{(1)}{}_{(1)}{}^{(2)}) \\
& \qquad \gamma^{-1} (p_{(1)}^{(1)}{}_{(-2)}, p_{(1)}^{(2)}{}_{(-2)}) \gamma (p_{(1)}^{(1)}{}_{(-1)}, p_{(1)}^{(2)}{}_{(-1)}) \diamond_B p_{(1)}^{(2)}{}_{(1)}^{(2)} \otimes q \\
&= p_{(0)} \otimes p_{(1)}^{(1)}{}_{(1)} \diamond_B p_{(1)}^{(2)}{}_{(1)} \otimes q \\
&= \Delta(p \otimes q).
\end{align*}
$$

□

Let $H$ be a Hopf algebra and $B \subseteq P$ be a right $H$-Galois extension. If $\sigma : H \otimes H \to k$
is a 2-cocycle, by [20] $B \subseteq P_\sigma$ is a $H^\sigma$-Galois extension with the original $H$-coaction on
$P$ and a twisted product

$$
p \cdot_{\sigma} q = p_{(0)} q_{(0)} \sigma^{-1}(p_{(1)}, q_{(1)}),
$$

for any $p, q \in P$. By [1], if $(H, K, P, B)$ is as above, $\gamma$ is a 2-cocycle on $K$ and $\sigma$ is
a 2-cocycle on $H$, then $\gamma(P_\sigma) = (\gamma P)_\sigma =: \gamma P_\sigma$. Moreover, $\gamma B \subseteq \gamma P_\sigma$ is a $H^\sigma$-Galois
extension. As a result of Lemma 6.5 and Theorem 6.7, we have

**Corollary 6.8.** If $(H, K, P, B)$ is as above, $\gamma$ is a 2-cocycle on $K$ and $\sigma$ is a 2-cocycle on $H$, then $L(\gamma P_\sigma, H^\sigma) \cong L(\gamma P, H) = L(P, H)^\Gamma$.

6.2. Examples.

6.2.1. *The SU(2) principal fibration.* Let $P = A(S^7)$, $B = A(S^4)$, $H = A(SU(2))$ and $K = A(\mathbb{T}^2)$. More precisely, $P = A(S_\theta^7)$ is generated by elements $z_a, z_a^*$, $a = 1, \dots, 4$ with the spherical relation $\sum_a z_a^* z_a = 1$. The Hopf algebra is $A(SU(2))$ is an unital complex $*$-algebra generated by $\omega_1, \bar{\omega}_1, \omega_2, \bar{\omega}_2$ subject to the relation $\omega_1 \bar{\omega}_1 + \omega_2 \bar{\omega}_2 = 1$. The coproduct, counit and antipode is given by:

$$
\Delta : \begin{pmatrix} \omega_1 & -\bar{\omega}_2 \\ \omega_2 & \bar{\omega}_1 \end{pmatrix} \mapsto \begin{pmatrix} \omega_1 & -\bar{\omega}_2 \\ \omega_2 & \bar{\omega}_1 \end{pmatrix} \otimes \begin{pmatrix} \omega_1 & -\bar{\omega}_2 \\ \omega_2 & \bar{\omega}_1 \end{pmatrix},
$$

with counit $\varepsilon(\omega_1) = \varepsilon(\bar{\omega}_1) = 1$, $\varepsilon(\omega_2) = \varepsilon(\bar{\omega}_2) = 0$ and antipode $S(\omega_1) = \bar{\omega}_1$, $S(\omega_2) = -\omega_2$. If we denote $A(S^7)$ and $A(SU(2))$ by matrix-valued function by

$$
\Psi = \begin{pmatrix} z_1 & -z_2^* \\ z_2 & z_1^* \\ z_3 & -z_4^* \\ z_4 & z_3^* \end{pmatrix}, \quad \omega = \begin{pmatrix} \omega_1 & -\bar{\omega}_2 \\ \omega_2 & \bar{\omega}_1 \end{pmatrix},
$$

then the right coaction can be written as $\delta(\Psi) = \Psi \otimes \omega$. This means in components
$\delta(\Psi_{ik}) = \Psi_{ij} \otimes \omega_{jk}$. The algebra $A(S^4)$ generated by $\zeta_1 = z_1 z_3^* + z_2^* z_4$, $\zeta_2 = z_2 z_3^* - z_1^* z_4$
and $\zeta_0 = z_1 z_1^* + z_2^* z_2 = 1 - z_3 z_3^* - z_4^* z_4$ is the subalgebra of coinvariant. Moreover,
$A(S^4) \subseteq A(S^7)$ is a right $A(SU(2))$-Galois extension. Let $t_i, t_i^*, i = 1, 2$ be the generators
of $A(\mathbb{T}^2)$, the left $K$-coaction of $P$ is given by

$$
K \delta(z_i) = \tau_i \otimes z_i, \quad K \delta(z_i^*) = \tau_i^* \otimes z_i^*,
$$
where $\tau_i = (t_1, t_1^*, t_2, t_2^*)$. The left $K$-coaction on $B$ is given by

$$ {}^K\delta(\zeta_1) = t_1 t_2^* \otimes \zeta_1, \quad {}^K\delta(\zeta_2) = t_1^* t_2^* \otimes \zeta_1, \quad {}^K\delta(\zeta_0) = 1 \otimes \zeta_0. $$

