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

[Oxymoron]

I have a C*-algebra A with a unit 1.

QUESTION 1: If \alpha\,:\,A\rightarrow A is an automorphism then does this mean that \alpha is also a *-isomorphism?

Now, I also have a positive functional f on A such that f(\alpha(a)) = f(a) for all a. Then I construct the following subspace of A:

N = \{a\,:\,f(a^*a)=0\}

QUESTION 2: If I quotient A by N to form the quotient space A/N then does this mean that my automorphism \alpha 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

L\,:\,A/N \rightarrow A/N

which is isometric. So perhaps, by quotienting A by N we have a subspace of a C*-algebra such that each element satisfies f(a^*a) \neq 0 then if we have an automorphism \alpha\,:\,N\rightarrow N we induce a isometric linear transformation L? I am not sure.

 

[Oxymoron]

QUESTION 3: I think that if f is a positive functional then the theorem by Gelfand and Naimark (1943) says then there is a representation \pi on the Hilbert space \mathcal{H}_f and a vector h\in\mathcal{H} such that

f(a) = (\pi(a)h\,|\,h) \quad\quad \forall\,A\in A

So, if \pi\,:\,A\rightarrow B(\mathcal{H}) is a representation and h \in \mathcal{H} then f(a) = (\pi(a)h\,|\,h) defines a positive functional on A! - And what's more, ALL positive functionals arise in this way! ...amazing.

 

[matt grime]

Answer 1: the heuristic argument would be: if any automorphism of a C* algebra as a vector space were a *-automorphism why would we define a *-automorphism differently? So try looking for an auto morphism that is not a *-map.

Answer 2: you should just go and think about linear algebra again, in particular the isomoprhism theorems. In any case as long as alpha maps N to N then it induces a map on the quotient.

"If I quotient A by N to form the quotient space A/N then does this mean that my automorphism goes from N to N instead of A to A?"

 

[Oxymoron]

I don't see why, if \alpha was originally an automorphism on A, then once I quotient A by N, the automorphism is now on N!? Shouldnt it be from A/N to A/N? Unless, of course, it does?...and it is also on N.

 

[matt grime]

I am not entirely clear what it is you're asking. But let me try to put in the thing that might help you understand.

It is the isomorphism theorems.

take M:V-->V a linear map. If M preserves a subspace W then M induces a map V/W to V/W by M[x]=[Mx] where [x] is a coset in V/W

Of course given any W<V, I can try to define a map from V/W to V/W by M[x]=[Mx] and if this is a well defined linear map on V/W to V/W it must send W to W since it must send the coset [0] to [0]. Of course proving this is usually just the same as proving directly that M maps W into W.

The problem would be if Mw was not in W for some w for then we would have

[0]=[M0]=M[0]=M[w]=[Mw]=/=[0]

 

[Oxymoron]

You seem to like turning functional calculus questions into vector space questions. I wish I could make the same analogies on my own, it seems to work for you. It looks like it makes this stuff a little easier to understand.

Im not too sure of your notation here on the right hand side: "M[x] = [Mx]". You say [x] is a coset in V/W. Does this mean that [x] is also an equivalence class of x in V? So what is [Mx]?

If you want to speak in terms of vector spaces then W is an ideal isn't it?

 

[Oxymoron]

Answer 2: you should just go and think about linear algebra again, in particular the isomoprhism theorems. In any case as long as alpha maps N to N then it induces a map on the quotient.

I know that alpha maps A to A and that it preserves adjoints. You say that as long as alpha maps N to N then it induces a map U from A/N to A/N. Quotienting A by N to form a subspace A/N of A, do I get a *-isomorphism fo free? If so, is it related to alpha at all? This probably doesn't make much sense... I have a *-isomorphism from A to A. I quotient A by N. Can I form a *-isomorphism from A/N to A/N? Is it related to alpha in any way? Can I form a *-isomorphism from N to N? Is it related to alpha in any way?

 

[matt grime]

The reason why I keep turning this into vector space stuff is because to some extent that is all that this is: we aren't really discussing any of the analytic properties like convergence of sums of elements. Just think about the additive nature of the question first, then you need to look at the algebraic nature (ie is the quotient actually an algebra rather than just a vector space). This question is just an example of the isomorphism theorems of linear algebra for vector spaces with extra structure.

Just take the vector space structure for now: [x] is an element in the quotient, the coset x+W, and [Mx] is the coset Mx+W. Given a map on V you define an induced map on the cosets, as long as M maps W to W this is well defined (i.e. take two different elements in the same coset and the induced map sends them to the same coset). From what you write I don't think you understand quotient spaces, just as vector spaces. Which is why I'm talking about those and ignoring the algebra part of the question for now.

