Does an Automorphism on a C*-Algebra Induce a *-Isomorphism?

[Oxymoron]

The big question I must now ask is: If L is isometric does it extend to an isometry, U, on the completion $\mathcal{H}_f$?

 

[matt grime]

Let me now try to make contact with the question. We know that $\alpha\,:\,A\rightarrow A$ is an automorphism of the C*-algebra A. Now define the subspace N by $N=\{a\,:\,f(a^*a)=0\}$, where $f(\alpha(a)) = f(a)$. Then $\alpha(N)$ is, in general, some subset of A.

However, if $\alpha(N) \subset N$ then we can define a linear map $L\,:\,A/N \rightarrow A/N$ by simply using, say

$$L(a+N) = \alpha(a) + N$$

The reason why I can see this working only if $\alpha$ maps N inside itself

I have repeatedly said that alpha *has* to map N into itself, it just does. For the map to even make sense at the level of vector spaces this must be what happens (and A, N and A/N are all vector spaces, alpha is a map of vector spaces).

That is, the formula

$$L(a+N) := \alpha(a) + N$$

is only well-defined in the situation where alpha maps N back inside itself.

Yes, we've been over this many times.

 

[matt grime]

The big question I must now ask is: If L is isometric does it extend to an isometry, U, on the completion $\mathcal{H}_f$?

What are L, H_f, etc? I mean what is a precise statement of the exact question with all the terms explained in one single post?

 

[Oxymoron]

Suppose $\alpha$ is an automorphism of a C*-algebra A with 1, and that f is a positive functional on A such that $f(\alpha(a))=f(a)$ for all a in A. Show that $\alpha\,:\,A\rightarrow A$ induces a unitary operator $U\,:\,\mathcal{H}_f\rightarrow\mathcal{H}_f$ such that $\pi_f(\alpha(a)) = U\pi_f(a)U^*$ for a in A.

Hint from lecturer: To get $\mathcal{H}_f$, we form the quotient of A by the subspace $N=\{a\,:\,f(a^*a)=0\}$. Check that $\alpha\,:\,N\rightarrow N$, and hence induces a linear transformation $L\,:\,A/N\rightarrow A/N$. Now check that L is isometric, hence extends to an isometry U on the completion $\mathcal{H}_f$, and that U is invertible, hence unitary.

$\mathcal{H}_f$ is a Hilbert Space.
$\pi$ is a representation of the form $\pi\,:\,A\rightarrow B(\mathcal{H})$.

 

[matt grime]

So, you are not forming the C* quotient after all, then, just the vector space quotient. N is not necessarily a two sided ideal. A/N is just a vector space (non-complete, otherwise it wouldn't make sense to take its completion). I should hve checked this before, but it seemed more important to get across what a quotient was in general. If N were an ideal, A/N would be a C* algebra and complete already, right? C* algebras are banach spaces are they not? (It is years since I did this, so do not take my word for it).

Anyway, what part is troubling you? alpha maps N to N (you checked?), hence maps A/N to A/N, L is isometric (you checked?), hence it extends to an isometry on the completion of A/N which I is H_f, right... U is invertible, again seems clear, so where is the problem?

But what is pi exactly? Is it pi(a)[x]=[ax]? Or something else?

 

[Oxymoron]

So, you are not forming the C* quotient after all, then, just the vector space quotient. N is not necessarily a two sided ideal. A/N is just a vector space (non-complete, otherwise it wouldn't make sense to take its completion). I should hve checked this before, but it seemed more important to get across what a quotient was in general. If N were an ideal, A/N would be a C* algebra and complete already, right? C* algebras are banach spaces are they not? (It is years since I did this, so do not take my word for it).

Yes, all C*-algebras are Banach algebras. C*-algebras just have the * law of composition, which makes them a little but more specific than Banach algebras.

Anyway, what part is troubling you? alpha maps N to N (you checked?)...

To be honest I don't know how I am meant to check such a thing. I mean it is one thing to say that I have such a map and another to show that it exists. Would I start by picking some element of N, n say, and show that alpha maps it to some element m in N?

...hence maps A/N to A/N...

and I am not going to ask you to show me this again, am I?!

L is isometric (you checked?), hence it extends to an isometry on the completion of A/N which I is H_f, right... U is invertible, again seems clear, so where is the problem?

I have not checked that L is isometric. To be honest once again, I am not sure how to. I would guess that I would need to use the norm somewhere in the proof - somehow show that the map L preserves the norm. In fact, this might be fairly easy to show, I seem to remember a corollary dealing with isometric mappings, Ill have to get back to you on that...

