Abstract
The fundamental notion of separability for commutative algebras was interpreted in categorical setting where also the stronger notion of heavily separability was introduced. These notions were extended to (co)algebras in monoidal categories, in particular to cowreaths. In this paper, we consider the cowreath \( \left( A\otimes H_{4}^{op}, H_{4}, \psi \right) \), where \(H_{4}\) is the Sweedler 4-dimensional Hopf algebra over a field k and \(A=Cl(\alpha , \beta , \gamma )\) is the Clifford algebra generated by two elements G, X with relations \(G^{2}=\alpha \), \(X^{2}=\beta \) and \(XG+GX=\gamma \), \( (\alpha , \beta , \gamma \in k \)) which becomes naturally an \(H_{4}\)-comodule algebra. We show that, when \(\textrm{char}\left( k \right) \ne 2, \) this cowreath is always separable and h-separable as well.
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1 Introduction
The general theory of separable algebras over commutative rings was introduced by Auslander and Goldmann [1] . This fundamental concept has many applications in almost all areas of algebra (cf. [10]). The classical notion of separability for algebras over commutative rings can be interpreted in categorical terms. This was done in [13] where the notion of separable functor was introduced and investigated. Recently, motivated by an example related to the tensor algebra, in [2], a stronger version of separable functor, called heavely separable, was introduced and investigated.
The notion of separability can be extended to (co)algebras in monoidal categories (see [6]). In [12], we investigated h-coseparable coalgebras in monoidal categories. In particular, we showed that, for a (strict) monoidal category \(\mathcal {C}\) with \(\otimes \)-generator unit object, the forgetful functor \(F:\mathcal {C }^{C}\rightarrow \mathcal {C}\) is (heavily) separable if and only if the coalgebra C is (heavily) coseparable i.e. there exists a morphism \( B:C\otimes C\rightarrow 1\) in \(\mathcal {C}\) such that conditions
hold.
Given an algebra A in a monoidal category \(\mathcal {C}\) one can construct a new category \(\mathcal {T}_{A}^{\#}\). The objects of this category are pair \((X, \psi )\), where X is an object in \(\mathcal {C}\) and \(\psi :X\otimes A\rightarrow A\otimes X\) is a morphism in \(\mathcal {C}\) compatible with algebra structure of A. A triple \((A, X, \psi )\) is called a cowreath in \( \mathcal {C}\) if \(\left( X, \psi \right) \) a coalgebra in \(\mathcal {T} _{A}^{\#} \). This is a kind of generalized entwined structure (as introduced by Brzezinski in [3] in order to develop Galois theory for coalgebras) and its module category, the category \(\mathcal {C}\left( \psi \right) _{A}^{X}\) of entwined modules over cowreath, include other well-know structures such that relative Hopf modules, Doi–Hopf modules and Yetter-Drinfeld modules. This structure is fundamental for developing a Galois theory for quasi-Hopf algebras (see details in [8]). In [9], separable functors for the category of Doi–Hopf modules were studied and generalizations for cowreath have been given in [4, 5] and for Galois cowreath in [7]. Then the h-separability was considered in [12]. We proved ( [12, Theorem 3.6]) that for a cowreath \(\left( A, X, \psi \right) \) in \(\mathcal {C}\), the h-separability of the forgetful functor \(F:\mathcal {C}^{C}\rightarrow \mathcal {C}\) is equivalent to the h-separability of the coalgebra \(\left( X, \psi \right) \) in \(\mathcal {T}_{A}^{\#}\).
In this paper, we continue the study of cowreaths constructed on two-sided Hopf modules (see [5]). Namely, let H be a Hopf algebra over a field k and let A be a right H-comodule algebra. Thus \(A\otimes H^{op}\) is a right \(H\otimes H^{op}\)-comodule algebra and H is a \(H\otimes H^{op}\)-module coalgebra and we can construct a cowreath \((A\otimes H^{op}, H) \) in the category of k-vector space. In [12, Theorem 5.1], we determined the separability and h-separability of a cowreath \((A\otimes H^{op}, H, \psi )\).
In [11], A. Masuoka classified cleft extensions for the \(H=H_{4}\) the Sweedler 4-dimensional Hopf algebra over a field k. Naturally, the k-algebra \(A=Cl(\alpha , \beta , \gamma )\) generated by two elements G, X and relations \(G^{2}=\alpha \), \(X^{2}=\beta \) and \(XG+GX=\gamma \), a generalization of the Clifford algebra appeared (note that the usual 4 -dimensional Clifford algebra is one of this type).
We shown that the cowreath \(\left( A\otimes H^{op}, H, \psi \right) \) is always separable and h-separable. Namely in Theorem 1 (resp. in Theorem 2) we determine all the conditions the bilinear form B should satisfy.
2 Preliminaries
In the following we will adopt all definitions and notations in [4] except for the unit of an algebra A, that we will denote by \(u:{\underline{1}}\rightarrow A\).
Let \(\mathcal {C}\) be a (strict) monoidal category and let \(\left( A, m, u\right) \) be an algebra in \(\mathcal {C}\). Recall that a (right) transfer morphism through A is a pair \(\left( X, \psi \right) \) with \(X\in \mathcal {C}\) and
such that
The category \(\mathcal {T}_{A}^{\#}\) has objects the right transfers. A morphism \(f:X\rightarrow Y\) in \(\mathcal {T}_{A}^{\#}\) is a morphism \( f:X\rightarrow A\otimes Y\) in \(\mathcal {C}\) such that
The composition of two morphisms \(f:X\rightarrow Y\) and \(g:Y\rightarrow Z\) in \(\mathcal {T}_{A}^{\#}\) is
and
The tensor product of \(\left( X, \psi _{X}\right) \) and \(\left( Y, \psi _{Y}\right) \) is
The tensor product of \(f:X\rightarrow X^{\prime }\) and \(g:Y\rightarrow Y^{\prime }\) is
The unit object is
Recall also that a cowreath in \(\mathcal {C}\) is a triple \(\left( A, X, \psi \right) \) where A is an algebra in \(\mathcal {C}\) and \(\left( X, \psi \right) \) is a coalgebra in \(\mathcal {T}_{A}^{\#}.\) This means that \(\left( X, \psi \right) \in \mathcal {T}_{A}^{\#}\) and there are morphisms
such that
We also recall that an entwined module over a cowreath \(\left( A, X, \psi \right) \) is a pair \(\left( M, \rho :M\rightarrow M\otimes X\right) \) where \(\left( M, \mu \right) \in \mathcal {C}_{A}\) satisfying
A morphism between entwined modules is a A-linear morphism \(f:M\rightarrow N\) such that \(\left( f\otimes X\right) \circ \rho =\rho \circ f.\)
The category of entwined modules will be denoted by \(\mathcal {C}\left( \psi \right) _{A}^{X}.\)
Let H be a Hopf algebra, A a right H-comodule algebra. \(A\otimes H^{op}\) is a right \(H\otimes H^{op}\)-comodule algebra with
H is a right \(H\otimes H^{op}\)-module coalgebra via
and H can be seen as a coalgebra in \({\mathcal {M}}_{H\otimes H^{op}}\)=\(_{H} {\mathcal {M}}_{H}.\) We consider
Then \(\left( H, \psi \right) \in \mathcal {T}_{A\otimes H^{op}}^{\#}\) and \( \left( H, \psi \right) \) is a coalgebra in \(\mathcal {T}_{A\otimes H^{op}}^{\#} \) via
The category of Doi–Hopf modules \({\mathcal {M}}(H\otimes H^{op})_{A\otimes H^{op}}^{H}\) is isomorphic to the category \(_{H}{\mathcal {M}}_{A}^{H}\) of two-sided (H, A)-bimodules over H.
