unitary equivalence for C*-algebras
John Baez
>Hi, John,
>
>Feel free to cc: your answer to s.p.r. if you like.
>
>I am reading BSZ's glossary entry on C^*-algebras. It says:
>
>"Any abstract C^*-algebra is isomorphic to some concrete C^*-algebra
>which, however, is in general not unique spatially in that, if two
>concrete C^*-algebras are *-algebraically isomorphic, they are in
>general not unitarily equivalent (although the norms automatically
>correspond)."
>
>The spectral radius is defined entirely algebraically, so this allows
>one to define the norm algebraically and explains why *-isomorphisms
>preserve the norm. Now, polarization allows one to define an inner
>product from the norm in terms of algebraic operations only, so any
>*-isomorphism should be unitary! What am I missing?
>
>Regards,
>
>Miguel
Your explanation of why *-isomorphisms between C*-algebras
are norm-preserving is correct, assuming you know what
you're doing - e.g. assuming you know that the spectral radius
of an element in a noncommutative C*-algebra needn't equal
its norm, and how to deal with this.
But if someone hands you a C*-algebra you can't "define an
inner product from the norm". C*-algebras don't even have
inner products! Of course you probably meant that you can
recover the inner product in a Hilbert space from its norm.
But that's a different matter, and it doesn't help at all here.
Say someone hands you two C*-algebras A and A' together with
a *-isomorphism f: A -> A'. Then f preserves the C*-algebra
norms - that's what we meant by saying "the norms automatically
correspond".
But now suppose they also tell you A is a C*-subalgebra of the
bounded operators on a Hilbert space H and A' is a C*-subalgebra
of the bounded operators on a Hilbert space H', and ask you if
A and A' are unitarily equivalent. In other words, they ask if
you can find a unitary operator u: H -> H' that "implements" the
isomorphism f: A -> A', meaning:
f(a) = uau^{-1}
In general, you can't do it! This is what we meant by "in general,
they are not unitarily equivalent".
In general, you can't do it even if A = A' and H = H'. This
is one reason dynamics as described by C*-algebra automorphisms is
more general than dynamics as described by unitary operators.
Miguel Carrion
Every abelian C* algebra is isomorphic to the algebra C(K) of complex
continuous functions on the compact space K, and the norm corresponds
to the topology of uniform convergence.
Now, when K is not compact, the right generalization seems (to me) to
be the continuous functions on K with the compact-open topology
(uniform convergence on compact sets). This is not a normed topology,
but it is given by a family of seminorms, so C(K) is not a
C*-algebra. Is it some other famous kind of algebra?
Is this generalization "obviously" useless?
And now back to BSZ and defining norms completely algebraically...
In article <bem7as$ntj$1...@glue.ucr.edu>, John Baez <ba...@galaxy.ucr.edu> wrote:
>My grad student Miguel Carrion in the UCR math department wrote:
>
>>
>>I am reading BSZ's glossary entry on C^*-algebras. It says:
>>
>>"Any abstract C^*-algebra is isomorphic to some concrete C^*-algebra
>>which, however, is in general not unique spatially in that, if two
>>concrete C^*-algebras are *-algebraically isomorphic, they are in
>>general not unitarily equivalent (although the norms automatically
>>correspond)."
>>
>>The spectral radius is defined entirely algebraically, so this allows
>>one to define the norm algebraically and explains why *-isomorphisms
>>preserve the norm.
>>
>
>Your explanation of why *-isomorphisms between C*-algebras
>are norm-preserving is correct, assuming you know what
>you're doing - e.g. assuming you know that the spectral radius
>of an element in a noncommutative C*-algebra needn't equal
>its norm, and how to deal with this.
For normal elements, that is, those satisfying
aa^*=a^*a,
the norm equals the spectral radius. I can then use the decomposition
of an arbitrary a as
a = b + ic
with self-adjoint b,c to define
|a|^2 = |b|^2 + |c|^2
purely algebraically.
>But if someone hands you a C*-algebra you can't "define an inner
>product from the norm". C*-algebras don't even have inner products!
>Of course you probably meant that you can recover the inner product
>in a Hilbert space from its norm.
No, I meant that the C*-algebra had an inner product... Duh!
>But now suppose they also tell you A is a C*-subalgebra of the
>bounded operators on a Hilbert space H and A' is a C*-subalgebra
>of the bounded operators on a Hilbert space H', and ask you if
>A and A' are unitarily equivalent. In other words, they ask if
>you can find a unitary operator u: H -> H' that "implements" the
>isomorphism f: A -> A', meaning:
>
>f(a) = uau^{-1}
Can we do this abstractly without reference to a Hilbert space? Let me
try:
If f : A -> A' is a *-algebra isomorphism, can one find unitary
(uu^*=1=u^*u) elements u in A and u' in A' such that u'f(a) = f(au)?
Is this equivalent to the representation-dependent notion of
equivalence?
Regards,
Miguel
ba...@math.ucr.edu
Miguel Carrion <mig...@math-cl-n02.math.ucr.edu> wrote:
>Another question about C* algebras...
Okay! I've taken so long to reply that I'll cc you,
in case you've given up reading s.p.r. every day and pining away
at your computer as you wait for an answer that never comes.
>Every abelian C* algebra is isomorphic to the algebra C(K) of complex
>continuous functions on the compact space K, and the norm corresponds
>to the topology of uniform convergence.
Yup!
And as you probably know, the compact space K might as well
be Hausdorff, too - if not, C(K) is isomorphic to C(K') for
some compact Hausdorff space K', the "Hausdorffication" of K.
>Now, when K is not compact, the right generalization seems (to me) to
>be the continuous functions on K with the compact-open topology
>(uniform convergence on compact sets). This is not a normed topology,
>but it is given by a family of seminorms, so C(K) is not a
>C*-algebra. Is it some other famous kind of algebra?
Umm, not famous enough for *me* to have heard of it.
>Is this generalization "obviously" useless?
Dunno. The power and beauty of C*-algebras, and the lesser
but still not inconsiderable power and beauty of Banach *-algebras,
makes most of us reluctant to invest much energy in more general
topological *-algebras. Though there *are* people who work on
such things.
What I would instinctively do, confronted with a noncompact but
still locally compact Hausdorff space that someone misleadingly called "K",
would be to form the C*-algebra of continuous functions on K
that vanish at infinity. This is a *nonunital* commutative C*-algebra.
By a slight generalization of the Gelfand-Naimark theorem,
and the category of nonunital commutative C*-algebras
is (contravariantly) equivalent to that of locally compact
Hausdorff spaces. So, if you're willing to accept
*local* compactness, you might as well use this C*-algebra.
>And now back to BSZ and defining norms completely algebraically...
Okay....
>In article <bem7as$ntj$1...@glue.ucr.edu>, John Baez
><ba...@galaxy.ucr.edu> wrote:
>>Your explanation of why *-isomorphisms between C*-algebras
>>are norm-preserving is correct, assuming you know what
>>you're doing - e.g. assuming you know that the spectral radius
>>of an element in a noncommutative C*-algebra needn't equal
>>its norm, and how to deal with this.
>For normal elements, that is, those satisfying
>
>aa^*=a^*a,
>
>the norm equals the spectral radius. I can then use the decomposition
>of an arbitrary a as
>
>a = b + ic
>
>with self-adjoint b,c to define
>
>|a|^2 = |b|^2 + |c|^2
>
>purely algebraically.
That's fine so far. But right now I'm forgetting how, or whether,
can define the norm purely algebraically for "abnormal" operators.
Without knowing how to do this, I don't remember why C*-algebra
*-isomorphisms should automatically preserve the norm! What
you've done is sufficient to handle the commutative case, where
all elements are normal...
I was mixed up: I thought I knew a general formula for
the norm in terms of the spectral radius, but actually I
know a general formula for the spectral radius in terms
of the norm! For any element a of a C*-algebra, its
spectral radius is the lim sup of the nth root of |a^n|.
>>But now suppose they also tell you A is a C*-subalgebra of the
>>bounded operators on a Hilbert space H and A' is a C*-subalgebra
>>of the bounded operators on a Hilbert space H', and ask you if
>>A and A' are unitarily equivalent. In other words, they ask if
>>you can find a unitary operator u: H -> H' that "implements" the
>>isomorphism f: A -> A', meaning:
>>
>>f(a) = uau^{-1}
>Can we do this abstractly without reference to a Hilbert space?
I don't think we can do THIS - I think this concept depends
crucially on how the C*-algebras are represented as subalgebras of
bounded operators on Hilbert spaces. But we can do something
ELSE:
>Let me try:
>
>If f : A -> A' is a *-algebra isomorphism, can one find unitary
>(uu^*=1=u^*u) elements u in A and u' in A' such that u'f(a) = f(au)?
>Is this equivalent to the representation-dependent notion of
>equivalence?
I doubt it, though I haven't seen this concept
before, so I can't rattle off a counterexample.
But your concept reminds me of another, which is
widely used. Given a *-automorphism of a C*-algebra
A, say f: A -> A, we say it is "inner" if there is
a unitary element u of A such that
f(a) = uau^{-1}
For a commutative C*-algebra, only the identity automorphism
is inner. But for the bounded operators L(H) on a Hilbert space H
there are lots of inner automorphisms and indeed physicists
rarely consider any others. I don't remember if they are
*all* inner in this case - if so, that would explain it. :-)
It's certainly true that if H is finite-dimensional, all
*-automorphisms of L(H) are inner.