Now this is where I struggle: "...extends to an isometry on the completion H_f" What does "extends" mean in this context!? I have done no work on extension theorems so surely it has nothing to do with that. Ugh, I don't like all these "words", I think I am meant to know what all this stuff means but I dont.

I think I know what "completion" means though. When I did topology (I like topology) complete meant that every Cauchy sequence in the set converged within that set. Now, I keep getting told that complete, in the analytic sense means that it's "span is dense". I am not claiming to know what this means, but when I think of "dense" I think of "approximable" or "can be approximated to polynomials" or something like that.

The hint says show that if L is isometric then it "extends" to an isometry U on the completion H_f. I guess H_f is meant to be the completion of A/N. But in my language this means that H_f (being complete) contains A/N as a dense subspace! Does this sound right to you?

But what is pi exactly? Is it pi(a)[x]=[ax]? Or something else?

All I am told is that it is the GNS-representation associated to f. (I think GNS stands for Gelfand, Naimark, and...Segal? Not sure). It comes from this theorem in 1943:

Let f be a positive functional on a C*-algebra A with 1. Then there is a representation pi_f on a Hilbert space H_f and a vector h_f in H such that

$$\{\pi_f(a)h_f\,:\,a\in A\}$$

is dense in H_f, and

$$f(a) = (\pi(a)h_f\,|\,h_f) \quad \forall\,a \in A$$.

 

[matt grime]

To be honest I don't know how I am meant to check such a thing.

yes you do, you've been doing this kind of thing for ages now (or you wouldn't be doing a course on C* algebras!).

You just need to show that if n is in N so is alpha(n). What is the characterisation of N? The n such that f(n*n)=0. So what is f(alpha(n)*alpha(n)), recalling that alpha is a (*-) automorphism and f(\alpha(x))=f(x) for all x?

I have not checked that L is isometric. To be honest once again, I am not sure how to. I would guess that I would need to use the norm somewhere in the proof - somehow show that the map L preserves the norm.

what is the definition of isometric? Use it.

Now this is where I struggle: "...extends to an isometry on the completion H_f" What does "extends" mean in this context!? I have done no work on extension theorems so surely it has nothing to do with that. Ugh, I don't like all these "words", I think I am meant to know what all this stuff means but I dont.

you must have done a course on analysis where they explain completions. The fundamental result about completions is that any continuous map f on X extends to a continuous map on the completion of X. The completion of X is defined purely in terms of cauchy sequences of elements of X so you can define f on the completion too because you can use it on the cauchy sequences.

You should definitely look up the GNS result and see how it works (the proof of it will show you what H_f is, and how the action is defined).

 

[Oxymoron]

$$f(\alpha(n)^*\alpha(n)) = f(\alpha(n)^*)f(\alpha(n))$$
$$\quad\quad = f(\alpha(n))^*f(\alpha(n))$$
$$\quad\quad = f(n)^*f(n)$$
$$\quad\quad = f(n^*)f(n)$$
$$\quad\quad = f(n^*n) =0$$

 

[matt grime]

f is only a linear functional, it is not a (*)-homomorphism so f(xy) is not in general f(x)f(y). To see why, if f is a linar functional so is 2f, and then yo'ure claiming that 2=4. alpha is a *-homomorphism, though.

 

[Oxymoron]

You should definitely look up the GNS result and see how it works (the proof of it will show you what H_f is, and how the action is defined).

I did, and it goes on and on about cyclic vectors (which, incidentally I need to know about for the next question!) The proof for the GNS theorem is absolutely huge. It's split up into 4 sections and requires 3 lemmas! (and ends with "...so this is easy". Which means "It really took me two days and 40 blank pages and 2 pens to come to this complication proof which I present to you now severly compressed and lacking any indication of how I came up with my ideas).

 

[matt grime]

But you do not necessarily need to understand all of the proof, you just need to know what H_f is and how A acts on it.

 

[Oxymoron]

f is only a linear functional, it is not a (*)-homomorphism so f(xy) is not in general f(x)f(y). To see why, if f is a linar functional so is 2f, and then yo'ure claiming that 2=4. alpha is a *-homomorphism, though.

Ah, but surely I can say that

$$f(\alpha(n)^*\alpha(n)) = f(\alpha(n^*)\alpha(n))$$

since alpha is a*-automorphism. Then

$$= f(\alpha(n^*n)) = f(n^*n)$$

since f(a(n)) = f(n). Then we have it equal zero.

 

[matt grime]

http://www.math.psu.edu/roe/520/home.html

read these notes, they're very good.

 

[Oxymoron]

But you do not necessarily need to understand all of the proof, you just need to know what H_f is and how A acts on it.