A Casimir morphism consists of a k-linear map \(B:H\otimes H\rightarrow A\otimes H^{op}\) with the following properties:
1. Casimir condition
2. Morphism condition
i.e. B is a morphism in \(\mathcal {T}_{A\otimes H^{op}}^{\#}\):
and we consider also the condition
3. Normalized condition
We denote by \(H_{4}\) the Sweedler 4-dimensional Hopf algebra. It is generated by g, x with the relations \(g^{2}=1_H\), \(x^{2}=0\) and \(xg=-gx\). The coalgebra structure is given by \(\Delta (g)=g\otimes g, \Delta (x)=x\otimes g+1_H\otimes x, \epsilon (g)=1\) and \(\epsilon (x)=0\) and the antipode \(S(g)=g, S(x)=gx\). It is well-know that \(t=x+gx\) (resp. \(r=x-gx\)) is left (resp. right) integral in \(H_{4}\).
We consider the Clifford algebra \(A=Cl(\alpha , \beta , \gamma )\) generated by G, X with the relations \(G^{2}=\alpha \), \(X^{2}=\beta \) and \(XG+GX=\gamma \) . This Clifford algebra is a right \(H=H_{4}\)-comodule algebra via \( 1_{A}\rightarrow 1_{A}\otimes 1_{H}, G\rightarrow G\otimes g, X\rightarrow X\otimes g+1_{A}\otimes x\).
The comultiplication on \(H_{4}\) can be written:
Let \(H=H_{4}\) and \(A=Cl(\alpha , \beta , \gamma )\). The H-coaction on A is given by:
The bilinear form \(B:H\otimes H\rightarrow A\otimes H\) is given \(h\otimes h^{\prime }\rightarrow B(h, h^{\prime })\) where in term of bases
for \(0\le i, j, k, l\le 1\).
3 Normalized Condition
The normalized condition is
Hence
4 Casimir Condition
The Casimir condition is
Thus we get
Now we compute \(\psi (g^{i}x^{j-a}\otimes G^{m}X^{n}\otimes g^{p}x^{q})\). We have
and
Hence
Finally the left side of the equation is:
where \(\alpha = ic+(c+j-a)(m+n-b)\).
The right side of the equation is:
The final equation is:
Now we are going to consider the sixteen occurrences of \(\left( i, j, k, l\right) \) which will be subdivided in groups sharing the same i and j.
4.1 Case \(i=1, j=0\)
The left side of the equation is
where \(\alpha =c+c(m+n-b)\). The terms with third component \(1_{H}\) are in the sum
with \(c+b=0\) i.e. \(c=b=0\) and \(m+n+p+q=1, 3\). Then \(\alpha =0\) and we obtain
The terms with third component g are in the sum
with \(c+b=0\) and \(m+n+p+q=0, 2, 4\). Then \(\alpha =0\) and we obtain
The terms with x as the third component are in the sum
with \(c+b=1\) and \(m+n+p+q= 0, 2, 4\). The sum simplifies to
there are two cases:
\(\bullet \) \(b=1\), \(c= 0\), then \(n=1\) and the sum \(m+p+q= 1, 3\). Therefore the sum simplifies
\(\bullet \) \(b=0\), \(c=1\), then \(q=1\) and the sum \(m+n+p= 1, 3\). Therefore the sums simplifies
Hence we get
The sums of the terms with gx in the third component appear with \(c+b=1\) and \(p+q+m+n=1, 3\),
There are two cases:
\(\bullet \) \(b=1\) and \(c=0\), then \(n=1\) and \(\alpha =0\).
we obtain
\(\bullet \) \(c=1\) and \(b=0\), then \(q=1\) and \(\alpha =1+m+n\).
and we get
Therefore the terms in gx in the third component are
4.1.1 \(i=1, j=k=l=0\)
This corresponds to \(B(g\otimes 1_H).\)
where \(\alpha = c+c(m+n-b)\).
By looking at the terms with \(1_{H}\) in the third component and using (13) we obtain
By considering the terms with g in the third component and using both (14) and (17) we realize that they are already zero. The same holds for terms with x in the third component. For the terms with gx in the third component we get, in view of (16)
Finally, we obtain
4.1.2 \(i=1, j=0, k=0, l=1\)
This corresponds to \(B(g\otimes x)\).
where \(\alpha = c(1+m+n-b)\). By looking at the terms with \(1_{H}\) in the third component and using (13) we obtain
Using this relation a straightforward computation, similar to the previous one, gives us:
4.1.3 \(i=1, j=0, k=1, l= 0\)
This correspond to \(B\left( g\otimes g\right) \) we now that it has to be \( 1_A\otimes 1_H\) by (10) .
4.1.4 \(i=1, j=0, k=1, l=1\)
This correspond to \(B(g\otimes gx)\).
where \(\alpha =c(1+m+n-b)\). By considering the terms with g in the third component and using (14) we get
By using both (13), (21) , (15), (16) and (10), a straightforward computation shows that
4.2 Case \(i=1, j=1\)
The left side of the equation is
where \(\alpha =c+(c+1-a)\left( m+n-b\right) \) i.e.
where \(\alpha =c+(c+1)(m+n-b)\) and \(\beta =c+c(m+n-b)\). We consider terms with \(1_{H}\) in the third component: they are in the sum
with \(c+b=0\) and \(\overline{p+q-c+1+m+n-b}=0\), i.e. \(b=c=0\) and \(\overline{ p+q+m+n}=1\). The sum becomes
We consider terms with g in the third component. They are in
\(\beta =c+c(m+n-b), \) with \(c+b=0\) and \(\overline{p+q+1+m+n}=1\), i.e. \( \overline{p+q+m+n}=0\), then
We consider the terms with x in the third component, we have two summands:
(I) in
with \(c+b=0\) and \(\overline{p+q+1+m+n}=0\), i, e, \(\overline{p+q+m+n}=1\), i.e.
(II) and in
with \(c+b=1\).
Two cases arise:
\(\bullet \) \(b=1, \) \(c=0\), hence \(n=1\) and \(\overline{p+q+m}=1\), we get
\(\bullet \) \(b=0, \) \(c=1\), hence \(q=1\) and \(\overline{p+m+n}=1\), we get
Thus we obtain from (II):
In conclusion, we get
We consider terms with gx in the third component: they are in the two following summands.
(I) In
with \(c+b=0\) and \(\overline{p+q+1+m+n}=1\), i.e. \(b=c=0\) and \(\overline{ p+q+m+n}=0\), we get
(II) and in
with \(c+b=1\) and \(\overline{p+q-c+1+m+n-b}=1\), i.e. \(\overline{p+q+m+n}=1\). We have two cases
(i) \(b=1\) and \(c=0\), then \(n=1\), we obtain
(ii) \(b=0\) and \(c=1\), then \(q=1\), we obtain
Therefore we obtain
4.2.1 \(i=1, j=1, k=0, l=0\)
This corresponds to \(B(gx\otimes 1_H)\).