Define $\gamma: K \otimes K \to k$ a 2-cocycle which is given by

$$ \gamma(t_i, t_j) = e^{i\pi\Theta_{ij}}, \quad \Theta_{ij} = \frac{1}{2} \begin{pmatrix} 0 & \theta \\ \theta & 0 \end{pmatrix}. \qquad (6.3) $$

By [1], $\gamma P = A(S_\theta^7)$ and $\gamma B = A(S_\theta^4)$. It is given in [18] that $A(S_\theta^4) \subseteq A(S_\theta^7)$ is a right $A(SU(2))$-Galois extension. More precisely, $A(S_\theta^7)$ is generated by elements $z_a, z_a^*$, $a = 1, \dots, 4$, subject to relations

$$ z_a z_b = \lambda_{ab} z_b z_a, \quad z_a z_b^* = \lambda_{ba} z_b^* z_a, \quad z_a^* z_b^* = \lambda_{ab} z_b^* z_a^*, \qquad (6.4) $$

and with the spherical relation $\sum_a z_a^* z_a = 1$, where $\lambda_{ab} = e^{2\pi i \theta_{ab}}$ and $(\theta_{ab})$ a real antisymmetric matrix given by

$$ \lambda_{ab} = \begin{pmatrix} 1 & 1 & \bar{\mu} & \mu \\ 1 & 1 & \mu & \bar{\mu} \\ \mu & \bar{\mu} & 1 & 1 \\ \bar{\mu} & \mu & 1 & 1 \end{pmatrix}, \quad \mu = \sqrt{\lambda} \quad \text{or} \quad \theta_{ab} = \frac{\theta}{2} \begin{pmatrix} 0 & 0 & -1 & 1 \\ 0 & 0 & 1 & -1 \\ 1 & -1 & 0 & 0 \\ -1 & 1 & 0 & 0 \end{pmatrix}. \qquad (6.5) $$

The generators of $A(S_\theta^4)$ has the relations

$$ \zeta_1 \zeta_2 = \lambda \zeta_2 \zeta_1, \quad \zeta_1 \zeta_2^* = \bar{\lambda} \zeta_2^* \zeta_1, \quad \zeta_1^* \zeta_1 + \zeta_2^* \zeta_2 = \zeta_0(1 - \zeta_0). $$

It is given by [12] that $L(A(S_\theta^7), A(SU(2)))$ has generator $p \otimes 1, 1 \otimes q$ and $V := \Psi \otimes \Psi^\dagger$, where

$$ p = \Psi \cdot \Psi^\dagger = \begin{pmatrix} \zeta_0 & 0 & \zeta_1 & -\bar{\mu}\zeta_2^* \\ 0 & \zeta_0 & \zeta_2 & \mu\zeta_1^* \\ \zeta_1^* & \zeta_2^* & 1-\zeta_0 & 0 \\ -\mu\zeta_2 & \bar{\mu}\zeta_1 & 0 & 1-\zeta_0 \end{pmatrix}, \quad q = \Psi \cdot_{op} \Psi^\dagger = \begin{pmatrix} \zeta_0 & 0 & \bar{\mu}\zeta_1 & -\zeta_2^* \\ 0 & \zeta_0 & \mu\zeta_2 & \zeta_1^* \\ \mu\zeta_1^* & \bar{\mu}\zeta_2^* & 1-\zeta_0 & 0 \\ -\zeta_2 & \zeta_1 & 0 & 1-\zeta_0 \end{pmatrix}. \qquad (6.6) $$

By Theorem 6.7, we have

**Lemma 6.9.** Let $P = A(S^7)$, $B = A(S^4)$, $H = A(SU(2))$, $K = A(\mathbb{T}^2)$ and $\gamma$ be the 2-cocycle given in (6.3). Then $L(A(S^7), A(SU(2))) = L(A(S_\theta^7), A(SU(2)))$, where $\Gamma$ is an invertible left 2-cocycle given by Lemma 6.6.