Hey! Wait a minute. There's something funny about your concept!
Since f is an isomorphism you're saying u'f(a) = f(a)f(u).
Taking a = 1 this says u' = f(u). So you're really just saying
the image of f commutes with f(u). I don't think this has
anything to do with u and u' "implementing" the isomorphism f.
Arkadiusz Jadczyk
On Fri, 25 Jul 2003 14:52:44 +0000 (UTC), ba...@math.ucr.edu wrote:
>In article <beq281$1ig$1...@glue.ucr.edu>,
>Miguel Carrion <mig...@math-cl-n02.math.ucr.edu> wrote:
(snip)
>>If f : A -> A' is a *-algebra isomorphism, can one find unitary
>>(uu^*=1=u^*u) elements u in A and u' in A' such that u'f(a) = f(au)?
(snip)
>Hey! Wait a minute. There's something funny about your concept!
Indeed there is something funny! Take u=I and u'=I and the answer is
"yes"!!!
ark
--
Arkadiusz Jadczyk
http://www.cassiopaea.org/quantum_future/homepage.htm
--
Miguel Carrion
I want to drive this thread in the direction of just how much analytic
information can be extracted from purely algebraic data. The reason
this may be relavant to physics is that when we are quantizing an
algebra of observables it doesn't necessarily carry norms or other
analytical information. In particular, you might have to dramatically
manipulate the algebra to make it support a C^* structure.
In article <bfrg7s$5f3$1...@glue.ucr.edu>, <ba...@math.ucr.edu> wrote:
>In article <beq281$1ig$1...@glue.ucr.edu>,
>Miguel Carrion <mig...@math-cl-n02.math.ucr.edu> wrote:
>
>>Every abelian C* algebra is isomorphic to the algebra C(K) of complex
>>continuous functions on the compact space K, and the norm corresponds
>>to the topology of uniform convergence.
>
It turns out that every abelian *-algebra is isomorphic to a
*-subalgebra of C(K) with Hausdorff K, so that the compact-open
topology on C(K) can be used to topologize the original *-algebra
entirely from the algebraic data. Moreover, if K is sigma-compact,
then this compact-open topology is metrizable (so C(K) is an F-space
--- if you close it you get a Frechet space). If K is compact, you get
a C^*-algebra. All of this from algebraic data only! So it would be
nice if you could then apply a modified version of the GNS theorem to
it.
>
>The power and beauty of C*-algebras, and the lesser
>but still not inconsiderable power and beauty of Banach *-algebras,
>makes most of us reluctant to invest much energy in more general
>topological *-algebras. Though there *are* people who work on
>such things.
Hmmm, power and beauty. Are you talking about the Hahn-Banach theorem,
which holds for any locally convex topological vector space? Are you
talking about the uniform-boundedness principle, the open mapping
theorem or the closed graph theorem, all of which hold for F-spaces?
Since none of these are special to Banach algebras, I guess you must
be talking about the spectral theory of bounded operators. But a large
number of operators of interest in physics are unbounded! And, if I
can embed any algebra in C(K) with the compact-open topology, I can
extend the functional calculus to any abelian algebra simply by using
C(K) to define the effect of applying a continous function to any
element of the algebra.
I am still having trouble with nonabelian algebras, though.
>What I would instinctively do, confronted with a noncompact but still
>locally compact Hausdorff space that someone misleadingly called "K",
>would be to form the C*-algebra of continuous functions on K that
>vanish at infinity. This is a *nonunital* commutative C*-algebra.
In other words, you would compactify K by a single point, likely
destroying any natural smooth structure that K may have. At least you
could look at functions tending to a constant at infinity, so you have
a unital C^*-algebra
Another thing you can do is restrict C(K) to compact subsets of K,
obtaining a C^*-algebra for each compact subset. So you get a family
of C^*-algebras which can be nicely put together. This is the second
time I come up with such a structure (the first wasn't on s.p.r), so
the issue seems to be whether one can get some physics out of this
idea of "multiple C^*-algebras".
>By a slight generalization of the Gelfand-Naimark theorem,
>and the category of nonunital commutative C*-algebras
>is (contravariantly) equivalent to that of locally compact
>Hausdorff spaces. So, if you're willing to accept
>*local* compactness, you might as well use this C*-algebra.
Is there a Gelfand-Naimark theorem for algebras of unbounded
operators?
Going back to the compact-open topology on C(K), you can define GNS
"states" on them which happen to be states on the C^* algebra
associated to some compact subset of K.
The idea is to take quotients of a locally convex *-algebra to obtain
various C^*-algebras and then use facts about those C^*-algebras to
prove facts about the locally convex *-algebra. This is essentially
how one would use the Stone-Weierstrass theorem (which is essentially
about Banach *-algebras) to prove that any continuous function on R
can be approximated by polynomials uniformly on compact sets.
>>And now back to BSZ and defining norms completely algebraically...
>
>>In article <bem7as$ntj$1...@glue.ucr.edu>, John Baez
>><ba...@galaxy.ucr.edu> wrote:
>
>>>Your explanation of why *-isomorphisms between C*-algebras
>>>are norm-preserving is correct, assuming you know what
>>>you're doing - e.g. assuming you know that the spectral radius
>>>of an element in a noncommutative C*-algebra needn't equal
>>>its norm, and how to deal with this.
>
>>For normal elements, that is, those satisfying
>>
>>aa^*=a^*a,
>>
>>the norm equals the spectral radius. I can then use the decomposition
>>of an arbitrary a as
>>
>>a = b + ic
>>
>>with self-adjoint b,c to define
>>
>>|a|^2 = |b|^2 + |c|^2
>>
>>purely algebraically.
>
>That's fine so far. But right now I'm forgetting how, or whether,
>can define the norm purely algebraically for "abnormal" operators.
Doesn't my formula take care of non-normal operators? I thought
*every* element of a *-algebra can be decomposed as a=b+ic, not just
the normal ones. In fact,
a^* a = b^2 + i [b,c] + c^2
a a^* = b^2 - i [b,c] + c^2
so a need not be normal, and I am wanting to define
|a|^2 = |b|^2 + |c|^2
Will the [b,c] terms prevent this from being a norm?
>>>But now suppose they also tell you A is a C*-subalgebra of the
>>>bounded operators on a Hilbert space H and A' is a C*-subalgebra
>>>of the bounded operators on a Hilbert space H', and ask you if
>>>A and A' are unitarily equivalent. In other words, they ask if
>>>you can find a unitary operator u: H -> H' that "implements" the
>>>isomorphism f: A -> A', meaning:
>>>
>>>f(a) = uau^{-1}
>
>>Can we do this abstractly without reference to a Hilbert space?
>
>I don't think we can do THIS - I think this concept depends
>crucially on how the C*-algebras are represented as subalgebras of
>bounded operators on Hilbert spaces. But we can do something
>ELSE:
>
Ok, the something ELSE I tried last time won't work. How about this?
let H and H' be the Hilbert spaces that appear in the proof of the GNS
theorem. Are all unitariyl equivalent C^*-algebras equivalent with
respect to these special H and H'?
Regards,
Miguel
Daniel Grubb
>It turns out that every abelian *-algebra is isomorphic to a
>*-subalgebra of C(K) with Hausdorff K, so that the compact-open
>topology on C(K) can be used to topologize the original *-algebra
>entirely from the algebraic data. Moreover, if K is sigma-compact,
>then this compact-open topology is metrizable (so C(K) is an F-space
>--- if you close it you get a Frechet space). If K is compact, you get
>a C^*-algebra. All of this from algebraic data only! So it would be
>nice if you could then apply a modified version of the GNS theorem to
>it.
You have to be careful doing this though. If your original algebra
*did* have a norm or topology, this proceedure can change it. An
example is given by L_1 (R) under convolution. This is embedded as
a *-subalgebra of C_0(R), but the norms changed in the process.
baez wrote:
>That's fine so far. But right now I'm forgetting how, or whether,
>can define the norm purely algebraically for "abnormal" operators.
>Without knowing how to do this, I don't remember why C*-algebra
>*-isomorphisms should automatically preserve the norm! What
>you've done is sufficient to handle the commutative case, where
>all elements are normal...
Yes, for C* algebras, you can define the norm purely algebraically.
First do it through the spectrum for self-adjoint elements and then
use the C* algebra condition that |a*a|=|a|^2 for the general case.
This fails for a general Banach *-algebra.
>Hmmm, power and beauty. Are you talking about the Hahn-Banach theorem,
>which holds for any locally convex topological vector space? Are you
>talking about the uniform-boundedness principle, the open mapping
>theorem or the closed graph theorem, all of which hold for F-spaces?
>Since none of these are special to Banach algebras, I guess you must
>be talking about the spectral theory of bounded operators. But a large
>number of operators of interest in physics are unbounded! And, if I
>can embed any algebra in C(K) with the compact-open topology, I can
>extend the functional calculus to any abelian algebra simply by using
>C(K) to define the effect of applying a continous function to any
>element of the algebra.
Again, you have to be careful of this. The usual function calculus
on a C*algebra gives elements of the C*algebra back. For unbounded
normal operators, you get potentially unbounded operators back. If
you try this proceedure with a general *-algebra, even if it is a
Banach *-algebra, you will not necessarily get elements of the
algebra back when applying a general continuous function. You *are*
guaranteed an analytic function calculus, though.