By (12) we know that \(B(gx\otimes 1_{H})=-B(g\otimes gx)\)
4.2.2 \(i=1, j=1, k=0, l=1\)
In this case we are computing \(B(gx\otimes x)\).
i.e.
where \(\alpha = c+(c+1)(m+n-b)\) and \(\beta = c+c(m+n-b)\).
By considering the terms with \(1_{H}\) in the third component, in view of (23) we get
By considering the terms with g in the third component, in view of (24) and (27) we do not get any further relation.
The terms with gx in the third component give us, in view of (26),
A straightforward computation shows that
4.2.3 \(i=1, j=1, k=1, l=0\)
This corresponds to \(B(gx\otimes g).\) We have
i.e.
where \(\alpha = c+(c+1)(m+n-b)\) and \(\beta = c+c(m+n-b)\).
By considering the terms with \(1_{H}\) in the third component, in view of (23) we get
A straightforward computation then shows
4.2.4 \(i=1, j=1, k=1, l=1\)
This case corresponds to \(B(gx\otimes gx)\).
i.e.
where \(\beta = c+c(m+n-b)\) and \(\alpha = c+(c+1)(m+n-b).\)
We consider the terms with g in the third component and by (24) we get
As far as the terms with x in the third component, in view of (25), we get
By using (23), (32) and (26), a straightforward computation shows
4.3 Case \(i=0, j=0\)
The left side of the equality is
The terms with \(1_{H}\) in third position appear as
The terms with g in third position are
Now we consider the terms with x in third position. In this case \(c+b=1\), and \(p+q+m+n=1, 3\). We have two subclasses \(\bullet \) \(c=1\) and \(b=0\) and \(q=1\). The sum is:
\(\bullet \) \(b=1\) and \(c=0\), so \(n=1\). The sum is:
In conclusion the term with x in third position are
Now, consider the terms with gx in third position. Then
Thus \(c+b=1\) i.e. \(b=1\) and \(c=0\) or \(c=1\) and \(b=0\).
If \(b=1\) and \(c=0\), then \(n=1\) and \(m+p+q=1, 3\), the sum is:
\(B(1_H\otimes g^kx^l;GX, 1_H)G\otimes 1_H\otimes gx+B(1_H\otimes g^kx^l;X, g)1_A\otimes g\otimes gx+ B(1_H\otimes g^kx^l;X, x)1_A\otimes x\otimes gx+B(1_H\otimes g^kx^l;GX, gx)G\otimes gx\otimes gx.\)If \(c=1\) then \(q=1\), \(m+n+q=1, 3\), the sum is \(-B(1_H\otimes g^kx^l;G, x)G\otimes 1_H\otimes gx-B(1_H\otimes g^kx^l;X, x)X\otimes 1_H\otimes gx+B(1_H\otimes g^kx^l;1_A, gx)1_A\otimes g\otimes gx+B(1_H\otimes g^kx^l;GX, gx)GX\otimes g\otimes gx.\)
Thus we get
4.3.1 \(i=0, j=0, k=0, l=0\)
This corresponds to \(B(1_H, 1_H)\) which is \(1_A\otimes 1_H\) by (9) .
4.3.2 \(i=0, j=0, k=0, l=1\)
This correspond to \(B(1_H\otimes x).\) The equation is:
since, by (9) \(B(1_{H}\otimes 1_{H};G^{m}X^{n}, g^{p}x^{q})=0\) except for \(B(1_{H}\otimes 1_{H};1_{A}, 1_{H})=1\). We get
By using (35) we get
By using (36), (41), (41) and (39) a straightforward computation shows
4.3.3 \(i=0, j=0, k=1, l=0\)
This correspond to \(B(1_H\otimes g)\). The equation is:
By using (35), (28) and (39) a straightforward computation shows
4.3.4 \(i=0, j=0, k=1, l=1\)
We compute \(B(1_H\otimes gx)\). We have to solve the following equation:
where \(\alpha = c(m+n-b)\).
By using (33), (40) and (28) a straightforward computation shows
4.4 Case \(i=0, j=1\)
The left side of the equality is
where \(\alpha =(c+1)(m+n-b)\) and \(\beta =c(m+n-b)\). The terms with third component \(1_{H}\) are in
with \(c+b=0\) (i.e. \(c=b=0\)) and \(p+q+m+n= 0, 2, 4\). Then it simplifies to:
The terms with third component g are in
with \(c+b=0\) (i.e. \(c=b=0\)) and \(p+q+m+n=1, 3\). Then it simplifies to:
The terms in the left side with third component x are either in
with \(c+b=0\) (i.e. \(c=b=0\)) and \(p+q+m+n= 0, 2, 4\) and we get
or in
with \(c+b=1\) and \(p+q+m+n=1, 3\), i.e. we have again two cases:
\(\bullet \) \(b=1\) and \(c=0\) so \(n=1\), and \(m+p+q=0, 2\) and since \(\beta =0\) the sum is
\(\bullet \) \(c=1\) and \(b=0\) so \(q=1\) and \(m+p+q=0, 2\) and the sum is
Thus, we obtain
The terms with third component gx are either in
with \(c+b=0\) and \(p+q+m+n=1, 3\). Thus in
or in
with \(c+b=1\) and \(p+q+m+n=0, 2, 4\). Two cases arise:
\(\bullet \) \(b=1\) and \(c=0\), then \(n=1\). We get
\(\bullet \) \(b=0\) and \(c=1\), then \(q=1\). We get
Thus we get
4.4.1 \(i=0, j=1, k=0, l=0\)
This corresponds to \(B(x\otimes 1_H)\)
where \(\alpha = (c+1)(m+n-b)\) and \(\beta = c(m+n-b)\). In view (46) we obtain
By using (49), (17), (49) and (48), a straightforward computation shows
4.4.2 \(i=0, j=1, k=0, l=1\)
We compute \(B(x\otimes x).\)
where \(\alpha =(c+1)(m+n-b)\) and \(\beta =c(m+n-b)\). Concerning the terms with third component \(1_{H}, \) in view of (45), we get
By using (51), (19) and (47), a straightforward computation shows
4.4.3 \(i=0, j=1, k=1, l=0\)
This corresponds to \(B(x\otimes g)\overset{(11)}{=} -B(1_H\otimes x).\) See (42) .
4.4.4 \(i=0, j=1, k=1, l=1\)
This corresponds to \(B\left( x\otimes gx\right) .\)
where \(\alpha =(c+1)(m+n-b)\) and \(\beta =c(m+n-b)\). In view of (46), we get
By using (53), (21) and (48), a straightforward computation shows
Proposition 1
Let \(B:H\otimes H\rightarrow A\otimes H^{op}\) a bilinear form. Then B is normalized and satisfies Casimir condition if and only if the equalities (9), (43), (42), (44) , (18), (10), (20), (10) , (22), (50), (11), (52) , (54), (12), (31) and (29) hold.
5 Morphism Condition
Now, we investigate when \(B:H\otimes H\rightarrow A\otimes H^{op}\) is a morphism in \(\mathcal {T}_{A\otimes H^{op}}^{\sharp }\). i.e.
This can be written as
This equality can be split in
In fact assume that (55) and (56) hold. Then we have
Now it is easy to show that if the equality (55) is true for \(a, b\in A\), then it is true for the product. Similarly if the equality (56) is true for \(s, t\in H, \) then it is true for the product. Therefore we only need to compute equality (56) for \(1_{A}\otimes g\) e \(1_{A}\otimes x\) and equality (55) for \(G\otimes 1_H\) and \(X\otimes 1_H\).
