# Seminar on Banach Algebras

This seminar was held in #mathematics on EFnet on 21st of May 2005 at 15:00 EDT by Kit.

It provided a basic introduction to the theory of commutative Banach algebras, focusing especially on the ideas of automatic continuity.

## Seminar Log

20:01 <@Kit> ok then.
20:01 <@Kit> Shall we give people a few minutes to make their way in, or shall I just start now?
20:01 < landen> start now
20:02 <@Kit> ok then.
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20:03 <@Kit> The study of Banach algebras is one of the big areas in modern functional analysis. It draws on a lot of theory from algebra proper, but the analysis shores up various aspects of the theory that aren't algebraically nice' in order to make them well behaved.
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20:03 <@Kit> Basically a banach algebra is a banach space (a complete normed space) with an algebra structure.
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20:04 <@Kit> That is we have an associative bilinear multiplication on it
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20:04 -!- mode/#mathematics [+l 62] by Evariste
20:04 < [CODE]XXX> +.v
20:04 <@Kit> That is it's a ring where the multiplication commutes with complex multiplication
20:04 <@Kit> So (za)b = a(zb) = z(ab)
20:05 <@Kit> In addition you also want the multiplication to behave well with respect to the norm
20:05 <@Kit> Specifically you want |ab| <= |a| |b|
20:05 <@Kit> We will also assume that it is unital. That it has an identity as a ring, and |1| = 1
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20:05 < gzl> !
20:05 <@Kit> (You can add in a unit to any non-unital banach algebra, and if |1|!= 1 you can replace it with an eequivalent norm such that it is)
20:05 <@Kit> gzl: Yes?
20:06 < gzl> do you only consider algebras over C?
20:06 <@Kit> I was going to mention that in a moment. :)
20:06 < gzl> sorry. :)
20:06 <@Kit> The definitions work over R
20:06 <@Kit> But the theory over C is much much better
20:06 <@Kit> In almost everything I say tonight we will be working over C
20:06 <@Kit> I'll occasionally make notes along the lines of look, this fails if we replace C by R'
20:07 -!- mode/#mathematics [+l 64] by Evariste
20:07 <@Kit> We've basically got two canonical examples of banach algebras.
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20:07 <@Kit> If X is a compact hausdorff space then the set of all continuous functions from X to C is a banach algebra with the uniform norm
20:07 <@Kit> We denote this by C(X)
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20:08 <@Kit> (multiplication and addition are pointwise)
20:08 <@Kit> If V is any banach space then End(V) is a banach algebra with composition and the operator norm.
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20:08 <@Kit> The first is commutative, the second is not.
20:08 -!- mode/#mathematics [+o Tolaria] by Thwash, bananach
20:08 -!- mode/#mathematics [+o Tolaria] by Evariste
20:08 <@Kit> We're not even going to look at the noncommutative theory of banach algebras tonight, cause it's hard. :)
20:08 <@Kit> Some of what I say will work there, much will not.
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20:09 -!- mode/#mathematics [+o NullPtr] by Evariste, bananach, Thwash
20:09 <@Kit> A reminder: If we have a closed subspace of a banach space, we can put a norm on the quotient by that subspace to make it into a banach space.
20:09 <@Kit> Specifically ||x + E|| = inf {||x + y|| : y \in E }
20:10 <@Kit> Exercise: If E is in fact a closed *ideal*, then this is an algebra norm on the quotient.
20:10 <@Kit> Anyway, those are the basic definitions. Any questions before I move on to some proper theory? :)
20:10 <@Kit> ok. Guess not.
20:11 <@Kit> Anyway, the basic reason why banach algebras are cool is that there's a huge amount of interplay between topological and algebraic structure.
20:11 <@Kit> For example we have the following result:
20:11 <@Kit> Let A be a banach algebra. We denote the set of invertible elements by G(A). G(A) is open.
20:12 <@Kit> (This isn't true in a general normed algebra, but I can't remember an example offhand)
20:12 -!- NullPtr [nullptr@cpe-70-112-94-63.austin.res.rr.com] has quit [Ping timeout: 186 seconds] 20:12 <@Kit> We first prove the following lemma: If |x| < 1 then 1 + x is invertible.
20:12 <@Kit> Proof: (1 + x)^{-1} = sum x^n
20:12 <@Kit> This converges because |x^n| <= |x|^n
20:12 <@Kit> So it converges absolutely
20:12 <@Kit> And because A is complete, absolute convergence => convergence.
