let A be an algebra and A1 the unitization of A, then multiplication algebra which is denoted by M(A) is algebra generated by identity operator and left multiplication and right multiplication also for the algebra A1, we have M(A1)
it is clearly that M(A) is subset of M(A1) when we have M(A)=M(A1)?
I am not an expert in this area. Maybe because of this the sentence that "M(A) is subset of M(A1)" is not so clear for me. Formally speaking, the elements of M(A) are operators on A, and the elements of M(A1) are operators on A1, so they are of different nature. When you are speaking about the inclusion of M(A) into M(A1) you maybe mean some canonical extension of operators defined on A to operators defined on A1. There is a little difficulty with this extension. Namely, if A has its own unit e, then in the unitarization one adds a new formal unit e1, so e is no more the init of the unitarization A1. Now, in this case the identity operator I_A on A is the same as the operator of multiplication by the element e. How should be I_A extended to be an element of M(A1)? There are two equally natural ways: 1) we can extend I_A to be the unitary operator I_{A1} on A1, or 2) we can extend I_A to be the operator of multiplication by e on A1, which is not the same. So what is the "canonical" inclusion of M(A) into M(A1)?
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