6.2.2. Quantum homogeneous spaces of quantum groups $SO_\theta(2n + 1, \mathbb{R})[9, 33]$. Let $P = A(SO(2n + 1, \mathbb{R}))$, $B = A(S^{2n})$, $H = A(SO(2n, \mathbb{R}))$ and $K = A(\mathbb{T}^n)$. We first define $A(M(2n, \mathbb{R}))$ being a bialgebra generated by elements $\mathbf{a} = (a_{jk})$, $\mathbf{b} = (b_{jk})$, $\mathbf{a}^* = (a_{jk}^*)$, $\mathbf{b}^* = (b_{jk}^*)$, $i, j, k = 1, \dots, n$. In matrix notation $A(M(2n, \mathbb{R}))$ has coproduct and counit given by

$$ M = (M_{JK}) = \begin{pmatrix} \mathbf{a} & \mathbf{b} \\ \mathbf{b}^* & \mathbf{a}^* \end{pmatrix}, \quad \Delta(M) = M \otimes M, \quad \varepsilon(M) = 1_{2n}. \qquad (6.7) $$

The Hopf algebra $A(SO(2n, \mathbb{R}))$ is given by the quotient of $A(M(2n, \mathbb{R}))$ by a Hopf ideal $I_Q$ which is given by

$$ I_Q = \langle M^t QM - Q, MQM^t - Q, \det(M) - 1 \rangle, \quad Q = \begin{pmatrix} 0 & 1_n \\ 1_n & 0 \end{pmatrix} = Q^{-1}. \qquad (6.8) $$

Similarly, $A(M(2n + 1, \mathbb{R}))$ is generated by $\mathbf{a} = (a_{jk})$, $\mathbf{b} = (b_{jk})$, $\mathbf{a}^* = (a_{jk}^*)$, $\mathbf{b}^* = (b_{jk}^*)$, and $u_i, u_i^*, v_i, v_i^*, i, j, k = 1, \dots, n$ and a hermitian scalar $x$. In matrix notation the coproduct and counit are given by

$$ N = \begin{pmatrix} \mathbf{a} & \mathbf{b} & \mathbf{u} \\ \mathbf{b}^* & \mathbf{a}^* & \mathbf{u}^* \\ \mathbf{v} & \mathbf{v}^* & x \end{pmatrix}, \quad \Delta(N) = N \otimes N, \quad \varepsilon(N) = 1_{2n+1}. \qquad (6.9) $$
The Hopf algebra $A(SO(2n + 1, \mathbb{R}))$ is given by the quotient of $A(M(2n + 1, \mathbb{R}))$ by a
Hopf ideal $J_Q$, which is given by

$$
J_Q = \langle N^t QN \rangle = Q, \quad NQN^t = Q, \quad \det(N) = 1 \rangle, \quad Q = \begin{pmatrix} 0 & 1_n & 0 \\ 1_n & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix} = Q^{-1}. \tag{6.10}
$$

The Hopf algebra $K$ is generated by $t_i, t_i^*$ subject to the relation $\sum_i t_i t_i^* = 1$. The coproduct and counit are

$$
\Delta(T) = T \otimes T, \quad \varepsilon(T) = 1_{2n}, \quad T := \mathrm{diag}\{t_1, \dots, t_n, t_1^*, \dots, t_n^*\}.
$$

There is a surjective Hopf algebra morphism $\pi : P \to H$ given by

$$
\pi : \begin{pmatrix} \mathbf{a} & \mathbf{b} & \mathbf{u} \\ \mathbf{b}^* & \mathbf{a}^* & \mathbf{u}^* \\ \mathbf{v} & \mathbf{v}^* & x \end{pmatrix} \mapsto \begin{pmatrix} \mathbf{a} & \mathbf{b} & 0 \\ \mathbf{b}^* & \mathbf{a}^* & 0 \\ 0 & 0 & 1 \end{pmatrix} \qquad (6.11)
$$

Therefore, $P$ is a right $H$-comodule algebra with coaction given by $\delta^H = (\mathrm{id} \otimes \pi) \circ \Delta : P \to P \otimes H$. The algebra $B = A(S^{2n})$ is generated by $u_i, u_i^*$ and $x$ subject to the relation $\sum_i 2u_i u_i^* + x^2 = 1$. Moreover, $B = A(S^{2n}) \subseteq P$ is a $H$-Galois extension. There is another surjective Hopf algebra morphism $\pi' : H \to K$ given by

$$
\pi' : a_{ij} \mapsto \delta_{ij} t_i, \quad a_{ij}^* \mapsto \delta_{ij} t_i^*, \quad b_{ij} \mapsto 0, \quad b_{ij}^* \mapsto 0.
$$

Therefore, $P$ is a left $K$-comodule algebra with coaction given by $\delta^H = (\pi' \circ \pi \otimes \mathrm{id}) \circ \Delta : P \to K \otimes P$.