20:13 <@Kit> We can rephrase this by saying that if |x - 1| < 1 then x is invertible
20:13 -!- landen [sizzle@adsl-216-103-91-31.dsl.snfc21.pacbell.net] has quit [Ping timeout: 268 seconds] 20:13 <@Kit> And fiddling around with some algebra then shows you that if y is invertible and |x - y| < |y^{-1}|^{-1}, we have x invertible.
20:14 <@Kit> (Caution: |y^{-1}|^{-1} is not in general the same as |y|)
20:14 <@Kit> Which gives us the result that G(A) is open. We'll need this later.
20:14 <@Kit> These simple results now let us develop one of the most important tools in banach algebra theory
20:15 <@Kit> (It may not be obvious that it's important yet. Hopefully you'll be more convinced later)
20:15 <@Kit> Let A be a banach algebra and a \in A. We define the spectrum of A to be the set { z \in C : z1 - a is *not* invertible }
20:15 <@Kit> err. spectrum of a
20:15 <@Kit> We'll write this sp(a)
20:16 <@Kit> Lemma: sp(a) is closed and bounded.
20:16 <@Kit> It's closed: it's the inverse image of G(A)^c under the map z -> z1 - a
20:16 <@Kit> And this map is clearly continuous
20:17 <@Kit> It's bounded: Suppose |z| > |a|. Then z1 - a = z(1 - z^{-1}a)
20:17 <@Kit> And |z^{-1}a| < 1
20:17 <@Kit> So this is invertible
20:17 <@Kit> We've actually proved somethign stronger: It's contained in the closed disk of radius |a|
20:18 <@Kit> If we define the spectral radius r(a) = sup \{ |z| : z \in sp(a) } this can be restated as r(a) <= |a|
20:18 <@Kit> Now. Big result coming up.
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20:18 <@Kit> In a general complex algebra, sp(a) can be empty
20:18 -!- mode/#mathematics [+o NullPtr] by bananach, Thwash
20:18 <@Kit> For example C[x]
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20:19 <@Kit> sp(x) is empty
20:19 <@Kit> err
20:19 <@Kit> Sorry
20:19 <@Kit> C(x)
20:19 <@Kit> (Rational functions, not polynomials)
20:19 -!- mode/#mathematics [+o NullPtr] by Evariste
20:19 <@Kit> Because z1 - x is not invertible iff it's zero
20:19 <@Kit> And that never happens
20:19 <@Kit> In a Banach algebra, this can't happen.
20:20 <@Kit> It may not look it, but this is a Big Thing. :)
20:20 <@Kit> Proof:
20:20 <@Kit> Let a \in A
20:20 <@Kit> Suppose sp(a) = \emptyset
20:20 <@Kit> Let f : A -> C be a continuous linear functional of norm 1
20:21 <@Kit> Define the map \phi : C -> C by \phi(z) = f( (z1 - a)^{-1} )
20:22 <@Kit> Note that |\phi(z)| <= |(z1 - a)^{-1}| = |z|^{-1} |(1 - a z^{-1})^{-1}|
20:22 <@Kit> Which tends to zero as |z| -> inf
20:22 <@Kit> I leave it as an exercise for the reader that \phi is an analytic function
20:22 <@Kit> (It will be in the notes)
20:22 <@Kit> So what we have is a bounded analytic function on all of C
20:23 <@Kit> Which by liouville's theorem can't possibly happen unless it's constant
20:23 <@Kit> And because it tends to zero it must then be constantly zero
20:23 <@Kit> So for every continuous linear functional of norm 1 and every z \in C we have f( (z1 - a)^{-1} ) = 0
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20:23 <@Kit> But by the hahn-banach theorem this implies that for every z \in C we have (z1 - a)^{-1} = 0, as norm 1 functionals seperate points.
20:23 <@Kit> But this is patently nonsense
20:24 <@Kit> As 0 is not invertible
20:24 <@Kit> So the spectrum couldn't have been empty in the first place
20:24 <@Kit> This then gives us the following important theorem, related to what I was saying about C(X)
20:24 <@Kit> Suppose A is a banach algebra in which every non-zero element is invertible
20:24 <@Kit> Then A = C
20:25 <@Kit> Because let a \in A. Then sp(A) is non-empty, so for some z we have z1 - a is not invertible
20:25 <@Kit> But this can only happen if z1 - a = 0
20:25 <@Kit> So a = z1
20:25 <@Kit> Thus the map z -> z1 is a banach algebra isomorphism between C and A
20:26 <@Kit> This is called the Gelfand-Mazur theorem
20:26 <@Kit> And it has some rather important consequences about the ideal structure of commutative algebras.