Once you understand the nature of what a quotient is, and what maps on the large space do to the quotient, then you can start adding in extra things like: is the quotient actually a C* algebra? (N needs to be an ideal as well now because we need to define [x][y] as [xy], which in terms of cosets looks like (x+N)(y+N)=(xy+N), and that is only going to happen if N is closed under adding elements of N (ie a vector subspace) and if xN and yN are in N too, i.e. N is an ideal.In this case, alpha is a *-iso on A, it restricts to a *-iso on N, and hence passes to a *-iso on the quotient. The isos are just the restriction and quotient of alpha, so they are intimately related to alpha.

 

[Oxymoron]

Ok, fair enough. Let's look at vector spaces. I think you are right, I am taking my knowledge of quotient spaces for granted because I was hoping I didnt need to know the details for this question. As a result of what you said, I got out my algebra book and refreshed myself.

First of all, the equivalence class to which x belongs (written [x]) is x + V, as you have said. This is a coset of V? If so, then if I add two of them together, say [x] and [y], then I get

[x] + [y] = (x+V)+(y+V) = (x+y)+V

Im not sure (or I can't remember) if the coset [x] obeys the vector space axioms implying that it is in fact a vector space of cosets of V. I am not too sure on this. I think it is tho...

If so, the vector space of cosets of V (denoted W) is called the quotient space of V by W, and it is written as V/W. So the quotient space V/W is actually a vector space in its own right.

QUESTION: Can I say that V/W is the complement of W?

 

[matt grime]

There is no such thing as 'the' complement of W. Think in two dimensions. Pick a line through the origin. There are uncountably many complementary subspaces to this line.

Let's fix notation W is a subspace of V, [x]=x+W, and is a coset of W. Remember this is thinking about V/~ where ~ is an equivalence relation: x~y if x-y is in , [x] is the equivalence class of x under ~

The set of equivalence classes of ~ is a vector space over the same field as V and [x]+[y]=[x+y], k[x]=[kx] for k a scalar.

 

[Oxymoron]

take M:V-->V a linear map. If M preserves a subspace W then M induces a map V/W to V/W by M[x]=[Mx] where [x] is a coset in V/W.

"If M preserves a subspace W". Does this mean that if W is a subspace in V then M(W) is also a subspace in W? So if M preserves subspaces, how does M induce a map V/W to V/W? I mean, what if M did not preserve subspaces? Why does it have to preserve subspaces for it to induce a map? And that's another thing that I don't really understand: "M induces a map". Is this just a way of saying that "because of the existence of M there exists a map..."?

 

[matt grime]

M does not preserve (all) subspaces (only the multiples of the identity map do that) it must just preserve W, which as you correctly guess means that M(W) is in W.

And I told you how it induces a map on V/W, and I explained why it is necessary for M to map W into W for this to be a well defined map. See post 5. But here it is again:

Let us by abuse of notation write M[x] to be the map

M[x]=[Mx]

this is a map of cosets and if M(W) is in W it is a well defined map: for any coset [x], [Mx] is another coset. If M did not preserve W then there is a w in W with Mw not in W, but then the cosets [0] and [w] are the same element in V/W, and [M0]=[0] is not the same coset as [Mw] since the zero coset contains exactly the things in W. Thus it is not a well defined map on the set of cosets if it does not preserve W. This should all have been taught to you in linear algebra.

I say M induces the map because I am using M to define the map. This is a common usage in maths for the term induce.

 

[Oxymoron]

In case you are wondering why I asked, I am trying to show that \alpha\,:\,A\rightarrow A, the automorphism we are talking about, induces a unitary operator U\,:\,\mathcal{H}_f \rightarrow \mathcal{H}_f, where f is a positive functional on A such that f(\alpha(a)) = f(a). But more on this when we get to it. By the way, do you have any "vector space analogies" for the Hilbert space? Or is it just another vector space to you?

And I told you how it induces a map on V/W, and I explained why it is necessary for M to map W into W for this to be a well defined map. See post 5.

Ah, you did too. Sorry about that.

this is a map of cosets and if M(W) is in W it is a well defined map: for any coset [x], [Mx] is another coset. If M did not preserve W then there is a w in W with Mw not in w, but then the cosets [0] and [w] are the same element in V/W, but [M0]=[0] is not the same coset as [Mw] since the zero coset contains exactly the things in W.