5.1 Case \(1_{A}\otimes g\)
We have
that is equivalen to
Now (58) can be rewritten as
Thus, equation (57) is equivalent to
Now, using the forms of the elements given in Proposition 1 it is straightforward to prove that these conditions hold.
5.2 Case \(1_{A}\otimes x\)
We have
From (61) we get
Thus
The left hand side of (62) is
The first summand of the right hand side of (62) is
Thus we get
which is equivalent to
and
Note that if these equalities hold for some \(h, h^{\prime }\) they do hold also for gh and \(gh^{\prime }.\) In fact we have
Therefore we have to check the previous equality for the six elements: \(B(1_H\otimes g), B(1_H\otimes x), B(1_H\otimes gx), B(x\otimes 1_H), B(x\otimes x )\) and \(B(x\otimes gx)\) mentioned before.
5.2.1 (a) \(h=1_H, h^{\prime }=g\)
In this case (63) and (64) become:
By using (50), (44), (66) , (65) and (59), we obtain
Now
so that we get
5.2.2 (b) \(h=1_H, h^{\prime }=x\)
In this case (63) and (64) become:
By using (54), (69), (70), (59), (60) and (11) we obtain
5.2.3 (c) \(h=1_H, h^{\prime }=gx\)
In this case (63) and (64) become:
By using (52), (73), (72) , (67), (59) and (60) we obtain
Cases \(h=x, h^{\prime }=1_H;\) \(h=x, h^{\prime }=x\) and \(h=x, \) \( h^{\prime }=gx\) give us no more information. In conclusion we get the new following form of the last three elements (67), (71) and (74)
5.3 Case \(G\otimes 1_{H}\)
We have
Now, by computing the left side of (75), we get
Now we compute the right side of (75)
Comparing both sides we obtain for any f in the basis of H.
Assume that (76) holds for some \(h, h^{\prime }.\) Let us prove it holds also for gh and \(gh^{\prime }.\)
Assume that (77) holds for some \(h, h^{\prime }.\) Let us prove it holds also for gh and \(gh^{\prime }.\)
Assume that (78) holds for some \(h, h^{\prime }.\) Let us prove it holds also for gh and \(gh^{\prime }.\)
Assume that (79) holds for some \(h, h^{\prime }.\) Let us prove it holds also for gh and \(gh^{\prime }.\)
Thus we have to consider only the usual six cases.
5.3.1 (a) \(h=1_H, h^{\prime }=g\)
In view of (43) we proceed as follows.
Equality (77) for \(f=g\) and \(f=x\) gives us
1) \(B(1_H\otimes g, 1_A, g)+\gamma B(1_H\otimes g, GX, g)=B(g\otimes 1_H, 1_A, g)\overset{(59) }{=}B(1_H\otimes g, 1_A, g)\) Hence
2) \(B(1_H\otimes g, 1_A, x)+\gamma B(1_H\otimes g, GX, x)=B(g\otimes 1_H, 1_A, x) \overset{(60)}{=}-B(1_H\otimes g, 1_A, x)\) since \( B(1_H\otimes g, GX, x)\overset{(43)}{=}0\), we get
Now (76) for \(f=1_H\) gives us
3) \(\alpha B(1_H\otimes g, G, 1_H)+\gamma B(1_H\otimes g, X, 1_H)=\alpha B(g\otimes 1_H, G, 1_H)\overset{(59) }{=}\alpha B(1_H\otimes g, G, 1_H)\) Thus we get, in view of (43),
Consider now (79) for \(f=1_H.\)
4) \(-B(1_H\otimes g, X, 1_H)=B(g\otimes 1_H, X, 1_H)\overset{(59)}{=}B(1_H\otimes g, X, 1_H)\) so that \(2B(1_H\otimes g, X, 1_H)=0\) which means, in view of (43), \( 2B(1_H\otimes g, 1_A, x)=0\) which is true by (81) . And now (76) for \(f=gx\)
5) \(\alpha B(1_H\otimes g, G, gx)+\gamma B(1_H\otimes g, X, gx)=\alpha B(g\otimes 1_H, G, gx)=\overset{(60) }{=}-\alpha B(1_H\otimes g, G, gx) \) Thus \(2\alpha B(1_H\otimes g, G, gx)+\gamma B(1_H\otimes g, X, gx)=0, \) but \(B(1_H\otimes g, X, gx)=0\) therefore
From (78) for \(f=g\) we obtain 6) \(-\alpha B(1_H\otimes g, GX, g)=\alpha B(g\otimes 1_H, GX, g)\overset{(59)}{=}\alpha B(1_H\otimes g, GX, g)\) Thus \(2\alpha B(1_H\otimes g, GX, g)=0\) by the form of the element \(2\alpha B(1_H\otimes g, G, gx)=0\). This is 5).
Following a procedure analogous to that of case a) we get
5.3.2 (b) \(h=1_H, h^{\prime }=x\)
We get
(1) \(B(1_H\otimes x, 1_A, x)+\gamma B(1_H\otimes x, GX, x)=-B(g\otimes gx, 1_A, x)\overset{ (60) }{=} B(1_H\otimes x, 1_A, x)\) i.e. \(\gamma B(1_H\otimes x, GX, x)=0\) this is already known in view of (42) .
(2) \(B(1_H\otimes x, 1_A, g)+\gamma B(1_H\otimes x, GX, g)=-B(g\otimes gx, 1_A, g)\overset{ (59) }{=} -B(1_H\otimes x, 1_A, g)\)
(3) \(\alpha B(1_H\otimes x, G, 1_H)+\gamma B(1_H\otimes x, X, 1_H)=-\alpha B(g\otimes gx, G, 1_H)\) \(\alpha B(1_H\otimes x, G, 1_H)+\gamma B(1_H\otimes x, X, 1_H)=-\alpha B(1_H\otimes x, G, 1_H)\) \(2\alpha B(1_H\otimes x, G, 1_H)+\gamma B(1_H\otimes x, X, 1_H)=0\) \(2\alpha B(1_H\otimes x, G, 1_H)+\gamma [1-B(1_H\otimes x;1_A, x)]=0\)
The following equalities are already known.
(4) \(\alpha B(1_H\otimes x, G, gx)+\gamma B(1_H\otimes x, X, gx)=-\alpha B(g\otimes gx, G, gx)\) \(\alpha B(1_H\otimes x, G, gx)+\gamma B(1_H\otimes x, X, gx)=\alpha B(1_H\otimes x, G, gx)\)
\(\gamma B(1_H\otimes x, X, gx) =0\)
(5) \(B(1_H\otimes x, X, 1_H)=B(g\otimes gx, X, 1_H), \) \( B(1_H\otimes x, X, 1_H)=B(1_H\otimes x, X, 1_H)\)
(6) \(\alpha B(1_H\otimes x, GX, g)=\alpha B(g\otimes gx, GX, g)\)
5.3.3 (c) \(h=1_H, h^{\prime }=gx\)
We get
(1) \(B(1_H\otimes gx, 1_A, 1_H)+\gamma B(1_H\otimes gx, GX, 1_H)=-B(g\otimes x, 1_A, 1_H)\overset{ (59) }{=} -B(1_H\otimes gx, 1_A, 1_H)\) so that \(2B(1_H\otimes gx, 1_A, 1_H)+\gamma B(1_H\otimes gx, GX, 1_H)=0.\) By considering (44), we get
(2) \(B(1_H\otimes gx, 1_A, gx)+\gamma B(1_H\otimes gx, GX, gx)=-B(g\otimes x, 1_A, gx)\overset{ (60) }{=}B(1_H\otimes gx, 1_A, gx)\) Thus \(\gamma B(1_H\otimes gx, GX, gx)=0\) i.e.\(\gamma B(1_H\otimes g;G, gx)=0\) This is (80).
