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1. • CommentRowNumber1.
   • CommentAuthorMike Shulman
   • CommentTimeAug 13th 2012
   • PermaLink
   Author: Mike Shulman
   Format: MarkdownItexI 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]].

   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.

2. • CommentRowNumber2.
   • CommentAuthorUrs
   • CommentTimeAug 13th 2012
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   Author: Urs
   Format: MarkdownItexThanks!!

   Thanks!!

3. • CommentRowNumber3.
   • CommentAuthorzskoda
   • CommentTimeAug 14th 2012
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   Author: zskoda
   Format: MarkdownItexI 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.

   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.

4. • CommentRowNumber4.
   • CommentAuthorMike Shulman
   • CommentTimeAug 14th 2012
   • PermaLink
   Author: Mike Shulman
   Format: MarkdownItexYes, I agree. The monoidal-category version of picking a fixed argument would be picking a morphism $I\to X$, and then composing with the evaluation map $[X,Y]\otimes X\to Y$ to obtain a map $[X,Y]\to Y$. Do you have examples in mind that don't even live in monoidal categories in this way?

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

5. • CommentRowNumber5.
   • CommentAuthorzskoda
   • CommentTimeAug 15th 2012
   • (edited Aug 15th 2012)
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   Author: zskoda
   Format: MarkdownItexCoherent state vectors are in correspondence with evaluation functionals. Take a holomorphic line bundle $L$ with Hermitean scalar product and the Hilbert space of holomorphic sections of it. Every point in the total space $q$ defines a functional. Namely look at the evaluation map $ev_{p(q)}$ on the Hilbert space of sections at the projection $p(q)$ of $q$ in the base space of the bundle. Divide $q$ by this value $s(p(q))$ 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 $q$.

   Coherent state vectors are in correspondence with evaluation functionals. Take a holomorphic line bundle $L$ with Hermitean scalar product and the Hilbert space of holomorphic sections of it. Every point in the total space $q$ defines a functional. Namely look at the evaluation map $ev_{p(q)}$ on the Hilbert space of sections at the projection $p(q)$ of $q$ in the base space of the bundle. Divide $q$ by this value $s(p(q))$ 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 $q$.

6. • CommentRowNumber6.
   • CommentAuthorTodd_Trimble
   • CommentTimeAug 15th 2012
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   Author: Todd_Trimble
   Format: MarkdownItexWithout 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.)

   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.)

7. • CommentRowNumber7.
   • CommentAuthorMike Shulman
   • CommentTimeAug 15th 2012
   • PermaLink
   Author: Mike Shulman
   Format: MarkdownItexOkay, 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]]!

   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!

8. • CommentRowNumber8.
   • CommentAuthorTodd_Trimble
   • CommentTimeAug 16th 2012
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   Author: Todd_Trimble
   Format: MarkdownItexActually, 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 $p: E \to B$ be a holomorphic line bundle, and let $Sect(p)$ 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 $H \hookrightarrow Sect(p)$. Or maybe he had in mind a context for $B$ where every holomorphic section is automatically square-integrable, e.g., $B$ a compact space with finite measure, so that $H = Sect(p)$.) Presumably one could think of $Sect(p)$ as embedded in a suitable smooth function space $E^B$, e.g., if we were working in some convenient cartesian closed category of smooth spaces. I'll pretend that's the case. Let $q: 1 \to E$ be a point in the total space (assumed to be nonzero; see below), and consider the restricted evaluation map (where we evaluate at $p(q)$): $$Sect(p) \hookrightarrow E^B \stackrel{eval_{p(q)}}{\to} E.$$ Because we are dealing with sections, this map factors through the fiber $E_{p(q)} \hookrightarrow E$, so we have a map $Sect(p) \to E_{p(q)}$. In torsor-like fashion, we can "divide by (nonzero) $q$" to get an explicit isomorphism $E_{p(q)} \cong \mathbb{C}$. (So even though we evaluate at $p(q)$, the specific $q$ is needed to specify the functional to $\mathbb{C}$. Incidentally, Zoran said to divide $q$ by $s(p(q))$, but is that really the order? because it can frequently happen that $s(p(q)) = 0$ for line bundle sections. Besides, we should want $s(p(q))/q$, not $q/s(p(q))$, if we want to get something linear in sections $s$.) Anyway, we put all this together to get a functional $H \hookrightarrow Sect(p) \to \mathbb{C}$ on the Hilbert space $H$, which corresponds to some vector in the conjugate Hilbert space $H^\ast$, 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 $\mathbb{P}(H^\ast)$? If so, we shouldn't really care which nonzero $q \in E_{p(q)}$ 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!

   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 $p: E \to B$ be a holomorphic line bundle, and let $Sect(p)$ 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 $H \hookrightarrow Sect(p)$. Or maybe he had in mind a context for $B$ where every holomorphic section is automatically square-integrable, e.g., $B$ a compact space with finite measure, so that $H = Sect(p)$.)

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

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

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

   Anyway, we put all this together to get a functional $H \hookrightarrow Sect(p) \to \mathbb{C}$ on the Hilbert space $H$, which corresponds to some vector in the conjugate Hilbert space $H^\ast$, 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 $\mathbb{P}(H^\ast)$? If so, we shouldn’t really care which nonzero $q \in E_{p(q)}$ 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!

9. • CommentRowNumber9.
   • CommentAuthorUrs
   • CommentTimeAug 16th 2012
   • (edited Aug 16th 2012)
   • PermaLink
   Author: Urs
   Format: MarkdownItex> (Incidentally, is a "state vector" really supposed to mean a line, i.e., a point in the projective space $\mathbb{P}(H ^\ast)$? 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.

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

   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.

10. • CommentRowNumber10.
    • CommentAuthorTodd_Trimble
    • CommentTimeAug 16th 2012
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    Author: Todd_Trimble
    Format: MarkdownItex@Urs #9: yes, thanks. I'm happy to suppose that people in the context of #5 would be concerned with superposing states.

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