Again, if you start with L_1(R), you can not apply a generic continuous
function and get back an L_1 function. Perhaps an even better example is
little l1 under convolution. The natural 'K' is the circle, but a general
continuous function on the circle does not act on the subalgebra.
---Dan Grubb
Daniel Grubb
>>>Every abelian C* algebra is isomorphic to the algebra C(K) of complex
>>>continuous functions on the compact space K, and the norm corresponds
>>>to the topology of uniform convergence.
>>
>It turns out that every abelian *-algebra is isomorphic to a
>*-subalgebra of C(K) with Hausdorff K, so that the compact-open
>topology on C(K) can be used to topologize the original *-algebra
>entirely from the algebraic data.
There is a problem that there may be several different K so that
your algebra is isomo]rphic to a subalgebra of C(K). The most
natural K to try is that provided by the Gelfand theory since
it is defined in terms of multiplicative linear functionals. Even
in the case of a Banach *-algebra this may not work, however.
In fact, if you consider the algebra of complex measures on the
circle group under convolution, the Gelfand map is *not* a *-isomorphism!
In the case of C* algebras, it is, but the proof doesn't work in
the general case.
So the real question is to pick out, from algebraic data, a natural
K that works. For a Banach *-algebra, this K will be a subset
of the Gelfand space. For the case of the measure algebra, the
obvious candidate would be the integers using the Fourier transform
for the *-isomorphism. This isn't compact even though the algebra
is unital.
>>What I would instinctively do, confronted with a noncompact but still
>>locally compact Hausdorff space that someone misleadingly called "K",
>>would be to form the C*-algebra of continuous functions on K that
>>vanish at infinity. This is a *nonunital* commutative C*-algebra.
>In other words, you would compactify K by a single point, likely
>destroying any natural smooth structure that K may have. At least you
>could look at functions tending to a constant at infinity, so you have
>a unital C^*-algebra
The C* algebra of functions tending to a constant at infinity
is just the algebra obtained when you add a unit to the non-unital
algebra of functions vanishing at infinity. And that algebra is the
standard object to consider for locally compact, non-compact spaces.
--Dan Grubb
John Baez
Miguel Carrion <mig...@math-cl-n01.math.ucr.edu> wrote:
>I want to drive this thread in the direction of just how much analytic
>information can be extracted from purely algebraic data. The reason
>this may be relevant to physics is that when we are quantizing an
>algebra of observables it doesn't necessarily carry norms or other
>analytical information. In particular, you might have to dramatically
>manipulate the algebra to make it support a C* structure.
Right! Quantization is a dirty business.
>It turns out that every abelian *-algebra is isomorphic to a
>*-subalgebra of C(K) with Hausdorff K,
Oh yeah? I don't believe that. We can make C[x]/<x^n>
(the algebra of complex polynomials in x, modulo x^n)
into an abelian *-algebra by setting x* = x, and this
is not isomorphic to a subalgebra of the continuous functions
on any space. The reason is that you can't find any nonzero
function f such that f^n = 0.
Of course, I'm picking on you, because you posted this article
ages ago, and you've subsequently realized that you need to
stick some fine into your claim to make it true: your abelian
*-algebra had better not contain "nilpotents": that is, nonzero
elements x with x^n = 0.
One reason this fine print is interesting is that it excludes
the "functions on a 1-dimensional odd supervector space", which
is the algebra C[x]/<x^2>. This is the fermionic analog of
the "functions on a 1-dimensional even supervector space",
better known as "the functions on the line". You'd see a
combination of these two algebras showing up if you studied
the functions on a general supervector space, as would be
natural when studying the classical configuration space or
phase space for a system with both bosonic and fermionic
degrees of freedom.
We had a long thread here about the fact that you can't use
C*-algebras to study the classical limit of fermions the way
you can to study the classical limit of bosons. Greg Weeks
instigated this, and he wrote:
>Putting fermions into a
>classical context, they become zero. (Zero being the only classical value
>that anticommutes with itself.)
>
>More fancily, from an algebraic viewpoint: The spectrum of an observable X
>is the set of values x such that X - x is not invertible. But if X^2 = 0
>and x != 0, then X - x has the inverse (X + x)/(-x^2). So the spectrum of
>a classical fermion is just {0}.
Of course, you may wonder why I'm talking about this, because
fermions anticommute and right now we're talking about commutative algebras!
One reason is that I'm just having fun, but the other
is that C[x]/<x^2> is both commutative and anticommutative!
More precisely, the generator x both commutes and anticommutes
with itself.
Anyway, I'm digressing. You just got me thinking about
the dangers of mixing nilpotence and C*-algebras. No
nonzero normal element of a C*-algebra can be nilpotent!
This doesn't mean C*-algebras can't handle fermions, but
it does seem to mean they can't can't handle "classical fermions":
Grassman algebras as opposed to Clifford algebras.
>so that the compact-open
>topology on C(K) can be used to topologize the original *-algebra
>entirely from the algebraic data.
Does the original *-algebra become a topological *-algebra?
I.e., do the addition, scalar multiplication, product and
* operation become continuous in this topology? I sure hope so.
Of course, we should stick in the hypothesis that A had no
nilpotents. Actually, it'd be even cuter to show that
the "semisimplification" of A, that is, A modulo its radical
(= the ideal consisting of all nilpotents) is contained in C(K)
and becomes a topological *-algebra. After all, it's more fun
to watch the nilpotents automatically getting killed off by this
map:
A -> C(K) (K = the spectrum of A)
than to stick in an extra hypothesis saying they were never
there.
> Moreover, if K is sigma-compact,
>then this compact-open topology is metrizable (so C(K) is an F-space
>--- if you close it you get a Frechet space).
Hey, good, it's great to see you doing analysis. Even though
I don't talk about it much anymore, I used to really like it,
and I still sort of do.
But are you hinting that C(K) might not already be a Frechet space
when K is sigma-compact? What does its closure contain that it does not?
In your thesis, it would be nice to have a theorem saying that
at this point we can take the semisimplication of A and take
its closure, obtaining a Frechet *-algebra under the assumption
that its spectrum is sigma-compact. I bet someone has done
this already, but that's okay.
>If K is compact, you get
>a C^*-algebra. All of this from algebraic data only! So it would be
>nice if you could then apply a modified version of the GNS theorem to
>it.
You're looking for a GNS theorem for more general abelian *-algebras?
You might as well prove a GNS theorem for general *-algebras and forget
about abelianness. There's a pretty easy theorem like that and it's sort
of interesting - I forget if I've ever seen it anywhere except
in my own notes. But you should think about where you're trying to
go with all this stuff, and not go too far unless you have a good reason.
>>The power and beauty of C*-algebras, and the lesser
>>but still not inconsiderable power and beauty of Banach *-algebras,
>>makes most of us reluctant to invest much energy in more general
>>topological *-algebras. Though there *are* people who work on
>>such things.
>Hmmm, power and beauty. Are you talking about the Hahn-Banach theorem,
>which holds for any locally convex topological vector space? Are you
>talking about the uniform-boundedness principle, the open mapping
>theorem or the closed graph theorem, all of which hold for F-spaces?
No, those are just bland general principles. I'm talking
about the *interesting* stuff, like in here:
Ola Bratteli and Derek W. Robinson, Operator Algebras and
Quantum Statistical Mechanics, 2 volumes, Springer, Berlin, 1987-1997.
Richard V. Kadison and John R. Ringrose, Fundamentals of
the Theory of Operator Algebras, 4 volumes, Academic Press,
New York, 1983-1992.
I know it's obnoxious to throw 6 books at you like this,
but the point is, people have worked out a very detailed
and tight theory of C*-algebras (especially von Neumann
algebras) and their applications to physics. The theory
of more general topological *-algebras just hasn't reached
anywhere near that level of sophistication. So, working
with a topological *-algebra instead of a C*-algebra is
like using a hand axe when there are precision tools available.
The theory of more general *-algebras may in part be weaker
because fewer people have worked on them, but there's a *reason*
why fewer people have worked on them - they're a much more
flabby sort of structure! It's sort of like the theory of
general groups versus the theory of Lie groups versus the
theory of simple Lie groups: as we zoom in we get a much
more vivid theory.
But, you may also be interested in attempts to justify
the importance of C*-algebras on *physical* grounds. These
are not completely convincing but they're pretty interesting.
I like this reference a lot:
Gerard G. Emch, Algebraic Methods in Statistical Mechanics
and Quantum Field Theory, Wiley-Interscience, New York, 1972.
>Since none of these are special to Banach algebras, I guess you must
>be talking about the spectral theory of bounded operators. But a large
>number of operators of interest in physics are unbounded! And, if I
>can embed any algebra in C(K) with the compact-open topology, I can
>extend the functional calculus to any abelian algebra simply by using
>C(K) to define the effect of applying a continous function to any
>element of the algebra.
The unbounded operators of interest in quantum physics are
almost always self-adjoint or at least normal. Other unbounded
operators are usually horrible! So, one of the tasks of
mathematical physics is to distinguish between unbounded operators
that are really self-adjoint and ones that just look like
they are - namely, the symmetric ones. We then focus on the
self-adjoint ones.