(3) \(\alpha B(1_H\otimes gx, G, g)+\gamma B(1_H\otimes gx, X, g)=-\alpha B(g\otimes x, G, g)\overset{ (59) }{=} -\alpha B(1_H\otimes gx, G, g)\)
\(2\alpha B(1_H\otimes gx, G, g) +\gamma B(1_H\otimes gx, X, g)=0\)
(4) \(\alpha B(1_H\otimes gx, G, x)+\gamma B(1_H\otimes gx, X, x)=-\alpha B(g\otimes x, G, x)\overset{ (60) }{=} \alpha B(1_H\otimes gx, G, x)\)
so that we get \(\gamma B(1_H\otimes gx, X, x)=0\) i.e. \(\gamma B(1_H\otimes g;1_A, x)=0.\) This is true by (82).
(5) \(-\alpha B(1_H\otimes gx, GX, 1_H)\overset{ (78) }{=}\alpha B\left( g\otimes gxg, GX, 1_H\right) =-\alpha B\left( g\otimes x, GX, 1_H\right) \overset{ (59) }{=}-\alpha B\left( 1_H\otimes gx, GX, 1_H\right) \)
This is trivial.
(6) \(-B(1_H\otimes gx, X, g)=B(g\otimes gxg, X, g)=-B\left( g\otimes x, X, g\right) \overset{ (59) }{=}-B(1_H\otimes gx, X, g).\) Thus we obtain
\( -B(1_H\otimes gx, X, g)=-B(g\otimes x, X, g)\). This follows from (59).
(7) \(-B(1_H\otimes gx, X, x)=-B(g\otimes x, X, x)\). This is true in view of (44) and (20) .
(8) \(-\alpha B(1_H\otimes gx, GX, gx)=\alpha B(g\otimes gxg, GX, gx)=-\alpha B(g\otimes x, GX, gx)\overset{ (60) }{=}\alpha B(1_H\otimes gx, GX, gx)\)
so that we get \(2\alpha B(1_H\otimes gx, GX, gx)=0.\) Now, in view of (44), the coefficient of \(GX\otimes gx\) is \( B(1_H\otimes g;G, gx)\) and hence we obtain \(2\alpha B(1_H\otimes g, G, gx)=0.\) This is (83).
5.3.4 (d) \(h=x, h^{\prime }=1_H\)
By considering (50) we obtain the following.
(1) \(B(x\otimes 1_H, 1_A, 1_H)+\gamma B(x\otimes 1_H, GX, 1_H) \overset{ (77) }{=}-B(gx\otimes g, 1_A, 1_H) \overset{ (59) }{=}-B(x\otimes 1_H, 1_A, 1_H)\)
Hence we get \(2B(x\otimes 1_H, 1_A, 1_H)=-\gamma B(x\otimes 1_H, GX, 1_H)\) and, in view of (50) we obtain
\(2B(x\otimes 1_H, 1_A, 1_H)=-\gamma (B(x\otimes 1_H;G, x)+B(g\otimes 1_H;G, 1_H)) \overset{ (59) }{=}B(1_H\otimes g;G, 1_H).\) Using (67) we obtain (86) again.
(2) \(B(x\otimes 1_H, 1_A, gx)+\gamma B(x\otimes 1_H, GX, gx) \overset{ (77) }{=}-B(gx\otimes g, 1_A, gx) \overset{ (60) }{=}B(x\otimes 1_H, 1_A, gx)\)
Hence by (50), we obtain \(\gamma B(1_H\otimes g;G, gx)=0. \)This is (80)
(3) \(\alpha B(x\otimes 1_H, G, g)+\gamma B(x\otimes 1_H, X, g) \overset{ (76) }{=}-\alpha B(gx\otimes g, G, g)\overset{ (59) }{=}-\alpha B(x\otimes 1_H, G, g)\)
so that we get, by (50), \(2\alpha B(x\otimes 1_H, G, g)+\gamma (-B(x\otimes 1_H;1_A, gx)-B(g\otimes 1_H;1_A, g))=0.\) Since \( B(g\otimes 1_H;1_A, g)\overset{ (59) }{=}B(1_H\otimes g;1_A, g)\) we obtain
using (67) this becomes the already known (87) .
(4) \(\alpha B(x\otimes 1_H, G, x)+\gamma B(x\otimes 1_H, X, x) \overset{ (76) }{=}-\alpha B(gx\otimes g, G, x)\overset{ (60) }{=}\alpha B(x\otimes 1_H, G, x).\)
Thus we get \(\gamma B(x\otimes 1_H, X, x)=0\) and hence, using (50) , \(\gamma B(g\otimes 1_H;1_A, x)=0\) and since \(\gamma B(1_H\otimes g;1_A, x)\overset{ (60) }{=}-\gamma B(g\otimes 1_H;1_A, x)\) we get \(\gamma B(1_H\otimes g;1_A, x)=0\) which is true by (82).
(5) \(-B(x\otimes 1_H, X, x)\overset{ (79) }{=}-B(gx\otimes g, X, x)\overset{ (60) }{=}B(x\otimes 1_H, X, x)\)
and hence \(B(x\otimes 1_H, X, x)=0\) so that we get, using (50) , \(B(g\otimes 1_H;1_A, x)=0\) and hence \(B(1_H\otimes g;1_A, x) \overset{ (60) }{=}-B(g\otimes 1_H;1_A, x)=0\) which is true by (81).
(6) \(-\alpha B(x\otimes 1_H, GX, gx)\overset{ (78) }{=}-\alpha B(gx\otimes g, GX, gx)\overset{ (60) }{=}\alpha B(x\otimes 1_H, GX, gx)\) so that we get
\(\alpha B(x\otimes 1_H, GX, gx)=0\) and hence, using (50) , \(\alpha B(g\otimes 1_H;G, gx)=0\) which holds by (83) .
(7) \(-\alpha B(x\otimes 1_H, GX, 1_H)\overset{(78)}{=}-\alpha B(gx\otimes g, GX, 1_H)\overset{ (59) }{=}-\alpha B(x\otimes 1_H, GX, 1_H)\) which is trivial.
(8) \(-B(x\otimes 1_H, X, g)\overset{ (79)}{=}-B(gx\otimes g, X, g)\overset{ (59) }{=} -B(x\otimes 1_H, X, g).\)This is trivial.
5.3.5 (e) \(h=x, h^{\prime }=x\)
By using (74) we obtain the following.
(1) \(B(x\otimes x, 1_A, x)+\gamma B(x\otimes x, GX, x)\overset{ (77) }{=}B(xg\otimes xg, 1_A, x)=B(gx\otimes gx, 1, x)\overset{ (60) }{=}-B(x\otimes x, 1_A, x)\)
so that we get \(2B(x\otimes x, 1_A, x)+\gamma B(x\otimes x, GX, x)=0\) so that we get, by (74),
By using (67) and (60) and (74), this becomes
which is trivial.
