Talk:Trace class

message: Problem here with displayed formula syntax.  Charles Matthews (Now fixed)
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I believe the previous definition of trace class (in terms of A rather than |A|) was wrong; (gimme a week and I'll think of a counterexample)CSTAR 17:03, 23 Jun 2004 (UTC)

message: I just wrote nuclear operator when I found this article. (I had been looking for "trace-class", with the dash, and when I did not find it, I assumed there was no article :(  I was vaguely contemplating expanding the other article to deal more with Banach spaces, where trace-class can be defined, but with only with some care. linas 19:08, 18 September 2005 (UTC)
replies: text: I don't think they should be merged. There is a lot more relevant material that doesn't generalize.  For example Schatten p-classes (or Hilbert Schmidt operators in case p=2). Also Dixmier trace would not fit into any category related to nuclear operators.--CSTAR 20:11, 18 September 2005 (UTC)
replies: text: OK, Works for me, I'll remove the merge tag. linas 03:28, 19 September 2005 (UTC)
thread_title:  merge with [[Nuclear operator]] 

message: In finite dimensions, the trace of a linear map exists for any vector space, not just for Hilbert spaces.  It is defined in terms of the natural transformation from V*⊗W to L(V,W), which is an isomorphism for finite dim.  In infinite dimensional spaces there is no isomorphism, but there is still a natural injection.  I wonder if it makes sense to define the trace class operators as the image of this injection.  This has the advantage that it doesn't rely on the use of an inner product, nor any other topological features.  It makes sense for any vector space.  Another idea is to define the trace in terms of the Aut(V) invariant linear form, though this seems harder to me.  -lethe talk [ +] 01:57, 7 March 2006 (UTC)
replies: text: Hmm I can't quite tell what you are trying to say here; but it sounds like you are describing nuclear operator which is the same concept, but without the inner product. linas 14:49, 7 March 2006 (UTC)
replies: text: I don't know about nuclear operators, but I'm reading the article we have on it, and yes, it does seem similar.  So let's see, my definition is this: T:V→W is trace class if there exist w∈W and f∈V* such that T=f⊗w or T(v)=f(v)w, and if V=W, then take trT = f(w).  (OK, actually, not all elements of V*⊗W can be written in the form f⊗w, but the universal property of tensor products takes care of us here).  Now, I guess the definition of nuclear operator looks similar, but is just more explicit.  It looks like it decomposes a linear transformation into eigenspaces, making explicit use of an eigenbasis.  Lemme think about whether I can show my definition includes the nuclear operator def. -lethe talk [ +] 18:16, 7 March 2006 (UTC)

text:  Isn't what you are describing the class of finite rank operators V → W? Nuclear operators include these but not the other way around.--CSTAR 18:30, 7 March 2006 (UTC)
replies: text: So like, the shift operator T(en)=en+1 is not finite rank, but it can be written as ∑σn⊗en+1 which is in V*⊗W.  Or is it?  Maybe I have to take the completion of V*⊗W? -lethe talk [ +] 18:41, 7 March 2006 (UTC)
replies: text:  You can't use the algebraic tensor product (without completion).  That's why Grothendieck in his famous paper in the early 1950s introduced topological tensor products to define operators which were like "integral" operators with a kernel (noyau) An operator with a noyau is "nucléaire".--CSTAR 18:49, 7 March 2006 (UTC)
replies: text: Yes, now I've followed the trail trace class→ nuclear operator→ Fredholm kernel→ topological tensor product→nuclear space.  Thanks for the pointers.  Keep an eye on their talk pages for more! -lethe talk [ +] 19:01, 7 March 2006 (UTC)
replies: text: Also Schwartz kernel (aha still missing), Mercer's theorem--CSTAR 04:30, 8 March 2006 (UTC)

text: At the risk of being off topic here: I think writing the matrix element M_kk as < M e_k, e_k > is quite unnatural. (Btw, < .,. > looks to me more like a duality bracket (cf "bra-ket") than like an inner product! (Admittedly, the bracket notation is also used in that article, not without protests on the talk page...)) It becomes clear in the next section why this is done, but I would prefer putting this complicated writing only there (like, "Mimicking the matrix case, where Mkk could be written <M ek, ek>, we now...") — MFH:Talk 22:34, 4 September 2007 (UTC)
thread_title:  trace class without inner product 

message: Can somebody explain me what I get wrong in the following? Let's fix an orthonormal basis e_n in H (n=1,2,...). Let A_n be an operator which sends e_i to 0 for i=1,2,...,n-1,2n,2n+1,... and is identity for n other e_i. 
Now, sequence A_n
*consists of trace class operators
*converges to the 0 operator in the weak operator topology
*Tr(A_n) diverges to infinity
while the article claims "It [Trace] is continuous in the weak operator topology". Where lies the problem?
Sirix 22:13, 30 September 2007 (UTC)
replies: text: you seem to be right. although the article doesn't say this directly, it certainly implies it. i think the mistake was mine. thanks for catching it. Mct mht 23:16, 30 September 2007 (UTC)
thread_title: Is trace really continuous in the [[weak operator topology]]?

message: In the sentence
: It is continuous in the weak operator topology and satisfies properties 1 and 2 above.
I can't figure out what properties 1 and 2 are supposed to be. -- Walt Pohl (talk) 05:11, 14 February 2008 (UTC)
replies: text:  Neither can I. They were in some previous version of the article. Too many editors.--CSTAR (talk) 06:22, 14 February 2008 (UTC)
thread_title:  Antecedent 

message: The article employs (defines!) a compound notation "Tr |⋅|" before defining Tr or |⋅|. In fact, the absolute value of a bounded operator is never defined or linked in the article. Is it just sqrt(A* A)? A link would be great.
In any case, wouldn't it make more sense to separate the two equalities
:\|A\|_1= \sum_k \langle (A^*A)^{1/2} \, e_k, e_k \rangle 
and
:\|A\|_1= {\rm Tr}|A| 
using the first one to perform the actual definition of trace-class, followed by the definition of Tr wherever valid, and adding the second one afterwards as a remark?
Or the other way around, define Tr for nonnegative self-adjoint operators first, and use it to define \|\cdot\|_1
and trace-class.
The difficulty is that Tr for trace-class operators and Tr for nonnegative self-adjoint operators have roughly co-equal status. Neither is really logically prior, and the domains of definition are not nested; rather, they agree on the intersection of their domains of definition. It's just like taking integrals of integrable functions versus nonnegative measurable functions.
178.38.85.195 (talk) 19:52, 13 April 2015 (UTC)
thread_title:  Tr |A|