Once we know our unbounded operator A is self-adjoint we instantly know
how to reduce every problem about it to a problem involving bounded
normal operators - e.g. the strongly continuous one-parameter
unitary group exp(itA), or the Cayley transform (A+i)/(A-i),
or the resolvent 1/(A + it), or if A is bounded below, the
strongly continuous one-parameter semigroup exp(-tA).
Moreover, if we have a formula involving *more than one*
unbounded self-adjoint operator, like the Heisenberg relation:
[p,q] = -i
it is completely useless for rigorous work until we have
converted it into one involving bounded operators, like
the Weyl relations:
exp(itp) exp(isq) = exp(its) exp(isq) exp(itp)
The formula involving bounded operators will typically imply the one
for unbounded operators but not vice versa. That's why we
get the Stone-von Neumann uniqueness theorem for irreps of
the Weyl relations, but nothing comparable for the Heisenberg
relations, which have myriads of weird irreps *of no use to physics*.
So, what I'm saying is that as far as rigorous mathematical
quantum physics goes, the unbounded operators that matter are
the ones that you can describe easily using bounded operators,
and this description is more or less necessary for doing any
serious work. SO, we might as well focus on C*-algebras rather
than some more general *-algebras designed for working with
unbounded operators.
I think this philosophy is pretty good *except* for problems
of quantization where you start out with some algebra, perhaps
not even equipped with a *-structure, and not equipped with
a faithful representation as operators on a Hilbert space.
I think the stuff you're doing in your thesis is good for that.
BUT, starting from this position where you just have an algebra,
I think the goal should be perform manipulations to get a C*-algebra,
and then mainly work with that (and its affiliated self-adjoint
operators in various representations). And of course it'd be
nice to understand just what the "right" manipulations are and
why they are "right".
>I am still having trouble with nonabelian algebras, though.
You should prove a little GNS theorem for general *-algebras.
>>What I would instinctively do, confronted with a noncompact but still
>>locally compact Hausdorff space that someone misleadingly called "K",
>>would be to form the C*-algebra of continuous functions on K that
>>vanish at infinity. This is a *nonunital* commutative C*-algebra.
>In other words, you would compactify K by a single point, likely
>destroying any natural smooth structure that K may have.
No, that's what would happen if I looked at the functions
tending to a constant at infinity, getting a unital commutative
C*-algebra whose spectrum was the one-point compactification of K.
>Another thing you can do is restrict C(K) to compact subsets of K,
>obtaining a C^*-algebra for each compact subset. So you get a family
>of C^*-algebras which can be nicely put together. This is the second
>time I come up with such a structure (the first wasn't on s.p.r), so
>the issue seems to be whether one can get some physics out of this
>idea of "multiple C^*-algebras".
I guess I'd really want to think about this in the context of
physically interesting examples, not in lofty abstraction.
>>By a slight generalization of the Gelfand-Naimark theorem,
>>and the category of nonunital commutative C*-algebras
>>is (contravariantly) equivalent to that of locally compact
>>Hausdorff spaces. So, if you're willing to accept
>>*local* compactness, you might as well use this C*-algebra.
>Is there a Gelfand-Naimark theorem for algebras of unbounded
>operators?
Well, by now you have proved a whole slew of GN-like theorems
for various sorts of algebras. Btw, in your thesis it would
look a lot nicer if you codified these as theorems: first seeing
what you can do with an abelian algebra, then a semisimple one,
then a *-algebra, then a semisimple one, then going further under
more conditions until finally you reach the usual GN theorem.
This stuff can't be new, but I've never seen it anywhere,
especially not all lined up neatly in once place. And
if you explain its physical significance, that probably *will* be new.
I think I'll stop here! If there's anything I left unanswered
that you want me to answer, please remind me of it! Or, just
go on talking about stuff.
Arnold Neumaier
> I want to drive this thread in the direction of just how much analytic
> information can be extracted from purely algebraic data. The reason
> this may be relevant to physics is that when we are quantizing an
> algebra of observables it doesn't necessarily carry norms or other
> analytical information. In particular, you might have to dramatically
> manipulate the algebra to make it support a C^* structure.
In gr-qc/0303037, I introduced in Section 3 the concept of
a Euclidean *-algebra. I have plenty of examples of it from
physics, and believe that it is the right mathematical framework
for quantum field theory. I spent a lot of time to investigate
many applications in order to check that these axioms are exactly
the right ones needed for doing interesting physics within its
scope. For example, one can derive in this setting all of
equilibrium thermodynamics with very little additional input.
I am just polishing my write up of these things, and hope to have
a manuscript within a few weeks.
A Euclidean *-algebra contains unbounded operators;
essentially it axiomatizing the algebra of operators on a nuclear
space. I think it would be interesting to investigate whether a
kind of converse can be proved. Since there is an integral, one
can do GNS-like constructions to complete part of the algebra,
in a similar way as one can get from Schwartz space to the
corresponding Hilbert space. There should also be analogues of
Sobolev spaces, etc.
If you find the concept interesting, I could provide you with
suggestions of interesting problems and suitable lines of attack.
Arnold Neumaier
nobody
>The unbounded operators of interest in quantum physics are
>almost always self-adjoint or at least normal. Other unbounded
>operators are usually horrible!
What about the boson annihilation and creation operators? My
impression from things that you've said before is that they
are reasonably well-behaved, e.g., (a+)+ = a (of course, I
should ask to what domain that statement applies).
nobody
Miguel Carrion
Daniel Grubb <gr...@math.niu.edu> wrote:
>
>>It turns out that every abelian *-algebra is isomorphic to a
>>*-subalgebra of C(K) with Hausdorff K, so that the compact-open
>>topology on C(K) can be used to topologize the original *-algebra
>>entirely from the algebraic data.
>
>*did* have a norm or topology, this proceedure can change it. An
>example is given by L_1 (R) under convolution. This is embedded as
>a *-subalgebra of C_0(R), but the norms changed in the process.
Are the norms equivalent in that they define the same topology? I
forget what the natural norm of (L_1,*) is, but I expect that it
should be obtained by the following procedure:
1) start with the L_1 norm on L_1(R).
2) define M_f(g)=f*g for all f,g in L_1(R). For each f in L_1(R), M_f
is a bounded endomorphism of L_1(R), and M_f M_g = M_{f*g}.
3) M_f has an operator norm as an operator from L_1 to L_1. Use this
norm to topologize L_1(R)
Question: are the norms in 1) and 3) the same? Are they equivalent?
There's some homework for me...
>>if I can embed any algebra in C(K) with the compact-open topology, I
>>can extend the functional calculus to any abelian algebra simply by
>>using C(K) to define the effect of applying a continous function to
>>any element of the algebra.
>
>Again, you have to be careful of this. The usual function calculus
>on a C*algebra gives elements of the C*algebra back. For unbounded
>normal operators, you get potentially unbounded operators back. If
>you try this proceedure with a general *-algebra, even if it is a
>Banach *-algebra, you will not necessarily get elements of the
>algebra back when applying a general continuous function. You *are*
>guaranteed an analytic function calculus, though.
You're right, and I should have been a little clearer on this. I don't
expect in general that the algebras I start with will be stable under
the functional calculus, but the idea is that, if A is embedded as a
*-subalgebra of C(K), there may be some intermediate algebra B which
is the closure of A under analytic or continuous functions. If the
original algebra didn't have a topology we had to complete it already,
so I don't mind enlarging it further to support a functional
calculus. Eventually you want to get a Von Neumann algebra too, so
that you can talk about spectral decompositions (i.e., characteristic
functions).
The idea is that you start with an abelian "algebra of observables" A,
from it you construct its spectrum K, and imbed A as a *-subalgebra of
C(K) with the compact-open topology. You may then want to enlarge the
algebra of observables by taking its closure. This gives a natural
prescription for which "sequences of observables" have limits, and
what the limits should be. But now everything happens in K.
Here's a physically relevant example. Take an abstract 1-parameter
group: W(s)W(t)=W(s+t) for all real s and t. Define W(s)^* = W(-s),
and let A be the complex *-algebra generated by the W(s) with those
relations. This means my abstract 1-parameter group is the group of
unitary elements of A.
What is the spectrum of A? A complex linear functional F : A -> \C is
determined by its action on the generators: F(W(s))=f(s) for some
function f : \R -> \C. Now, for F to be a *-algebra homomorphism, it
must satisfy f(s)f(t)=f(s+t) and f(s)^*=f(-s), so f(s)=exp(i s theta)
for some real theta. The spectrum of A is, then, K=\R, and the
Gel'fand transform maps W(s) to S(theta)=exp(i s theta). All the
elements of A are linear combinations of bounded funcions on \R, so
they are all bounded and so K must be compact... Wait! \R is not
compact!
That is because the topology on the spectrum K is not the usual
topology on \R, but the weakest topology making all the exp(i s theta)
continuous. This is not much weaker than the usual topology, but it is
weak enoug to make \R compact! Other surprising properties of this
topology are that every open set has infinite Lebesgue measure.
I am about to get carried away and claim that this should be the
topology of the "physical real line", but I'll stop short of that and
instead go home and ponder the implications.
[As a side note, this seems like an important an interesting topology,
so I don't understand why I didn't learn about it in my point-topology
courses. We spent too much time on the Sorgenfrey line, and other
silly counterexamples by Polish mathematicians, and hardly any time on
counterintuitive but useful examples like this one. Grrr...]