(2) \(B(x\otimes x, 1_A, g)+\gamma B(x\otimes x, GX, g)\overset{ (77) }{=}B(xg\otimes xg, 1_A, g)=B(gx\otimes gx, 1_A, g)\overset{ (59) }{=}B(x\otimes x, 1_A, g)\)
\(B(x\otimes x, 1_A, g)+\gamma B(x\otimes x, GX, g)=B(x\otimes x, 1_A, g)\)
and hence
which holds by (74).
(3) \(\alpha B(x\otimes x, G, 1_H)+\gamma B(x\otimes x, X, 1_H) \overset{ (76) }{=}\alpha B(xg\otimes xg, G, 1_H)=\alpha B(gx\otimes gx, G, 1_H)\overset{)(60) }{=}\alpha B(x\otimes x, G, 1_H)\)
We obtain \(\gamma B(x\otimes x, X, 1_H)=0\). This follows from (74).
(4) \(\alpha B(x\otimes x, G, gx)+\gamma B(x\otimes x, X, gx) \overset{ (76) }{=}\alpha B(xg\otimes xg, G, gx)=\alpha B(gx\otimes gx, G, gx)\overset{ (60) }{ =}-\alpha B(x\otimes x, G, gx)\)
so that we get
\(2\alpha B(x\otimes x, G, gx)+\gamma B(x\otimes x, X, gx)=0\) and hence, by (74),
(5) \(-B(x\otimes x, X, 1_H)\overset{ (79)}{=}B\left( xg, xg, X, 1_H\right) =B(gx\otimes gx, X, 1_H)\overset{ (59) }{=}B(x\otimes x, X, 1_H)\)
so that we get \(2B(x\otimes x, X, 1_H)=0\) true by (74).
(6) \(-B(x\otimes x, X, gx)\overset{ (79) }{=}B(xg\otimes xg, X, gx)=B(gx\otimes gx, X, gx)\overset{ (60) }{=}-B(x\otimes x, X, gx)\)
Trivial.
(7) \(-\alpha B(x\otimes x, GX, g)\overset{\text {}(78) }{=}\alpha B(gx\otimes gx, GX, g)\overset{ (59) }{=}\alpha B(x\otimes x, GX, g)\)
so that we get \(2\alpha B(x\otimes x, GX, g)=0\) this is true by (74).
(8) \(-\alpha B(x\otimes x, GX, x)\overset{(78) }{=}\alpha B(gx\otimes gx, GX, x)\overset{ (60) }{=}-\alpha B(x\otimes x, GX, x)\)
This is trivial.
5.3.6 (f) \(h=x, h^{\prime }=gx\)
By considering (71) we obtain
5.4 Case \(X\otimes 1_{H}\)
We have
Now we compute the left side of (90).
Now, by computing first summand of the right side (90), we get
We can summarize the equality (90) in the following form:
Assume that (91) holds for some \(h, h^{\prime }.\) Let us prove it holds also for gh and \(gh^{\prime }.\)
Assume that (92) holds for some \(h, h^{\prime }.\) Let us prove it holds also for gh and \(gh^{\prime }.\)
Assume that (93) holds for some \(h, h^{\prime }.\) Let us prove it holds also for gh and \(gh^{\prime }.\)
Assume that (94) holds for some \(h, h^{\prime }.\) Let us prove it holds also for gh and \(gh^{\prime }.\)
Therefore we have to check the previous equality for the usual six elements.
5.4.1 (a) \(h=1_H, h^{\prime }=g\)
\(B(1_H\otimes g, 1_A, f)\overset{(91) }{=}B(g\otimes 1_H, 1_A, f)+\gamma B(g\otimes \) \(1_H, GX, f)+B(x\otimes 1_H;X, f)+B(1_H\otimes gx;X, f)\)
\(\beta B(1_H\otimes g, X, f)\overset{(93) }{=}\beta B(g\otimes 1_H, X, f)+\gamma B(g\otimes 1_H, G, f)+B(x\otimes \) \(1_H;1_A, f)+B(1_H\otimes gx;1_A, f)\)
\(\beta B(1_H\otimes g, GX, f)\overset{(94)}{=}-\beta B(g\otimes 1_H, GX, f)+B(x\otimes 1_H;G, f)+B(1_H\otimes \) gx; G, f)
\(B(1_H\otimes g, G, f)\overset{(92) }{=}-B(g\otimes 1_H, G, f)+B(x\otimes 1_H;GX, f)+B(1_H\otimes gx;GX, f)\)
Now using (59) and (60) we get
Now, in view of (43), we obtain the following.
(1) \(B(1_H\otimes g, 1_A, g)\overset{(95) }{=}B(1_H\otimes g, 1_A, g)+\gamma B(1\otimes g, GX, g)+B(x\otimes 1;X, g)+B(1\otimes gx;X, g)\)
so that, by using (67) and (44), we get
(2) \(B(1_H\otimes g, 1_A, x)=B(g\otimes 1_H, 1_A, x)+\gamma B(g\otimes 1_H, GX, x)+B(x\otimes 1_H;X, x)+B(1_H\otimes gx;X, x)\)
By means of (96), we get
By using this and (67) and (44) we get \(2B(1_H\otimes g, 1_A, x)=B(1_H\otimes g;1_A, x)+B(1_H\otimes g;1_A, x)\) which is trivial.
(3) \(B(1_H\otimes g, G, 1_H)\overset{(97)}{=}-B(1_H\otimes g, G, 1_H)+B(x\otimes 1_H;GX, 1_H)+B(1_H\otimes gx;GX, 1_H)\)
so that, by means of (67) and (44), we obtain
(4) \(B(1_H\otimes g, G, gx)\overset{(98)}{=}B(1_H\otimes g, G, gx)+B(x\otimes 1_H;GX, gx)+B(1_H\otimes gx;GX, gx) \)
so that \(B(x\otimes 1_H;GX, gx)+B(1_H\otimes gx;GX, gx)=0\). By using (67) and (44), we obtain \( 2B(1_H\otimes g, G, gx)=0\) which is (68).
(5) \(\beta B(1_H\otimes g, X, 1_H)\overset{(99) }{=}\beta B(1_H\otimes g, X, 1_H)+\gamma B(1_H\otimes g, G, 1_H)+B(x\otimes 1_H;1_A, 1_H)+B(1_H\otimes gx;1_A, 1_H)\)
so that, by using (67), we get
(6) \(\beta B(1_H\otimes g, GX, g)\overset{(99) }{=}-\beta B(1_H\otimes g, GX, g)+B(x\otimes 1_H;G, g)+B(1_H\otimes gx;G, g)\)
so that we get \(2\beta B(1_H\otimes g, GX, g)=B(x\otimes 1_H;G, g)+B(1_H\otimes gx;G, g).\)
Since \(B(1_H\otimes g, GX, g)\overset{(43)}{=} B(1_H\otimes g, G, gx)\), we obtain, by means of (67),
5.4.2 (b) \(h=1_H, h^{\prime }=x\)
By using (91) we get
Now, by using (59) and (60), we obtain
By means of (92), we obtain
Now, by (59), we have
In view of (42) we consider the following cases.
(1)
\(2B(1_H\otimes x, 1_A, g)\overset{(107) }{=}-\gamma B(1_H\otimes x, GX, g)-B(x\otimes gx;X, g)\)By using (71) we get \(2B(1_H\otimes x, 1_A, g)=-\gamma B(1_H\otimes x, GX, g).\) This is (84).
