As is known, multiplier Hopf algebras are a generalization of Hopf algebras[1-2]. Differently from Hopf algebra, the underlying algebra is no longer assumed to be a unit, but the product of algebra is non-degenerate. Some studies of multiplier Hopf algebras and their applications can be found in Refs.[3-6]. In Ref.[7], the author considered the module algebra as a regular multiplier Hopf algebra, and the theory of smash products was generalized to multiplier Hopf algebras from Hopf algebras. The definition and properties of diagonal crossed products were introduced in Ref.[8]. In this article, we will consider the bimodule algebra as a regular multiplier Hopf algebra and generalize the theory of diagonal crossed products to the multiplier Hopf algebra case.
1 Preliminaries
Throughout this article, let k be a fixed field of characteristic 0 (i.e., all algebraic systems are over k). In order to facilitate our computations, we always omit the summation symbol Σ.
In the following, we recall some definitions.
For an associative algebra A, A has a non-degenerate product with or without identity. We denote its multiplier algebra by M(A), and M(A) always contains a unit 1. In fact, M(A) can be characterized as the largest algebra with a unit, in which A is regarded as an essential two-side ideal. Clearly, A=M(A) if and only if A has a unit (see the appendix in Ref.[2] for details), which is similar to M(A⊗A).
A comultiplication on A is a homomorphism ΔA:A→M(A⊗A) such that ΔA(a)(1⊗b) and (a⊗1)ΔA(b) belong to A⊗A for all a,b∈A, and ΔA is coassociative in the sense that
(a⊗1⊗1)(ΔA⊗i)(ΔA(b)(1⊗c))=
(i⊗ΔA)((a⊗1)ΔA(b))(1⊗1⊗c)
(1)
for all a,b,c∈A, where i is the identity map.
A pair (A, ΔA), in which A is an algebra with a non-degenerate product and ΔA is a comultiplication on A, is called a multiplier Hopf algebra if there are two linear bijections 
which are defined as
We say that (A,ΔA) is regular if σΔA is again a comultiplication on A such that a pair (A,σΔA) is also a multiplier Hopf algebra, where σ is the flip.
Remark 1 1) The use of the Sweedler notation for regular multiplier Hopf algebra is discussed in Ref.[1]. Take a,b∈A as an example and consider ΔA(a)(1⊗b). If we choose e such that eb=b, where e is called the local unit (see Ref.[1], Proposition 2.2), then, ΔA(a)(1⊗b)=(ΔA(a)(1⊗e))(1⊗b)=a1⊗a2b.
2) By the definition of two linear maps 
Eq.(1) can be replaced by 
Definition 1[9] Let A be a regular multiplier Hopf algebra. Then, algebra B is called an A-bimodule algebra if
1) B is a unital left A-module and a unital right A-module such that (a·b)·a′=a·(b·a′);
2) a·bb′=(a1·b)(a2·b′);
3) bb′·a=(b·a1)(b′·a2) for all b,b′∈B and a,a′∈A.
2 Main Results
In this section, we will give the construction of the diagonal crossed product in the sense of multiplier Hopf algebras. More information can be found in Refs.[3-4].
Definition 2 Let A be a regular multiplier Hopf algebra and B an A-bimodule algebra. Then, the diagonal crossed product B#A built on B⊗A with multiplication is given as
(b⊗a)(b′⊗a′)=b(a1·b′·S-1(a3))⊗a2a′
(2)
for all b, b′∈B and a,a′∈A. Note that on the right side, each decomposition is well-covered.
We will further investigate the algebra B#A.
Lemma 1 Let A be a cocommutative multiplier Hopf algebra and B is a regular multiplier Hopf algebra such that B is an A-bimodule algebra. Then, the product in B#A is non-degenerate.
