- #1
Oxymoron
- 870
- 0
I have a C*-algebra A with a unit 1.
QUESTION 1: If [itex]\alpha\,:\,A\rightarrow A[/itex] is an automorphism then does this mean that [itex]\alpha[/itex] is also a *-isomorphism?
Now, I also have a positive functional f on A such that [itex]f(\alpha(a)) = f(a)[/itex] for all a. Then I construct the following subspace of A:
[tex]N = \{a\,:\,f(a^*a)=0\}[/tex]
QUESTION 2: If I quotient A by N to form the quotient space A/N then does this mean that my automorphism [itex]\alpha[/itex] goes from N to N instead of A to A? If not, could I possibly say that IF it does go from N to N THEN it induces a linear transformation
[tex]L\,:\,A/N \rightarrow A/N[/tex]
which is isometric. So perhaps, by quotienting A by N we have a subspace of a C*-algebra such that each element satisfies [itex]f(a^*a) \neq 0[/itex] then if we have an automorphism [itex]\alpha\,:\,N\rightarrow N[/itex] we induce a isometric linear transformation L? I am not sure.
QUESTION 1: If [itex]\alpha\,:\,A\rightarrow A[/itex] is an automorphism then does this mean that [itex]\alpha[/itex] is also a *-isomorphism?
Now, I also have a positive functional f on A such that [itex]f(\alpha(a)) = f(a)[/itex] for all a. Then I construct the following subspace of A:
[tex]N = \{a\,:\,f(a^*a)=0\}[/tex]
QUESTION 2: If I quotient A by N to form the quotient space A/N then does this mean that my automorphism [itex]\alpha[/itex] goes from N to N instead of A to A? If not, could I possibly say that IF it does go from N to N THEN it induces a linear transformation
[tex]L\,:\,A/N \rightarrow A/N[/tex]
which is isometric. So perhaps, by quotienting A by N we have a subspace of a C*-algebra such that each element satisfies [itex]f(a^*a) \neq 0[/itex] then if we have an automorphism [itex]\alpha\,:\,N\rightarrow N[/itex] we induce a isometric linear transformation L? I am not sure.