(2)
\(B(1_H\otimes x, 1_A, x)\overset{(108) }{=} B(1_H\otimes x, 1_A, x)+\gamma B(1_H\otimes x, GX, x)-B(x\otimes gx;X, x)\) using (71) we obtain \(\gamma B(1_H\otimes x, GX, x)=0\) which is trivial by (42).
(3)
\(B(1_H\otimes x, G, 1_H)\overset{ (110)}{=} B(1_H\otimes x, G, 1_H)-B(gx\otimes x;GX, 1_H)=B(1_H\otimes x, G, 1_H)-B(gx\otimes x;GX, 1_H)\overset{ (59) }{=}\)
\(=B(1_H\otimes x, G, 1_H)-B(x\otimes gx;GX, 1_H)\overset{(71) }{=}B(1_H\otimes x, G, 1_H)\) This is trivial.
(4) \(B(1_H\otimes x, G, gx)\overset{(111)}{= }-B(1_H\otimes x, G, gx)+B(gx\otimes x;GX, gx).\) Therefore we get
Now, by using (93), we get
(5) \(\beta B(1_H\otimes x, X, 1_H)\overset{(112) (59) }{=}-\beta B(1_H\otimes x, X, 1_H)-\gamma B(1_H\otimes x, G, 1_H)-B(x\otimes gx;1_A, 1_H)\)
\(2\beta B(1_H\otimes x, X, 1_H)=-\gamma B(1_H\otimes x, G, 1_H)-B(x\otimes gx;1_A, 1_H)\) and by (71) and (42) we get
(6) \(\beta B(1_H\otimes x, GX, f)\overset{(94)}{=}-\beta B(g\otimes xg, GX, f)+B(x\otimes xg;G, f)+B(1_H\otimes xx;G, f)=\beta B(g\otimes gx, GX, f)-B(x\otimes gx;G, f)\)so that
and hence \(\beta B(1_H\otimes x, GX, g)\overset{(114) }{=}\beta B(g\otimes gx, GX, g)-B(x\otimes gx;G, g)\overset{ (59) }{=}\beta B(1_H\otimes x, GXc, g)-B(gx\otimes x;G, g)\) so that we get \(B(gx\otimes x;G, g)=B(x\otimes gx;G, g)=0\) which is already known by (71).
5.4.3 (c) \(h=1_H, h^{\prime }=gx\)
By means of (91), we get
(1) By using (115), (59) , (74) and (44) we get
(2) By using (115), (60) , (74) and, (44) we obtain
In view of (92) we get
(3) \(B(1_H\otimes gx, G, g)\overset{(117)}{=} B(g\otimes x, G, g)-B(x\otimes x;GX, g)\overset{ (59) }{ =}B(1_H\otimes gx, G, g)-B(x\otimes x;GX, g)\)
and hence we get \(B(x\otimes x;GX, g)=0\) which is true in view of (74)
(4) \(B(1_H\otimes gx, G, x)\overset{(117)}{=} B(g\otimes x, G, x)-B(x\otimes x;GX, x)\overset{ (60) }{ =}-B(1_H\otimes gx, G, x)-B(x\otimes x;GX, x)\)
so that we get in view of (74)
\(\beta B(1_H\otimes gx, X, f)\overset{(93) }{=}=\beta B(g\otimes gxg, X, f)+\gamma B(g\otimes gxg, G, f)+B(x\otimes gxg;1_A, f)+B(1_H\otimes gxx;1_A, f)\)
\(=-\beta B(g\otimes x, X, f)-\gamma B(g\otimes x, G, f)-B(x\otimes x;1_A, f)\) so that we get
(5) In view of (119) (59) (44) and(74), we get
(6) By using (119), (60) and (74) we get
Now, in view of (94), we have
(7) \(\beta B(1_H\otimes gx, GX, 1_H)\overset{(122) }{=}\beta B(g\otimes x, GX, 1_H)-B(x\otimes x;G, 1_H)\overset{ (59) }{=}\beta B(1_H\otimes gx, GX, 1_H)-B(x\otimes x;G, 1_H)\)
so that we get \(B(x\otimes x;G, 1_H)=0\) which is true in view of (74)
(8) By using (122), (60), (74) and (44), we get
5.4.4 (d) \(h=x, h^{\prime }=1_H\)
By means of (91) we obtain
(1) By using (124), (59) , (74) and (67), we get
(2) In view of (124), (60), (67) and (74) we obtain
Now by using (92), (59) and (60) we get
and
in particular
(3)\(B(x\otimes x;GX, g)=0\) which is known.
(4)\(2B(x\otimes 1_H, G, x)=B(x\otimes x;GX, x)\) By (67) and (74) we get \(2B(1_H\otimes gx, G, x)=0\) which is (118). Now, by using (93) we get
In particular
(5) \(2\beta B(x\otimes 1_H, X, g) =-\gamma B(x\otimes 1_H, G, g)+ B(x\otimes x;1_A, g) \) By means of (67) and (74), we get
(6) \(\beta B(x\otimes 1_H, X, x)=\beta B(x\otimes 1_H, X, x)+\gamma B(x\otimes 1_H, G, x)+B(x\otimes x;1_A, x)\) i.e.
\(\gamma B(x\otimes 1_H, G, x)+B(x\otimes x;1_A, x)=0\)
By (67) and (74) , we get \(\gamma B(1_H\otimes gx, G, x)=0\) which is (121).
Now, by using (94), we get
and
In particular
(7) For \(f=1_H\)we get \(B(x\otimes x;G, 1_H)=0\) which is already known by (74) and
(8) for \(f=gx\) and by using (67) and (74) we get
5.4.5 (e) \(h=h^{\prime }=x.\)
By using (91), (59) and (60) we obtain
and
so that we get
(1) \(\gamma B(x\otimes x, GX, g)=0\) which is true in view of (74)
(2) \(2B(x\otimes x, 1_A, x)=-\gamma B(x\otimes x, GX, x)\) which is true in view of (74).
Now by using (92) we get
and
in particular
(3) \(2B(x\otimes x, G, 1_H)=0\) which is true in view of (74).
(4)\(B(x\otimes x, G, gx)=B(x\otimes x, G, gx)\) which is trivial. Now, in view of (93), (59) and (60) we get
and
In particular
(5) \(\gamma B(x\otimes x, G, 1_H)=0\). This is true in view of (74).
(6) \(2\beta B(x\otimes x, X, gx)=-\gamma B(x\otimes x, G, gx)\) By (74) we get \(2\beta [2B(1_H\otimes gx, 1_A, gx)-2B(1_H\otimes g, 1_A, g)]=-\gamma \left[ -2B(1_H\otimes gx, G, g) \right] \) i.e.
Now, by using (94), (59) and (60), we get
and
In particular
(7) \(\beta B(x\otimes x, GX, g)=-\beta B(x\otimes x, GX, g)\)which is true by (74) and
(8) \(\beta B(x\otimes x, GX, x)=\beta B(x\otimes x, GX, x)\) which is trivial.
5.4.6 (f) \(h=x\) \(h^{\prime }=gx\)
By using (91), (59) and (60) , we get
and
In particular
(1) for \(f=1_H\) we get \(\gamma B(x\otimes gx, GX, 1_H)=0\) which is true in view of (71).