Proof Suppose that bi#ai∈B#A and that (bi#ai)(b#a)=0 for all b∈B and a∈A. Then, according to the definition of the product in B#A and the non-degeneracy of the product in A, we have
bi(ai1·b·S-1(ai3))#ai2=0
Applying Δ and S, multiplying with a from the right and replacing b by a″b, we can obtain
bi(ai1a″·b·S-1(ai4))⊗S(ai2)a⊗ai3=0
for all b∈B and a,a′,a″∈A. Replacing a″ by S(ai2)a, we have
bi(ai1S(ai2)a·b·S-1(ai4))⊗ai3=0
That is
bi(a·b·S-1(ai2))⊗ai1=0
As that A is cocommutative, we can obtain
bi(a·b·S-1(ai1))⊗ai2=0
Applying Δ again, multiplying with a′ from the left and replacing b by ba″, we obtain
Replacing a″ by 
we can obtain
Hence, bi(a·b·a′)⊗ai=0. Note that B has a non-degenerate product and it is a unital left A-module and a unital right A-module, and we obtain bi⊗ai=0.
On the other hand, suppose that (b#a)(bi#ai)=0 for all b∈B and a∈A. Similar to the proof of Lemma 5.6 in Ref.[1], we can obtain bi⊗ai=0.
Moreover, we will construct a comultiplication for the diagonal crossed product such that it admits a structure of multiplier Hopf algebra. First, we give two maps as follows.
Let A and B be regular multiplier Hopf algebras such that B is an A-bimodule algebra. The multiplication of diagonal crossed product B#A is defined by two twist maps R:A⊗B→B⊗A via a⊗b
a1·b⊗a2 and R′:B⊗A→B⊗A via b⊗ab·S-1(a2)⊗a1. It can easily be checked that R and R′ are two bijections. R-1:B⊗A→A⊗B via b⊗aa2⊗S-1(a1)·b and R′-1:B⊗A→B⊗A via b⊗ab·a2⊗a1.
We note that the comultiplication ΔA in a multiplier Hopf algebra (A,ΔA) is determined by two linear bijections 
For all b,b′∈B and a,a′∈A, we define
○
○ R34○
○ (R-1)34○ (R′-1(b#a⊗b′#a′)=
b1#a1⊗b2(a2·b′·S-1(a4))#a3a′
where 
on the first and the third components, similar to other operators. Observe that a2 and a4 are covered by b′, a3 is covered by a′, and b2 is covered by (a2·b′·S-1(a4)).
○
○ R12○
○ (R-1)12○ (R′-1(b#a⊗b′#a′)=
Observe that a1 and a5 are covered by b, 
is covered by (S-1(a1)·b·a5), a2 and a4 are covered by [(S-1(a1)·b·a5)b′1] and a′1 is covered by a3.
Definition 3 For all b, b′,b″∈B and a, a′,a″∈A, we define
Δ#(b#a)(b″#a″⊗b′#a′)=
b1(a1·b″·S-1(a3))#a2a″⊗b2(a4·b′·S-1(a6))#a5a′
and
(b″#a″⊗b′#a′)Δ#(b#a)=
b2·
Lemma 2 For any b∈B and a∈A, Δ#(b#a) is a two-sided multiplier of B#A⊗B#A.
Proof According to the definition of the comultiplication[2], we only need to prove that the equation 
holds. For all b,b′,b″∈B and a,a′, a″∈A, we have
b2(a2·b′·S-1(a4))#a3a′
(b″#a″⊗1#1)(b1#a1⊗b2(a2·b′·S-1(a4))#a3a′)=
Lemma 3 The comultiplication Δ# is coassociative on B#A.
Proof To inquire if Δ# is coassociative in the sense of Definition 2.2 in Ref.[2], we have to check if the linear maps 
obey the following relationship:
For all b,b′,b″∈B and a, a′, a″∈A, we have
b12#a12⊗b2(a2·b″·S-1(a4))#a3a″)
and
b22(a22·b″·S-1(a4))#a3a″)
We can easily see that these two terms are equal by using 
on B⊗B⊗B.
Before we proceed to give the main result, we need the following definition and lemma.
Definition 4 Let A and B be regular multiplier Hopf algebras and B is an A-bimodule algebra. Then, B is an A-bimodule bialgebra if
(3)
ε#(a·b·a′)=εA(a)εB(b)εA(a′)
(4)
for all b,b′∈B and a, a′∈A. Observe that on the right side, all decompositions are well-covered.