We can define a 2-cocycle $\gamma$ of $K$ on the generators by

$$
\gamma(t_i, t_j) = e^{i\pi\theta_{ij}}, \quad \theta_{ij} = -\theta_{ji}. \tag{6.12}
$$

There is a 2-cocycle $\sigma$ on $H$ given by $\sigma(g, h) = \gamma(\pi'(g), \pi'(h))$ for any $g, h \in H$. By [1], $\gamma P_\sigma = A(SO_\theta(2n+1, \mathbb{R}))$, $H^\sigma = A(SO_\theta(2n, \mathbb{R}))$ and $\gamma B = A(S_\theta^{2n})$. It is given by [9, 33], $A(S_\theta^{2n}) \subseteq A(SO_\theta(2n+1, \mathbb{R}))$ is a $A(SO_\theta(2n, \mathbb{R}))$-Galois extension. More precisely, the generators $\mathbf{a} = (a_{jk})$, $\mathbf{b} = (b_{jk})$, $\mathbf{a}^* = (a_{jk}^*)$, $\mathbf{b}^* = (b_{jk}^*)$ of $A(SO_\theta(2n, \mathbb{R}))$ satisfy the relations:

$$
\begin{align}
a_{ij}a_{kl} &= \lambda_{ik}\lambda_{lj} a_{kl}a_{ij}, & a_{ij}b_{kl}^* &= \lambda_{ki}\lambda_{lj} b_{kl}^*a_{ij} \nonumber \\
a_{ij}b_{kl} &= \lambda_{ik}\lambda_{jl} b_{kl}a_{ij}, & a_{ij}a_{kl}^* &= \lambda_{ki}\lambda_{jl} a_{kl}^*a_{ij} \nonumber \\
b_{ij}b_{kl} &= \lambda_{ik}\lambda_{lj} b_{kl}b_{ij}, & b_{ij}b_{kl}^* &= \lambda_{ki}\lambda_{jl} b_{kl}^*b_{ij} \tag{6.13}
\end{align}
$$

together with their *-conjugated. The commutation relations of $A(SO_\theta(2n + 1, \mathbb{R}))$ is
given by

$$
N_{IJ} N_{KL} = \lambda_{IK} \lambda_{LJ} N_{KL} N_{IJ}, \quad (6.14)
$$

where $\lambda_{IK}$ is induced by the 2-cocycle $\gamma$ and $I, J, K, L = 1, \dots, 2n$. The generators $u_i, u_i^*, x$ of $S_\theta^{2n}$ satisfy (6.14):

$$
u_i u_j = \lambda_{ij} u_j u_i, \quad u_i^* u_j^* = \lambda_{ij} u_j^* u_i^*, \quad u_i u_j^* = \lambda_{ji} u_j^* u_i, \qquad (6.15)
$$

and $x$ central. The orthogonality conditions imply the sphere relation

$$
\sum_{j=1}^{n} 2u_j^* u_j + x^2 = 1.
$$

It is given by [12] that $L(A(SO_\theta(2n + 1, \mathbb{R})), A(SO_\theta(2n, \mathbb{R})))$ is generated by elements
$1 \otimes \mathbf{u}, 1 \otimes \mathbf{u}^*, 1 \otimes x$, and $\mathbf{u} \otimes 1, \mathbf{u}^* \otimes 1, x \otimes 1$, together with the entries of the matrix

$$
\mathbf{V} = \begin{pmatrix} \mathbf{a} & \mathbf{b} \\ \mathbf{b}^* & \mathbf{a}^* \\ \mathbf{v} & \mathbf{v}^* \end{pmatrix} \otimes \begin{pmatrix} \mathbf{a} & \mathbf{b} \\ \mathbf{b}^* & \mathbf{a}^* \\ \mathbf{v} & \mathbf{v}^* \end{pmatrix}^{\dagger},
$$

By Corollary 6.8, we have
**Lemma 6.10.** Let $P = A(SO_θ(2n + 1, \mathbb{R}))$, $B = A(S^{2n})$, $H = A(SO_θ(2n, \mathbb{R}))$, $K = A(\mathbb{T}^{2n})$ and $γ$ be the 2-cocycle on $A(\mathbb{T}^{2n})$ in (6.12). Then $L(A(SO(2n+1, \mathbb{R})), A(SO(2n, \mathbb{R})))^Γ \cong L(A(SO_θ(2n + 1, \mathbb{R})), A(SO_θ(2n, \mathbb{R})))$, where $Γ$ is an invertible left 2-cocyle given by Lemma 6.6.

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QUEEN MARY UNIVERSITY OF LONDON

Email address: x.h.han@qmul.ac.uk

UNIVERSITÉ BOURGOGNE EUROPE, CNRS, IMB UMR 5584, 21000 DIJON, FRANCE

Email address: peter.schauenburg@ube.fr