John Baez
Miguel Carrion <mig...@math-cl-n01.math.ucr.edu> wrote:
>I forget what the natural norm of (L_1,*) is,
It's, err, umm...
... the L_1 norm!
L_1 functions on the real line are the classic example of a
B*-algebra that's not a C*-algebra.
It's a Banach algebra by that theorem which bounds the L_1 norm of
f*g by the product of the L_1 norms of f and g. And it's a B*-algebra
because the L_1 norm of the complex conjugate of f equals that of f.
It's nonunital, though - unless we throw in a unit, namely
the Dirac delta at the origin!
>but I expect that it
>should be obtained by the following procedure:
>1) start with the L_1 norm on L_1(R).
>2) define M_f(g)=f*g for all f,g in L_1(R). For each f in L_1(R), M_f
> is a bounded endomorphism of L_1(R), and M_f M_g = M_{f*g}.
>3) M_f has an operator norm as an operator from L_1 to L_1. Use this
> norm to topologize L_1(R)
>Question: are the norms in 1) and 3) the same? Are they equivalent?
>There's some homework for me...
Well, you're making life hard on yourself, but this
question gets easier if you think of this as an example
of a more general question.
Someone hands Miguel a Banach algebra, A.
Dissatisfied with the original norm, Miguel
notes that any element of A acts by left
multiplication on A:
M_a: b |-> ab
so he can use the operator norm of M_a to put
a new norm on A.
He then asks if this new norm is equivalent to
the original one!
I claim they're obviously *equal* if the
Banach algebra is unital: |M_a| is never
bigger than |a|, but if there's a unit we have
|M_a 1| = |a| so |M_a| is at least |a|.
If the Banach algebra is nonunital, the
question gets trickier. But in the case of (L_1,*),
I think we can copy the above argument using a tall
skinny bump function as an approximate version
of "1", and still show the original norm on L_1
is equal to the operator norm.
There's a general concept of "approximate identity"
for Banach algebras, namely a sequence (or net)
of elements 1_n such that
1_n a -> a
and
a 1_n -> a
for all a. My claim is that bump functions give an
approximate identity for (L_1,*).
It's easy to see that whenever we have an approximate
identity on a Banach algebra, the original norm equals
Miguel's operator norm.
I forget if all nonunital Banach algebras have approximate
identities, but I seem to recall that all nonunital C*-algebras
do. There's stuff about that in Takesaki's book on operator
algebras. It's fun (and easy) to think about this issue for
commutative nonunital C*-algebras.
But we're drifting away from physics...
John Baez
nobody <nob...@polyester.richmond.edu> wrote:
>John Baez wrote:
>>The unbounded operators of interest in quantum physics are
>>almost always self-adjoint or at least normal. Other unbounded
>>operators are usually horrible!
>What about the boson annihilation and creation operators?
True, those are unbounded and unhorrible.
>My impression from things that you've said before is that they
>are reasonably well-behaved, e.g., (a+)+ = a (of course, I
>should ask to what domain that statement applies).
Yes, if you pick the right domains you get this equation.
This book will tell you which domain work:
http://math.ucr.edu/home/baez/bsz.html
if you've got the guts to download a 200-megabyte PDF file.
But, I'm trying to convince Miguel not to write his
thesis on some generalization of C*-algebras that
includes unbounded operator, so I wish you hadn't mentioned
this. :-)
Squark
ba...@galaxy.ucr.edu (John Baez) wrote in message news:<3f3483ec$2...@news.sentex.net>...
> ...That's why we
> get the Stone-von Neumann uniqueness theorem for irreps of
> the Weyl relations, but nothing comparable for the Heisenberg
> relations, which have myriads of weird irreps *of no use to physics*.
How does one construct those? I can't recall encountering them.
Best regards,
Squark
------------------------------------------------------------------
Write to me using the following e-mail:
Skvark_N...@excite.exe
(just spell the particle name correctly and change the
extension in the obvious way)
John Baez
Squark <fii...@yahoo.com> wrote:
>ba...@galaxy.ucr.edu (John Baez) wrote in message
>news:<3f3483ec$2...@news.sentex.net>...
>> ...That's why we
>> get the Stone-von Neumann uniqueness theorem for irreps of
>> the Weyl relations, but nothing comparable for the Heisenberg
>> relations, which have myriads of weird irreps *of no use to physics*.
>How does one construct those? I can't recall encountering them.
There are lots of different "useless" irreps of the Heisenberg
relations. Using sneaky constructions you can get examples
that have a *lot* of the same nice properties as the useful irrep -
but are still, in fact useless. My favorite is the crazy example
due to Ed Nelson, which is explained near the end of Reed & Simon's
first volume. But here's a much simpler example, just to get
you warmed up:
Let H = L^2[0,1] and let D be the subspace of H consisting of
all smooth functions vanishing in a neighborhood of the endpoints
of the interval [0,1]. D is a dense subspace of H. Define
linear operators
p,q: D -> D
as follows: q is multiplication by x, p is -i times differentiation
with respect to x.
Then p,q are densely defined symmetric unbounded operators on H
with
[p,q] = -i
on D. But they are not at all like the usual ones. In particular,
while both p and q admit self-adjoint extensions, the spectra of
these self-adjoint extensions are not the whole real line.
Of course p and q are not self-adjoint, so you might think
that's the problem. Or, you might think the problem is that
p admits a whole circle's worth of different self-adjoint
extensions. But that's the only problem, as fancier examples show.
Squark
> Let H = L^2[0,1] and let D be the subspace of H consisting of
> all smooth functions vanishing in a neighborhood of the endpoints
> of the interval [0,1]. D is a dense subspace of H. Define
> linear operators
>
> p,q: D -> D
>
> as follows: q is multiplication by x, p is -i times differentiation
> with respect to x.
Okay, that works.
> ...you might think the problem is that
> p admits a whole circle's worth of different self-adjoint
> extensions. But that's the only problem, as fancier examples show.
Did you mean to say "that's the only problem" i.e. there are no
"anomalous" representations with unique self-adjoint extension
or did you mean to say "that's not the only problem"?
Btw, this example is not that useless, it (in a way) corresponds
to a quantum particle in a 1D box. Or, to a quantum particle on
a circle, if we extend to H.
Daniel Grubb
>>*did* have a norm or topology, this proceedure can change it. An
>>example is given by L_1 (R) under convolution. This is embedded as
>>a *-subalgebra of C_0(R), but the norms changed in the process.
>Are the norms equivalent in that they define the same topology? I
>forget what the natural norm of (L_1,*) is, but I expect that it
>should be obtained by the following procedure:
>1) start with the L_1 norm on L_1(R).
>2) define M_f(g)=f*g for all f,g in L_1(R). For each f in L_1(R), M_f
> is a bounded endomorphism of L_1(R), and M_f M_g = M_{f*g}.
>3) M_f has an operator norm as an operator from L_1 to L_1. Use this
> norm to topologize L_1(R)
>Question: are the norms in 1) and 3) the same? Are they equivalent?
>There's some homework for me...
Yes, the norms are the same since L1 has approximate identities.
Clearly, ||M_f|| <= ||f||_1. The other inequality follows by considering
an approximate identity {g_n}. Then f*g_n->f in L1 while each
||g_n ||_1 =1.
It turns out that the Gelfand transform for the *-algebra L1
is equivalent to the Fourier transform which takes L1 to a
dense *-subalgebra of C_0(R). If the norms on the two sides
were the same, the image subalgebra couldn't be dense because
it would be complete. (Use Stone-Weierstrass to show density)
As for the measure algebra M(R), this is a unital *-algebra,
so the Gelfand map takes it to a subalgebra of C(K) for some
compact K. However, the image subalgebra is *not* a *-subalgebra.
There is more structure, though. For example, K can be made
into a semi-group. See Graham and McGehee 'Essays in Commutative
Harmonic Analysis' for a treament of more concerns than physicists
may want to know. On the other hand, M(R) *is* embedded in C(R) as
a *-subalgebra through the Fourier transform, although once again,
the norm changed dramatically.
>The idea is that you start with an abelian "algebra of observables" A,
>from it you construct its spectrum K, and imbed A as a *-subalgebra of
>C(K) with the compact-open topology. You may then want to enlarge the
>algebra of observables by taking its closure. This gives a natural
>prescription for which "sequences of observables" have limits, and
>what the limits should be. But now everything happens in K.
The problem is that the spectrum may not give an embedding as
a *-subalgebra. See above with the measure algebra. Also, there
may be several other embeddings that are as *-subalgebras, but
those may not give equivalent norms. Hence the limits may exist in
some embeddings but not in others.
>Here's a physically relevant example. Take an abstract 1-parameter
>group: W(s)W(t)=W(s+t) for all real s and t. Define W(s)^* = W(-s),
>and let A be the complex *-algebra generated by the W(s) with those
>relations. This means my abstract 1-parameter group is the group of
>unitary elements of A.
>What is the spectrum of A? A complex linear functional F : A -> \C is
>determined by its action on the generators: F(W(s))=f(s) for some
>function f : \R -> \C. Now, for F to be a *-algebra homomorphism, it
>must satisfy f(s)f(t)=f(s+t) and f(s)^*=f(-s), so f(s)=exp(i s theta)
>for some real theta. The spectrum of A is, then, K=\R, and the
>Gel'fand transform maps W(s) to S(theta)=exp(i s theta). All the
>elements of A are linear combinations of bounded funcions on \R, so
>they are all bounded and so K must be compact... Wait! \R is not
>compact!
