From: squark@my-deja.com Subject: Re: Classification of H*-algebras Date: Sun, 21 Jan 2001 17:43:57 GMT Newsgroups: sci.math.research In article <94a59u$h11$1@nntp.itservices.ubc.ca>, israel@math.ubc.ca (Robert Israel) wrote: > Sec. 4.10 of Rickart's book, "Banach Algebras", contains a complete > classification. The H*-algebra A is the direct sum of its annihilator > ideal Z = {u: ua = 0 for all a in A} and the orthogonal complement > of Z, which is a semi-simple H*-algebra. A semi-simple H*-algebra > is the orthogonal direct sum of its minimal-closed 2-sided ideals, > which are simple H*-algebras. A simple H*-algebra is *-isomorphic to > the algebra of Hilbert-Schmidt operators on a Hilbert space (Rickart > calls it a "full matrix algebra". Alright, I have what must be a stupid question then. Consider the *-algebra L^2(S^1) where S^1 is, of course, the circle. It is unital, commutative, and seems to be and H*-algebra by all means, where the inner product is the usual L^2(S^1) inner product. It has no annihilator ideal, and any full matrix algebra of a more than 1 dimensional Hilbert space is non-commutative. Thus, it must be a direct orthogonal sum of a lot of copies of the full matrix algebra of C, that is, C. This sum must be countable, as L^2(S^1) has countable dimension. This gives the algebra of square-summable complex sequences, with by place multiplication. The later, however, is not unital. Contradiction? Best regards, squark. Sent via Deja.com http://www.deja.com/ ============================================================================== From: Walter Kunhardt Subject: Re: Classification of H*-algebras Date: Sun, 21 Jan 2001 22:49:29 +0100 Newsgroups: sci.math.research On Sun, 21 Jan 2001 squark@my-deja.com wrote: > In article <94a59u$h11$1@nntp.itservices.ubc.ca>, > israel@math.ubc.ca (Robert Israel) wrote: > > Sec. 4.10 of Rickart's book, "Banach Algebras", contains a complete > > classification. The H*-algebra A is the direct sum of its annihilator > > ideal Z = {u: ua = 0 for all a in A} and the orthogonal complement > > of Z, which is a semi-simple H*-algebra. A semi-simple H*-algebra > > is the orthogonal direct sum of its minimal-closed 2-sided ideals, > > which are simple H*-algebras. A simple H*-algebra is *-isomorphic to > > the algebra of Hilbert-Schmidt operators on a Hilbert space (Rickart > > calls it a "full matrix algebra". > > Alright, I have what must be a stupid question then. Consider the > *-algebra L^2(S^1) where S^1 is, of course, the circle. It is unital, > commutative, and seems to be and H*-algebra by all means, where the > inner product is the usual L^2(S^1) inner product. Alas, it's not an algebra, since it's not closed under pointwise multiplication. Exercise: find a function f in L^2(S^1) such that f^2 is not in L^2(S^1) any more. > It has no > annihilator ideal, and any full matrix algebra of a more than 1 > dimensional Hilbert space is non-commutative. Thus, it must be a direct > orthogonal sum of a lot of copies of the full matrix algebra of C, that > is, C. This sum must be countable, as L^2(S^1) has countable dimension. > This gives the algebra of square-summable complex sequences, with by > place multiplication. The later, however, is not unital. Contradiction? > One more remark/question. If I understand correctly what R. Israel wrote, then no infinite-dimensional H*-algebra A is unital. My reasoning goes as follows: Either A contains the algebra of Hilbert-Schmidt operators on an infinite-dim. Hilbert space. Since the latter is not unital, A isn't unital either. Otherwise A is an infinite direct sum of finite-dimensional matrix algebras A_1 + A_2 + ..... (plus possibly the annihilator ideal). Then the only candidate for the unit in A would be the direct sum of the units in the summands A_j. But since there are infinitely many of them, that sum of units has infinte norm, i.e. it's not an element of A. Hence A is not unital. Is this correct? Regards, Walter Kunhardt ============================================================================== From: Walter Kunhardt Subject: Re: Classification of H*-algebras Date: Mon, 22 Jan 2001 12:36:54 +0100 Newsgroups: sci.math.research On Sun, 21 Jan 2001, I wrote: [quote of original post deleted --djr] What I wrote is correct as far as I can see, but nevertheless I should have added that L^(S^1) actually _is_ an H*-algebra. :-? The point is that the pointwise product of functions just isn't the right product here. More precisely, let's pretend L^(S^1) is actually L^2(T), where T = U(1) is the group of unit complex numbers. Then L^2(T) is _canonically_ isomorphic to the Hilbert space l^2(Z) , the unitary operator going from the first to the latter being the Fourier transform. (Of course, L^2(S^1) too is isomorphic to l^2(Z), but not in a natural way.) As you noticed, l^2(Z) is a (commutative, non-unital) H*-algebra when equipped with "placewise" multiplication. Now one can use the inverse Fourier transform to carry that product back to L^2(T), where one obtains the convolution of functions. Three remarks: - The convolution clearly exists (on the dense subspace of smooth functions, say) because T is a compact Lie group. - It should be possible to show that the convolution can be extended that dense subspace to all of L^2(T) and that it does not leave L^2(T). - One sees why L^2(T) is not unital: the candidate for a unit element would be the Dirac delta function sitting at the unit element of T, but that "function" is not square-integrable. Regards, Walter Kunhardt.