Lemma 4 A and B are the same as those in Definition 4. We denote the antipode of B (A, resp.) by SB (SA, resp.). Then,
a·SB(b)·a′=SB(a·b·a′)
The proof is straightforward.
Now, we can formulate the main result as follows.
Theorem 1 Let A be a cocommutative regular multiplier Hopf algebra and B is a regular multiplier Hopf algebra such that B is an A-bimodule bialgebra. Then, Δ# is a comultiplication on B#A such that (B#A,Δ#) is a regular multiplier Hopf algebra.
Proof According to Proposition 2.9 in Ref.[10], our proof is given as follows.
First, we can easily check that the diagonal crossed product B#A is an associative algebra, and we prove that it has a non-degenerate product by Lemma 1.
The comultiplication Δ# as a multiplier of B#A⊗B#A is coassociative by Lemma 3. We now show that Δ#:B#A→M(B#A⊗B#A) is a homomorphism. For all b,b′,b″∈B and a,a′,a″∈A, we have
Δ#(b#a)(Δ#(b′#a′)(1#1⊗b″#a″))=
Δ#(b(a1·b′·S-1(a3))#a2a′)(1#1⊗b″#a″)=
Δ#((b#a)(b′#a′))(1#1⊗b″#a″)
We can easily prove that the functions 
and R and R′ are bijective.
There is a counit ε# defined as
ε#(b#a)=εB(b)εA(a)
for all b∈B and a∈A. It can easily be checked that these two equations (ε#⊗i)(Δ#(b#a)(1#1⊗b′#a′))=(b#a)(b′#a′) and (i⊗ε#)((b′#a′⊗1#1) Δ#(b#a))=(b′#a′)(b#a) are valid. From Eq.(4), it easily follows that ε# is a homomorphism.
There is an invertible antipode S# defined as
S#(b#a)=R′oRo(SA⊗SB)oσ(b#a)=
SA(a3)·SB(b)·a1#SA(a2)∈B#A
For any b, b′∈B and a, a′∈A, we obtain
m#(S#⊗iB#iA)(Δ#(b#a)(1#1⊗b′#a′))=
m#(S#⊗iB#iA)(b1#a1⊗b2(a2·b′·S-1(a4))#a3a′))=
m#(S(a3)·S(b1)·a1#S(a2)⊗b2(a4·b′·S-1(a6))#a5a′)=(S(a5)·S(b1)·a1)(S(a4)·
(b2(a6·b′·S-1(a8)))·a2)#S(a3)a7a′=
S(a3)·(S(b1)b2(a4·b′·S-1(a6)))·a1#S(a2)a5a′=
εB(b)b′·S-1(a4)·a1#S(a2)a3a′=εB(b)εA(a)b′#a′=
εB#A(b#a)(b′#a′)
Similarly, we can obtain
m#(iB#iA⊗S#)((b′#a′⊗1#1)Δ#(b#a))=
εB#A(b′#a′)(b#a)
Using Lemma 4, we can easily prove that S# is an anti-homomorphism.
Finally, we can easily observe that the multipliers Δ#(b#a)(b′#a′⊗1#1) and (1#1⊗b′#a′)Δ#(b#a) are in B#A⊗B#A.
The proof is complete.
In Theorem 1, if the right action is trivial, we have the following corollary which is the main result of Ref.[7].
Corollary 1 Let A be a cocommutative multiplier Hopf algebra and B is a regular multiplier Hopf algebra such that B is an A-bimodule bialgebra. Then, Δ# is a comultiplication on B#A such that the smash product (B#A,Δ#) is a regular multiplier Hopf algebra.
In Ref.[10], for multiplier Hopf algebras, there is a natural notion of left and right invariance for linear functionals (called integrals in the theory of Hopf algebra). We now suppose that A and B are multiplier Hopf algebras which are the same as those in Theorem 1. Furthermore, we suppose that A and B have invariant functions. In the following proposition, we will give these integrals on (B#A,Δ#).
Proposition 1 A and B are the same as those in Theorem 1. Let φB(φA, resp.) be a left integral of B(A, resp.). Then, φB⊗φA is the left integral of B#A. The same statement yields for the right integral.
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