Emil, isn't that point of whole exercise that legal colors within a gamut get mapped to out of gamut in another system. Please think about it.
Joofa
There are several points:
1) A color space A is defined wrt a particular illuminant I. The gamut of allowed colors in that color space (call it G(A,I)) and implicitly referred to that illuminant, is a paralellepiped in XYZ space, three of whose corners are the R,G,B primaries for that color space.
2) Chromatic adaptation maps the parallelepiped of colors in the gamut defined for the reference illuminant, to another parallelepiped G(A',I') of colors referred to another illuminant I'. There are adapted primaries R',G',B'.
3) The adaptation map can be such that the XYZ coordinates of eg the blue primary of A lie outside the parallelepiped G(A',I'). We note this fact and continue.
4) The color 'white' does not have unique XYZ coordinates; it depends on the illuminant, W=W(I). Part of the map G(A,I)->G(A',I') is W(I)->W(I'). W(I) can also lie outside G(A',I').
5) A second color space B' might be defined wrt a different illuminant I', and determine a gamut parallelepiped G(B',I').
6) If we are to discuss a given image being formed by a particular illuminant I", each pixel should have a set of X,Y,Z coordinates determined by the spectral power distribution (SPD) of the illuminant and the surface reflectivities of the objects in the scene.
7) Then to compare the recording of that image in different color spaces A and B', we should be looking at the coordinatization of the colors represented by X,Y,Z as linear combinations of the respective primaries adapted to I" and which may or may not lie within G(A",I") and/or G(B",I"). One can if one wishes then transform the XYZ values to ones determined in the reference color spaces G(A,I) and/or G(B',I') by inverting the chromatic adaptation transform.
8 ) If a viewer wishes to render the image, to be viewed under yet another illuminant I''', then the XYZ values should be mapped by chromatic adaptation to that illuminant. I'''=I and I'''=I' are special cases.
9) There is no point in the above discussion of points (6-8) where it is necessary to discuss either point (3), or whether eg the blue primary of G(A,I) lies outside G(B',I'). The only issue of possible relevance is whether it lies outside of G(B,I). In the case of AdobeRGB and Prophoto, the answer is no, the blue primary of Adobe, properly adapted to the illuminant, lies within Prophoto. This is why I commented that it was a purely theoretical question, without practical import.
At least, this is my current understanding. I am happy to be further illuminated
by the experts.
