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    • CommentRowNumber1.
    • CommentAuthorMike Shulman
    • CommentTimeAug 13th 2012

    I created function application, so as to be able to link to it from fixed-point combinator. While adding links, I was motivated to expand a bit on function.

    • CommentRowNumber2.
    • CommentAuthorUrs
    • CommentTimeAug 13th 2012

    Thanks!!

    • CommentRowNumber3.
    • CommentAuthorzskoda
    • CommentTimeAug 14th 2012

    I see that entry evaluation map refers only to the notion in monoidal categroies. However the application of functions in some function space to a fixed argument is also an evaluation map in widely spread terminology.

    • CommentRowNumber4.
    • CommentAuthorMike Shulman
    • CommentTimeAug 14th 2012

    Yes, I agree. The monoidal-category version of picking a fixed argument would be picking a morphism IXI\to X, and then composing with the evaluation map [X,Y]XY[X,Y]\otimes X\to Y to obtain a map [X,Y]Y[X,Y]\to Y. Do you have examples in mind that don’t even live in monoidal categories in this way?

    • CommentRowNumber5.
    • CommentAuthorzskoda
    • CommentTimeAug 15th 2012
    • (edited Aug 15th 2012)

    Coherent state vectors are in correspondence with evaluation functionals. Take a holomorphic line bundle LL with Hermitean scalar product and the Hilbert space of holomorphic sections of it. Every point in the total space qq defines a functional. Namely look at the evaluation map ev p(q)ev_{p(q)} on the Hilbert space of sections at the projection p(q)p(q) of qq in the base space of the bundle. Divide qq by this value s(p(q))s(p(q)) what you can do because you are in a line bundle. The Riesz dual of this composition functional is the coherent state vector corresponding to qq.

    • CommentRowNumber6.
    • CommentAuthorTodd_Trimble
    • CommentTimeAug 15th 2012

    Without particularly wanting to discuss the question at #4, one also often wants to refer to evaluation of partially defined functions, which itself is a partially defined function. (My current interest in this will be evident shortly.)

    • CommentRowNumber7.
    • CommentAuthorMike Shulman
    • CommentTimeAug 15th 2012

    Okay, I didn’t understand very much of #5, but I gather that it is supposed to be a “yes” answer to #4. Everyone should feel free to add other possible meanings to evaluation map!

    • CommentRowNumber8.
    • CommentAuthorTodd_Trimble
    • CommentTimeAug 16th 2012

    Actually, I changed my mind: I’d like to discuss #5 for just a minute – I think it’s basically nothing more than a “restricted evaluation map” being called an “evaluation map”.

    I won’t pretend I followed #5 down to the last detail, but the general idea I got was this. Let p:EBp: E \to B be a holomorphic line bundle, and let Sect(p)Sect(p) be the space of holomorphic sections. (Zoran seemed to be going further and considering square-integrable holomorphic sections, presumably with respect to a measure on the base space, as forming a Hilbert space HSect(p)H \hookrightarrow Sect(p). Or maybe he had in mind a context for BB where every holomorphic section is automatically square-integrable, e.g., BB a compact space with finite measure, so that H=Sect(p)H = Sect(p).)

    Presumably one could think of Sect(p)Sect(p) as embedded in a suitable smooth function space E BE^B, e.g., if we were working in some convenient cartesian closed category of smooth spaces. I’ll pretend that’s the case. Let q:1Eq: 1 \to E be a point in the total space (assumed to be nonzero; see below), and consider the restricted evaluation map (where we evaluate at p(q)p(q)):

    Sect(p)E Beval p(q)E.Sect(p) \hookrightarrow E^B \stackrel{eval_{p(q)}}{\to} E.

    Because we are dealing with sections, this map factors through the fiber E p(q)EE_{p(q)} \hookrightarrow E, so we have a map Sect(p)E p(q)Sect(p) \to E_{p(q)}. In torsor-like fashion, we can “divide by (nonzero) qq” to get an explicit isomorphism E p(q)E_{p(q)} \cong \mathbb{C}. (So even though we evaluate at p(q)p(q), the specific qq is needed to specify the functional to \mathbb{C}. Incidentally, Zoran said to divide qq by s(p(q))s(p(q)), but is that really the order? because it can frequently happen that s(p(q))=0s(p(q)) = 0 for line bundle sections. Besides, we should want s(p(q))/qs(p(q))/q, not q/s(p(q))q/s(p(q)), if we want to get something linear in sections ss.)

    Anyway, we put all this together to get a functional HSect(p)H \hookrightarrow Sect(p) \to \mathbb{C} on the Hilbert space HH, which corresponds to some vector in the conjugate Hilbert space H *H^\ast, and this is basically the coherent state vector. (Incidentally, is a “state vector” really supposed to mean a line, i.e., a point in the projective space (H *)\mathbb{P}(H^\ast)? If so, we shouldn’t really care which nonzero qE p(q)q \in E_{p(q)} we use.)

    If this is the right idea, then we’re basically dealing with a restricted evaluation map, which I guess most mathematicians would be happy to call simply an evaluation map!

    • CommentRowNumber9.
    • CommentAuthorUrs
    • CommentTimeAug 16th 2012
    • (edited Aug 16th 2012)

    (Incidentally, is a “state vector” really supposed to mean a line, i.e., a point in the projective space (H *)\mathbb{P}(H ^\ast)?

    That’s the case as long as we are interested only in the value of the quantum observables in that single state: in this case the phase of the state drops out. But as soon as one is interested in forming the “superposition” of this state with another state, the phase does crucially matter, hence in that case we really must consider it as an element in the Hilbert space itself, instead of in its projectivization.

    • CommentRowNumber10.
    • CommentAuthorTodd_Trimble
    • CommentTimeAug 16th 2012

    @Urs #9: yes, thanks. I’m happy to suppose that people in the context of #5 would be concerned with superposing states.

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