Ok, I read the proof again and got some ideas from it:

* N is a vector subspace of A (which we already knew).
* We define a pairing on A/N by (a+N|b+N):=[a|b] = f(b*a) which is a well-defined inner product - which I easily show, just went through that proof. (inner products! That's Hilbert space territory! Oh oh)
* I found that the operator $\pi(a)$ is linear and bounded.
* A/N is a subspace of H
* pi_f (a*) = pi_f(a)*

*The vector space quotient is $$A/N := \{a+N\,:\,a\in A\}$$ which we already know, but could have been handy for me a lot earlier.

 

[matt grime]

What is the operator pi(a), precisely? (To satisfy my own curiosity. I would guess pi(a)[x]=[ax] as the obvious candidate). It should be clear that L (the thing alpha induces) is an isometry now.

 

[Oxymoron]

What is the operator pi(a), precisely? (To satisfy my own curiosity. I would guess pi(a)[x]=[ax] as the obvious candidate).

Im not 100% sure either. This is the first time I've ever encountered such operators (representations). I am not sure what pi(a)[x] means. pi is a linear operator acting on a. Then "multiplied" by [x]? I don't know.

 

[Oxymoron]

It should be clear that L (the thing alpha induces) is an isometry now.

This is all I know about L at the moment:

1) $$L\,:\,A/N\rightarrow A/N$$

2) L is induced by a *-isomorphism $\alpha\,:\,A\rightarrow A$.

Oh wow. I think that maybe L is a representation! Since it satisfies

$$L(a)(b+N) = \alpha(a)(b)+N = ab+N$$

So if I can show that L is bounded and linear then it must be an isometry.

 

[Oxymoron]

Also L(ab) = L(a)L(b) on A/N.

 

[matt grime]

Uh? Why are you defining L(a) like that? L is the map of A/N to A/N, it sends cosets to cosets L[a]=[alpha(a)], and is obviously an isometry:

(L([a],L)= f(alpha(a)*alpha(b)) = f(a*b) hence is an isometry.

 

[matt grime]

Im not 100% sure either. This is the first time I've ever encountered such operators (representations). I am not sure what pi(a)[x] means. pi is a linear operator acting on a.[\quote]

Just to clarify, you do realize (you've said it more than once) pi is a *-homomorphism from A to B(H_f), i.e. pi(a) is an element of B(H_f), thus pi(a) is a linear map on H_f (whcih is the completion of A/N).

 

[Oxymoron]

If $n\in N$ then $\alpha(a)(n) \in N$ for all $a \in A$. But we have

$$f((an)^*(an)) = f(n^*a^*an) = 0$$

which is in N. So $\alpha\,:\,N\rightarrow N$ gives $L\,:\,A/N\rightarrow A/N$ such that

$$L(a)(b+N) = \alpha(a)(b) + N = ab+N$$

Then, L is linear because alpha is. To show that it is bounded we have

$$\|L(a)(b+N)\|^2 = (ab+N\,|\,ab+N) = f(b^*a^*ab) = f((cb)^*cb) \geq 0$$

Thus

$$\|a\|^2(b+N\,|\,b+N) = \|a\|^2f(b^*b) \geq f(b^*a^*ab) = \|L(a)(b+N)\|^2$$

and L(a) is bounded with $\|L(a)\|\leq \|a\|$. Therefore L is an isometry.

 

[matt grime]

If $n\in N$ then $\alpha(a)(n) \in N$ for all $a \in A$.

Why? I see no reason for N to be an ideal.

But we have

$$f((an)^*(an)) = f(n^*a^*an) = 0$$

again, why would that be true? N is not an ideal, f is not multiplicative.

which is in N. So $\alpha\,:\,N\rightarrow N$ gives $L\,:\,A/N\rightarrow A/N$ such that

$$L(a)(b+N) = \alpha(a)(b) + N = ab+N$$

again, this doesn't make sense. What are you doing?

Then, L is linear because alpha is. To show that it is bounded we have

$$\|L(a)(b+N)\|^2 = (ab+N\,|\,ab+N) = f(b^*a^*ab) = f((cb)^*cb) \geq 0$$

Thus

$$\|a\|^2(b+N\,|\,b+N) = \|a\|^2f(b^*b) \geq f(b^*a^*ab) = \|L(a)(b+N)\|^2$$

and L(a) is bounded with $\|L(a)\|\leq \|a\|$. Therefore L is an isometry.

What has multiplication by L(a) got to do with anything? I don't understand what you're trying to accomplish by looking at this.

f is just a linear functional, it does not have any of these properties you are assigning to it.

I already explained why L is an isometry, and this isn't it.L maps A/N to A/N. So, take b+N in A/N, what does L map it to? Lb+N, that is all. It does not require you to pick and a and work out L(a)(b+N).