...and this proves that M must preserve the subspace W for it to induce the map M[x] = [Mx] - (is this the map from V/W to V/W?). I mean, M is the automorphism from V to V, right? So M acting on a coset [x] gives another coset [Mx]. Is this the correct definition of that terminology?

 

[matt grime]

We're just talking about vector spaces and linear maps right now, M does not have to be an automorphism, it just has to map W to W. Don't get hung up on the actual question you started with yet until you understand the basics here. M must preserve W to give a *well defined* map on the quotient. That is all. Something is well defined if it is independent of choices. Here you have to make a choice of coset representative to define the map. If you choose different representatives of the same coset it must give the same answer. What does that mean here? Suppose [x]=[y], then x-y is in W, for [Mx] to equal [My] it means that Mx-My is in W, or M(x-y) is in W. Since x-y is in W this can happen if and only if M(x-y) is in W, ie M must preserve W.

But I repeat there is no necessity for M to an automorphism.

If M is an automoprhism, so is the induced map on V/W (because it has an inverse, M^{-1} on V this passes to an inverse on the quotient).

Now, in your case, we're dealing with something more than just a vector space but the basics are still the same.

N must be both a a vector subspace just thinking of vector spaces alone, but it must be an ideal when one considers the algebra structure.Banach spaces, C* algebras and Hilbert spaces are more complicated than just vector spaces, but right now you're really only looking at their vector space structure. There is no harm, to begin with, of just thinking: what does this mean for a finite dimensional vector space over C? Which is after all examples of hilbert spaces. If you don't use the finite dimensionality at all in your thinking then you've got an idea of what is going on in general.

 

[Oxymoron]

We're just talking about vector spaces and linear maps right now, M does not have to be an automorphism, it just has to map W to W.

See, here I was thinking a map of the form M:V-->V is an automorphism! But it must also be an isomorphism (as well as being a map to itself). Right?

Ok, so a map M:V-->V which preserves the subspace W of V induces a well defined map on the quotient, V/W. If M is an automorphism then so is the induced map.

 

[Oxymoron]

Ok, N is a subspace. Membership for N means that an element a must satisfy f(a^*a)=0. So A/N = \{a\,:\,f(a^*a)\neq 0\}?

And you are saying that N is also an ideal?

 

[matt grime]

See, here I was thinking a map of the form M:V-->V is an automorphism! But it must also be an isomorphism (as well as being a map to itself). Right?

yes, an automorphism is an invertible 'morphism' in any place where this makes sense (morphisms are another term for maps, and useful one. in category theory we just use morphims to mean a map and then append words in front like auto and endo, mono and epi to mean different types of maps).

 

[matt grime]

Ok, N is a subspace. Membership for N means that an element a must satisfy f(a^*a)=0. So A/N = \{a\,:\,f(a^*a)\neq 0\}?

Elements of A/N are not elements of A! They are sets of elements of A, equivalence classes of elements of A. Just like we sorted out that V/W is *not* the complement of W. V/W is not even a subspace of V. What you've written is a subset of A, and A/N is not a subset of A. Take another analogy. Z/nZ modulo arithmetic, the elements of Z/nZ are not elements of Z, are they? (We might use the same symbols but this is an abuse of notation, 1 means [1] and so on, i.e. we pick a distinguished element from each equivalence class)

Just to clarify, N must be a two sided ideal: I made a mistake earlier:

(x+N)(y+N)=xy+xN+Ny+N

so N needs to be a lft ideal (xN<N) and a right ideal (Ny<N)

 

[Oxymoron]

Elements of A/N are not elements of A! They are sets of elements of A, equivalence classes of elements of A. What you've written is a subset of A, and A/N is not a subset of A. Take another analogy. Z/nZ modulo arithmetic, the elements of Z/nZ are not elements of Z, are they? (We might use the same symbols but this is an abuse of notation, 1 means [1] and so on, i.e. we pick a distinguished element from each equivalence class)

I see, I see. Very good. I seem to remember my algebra lecturer drumming this into my brain a couple of years ago. The drumming didnt help apparently. Of course, your Z/nZ example is a clear expression of my mistake. Indeed, they are sets!

Just like we sorted out that V/W is *not* the complement of W. V/W is not even a subspace of V.

Which mean that I cannot say that

A/N = \{a\,:\,f(a^*a)\neq 0\}

right? Instead, I must first pick a representative, any representative, of A, call it a. Then define an equivalence relation ~ on A by saying that a ~ b <=> a-b is in N, where N is a subspace of A. Then the quotient space A/N is defined as A/~: the set of all equivalence classes over A by ~. So the quotient space A/N is a set of equivalence classes. Now, what does this translate into for C*-algebras? Or is there more to consider first?