20:26 <@Kit> Note: This fails for R. For example C is a banach algebra over R which is not isomorphic to R
20:26 <@Kit> Similarly, i \in C has empty spectrum over R
20:27 <@Kit> ok. This is about as far as I'm going with the general theory. The rest will all be commutative stuff.
20:27 <@Kit> Any questions before I move on to that?
20:27 < azta> nope :)
20:27 <@Kit> ok then.
20:28 <@Kit> From now on when I say banach algebra' I really mean commutative banach algebra'
20:28 <@Kit> Some of what I say will work without that, but not an awful lot
20:28 <@Kit> Let A be a banach algebra. A character on A is a ring homomorphism f : A -> C 20:28 <@Kit> Sorry. Algebra homomorphism
20:29 <@Kit> We also want it to be C-linear
20:29 <@Kit> (I suppose we also want it to be non-zero if you don't believe ring homomorphisms send 1 to 1)
20:29 <@Kit> Note there is no continuity assumption here.
20:29 <@Kit> This gives us our first automatic continuity theorem, which motivates a lot of other stuff:
20:29 <@Kit> Let A be a banach algebra and f a character. Then f is continuous and |f| <= 1
20:30 <@Kit> Proof:
20:30 <@Kit> Suppose not and we have some a with |f(a)| > |a|
20:30 <@Kit> By scaling a we can assume that f(a) = 1 and |a| < 1
20:31 <@Kit> Ack. Temporary brain failure. One moment. :)
20:32 <@Kit> (I hate it when this happens)
20:33 <@Kit> ok. Lets see if I can think through this.
20:34 <@Kit> Define y = sum_{n >= 0} a^n
20:34 <@Kit> As before, this converges.
20:34 <@Kit> y = 1 + a + a^2 + ... = 1 + a(1 + a + ... )
20:34 <@Kit> Ah yes
20:34 <@Kit> So y = 1 + ay
20:34 <@Kit> So f(y) = 1 + f(a) f(y)
20:35 <@Kit> And f(a) = 1
20:35 <@Kit> So f(y) = 1 + f(y), and 1 = 0. Oops. Contradiction.
20:35 <@Kit> Phew. :)
20:35 < azta> :)
20:35 <@Kit> ok. By the isomorphism theorem, if f is a multiplicative linear functional then A / ker(f) = C
20:36 <@Kit> (This is a purely algebraic statement - no norms yet)
20:36 <@Kit> So ker(f) is a maximal ideal
20:36 <@Kit> We will in fact show that *all* maximal ideals arise this way.
20:36 <@Kit> Step 1: Maximal ideals are closed.
20:37 <@Kit> Proof: Let I be an ideal. Then cl(I) is an ideal. So either I = cl(I) and I is closed, or cl(I) = A
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20:37 <@Kit> So we just need to rule out the second possibility
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20:37 <@Kit> But we know that G(A) is open
20:37 <@Kit> And I is included in G(A)^c because it is a proper ideal
20:37 <@Kit> So cl(I) is included in G(A)^c
20:38 <@Kit> Thus cl(I) isn't the whole algebra, so it must be equal to I
20:38 <@Kit> (The first bit used that I is maximal of course)
20:38 <@Kit> So suppose we have some maximal ideal I
20:38 <@Kit> Then A / I is a field
20:38 <@Kit> And it's also a banach algebra
20:39 <@Kit> So by Gelfand-Mazur it's C
20:39 <@Kit> Thus the quotient map into A / I is a character f with ker(f) = I
20:39 <@Kit> So we have a correspondence between maximal ideals and characters.
20:39 <@Kit> This is another cool thing about banach algebras
20:39 <@Kit> In a general ring the fields that can arise this way are potentially rather complicated
20:40 <@Kit> So you need to understand lots of homomorphisms into fields, not just ones over the base field of the algebra
20:40 <@Kit> (Usual R, C example again)
20:40 <@Kit> This also gives us another way to characterise the spectrum.
20:40 <@Kit> Theorem: sp(a) = { f(a) : f a character }
20:41 <@Kit> We first show sp(a) <= { etc }
20:41 <@Kit> Suppose we have a character f
20:41 <@Kit> Then f( a - f(a) 1 ) = 0
20:41 <@Kit> So a - f(a) 1 \in ker(f)
20:41 <@Kit> Which is a proper ideal, so contains no invertible elements
20:41 <@Kit> So f(a) \in sp(a)
20:42 <@Kit> In the other direction, suppose z1 - a is not invertible
20:42 <@Kit> Then it is contained in some proper ideal, and so in some maximal ideal.
20:42 <@Kit> Say I
20:42 <@Kit> If we let ker(f) = I then f(z1 - a) = 0
20:42 <@Kit> So z = f(a)
20:42 <@Kit> This proves the equality.
