1. Jun 5, 2006

### 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? Im not sure.

2. Jun 5, 2006

### 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.

3. Jun 5, 2006

### 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?"

4. Jun 5, 2006

### Oxymoron

I dont 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.

5. Jun 5, 2006

### 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]

6. Jun 5, 2006

### 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?

Last edited: Jun 5, 2006
7. Jun 5, 2006

### Oxymoron

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 doesnt 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?

8. Jun 5, 2006

### 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.

Last edited: Jun 5, 2006
9. Jun 5, 2006

### 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 cant remember) if the coset [x] obeys the vector space axioms implying that it is in fact a vector space of cosets of V. Im 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?

10. Jun 5, 2006

### 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.

Last edited: Jun 5, 2006
11. Jun 5, 2006

### Oxymoron

"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 dont really understand: "M induces a map". Is this just a way of saying that "because of the existence of M there exists a map..."?

12. Jun 5, 2006

### 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.

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13. Jun 5, 2006

### Oxymoron

In case you are wondering why I asked, Im 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?

Ah, you did too. Sorry about that.
...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?

14. Jun 5, 2006

### 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.

Last edited: Jun 5, 2006
15. Jun 5, 2006

### Oxymoron

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.

16. Jun 5, 2006

### 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?

Last edited: Jun 5, 2006
17. Jun 5, 2006

### matt grime

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).

18. Jun 5, 2006

### matt grime

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)

Last edited: Jun 5, 2006
19. Jun 5, 2006

### Oxymoron

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!

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?

Last edited: Jun 5, 2006
20. Jun 5, 2006

### 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).

21. Jun 6, 2006

### 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?

22. Jun 6, 2006

### 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?

23. Jun 6, 2006

### 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.

Last edited: Jun 6, 2006
24. Jun 6, 2006

### 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:

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

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 Im not sure if it relates to anything we have been discussing here. Although it does looks half-relevant.

Last edited: Jun 6, 2006
25. Jun 6, 2006

### 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).