The problem is that the function f(s) may not be continuous. This
means that f(s) may not be of the form you claim. In fact, the whole
collection of such f is called the Bohr compactification of \R, bR.
That's right, Bohr. However, it's Harold Bohr, not Niels (they were
brothers). Now, bR is a compact group in which \R is embedded
continuously as a dense subgroup.
>That is because the topology on the spectrum K is not the usual
>topology on \R, but the weakest topology making all the exp(i s theta)
>continuous. This is not much weaker than the usual topology, but it is
>weak enoug to make \R compact! Other surprising properties of this
>topology are that every open set has infinite Lebesgue measure.
Not quite true. The topology you are talking about is that induced from
bR to \R. It is not a compact topology on \R! But, again, bR *is* compact.
There is a still larger compact space in which \R is naturally embedded,
this time preserving the topology. That is the Cech compactification.
\beta \R. It is *much* larger, but is the space you obtain when
you use the Gelfand transform of the uniutal *-algebra, C(R), of
bounded continuous functions on \R. This is typical Polish topology
stuff. ;)
I hope this clarifies things some.
--Dan Grubb
C. M. Heard
> Of course p and q are not self-adjoint, so you might think
> that's the problem. Or, you might think the problem is that
> p admits a whole circle's worth of different self-adjoint
> extensions. But that's the only problem, as fancier examples show.
I think you meant to say "that's NOT the only problem."
//cmh
P.S. Thanks for the cute example.
Miguel Carrion
In article <3f3483ec$2...@news.sentex.net>, John Baez <ba...@galaxy.ucr.edu> wrote:
>In article <bfuri2$pi7$1...@glue.ucr.edu>,
>Miguel Carrion <mig...@math-cl-n01.math.ucr.edu> wrote:
>
>>so that the compact-open topology on C(K) can be used to topologize
>>the original *-algebra entirely from the algebraic data.
>
>Does the original *-algebra become a topological *-algebra? I.e., do
>the addition, scalar multiplication, product and * operation become
>continuous in this topology? I sure hope so.
Yes, basically because the *-algebra A is imbedded as a subalgebra of
C(K), and C(K) is a topological *-algebra with the compact-open
topology.
>Hey, good, it's great to see you doing analysis. Even though I don't
>talk about it much anymore, I used to really like it, and I still
>sort of do.
Like I once told my old professor Fernando Bombal, if someone had told
me when I was learning functional analysis from him that I'd be doing
so much analysis three years down the line, I wouldn't have beleved
them :-)
>But are you hinting that C(K) might not already be a Frechet space
>when K is sigma-compact? What does its closure contain that it does
>not?
When K is locally compact C(K) is closed in the compact-open
topology. What I meant is that A is an F-space and that its closure is
a Fr'echet space. The interesting thing is that, when K is locally
compact, then A is actually dense in C(K).
Now I am a little puzzled. Consider the C*-algebra of bounded
continuous complex functions on R with the sup norm. The spectrum is
supposed to be compact, but R is not compact! Is the spectrum of this
C^*-algebra different from R, or is R being given a topology different
from the usual one that makes it compact? I have a hunch: that the
spectrum contains an "invariant mean" which compactifies R by a point
at infinity.
>So, what I'm saying is that as far as rigorous mathematical
>quantum physics goes, the unbounded operators that matter are
>the ones that you can describe easily using bounded operators,
>and this description is more or less necessary for doing any
>serious work. SO, we might as well focus on C*-algebras rather
>than some more general *-algebras designed for working with
>unbounded operators.
My problem with this is that in probability theory (which is to
abelian C*-algebras what quantum theory is to non-abelian ones) not
every sensible function has finite expectation value with respect to
every sensible probability distribution, but GNS states on an abelian
C^*-algebra are precisely that: bounded expectations. So already in
probability theory there is a need for densely-defined GNS states on
algebras which I want to call "locally C*" (because they have locally
compact spectrum!), for which continuous GNS states correspond to
compactly-supported probability distributions. Of course, if you think
that "sensible" functions are polynomials and that "sensible"
distributions have finite variance, then C^*-algebras are the
"sensible" setting. So I guess ultimately that's why I think
C^*-algebras are too rigid.
Regards,
Miguel
backdoorstudent
ba...@galaxy.ucr.edu (John Baez) wrote in message news:<bh7h33$drr$1...@glue.ucr.edu>...
>
> Yes, if you pick the right domains you get this equation.
> This book will tell you which domain work:
>
> http://math.ucr.edu/home/baez/bsz.html
>
> if you've got the guts to download a 200-megabyte PDF file.
>
Just in case you don't know, that file can easily be compressed to
take up only 53.4 Mb using a utility such as PDF Shrink
(http://www.apago.com/pdfshrink/) or (PDF Enhancer
http://www.pdfsages.com/enhancer.html) with negligible degradation in
quality.
gr...@math.niu.edu
was stopped.
> Now I am a little puzzled. Consider the C*-algebra of bounded
> continuous complex functions on R with the sup norm. The spectrum is
> supposed to be compact, but R is not compact! Is the spectrum of this
> C^*-algebra different from R, or is R being given a topology different
> from the usual one that makes it compact? I have a hunch: that the
> spectrum contains an "invariant mean" which compactifies R by a point
> at infinity.
Let me try to clarify what is happening. This will consist of a bit
of an overview of the Gelfand theory, so bear with me.
Suppose we start with a commutative Banach algebra A. We can look
at the space of all continuous homomorphisms of A into the complex
numbers, C. This space is also in one-to-one correspondence with
the collection of maximal modular ideals in A. Call this space of
continuous homomorphisms D. It is also called the structure space of
A.
Now D can be given a topology. We use the weakest topology that makes
all evaluation maps a^:D-->C continuous, where a^(L)=L(a) for a in A
and L in D. This topology is always locally compact and is even compact
if the algebra A has an identity. The map that send the element a of A
to the continuous function a^ on D is called the Gelfand transform.
It turns out that a^ is always in C_0 (D), the continous functions
vanishing at infinity on D.
There are several difficulties that can occur in general. First is that
the Gelfand transform may not be one-to-one. When it is, we say the
algebra A is semi-simple. What we can see is that the kernel of the Gelfand
map is simply the intersection of all the maximal modular ideals in A.
Since any nilpotent element will be in all of these, there cannot be any
nilpotent (or even topologically nilpotent) elements in a semi-simple
algebra.
Now, suppose that A is actually a *-algebra. It is quite possible that
the Gelfand map is not a *-homomorphism, even when A is semi-simple.
In fact, if M(R) is the measure algebra on the real line under convolution,
we can create a * operation on M(R) but the Gelfand map is definitely
---not--- a *-homomorphism! The key word here is that M(R) is asymmetric.
Next, we have nice algebras like L1(R). For this algebra, the Gelfand transform
is essentially the Fourier transform, so is a one-to-one *-homomorphism.
However, the image is not all of C_0(R), it is just a dense subalgebra.
Next, there are C* algebras, where the Gelfand map is a one-to-one
*-homomorphism -onto- C_0(D). These a particularly nice algebras!
Now, to answer your question. If we start with the collection of all bounded
functions on R, i.e. C(R), the structure space is compact since C(R)
has an identity. However, we do not get D=R as sets! The space D turns out
to be what is called the Stone-Cech compactification of R. This is a
BIG space, in which R sits as a dense set, but one of relatively small
cardinality. The space D has the same cardinality as the power set of R!
There are a lot of points in the remainder!!! The topology of R as a subset
of D is the usual one however.
In fact, if we start with any Tychonoff space X, and consider C(X), the
structure space is the Stone-Cech compactification of X.
I hope this helps.
Now for some criticism of your overall strategy:
If you start with a *-algebra that is semi-simple (this can be given
a purely algebraic definition), the Gelfand map may not give a *-embedding
of A into some C(D). However, the Gelfand map is the natural one from
a purely algebraic perspective. On the other hand, there may be another
space D' and another map from A into C(D') that is a *-homomorphism.
It is not at all clear to me how to find D' via purely algebraic means
however. From the example of M(R), I suspect that D' may not be compact
even if A is unital (assuming such a D' exists).
--Dan Grubb
John Baez
Squark <fii...@yahoo.com> wrote:
>ba...@galaxy.ucr.edu (John Baez) wrote in message
>news:<bhf7dr$67k$1...@glue.ucr.edu>...
>> Let H = L^2[0,1] and let D be the subspace of H consisting of
>> all smooth functions vanishing in a neighborhood of the endpoints
>> of the interval [0,1]. D is a dense subspace of H. Define
>> linear operators
>>
>> p,q: D -> D
>>
>> as follows: q is multiplication by x, p is -i times differentiation
>> with respect to x.
... thus obtaining a pair of operators p and q satisfying
[p,q] = -i
which are not at all like the momentum and position operators
we know and love! Shudder shudder, gasp gasp!
>> ...you might think the problem is that
>> p admits a whole circle's worth of different self-adjoint
>> extensions. But that's the only problem, as fancier examples show.
>Did you mean to say "that's the only problem" i.e. there are no
>"anomalous" representations with unique self-adjoint extension
>or did you mean to say "that's not the only problem"?