20:43 <@Kit> We now introduce another key notion:
20:43 <@Kit> That of a semisimple banach algebra.
20:43 <@Kit> Let A be a banach algebra. We define the Jacobson radical, J(A) to be the intersection of all maximal ideals.
20:43 <@Kit> We now have several good ways of characterising this
20:44 <@Kit> x \in J(A) iff f(x) = 0 for every character
20:44 <@Kit> Because maximal ideals are kernels of characters
20:44 <@Kit> So x \in J(A) iff r(x) = 0, by the relation between the spectrum and characters
20:45 <@Kit> So x \in J(A) iff z1 + x is invertible for every z != 0
20:45 <@Kit> An algebra is semisimple if J(A) = 0
20:45 <@Kit> Example: C(X) is semisimple.
20:45 <@Kit> Proof: Let f \in C(X) be non-zero.
20:46 <@Kit> Then there exists z with f(z) != 0
20:46 <@Kit> The map \phi_z : g -> g(z) is a character on C(X) with \phi_z(f) != 0
20:46 <@Kit> Sorry. I should have given that as an example when I first introduced characters.
20:47 <@Kit> In fact all characters on C(X) are of this form. I leave this as an exercise for the dedicated reader. :)
20:47 <@Kit> I'm going to prove the following automatic continuity theorem:
20:47 <@Kit> Let A, B be banach algebras with B semisimple. Let f : A -> B be an algebra homomorphism. Then f is continuous.
20:48 <@Kit> This should be considered Cool. It's a purely algebraic statement, from which we are deducing a purely topological one.
20:48 <@Kit> (Modulo the bit of analysis in the definition of banach algebra)
20:48 <@Kit> For this I'm going to introduce something called the seperating subspace.
20:48 <@Kit> Definition:
20:48 <@Kit> Let A, B be banach spaces (not neccesarily algebras), let f : A -> B be linear.
20:48 <@Kit> Recall the closed graph theorem.
20:49 <@Kit> The graph of f is the subset of A (+) B defined by { (x, f(x)) : x \in A }
20:49 <@Kit> f is continuous iff the graph of f is closed.
20:49 <@Kit> Let G_f be the graph of f. f is continuous iff cl(G_f) = G_f
20:50 <@Kit> The seperating subspace of f is S_f = { y \in B : (0, y) \in cl(G_f) }
20:50 <@Kit> The closed graph theorem can be restated as saying that f is continuous iff S_f = {0}
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20:51 <@Kit> Because if (x, y) \in cl(G_f) then y - f(x)) \in S_f
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20:52 -!- mode/#mathematics [-m] by Kit
20:52 <@Kit> I'm going to show that if f is an algebra homomorphism between banach algebras then S_f <= J(B)
20:52 <@Kit> This will then prove the automatic continuity theorem.
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20:52 <@Kit> Because if B is semisimple then J(B) = 0 so S_f = 0
20:53 <@Kit> Ok. So let A, B, D be banach algebras and let f : A -> B be a homomorphism. Let g : B -> D be a *continuous* homomorphism.
20:53 <@Kit> If gf is continuous then we must have S_f <= ker g
20:54 <@Kit> Because G_{gf} = (id (+) g) G_f
20:54 <@Kit> (id (+) g is the direct sum of the two maps: identity on the first factor, g on the second)
20:55 <@Kit> So by continuity we have cl( G_{gf} ) >= (id (+) g) cl(G_f)
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20:55 <@Kit> So g(S_f) <= S_{gf}
20:55 <@Kit> Thus if S_f contains an element not in ker(g) then S_{gf} contains a non-zero element
20:56 -!- mode/#mathematics [+l 66] by Evariste
20:56 <@Kit> So gf is not continuous
20:56 <@Kit> Surprisingly, the result is now almost immediate:
20:56 <@Kit> Let f : A -> B, etc.
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20:56 <@Kit> And let g now be a character on B
20:56 < pkrumins> arghhh!! i missed everything!
20:56 <@Kit> Then gf is a character on A
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20:56 <@Kit> So it's continuous
20:57 <@Kit> Thus S_f <= ker(g)
20:57 <@Kit> And as we've shown, all the maximal ideals arise as ker(g)
20:57 <@Kit> So S_f is contained in every maximal ideal, and so in J(B)
20:57 <@Kit> Ta da
20:58 <@Kit> We have now proved the basic example of a non-trivial automatic continuity question. :)
20:58 < azta> so simple :)
20:58 <@Kit> These sorts of questions motivate a lot of modern banach algebra theory. There are some Big unsolved problems related to them.
20:58 <@Kit> Mostly in the noncommutative case.