(2) for \(f=gx\) we get, by using (71), we get
Now, in view of (94) and (60) we get \(\ \beta B(x\otimes gx, GX, f)=\beta B(x\otimes gx, GX, f)\) for \(f=x, gx\), which is trivial.
5.5 The List of Equalities
Thus we obtained the following equalities: (80), (81), (82), (83) , (83), (84), (85) , (86), (87), (88) , (89), (101), (103), (68), (104), (105) , (113), (116), (118) , (120), (121), (123) , (125), (126), (127) , (128), (131) and (132) .
5.6 Simplifications for Char(k) \(\ne 2\)
The reader can check that, by assuming char(k) \(\ne 2\) we obtain the following list which we relabel as follows.
5.7 The New Form of The Six Elements Using Last List
In the following, we set: \({\textbf{A}}=B(1_H\otimes g;1_A, g), \) \({\textbf{B}} =B(1_H\otimes g;G, 1_H), {\textbf{C}}=B(1_H\otimes x;G, 1_H), {\textbf{D}} =B(1_H\otimes x;1_A, x), {\textbf{E}}=B(1_H\otimes x, GX, g).\) By using the above equalities we get the new form of the six elements.
6 The Separability Result
Theorem 1
Let \(A=Cl(\alpha , \beta , \gamma )\) and \(H=H_{4}.\) Assume that \(\textrm{char}\left( k \right) \ne 2\). Then the cowreath \( (A\otimes H^{op}, H)\) is separable with respect to any bilinear form satisfying (142), (143), (144), (145), (146) and (147) and (139), (141).
Proof
In view of [4, Proposition 7.4], we have to find a bilinear form
which is a Casimir morphism satisfying also the normalized condition. In view of Proposition (1) and of the morphism condition, since the last form of the six elements (142), (143), (144) , (145), (146) and (147) was obtained by using all the equalities in 5.6 except (139) and (141) we rewrite these two equalities remaining equalities.
Note that this system has always a solution, namely \({\textbf{C}}=0\) and \( {\textbf{D}}=1.\)
Thus by means of any k-linear map \(B:H\otimes H\rightarrow A\otimes H^{op}\) satisfying all these equalities, the cowreath \((A\otimes H^{op}, H)\) is separable. \(\square \)
7 h-Separability
Now, we are going to investigate the h-separability as introduced in [12, Theorem 5.1]. We still assume \(\textrm{char}\left( k\right) \ne 2.\) Equation [12, (3)] in our case reduces to
for all \(h, h^{\prime }, h^{\prime \prime }\in H_{4}\).
Thus possible values of \(h^{\prime }\) in the bases we obtain four equations:
We claim that (152) and (154) follows from (151) and (153). Multiplying equation (151) by \((1_A\otimes g)\) on the left and on the right we get
By equation (58) we obtain
Since any \(h\in H\) can be written as \(h=g\left( gh\right) \), we deduce that (152) can be obtained from (151).
Similarly, multiplying equation (153) by \((1_A\otimes g)\) on the left and on the right we get
so that, by equation (58), we obtain
and we get (154).
7.1 The First Equation
Here we will analyze the several occurrences of equation (151).
Whenever either \(h=1_H\) or \(h^{\prime \prime }=1_H, \) in view of (9), the equation is trivially satisfied.
7.1.1 \(h=g, \) \(h^{\prime \prime }=g\)
Therefore we obtain
7.1.2 \(h=g, \) \(h^{\prime \prime }=x\)
Proceeding as in the previous case we get, by using (142) , (143) we obtain the first side
and by using (144) we get the second side
In conclusion we get the following equality
which gives us
7.1.3 \(h=g, \) \(h^{\prime \prime }=gx\)
Proceeding as in the previous case, by using (142) , (144) we obtain the first side of the equation
and by using (143) we get the second side
In conclusion we get the following equality
so that, in view of (156) gives us
and hence, by (156), we get
7.1.4 \(h=x, h^{\prime \prime }=g\)
so that we get the first side
and by using (143) we get
so that in view of (162), (163), (165) and (166) we get no new information.
7.1.5 \(h=x, h^{\prime \prime }=x\)
We get, by using (143), (145) and (165), the first side
so that, in view of (146), (159), (160) , (158), (157) (165) we get
7.1.6 \(h=x, h^{\prime \prime }=gx\)
In view of (145) and (144) the first side is
so that, by using (147) and (165) we get
7.1.7 \(h=gx, h^{\prime \prime }=g\)
By (58) we get
so that by (11) we get
and hence the first side is, by using (165) and (159),
and the second side is
so that we get in view of (158), (160), (157) and (165) we do not get anything new.
7.1.8 \(h=gx, h^{\prime \prime }=x\)
By (58) we get
and hence, by using (165) and (159), the first side is
In view of (147) and (165) second side is 0 so that we obtain
7.1.9 \(h=gx, h^{\prime \prime }=gx\)
By using (58), (11) and (146) we obtain
In view of (143), (144) and (165) we obtain
Hence, by using (160), (158), (167) and (165), we obtain \(\alpha \textbf{BC}+\frac{\gamma }{2}{\textbf{B}}=0\) which follows from (157) and (165).
Remark 1
By using (165) we get the new form of the following elements.
By using (159), (160), (158) and (165) we obtain
7.2 The Third Equation
Here, by using the results above, we show how (153) can be deduced from (151) . By using (151) , we get
now we compute
From this we deduce (153).
7.3 The Main Result
Theorem 2
Let \(A=Cl(\alpha , \beta , \gamma )\) and \(H=H_{4}.\) Assume that \(\textrm{char}\left( k \right) \ne 2\). Then the cowreath \( (A\otimes H^{op}, H)\) is always h-separable. This happens whenever the bilinear form \(B:H\otimes H\rightarrow A\otimes H^{op}\) satisfies (142), (143), (144), (145), (146) and (147) and (139), (141).
Proof
We have already seen that (150) is equivalent to (151) . In view of (165), we know that \({\textbf{E}}=0\), by taking in account of this we collect here all the equalities we have obtained in (7.1) .
Moreover, we have to add the two separability equalities
and
Assume \({\textbf{B}}=0\). In this case we get that all the equalities above reduce to the following.
and we have a solution. \(\square \)
Remark 2
In contrast note that if \({\textbf{A}}=0, \) we get
Hence \(\alpha \in \left( k^{\times }\right) ^{2}\) and \(\mathbf {B\ne }0\) so that from
we get
so that (163) can be deduced from (155) . By substituting \({\textbf{C}}\) inside (169), we obtain \(4\alpha \beta -\gamma ^{2}=0.\) Therefore we are exactly in the situation of [12, Theorem 6.1].
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Communicated by Uwe Kaehler.
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This paper was written while the first author was a member of the “National Group for Algebraic and Geometric Structures and their Applications” (GNSAGA-INdAM) and was partially supported by MIUR within the National Research Project PRIN 2017. The second author was partially supported by FEDER-UAL18-FQM-B042-A, PY20-00770 from Junta Andalucía and by research project PID2020-113552GB-I00 from MICIN.
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Menini, C., Torrecillas, B. Separable Cowreaths Over Clifford Algebras. Adv. Appl. Clifford Algebras 33, 19 (2023). https://doi.org/10.1007/s00006-023-01258-y
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DOI: https://doi.org/10.1007/s00006-023-01258-y