Yikes! I meant to say that's NOT the only problem!!!
Indeed, you CAN cook up bizarre solutions of
[p,q] = -i
where both p and q are self-adjoint, but are not unitarily
equivalent to the p and q we know and love.
And this is why we need to use the exponentiated version
of the relation [p,q] = -i if we want to state a useful
version of the Stone-von Neumann uniqueness theorem.
>Btw, this example is not that useless, it (in a way) corresponds
>to a quantum particle in a 1D box.
In a way, but only in a misleading sort of way. There's no good
momentum operator for a particle in a box because there's
no translation symmetry. This is related to the fact that the
p operator described above doesn't have a unique self-adjoint
extension.
>Or, to a quantum particle on a circle, if we extend to H.
This is more relevant. Indeed, the p operator described
above has a whole circle's worth of self-adjoint extensions.
Each one generates a different one-parameter unitary group
on L^2[0,1] = L^2(S^1), each giving our particle a different
"phase shift" when we translate it all the way around the circle S^1.
We've talked about this before. You may remember:
..................................................................
From: ba...@galaxy.ucr.edu (John Baez)
Newsgroups: sci.physics.research
Subject: Re: Particle in a box
Date: Fri, 11 Oct 2002 01:34:30 +0000 (UTC)
In this article we shall see how many ways a quantum particle
can bounce off a wall.
In article <200210071527...@mscan1.ucar.edu>,
nobody <nob...@localhost.ucar.edu> wrote:
>mandro wrote:
>> I'm wondering
>> a) In how many ways can I describe mathematically a particle in a box.
Excellent question!
If you read the first volume of Reed and Simon, the one called
Functional Analysis, you'll learn a lot about this. Also the
second volume, Fourier Analysis and Self-Adjointness, especially
section X.1. I'll say a bit here, but no self-respecting mathematical
physicist can do without Reed and Simon.
>If you are restricting yourself to a 1-dimensional box with impenetrable
>sides such that the particle bounces off the sides (as you describe below)
>then there is, I think, just one self-adjoint Hamiltonian to describe it
>(modulo unitary equivalences of course).
As we'll see, it depends a bit on what you mean by "bounce"!
>> b)I have seen the standard textbook derivation on R with an infinite
>> potential outside [a, b], but, I'm wondering can I do the same in
>> L^2([a,b]) just using a different momentum operator?
There's not really any good momentum operator on L^2([a,b]).
More precisely, if we start with the operator
P = -i d/dx
defined on the domain of smooth functions vanishing at the endpoints
of the interval [a,b], this operator is symmetric but not self-adjoint.
It has a circle's worth of self-adjoint extensions, one for each
unit complex number z:
|z| = 1.
Let's call these P(z).
Each of these operator P(z) generates a one-parameter unitary group U(t).
When you apply U(t) to a function, it just translates the function -
at least when t is small and your function vanishes near the endpoints
of the interval.
However, as t becomes larger, your function gets translated to the right
until it "hits the right endpoint". Then it "wraps around" and reappears
at the left endpoint multiplied by the number z!
In short, we get self-adjoint operators that look and feel like
-i d/dx only if we assume "twisted periodic boundary conditions",
where the "twist" is this unit complex number z.
These are not particularly relevant to a particle in a box.
Part of the point is that momentum is not conserved for a particle
in a box. Naively it seems the momentum operator -i d/dx commutes
with the Hamiltonian - d^2/dx^2, which would imply conservation of
momentum! A paradox? No: it's just that there is no good momentum
operator.
>> By the way when
>> I say particle in a box I mean one that bounces. In short, is there
>> a formalism on L^([0, 1]) that could define the particle bouncing off
>> walls, and what would H and p be in that case--esp what would be the
>> domains.
I'm really glad you said the word "domains" - because this is one of
those physics questions you can only understand if you realize
the importance of PICKING A DOMAIN for your would-be self-adjoint
operator! All the operators P(z) have different domains, though
all their domains contain that of the operator P.
Anyway, there's no good momentum operator "-i d/dx" for a particle in
a box, but there is at least one good Hamiltonian "-d^2/dx^2".
>Actually, I discussed this question briefly back in July in article
><ahmkib$br1$1...@localhost.localdomain>. The answer is that you can
>indeed to this. Here is one route. Note that units are chosen so
>that h_cross = 1.
I hear that Dirac always pronounced it "h cross". I always
say "h bar". So, I'm guessing you're British.
>Take the Hilbert space to be L^2 [a, b] and take
>the Hamiltonian H: f -> (-1/2m ) f'' (i.e. a
>free particle Hamiltonian). This mapping is defined
>for functions f with absolutely continuous first
>derivative and second derivative in L^2 [a, b].
>But it's not self-adjoint over that whole domain.
>To make the operator self-adjoint, we can restrict
>the domain in (at least) these two different ways:
>
>(i) require that f(b) = 0 and f(a) = 0;
>(ii) require that f(b) = f(a) and f'(b) = f'(a).
Right! There are a bunch of other self-adjoint
restrictions of this operator, as well. Even for the
particle on the half-line [0,infinity) there is a whole
circle's worth of self-adjoint operators deserving the
name "-d^2/dx^2". Here there should be even more.
The point is that you want the particle to bounce off
both endpoints of your interval, not wrap around as in
case (ii). Case (i) makes the particle bounce, but it
isn't the only way to get a "bounce". You can have it
bounce back with any phase shift you like! Case (i)
gives a phase shift of -1. This corresponds to a "hard
wall potential". So, this is the right answer for most
purposes. But there are other options.
Note: here you are starting with a large domain and looking
for ways to restrict the operator to smaller domains to
make it self-adjoint. I prefer to start with a small
domain and look for self-adjoint extensions. To some
extent it's a matter of taste. But if you use the method
I'm choosing, von Neumann's powerful theory of "deficiency
indices" lets you completely classify the self-adjoint
extensions! You can read about this in section X.1 of
the second volume of Reed and Simon.
I don't know a comparable theory for restrictions.
John Baez
Miguel Carrion <mig...@math-cl-n01.math.ucr.edu> wrote:
>When K is locally compact C(K) is closed in the compact-open
>topology. What I meant is that A is an F-space and that its closure is
>a Fr'echet space. The interesting thing is that, when K is locally
>compact, then A is actually dense in C(K).
Okay. By the way, I hope you really think a bunch about what
happens here in that example where A = L^1(R) with its convolution
product.
>Now I am a little puzzled. Consider the C*-algebra of bounded
>continuous complex functions on R with the sup norm. The spectrum is
>supposed to be compact, but R is not compact! Is the spectrum of this
>C^*-algebra different from R, or is R being given a topology different
>from the usual one that makes it compact?
Neither! The spectrum of this algebra is the Stone-Cech compactification
of R, more or less by definition.
Though there are other ways to define the Stone-Cech compactification
of a locally compact space K, and you should learn some,
the way I like to do it is to take the spectrum of the algebra of bounded
continuous functions on K.
If instead we use the algebra of bounded continuous functions that
approach a constant at infinity, we get the one-point compactification.
The Stone-Cech compactification is the "biggest possible" one,
while the one-point compactification is the "smallest possible" one.
Do you know the variant of this trick that gives the Bohr
compactification of R?
How about the 2-point compactification of R (= the closed unit interval)?
(I'll email this to you as well as posting it....)
While I'm raising puzzles: what happens to your machinery
if we start with the algebra of all rational functions of n
variables? What's the spectrum of this algebra?
Miguel Carrion
Daniel Grubb <gr...@lola.math.niu.edu> wrote:
>
>The problem is that the function f(s) may not be continuous. This
>means that f(s) may not be of the form you claim. In fact, the whole
>collection of such f is called the Bohr compactification of \R, bR.
>That's right, Bohr. However, it's Harold Bohr, not Niels (they were
>brothers). Now, bR is a compact group in which \R is embedded
>continuously as a dense subgroup.
I seem to recall that if f(s+t)=f(s)f(t) is a real function and it is
not continuous, then it is not measurable, and that the
counterexamples use the axiom of choice (we're using Tychonoff's
theorem all over the place already). Are we saying that bR consists
mostly of non-measurable functions? Of course when discussing the
complex homomorphisms we are at the level of algebra, not of analysis,
so that's not necessarily a problem.
>Not quite true. The topology you are talking about is that induced from
>bR to \R. It is not a compact topology on \R! But, again, bR *is* compact.
Sorry, I was being careless again.
>I hope this clarifies things some.
Yes, thanks a lot for all your postings.
Miguel
Miguel Carrion
>In article <bh66v2$394$1...@glue.ucr.edu>,
>Miguel Carrion <mig...@math-cl-n01.math.ucr.edu> wrote:
>
>Though there are other ways to define the Stone-Cech compactification
>of a locally compact space K, and you should learn some,
>the way I like to do it is to take the spectrum of the algebra of bounded
>continuous functions on K.
>
>If instead we use the algebra of bounded continuous functions that
>approach a constant at infinity, we get the one-point compactification.
>
>The Stone-Cech compactification is the "biggest possible" one,
>while the one-point compactification is the "smallest possible" one.
>
>Do you know the variant of this trick that gives the Bohr
>compactification of R?
I think it is the algebra generated by the imaginary exponentials f(x)
= exp (ikx) for all real k.