20:58 <@Kit> But we don't have time for that today. :)
20:58 <@Kit> Any questions?
20:59 <@Kit> pkrumins: Sorry you missed it. Log is in the topic.
20:59 < azta> I have a million questions, but I think I will go over what you've said again, first :)
21:00 <@Kit> Hopefully that's because it's interesting rather than incomprehensible. :)
21:00 < azta> It was
21:00 < azta> Thanks a lot Kit
21:00 < pkrumins> ok, this is it?
21:01 <@Kit> Yep.
21:01 * dioid claps
21:01 < azta> :)
21:01 <@Kit> I got complaints about the length of my previous seminars, so I've kept this one short. :)
21:02 < azta> almost exactly 1 hour.
21:02 <@Galois> did someone mention that semisimple is equivalent to a direct sum of simple modules?
21:02 <@Galois> algebras, whatever
21:02 * vitriol claps
21:02 <@Kit> Galois: Hmm. Is that really true for a general ring?
21:02 <@Galois> depends on what ring means?
21:02 <@Kit> ok. Say any commutative ring with 1
21:03 <@Galois> think so
21:03 <@Kit> I thought there was some sort of finitness condition on it. I'm probably wrong though
21:03 < vitriol> the definition of semi-simple is that it's a sum of simple modules. in a ring this also implies that its jacobson radical is 0
21:03 < vitriol> and also in a ring it implies that its a finite sum of simple modules
21:03 <@Kit> Hmm.
21:03 < vitriol> so that might be the finiteness condition you're referring to
21:04 <@Kit> I meant more noetherian rings.
21:04 <@Galois> this theorem about A / ker(f) always equalling C is an analytic analogue of the nullstellensatz
21:04 <@Kit> Oh. I'm being silly. What I was thinking of isn't true.
21:04 <@Kit> e.g. C(X) is the sum of any two of its maximal ideals
21:05 <@Kit> So yeah, that makes more sense in this case than I thought it did. :)
21:05 < vitriol> fortunately it's true for all rings. also i think semi-simple is equivalent to artinian and J(R) = 0
21:05 < vitriol> for rings
21:06 < pkrumins> Kit, when's the next lecture?
21:06 <@Kit> pkrumins: I don't have any more planned at this point in time. :)
21:06 <@Kit> I think vitriol was going to do one after the exams
21:06 <@Kit> (Any comments vitriol? :) )
21:06 < vitriol> yeah hopefully if i have time :)
21:07 <@Kit> ok. So the definition of semisimple I gave isn't a proper algebraist's semisimple? Sorry about that. :)
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21:08 <@Kit> So have I managed to convince any of you algebraists that banach algebras are cool? :)
21:08 < vitriol> did you define semi-simple as J(R) = 0 ?
21:08 <@Kit> Yes
21:09 < vitriol> that's usually semi-primitive i believe, but if the ring is artinian i think it's ok
21:09 < azta> http://snipurl.com/f1zz
21:10 < vitriol> banach algebras do seem nice. some of it bore resemblance to analysis II example sheets
21:10 < azta> html with joins/parts/botmsgs removed
21:10 <@Kit> Banach algebras are very rarely artinian or noetherian
21:10 <@Kit> I think if X is infinite then C(X) is never noetherian or artinian
21:10 < vitriol> oh i see
21:11 * vitriol almost never deals with non-noetherian rings!
21:11 < vitriol> evil things
21:11 <@Kit> Actually, that's one of the reason banach algebras are cool. :)
21:11 <@Kit> `Banach' can be thought of as a finiteness condition
21:11 <@Kit> You can prove things about Banach algebras that are rather close to the things you can prove about noetherian algebras
21:12 <@Kit> Despite the fact that they're not noetherian
21:12 < vitriol> ah ok, that's quite interesting
21:12 <@Kit> I'm not quite confident enough on the noetherian side of things to support that claim. :)
21:12 <@Kit> But I have heard it claimed by people who know a good deal more about both than I do
21:12 <@Kit> (e.g. David Loeffler)
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21:18 <@Kit> Hmm. In retrospect I don't think this really needs notes.
21:18 <@Kit> Would anyone be heartbroken if I didn't provide them? :)
21:18 < azta> ;(
21:18 <@Kit> I'm planning to produce a big wodge of notes over the next year or so
21:18 <@Kit> Which will include notes on banach algebras
21:19 <@Kit> But I think most of today's talk stands fairly well on its own
21:20 < azta> It does
21:20 < vitriol> i think revision is more pressing :)
21:21 < gzl> should I upload the log?
21:21 <@Kit> vitriol: Indeed. :)