 

[matt grime]

As long as N is a two sided ideal the quotient vector space inherits a multiplication from A, and a * map from A. All is fine there (though I wouldn't say anything about picking a representative of A, you are picking, if you really must, representaives of equivalence classes, though there is no harm in talking purely about equivalence classes and not representatives).

 

[Oxymoron]

So if I quotient any vector space A by a two-sided ideal, the quotient space A/N inherits the binary operation from A? I pretty sure I grasp that idea. And from what we have been talking about (i.e. Post #14) there is an induced map A/N to A/N - and if the corresponding map M:A-->A is an automorphism then so is L:A/N --> A/N. Is this right?

 

[Oxymoron]

So before we saw that if we had a map M:V-->V (which preserved subspaces) and we quotiented the V by some subspace W, then we automatically induce another map L:V/W-->V/W.

Now certainly, if we change this scenario slightly, and instead of W being some subspace of V it is instead a two-sided ideal of V, the induced map is the same. However, would M:V-->V have to preserve ideals as well? Or is that inherent in making M preserve subspaces?

 

[matt grime]

Vector spaces do not have two sided ideals. Algebras have two sided ideals. You need a multiplication of elements.

You still seem to think that M preserves subspaces/ideals. It does not. That implies it preserves *all* subspaces/ideals. It needs to only preserve one subspace/ideal that in question. It just has to map whatever you're quotienting about by to itself, whatever that object is in whatever sense of quotient this is. It is just the isomorphism theorem. It works for quotient vector spaces, quotient groups, quotient rings, quotient algebras, quotient lie algebras, quotient modules, any quotient thing.

 

[Oxymoron]

When I did algebra I was introduced to the idea of an ideal via rings and ring homomorphisms. I was taught the following Theorem:

\mbox{Let } \phi\,:\,R\rightarrow R&#039; \mbox{ be a ring homomorphism with kernel } H\mbox{. Then the additive cosets of }H\mbox{ form the quotient ring }R/H\mbox{ whose binary}
\mbox{ operations are defined by choosing representatives. That is, the sum of two cosets is defined by }(a+H)+(b+H) = (a+b) + H\mbox{ and the}
\mbox{product of cosets is defined by }(a+H)(b+H) = ab+H\mbox{. Also, the map }M\,:\,R/H \rightarrow \phi[R]\mbox{ defined by }M(a+H) = \phi(a)\mbox{ is an isomorphism.}

Is this the same theorem we are talking about here?

Because, after this, we defined an ideal to be the additive subgroup N of a ring R satisfying

aN \subseteq N \quad \mbox{and} \quad Nb \subseteq N \quad \forall\,a,b \in R

and after this we defined the fundamental homomorphism theorem:

\mbox{Let }N\mbox{ be an ideal of a ring }R.\mbox{ Then }y\,:\,R\rightarrow R/N\mbox{ given by }y(x) = x + N\mbox{ is a ring homomorphism.}

All this stuff I can prove, but I am not sure if it relates to anything we have been discussing here. Although it does looks half-relevant.

 

[matt grime]

That is part of it.

A C* algebra is just a ring. You are just forming a ring theoretic quotient, it so happens the C* algebra carries extra structure (it is an algebra, ie a vector space, and has a * operation too), and so does the subalgebra (which is a vector space, a two sided ideal, and a * algebra in its own right) you're quotienting out by, and it can be shown that the quotient therefore inherits the extra structure too (ie it is also a vector space, and therefore an algebra has a * map too).

 

[Oxymoron]

Is that why the map \alpha\,:\,A\rightarrow A induces a map L\,:\,A/N\rightarrow A/N?

 

[matt grime]

Erm, haven't we established this already? Just define it, using alpha, like I cited in posts 5, 8, 12, 14, and referred to all the time: L[a] = [\alpha(a)], that is the whole point of what I have said about 5 times now: if M (or alpha) is a map from V to V (or A to A) and it sends some W to W (or N to N) then it acts in a well defined manner on the cosets of V/W (or A/N). Please don't ask me to re-explain what well defined is. Please, I urge you to reread all the posts so far so I don't have to say the same thing for a sixth time.

 

[Oxymoron]

Erm, haven't we established this already? Just define it, using alpha, like I cited in posts 5, 8, 12, 14, and referred to all the time: L[a] = [\alpha(a)], that is the whole point of what I have said about 5 times now: if M (or alpha) is a map from V to V (or A to A) and it sends some W to W (or N to N) then it acts in a well defined manner on the cosets of V/W (or A/N). Please don't ask me to re-explain what well defined is. Please, I urge you to reread all the posts so far so I don't have to say the same thing for a sixth time.