>How about the 2-point compactification of R (= the closed unit interval)?
These are the continuous functions tending to possibly different
constants at positive and negative infinity.
>While I'm raising puzzles: what happens to your machinery
>if we start with the algebra of all rational functions of n
>variables? What's the spectrum of this algebra?
Rational functions form a field, so there can't be any complex
homomorphisms and the spectrum is empty! If f is a nonconstant
rational function, and h is a complex algebra homomorphism, then
f-h(f) is also a nonconstant rational function, hence invertible. Now,
algebra homomorphisms preserve invertibility, so h(f-h(f))=0, which is
a contradiction.
Regards,
Miguel
gr...@math.niu.edu
> In article <bhjm9u$3u8$1...@news.math.niu.edu>,
> Daniel Grubb <gr...@lola.math.niu.edu> wrote:
> >
> >The problem is that the function f(s) may not be continuous. This
> >means that f(s) may not be of the form you claim. In fact, the whole
> >collection of such f is called the Bohr compactification of \R, bR.
> >That's right, Bohr. However, it's Harold Bohr, not Niels (they were
> >brothers). Now, bR is a compact group in which \R is embedded
> >continuously as a dense subgroup.
>
> I seem to recall that if f(s+t)=f(s)f(t) is a real function and it is
> not continuous, then it is not measurable, and that the
> counterexamples use the axiom of choice (we're using Tychonoff's
> theorem all over the place already). Are we saying that bR consists
> mostly of non-measurable functions? Of course when discussing the
> complex homomorphisms we are at the level of algebra, not of analysis,
> so that's not necessarily a problem.
Yes, this depends on the Axiom of Choice, but so does the whole
Gelfand theory! In particular, to prove the compactness of the
structure space requires the Tychonoff product theorem, which is
equivalent to AC. Hmmmm...I guess we actually only need the product
theorem for Hausdorff spaces, so it isn't equivalent to the full
AC, but it's close.
Yes, bR is the collection of *all* homomorphisms of \R into the
circle group (i.e. continuous in the discrete topology). Most
of these functions are not Lebesgue measurable. However, it *is*
known that \R is measurable inside of bR.
---Dan Grubb
gr...@math.niu.edu
> In article <bi4ofa$6fm$1...@glue.ucr.edu>, John Baez <ba...@galaxy.ucr.edu> wrote:
> >In article <bh66v2$394$1...@glue.ucr.edu>,
> >Miguel Carrion <mig...@math-cl-n01.math.ucr.edu> wrote:
> >
> >Though there are other ways to define the Stone-Cech compactification
> >of a locally compact space K, and you should learn some,
> >the way I like to do it is to take the spectrum of the algebra of bounded
> >continuous functions on K.
> >
> >If instead we use the algebra of bounded continuous functions that
> >approach a constant at infinity, we get the one-point compactification.
> >
> >The Stone-Cech compactification is the "biggest possible" one,
> >while the one-point compactification is the "smallest possible" one.
> >
> >Do you know the variant of this trick that gives the Bohr
> >compactification of R?
>
> I think it is the algebra generated by the imaginary exponentials f(x)
> = exp (ikx) for all real k.
You do have to be careful because the Bohr compactification is not
a 'true' compactification. The topology of \R as a subset of bR is
not the usual topology. The topology as a subset of the Stone-Cech
compactification *is* the same as the usual topology.
--Dan Grubb
Miguel Carrion
I'd like to come back to the example of L_1 and the space of measures
on R, both with convolution.
Let's take L_1 first. It is not hard to see that the continuous
complex homomorphisms are precisely the Fourier integrals, and that
the resulting Gelfand map takes L_1 functions to bounded continuous
functions on R. Now, L_1 has two involutions:
____
f*(x) = f(x)
_____
f*(x) = f(-x)
In the first case, the only Fourier integral which is a *-homomorphism
is the integral w.r.t the constant function 1. Hence, the Gelfand
transform cannot be a *-homomorphism. In the second case, every
Fourier integral is a *-homomorphism, so the Gelfand transform is a
*-homomorphism.
I am wondering if the pathology ("asymmetry") that Dan Grubb mentioned
regarding the space of measures [i.e., that the Gelfand transform is
not a *-homomorphism but that the Fourier transform imbeds it into
C(R)] is related to this.
Regards,
Miguel
gr...@math.niu.edu
No, it is not related to this. L1(R) natually imbeds into M(R) by
sending f to fd\lambda where \lambda is the usual Lebesgue measure
on R. The involution chosen for M(R) extends your second involution
on L1(R). Almost nobody in harmonic analysis uses the first one.
What really occurs in much more subtle. With the 'correct' involution,
Fourier transform of measures provides a *-homomorphism into C(R).
However, the Gelfand structure space for M(R) is *much, much* larger
than just the collection of Fourier transforms. The problem is that
there are measures that are self-adjoint but whose Gelfand transform
is not real valued, even though their Fourier transform is.
In particular, there are self-adjoint probability measures \mu that
are 'independent power' measures in the sense that every convolution
power of \mu is singular with respect to every other convolution
power. The spectrum of such a measure turns out to be the whole
unit disk in the complex plane. This happens even though the Fourier
transform of \mu is real valued and even though \mu is self-adjoint
and a probability measure.
Another consequence is that there are non-invertible measures whose
Fourier transform is bounded away from 0.
There is a lot known about the measure algebra on a general locally
compact abelian group. A good, but probably out of date reference is
Colin C. Graham & O. Carruth McGehee, Essays in Commutative Harmonic
Analysis, Springer-Verlag 1979, Grundlehren der mathematischen Wissenschaften
238
Look at chapters 6 and 8, but don't expect an easy time.
---Dan Grubb
gr...@math.niu.edu
I know I responded once to this, but I have thought about it some more
and have a couple more insights for you. In the case of L1 with either
involution, the image of the Gelfand transform in C_0(R) is convolution
closed. In other words, for each f in L1, there is g such that the transform
of g is the conjugate of the transform of f. This essentially picks out
the second involution as the 'correct' one since it is the one that
gives the 'correct' transform.
If you look at M(R), however, things are different. If X is the Gelfand
structure space, the image of M(R) in C(X) is *not* conjugate closed.
There is a measure \mu such that the conjugate of it's Gelfand transform
is not the transform of *any* measure. This holds even though M(R) is
a *-algebra. What happens is that the Fourier transforms give a subset
of the structure space that determine *which* involution should be the
*-operation. But this *-operation does not, in fact, work.
--Dan Grubb
nobody
>In article <200308090619...@localhost.localdomain>,
>nobody <nob...@polyester.richmond.edu> wrote:
>
>>John Baez wrote:
>>
>>>The unbounded operators of interest in quantum physics are
>>>almost always self-adjoint or at least normal. Other unbounded
>>>operators are usually horrible!
>>
>>What about the boson annihilation and creation operators?
>
>True, those are unbounded and unhorrible.
>
>>My impression from things that you've said before is that they
>>are reasonably well-behaved, e.g., (a+)+ = a (of course, I
>>should ask to what domain that statement applies).
>
>This book will tell you which domain work:
>
>http://math.ucr.edu/home/baez/bsz.html
>
>if you've got the guts to download a 200-megabyte PDF file.
Fortunately, I won't have to do that. I was just given a nearly
mint-condition print copy of BSZ that a friend got for me from a
used book dealer in Kansas (via Amazon).
The bad news is, I don't know where in that book to find those
domains ... so I have to ask for a hint. Dr. Baez?
nobody
John Baez
nobody <nob...@localhost.localdomain> wrote:
>On Mon, 11 Aug 2003 John Baez wrote:
>>In article <200308090619...@localhost.localdomain>,
>>nobody <nob...@polyester.richmond.edu> wrote:
>>>John Baez wrote:
>>>>The unbounded operators of interest in quantum physics are
>>>>almost always self-adjoint or at least normal. Other unbounded
>>>>operators are usually horrible!
>>>What about the boson annihilation and creation operators?
>>True, those are unbounded and unhorrible.
>>My impression from things that you've said before is that they
>>>are reasonably well-behaved, e.g., (a+)+ = a (of course, I
>>>should ask to what domain that statement applies).
>>Yes, if you pick the right domains you get this equation.
>>This book will tell you which domain work:
>>http://math.ucr.edu/home/baez/bsz.html
>Fortunately, I won't have to do that. I was just given a nearly
>mint-condition print copy of BSZ that a friend got for me from a
>used book dealer in Kansas (via Amazon).
Cool! It's a rare book by now.
>The bad news is, I don't know where in that book to find those
>domains ... so I have to ask for a hint. Dr. Baez?
[Pulls book out of bookshelf, turns to index, looks up
"annihilation operator", sees first page they're referred to
is page 48....]
Check out the first theorem on page 49.
Any complex Hilbert space H has a Fock space K(H) on which
there are self-adjoint field operators phi(z) for all vectors
z in H, such that
[phi(z), phi(z')] = -i Im<z,z'>
on the domain of definition of the left-hand side. This is
an abstract formulation of the canonical commutation relations.
The annihilation and creation operators are then defined as
a(z) = (phi(z) + i phi(iz))/sqrt(2)
a*(z) = (phi(z) - i phi(iz))/sqrt(2)
with their domains being the intersection of the domains of
phi(z) and phi(iz). They are then each other's adjoint!