Wait a minute. We've been talking about vector spaces and subspaces, only now we are talking about algebras and ideals. I merely wanted to make sure the same things you've been speaking about (in 5, 8, 12 and 14). hold in this case. I guess, but I am not sure, there are analogous isomorphism theorems for algebras as there are for vector spaces?

Just out of curiosity, is there anything that holds in vector spaces/subspaces which does not hold in algebras/ideals? I mean, will anything we discuss in the vector space setting carry over to the algebra setting?

The way I see it (which could very well be incorrect), a vector space can be defined by an abelian group with a law of composition over some field. Then from vector spaces we can form algebras. Then a C*-algebra is just an algebra with the * law of composition. Then an algebra isomorphism is a bijective homomorphism and if there exists an isomorphism between two different algebras then they are isomorphic.

Now say I take a regular vector subspace N of my algebra A. If it is closed with respect to products then I can form left and right ideals in the usual way. Ideals don't need the * law of composition of vectors. Then a two-sided ideal (or simply an ideal) which is both left and right.

Now, let H:A-->A be an algebra homomorphism (Im not sure if this is automatically an isomorphism?). The kernel of H, kerH, is the subspace of A consisting of those vectors mapped to 0 in A. It just so happens that the kernal of any such map is an ideal. So here is an example of an ideal which I believe in some way reflects the problem at hand (not exactly though).

If N is a (two-sided) ideal of A then there is an algebra structure define on the quotient space A/N right? So a map P:A-->A/N, where A is mapped to [A] = A+N, is a homomorphism with kernel N! Although P:A-->A/N is not quite the same as M:A/N-->A/N - maybe it is and I don't know it!

If N is an ideal of A, then typical elements of A/N are cosets [A] = A+N. Then all we have to do is define an algebra structure on A/N by setting [A]=[AB]. (Note: I am attempting to use your concise notation, I hope I am not abusing it!). Then I can see why this is natural, because it is independent of the choice of representative - just like you said! In fact, exactly like you said. I am finally starting to understand this. By the way, when I am told something, I like to attempt to repeat it on my own to see if I really have understood it. That is why a lot of the stuff I write may be repeating some of the stuff you have said. Please don't be offended.

 

[matt grime]

Wait a minute. We've been talking about vector spaces and subspaces, only now we are talking about algebras and ideals. I merely wanted to make sure the same things you've been speaking about (in 5, 8, 12 and 14). hold in this case. I guess, but I am not sure, there are analogous isomorphism theorems for algebras as there are for vector spaces?

it is all the same principle: understand one quotient idea understand them all.

Now, let H:A-->A be an algebra homomorphism (Im not sure if this is automatically an isomorphism?)

of course not because...

The kernel of H, kerH,

might not be zero

If N is a (two-sided) ideal of A then there is an algebra structure define on the quotient space A/N right? So a map P:A-->A/N, where A is mapped to [A] = A+N, is a homomorphism with kernel N! Although P:A-->A/N is not quite the same as M:A/N-->A/N - maybe it is and I don't know it!

It cannot be the same map. The quotient map p:A->A/N does not need M to be defined to make sense, just look at the definition of the two maps: one sends x to [x], the other [y] to [My]. M maps from A to A and induces a map from A/N to A/N, the induced map M:A/N -> A/N doesn't even have the same domain as the map A->A/N.

If N is an ideal of A, then typical elements of A/N are cosets [A] = A+N. Then all we have to do is define an algebra structure on A/N by setting [A]=[AB]. (Note: I am attempting to use your concise notation, I hope I am not abusing it!).

A stood for the algebra, now it stands for a coset: that is not good.

 

[Oxymoron]

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 is that suppose n \in N but \alpha(n) \notin N. Let a\in A and write a^{\prime} for a-N. Then a ~ a' because a-a&#039; = n \in N. Therefore, a+N and a&#039;+N are the same element of a/N.

However, this is a problem because we have defined L by the formula above and since a+N = a&#039;+N \in a/N, if L is to be a linear map of A/N we must have L(a+N) = L(a&#039;+N). However, L(a+N) - L(a&#039;+N) = \alpha(a) + N - \alpha(a&#039;) + N = \alpha(n) + N, and since \alpha(n) \notin N, this is not the zero element of A/N.

That is, the formula

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

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

 

[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).