I'm reading into color theory and there were a few questions which I asked myself along the way, maybe you can put me forward to some source where I can find answers or give them directly.

The following paragraph contains some context, I'll try to summarize everything in purely mathematical terms below.

The human eye has three main types of color reception cells (one for reddish light, one for greenish light and one for blueish light). So every source of light (monochromatic, i.e. single wavelength, or not) induces a state in those cells, which can be described as a three tuple of non-negative, real numbers and it is assumed that this transition works semi-linearly, i.e. adding two sources of light induces the sum of the states induced by the single sources of light and same with multiplying by non-negative scalars. I call this transition $T:S\rightarrow{\bf R}^3_{\ge 0}$ where $S$ is the space of spectral power distributions. In the 1930s, as far as I understood, they didn't know exactly how $T$ looked like, so in order to build up a color space in accordance with human perception, they fixed three monochromatic sources of reg, green and blue light (they chose those wavelengths which were most easy to reproduce physically) and let a user mix it together in order to match the color of an arbitrary given monochromatic source of light. This was not always possible, so they allowed the user to choose negative values for the primaries, which meant that instead of adding the wavelength to the mixture, it was added to the given source. They succeeded to match every wavelength in the visible spectrum and took those values as a base of a first color space. Again you get a semi-linear transition $T':S\rightarrow{\bf R}^3$ but this time non-negative values are included, yet it's not surjective on ${\bf R}^3$. What they did then was to choose a linear isomorphism which takes $T'(S)$ to ${\bf R}^3_{\ge 0}$, so they can describe colors using non-negative values only.

So in pure terms, there is the semi-linear space $S$ of spectral power distributions (which one can safely think as ${\bf R}^n_{\ge 0}$ for some large $n$) and a semi-linear transformation $T:S\rightarrow{\bf R}^3_{\ge 0}$ which is in general not known. Also there are 'wavelengths' $p_1,p_2,p_3\in S$ and a semi-linear transformation $T':S\rightarrow{\bf R}^3$ such that for every $s\in S$ we have, denoting ${\sf max}(x,0)$ by $x^+$ and $(-x)^+$ by $x^-$, the following: \begin{equation*}\tag{*}T(\sum_i T'(s)^+_i p_i)=T(s+\sum_i T'(s)^-_i p_i)\end{equation*}

My questions related to this:

  1. Can we derive $T$ (or some of its properties) from $T'$ and (*)?

  2. Assuming we know $T$, are there other triples of wavelengths which make this procedure possible, and if yes,

  3. What are their properties purely in terms of $T$?

  4. How does $T'(S)$ look like (geometrically) and

  5. How should it look like so we can find a linear isomorphism taking it to ${\bf R}^3_{\ge 0}$?

My thoughts so far:

(1) Here it helps to think the codomain of $T$ as whole ${\bf R}^3$, then one can easily derive $T(s)=\sum_i T'(s)_i T(p_i)$ or equivalently $T=M\circ T'$ where $M$ is the $3\times 3$-matrix $(T(p_1),T(p_2),T(p_3))$. If $T(S)$ is $3$-dimensional, i.e. the set is not contained in a $2$-dimensional subspace (which I assume), then $M$ is full rank, hence $T'$ is indeed semi-linear and we have that vice versa every full rank matrix $M$ with $M(T'(S))\subseteq{\bf R}^3_{\ge 0}$ produces a permissible $T$ which satisfies (*) via $T=M\circ T'$.

(2), (3) Constructing $T'$ should be possible for every triplet $p_1,p_2,p_3$ such that the $T(p_i)$ are linearly independent.

(4) I realized that the semi-linearly closed subsets of ${\bf R}^n$ equal exactly the convex subsets which are closed under non-negative scalar multiplication. For semi-linearly closed subsets of ${\bf R}^n_{\ge 0}$ these correspond exactly with the convex subsets of the $(n-1)$-simplex (via projection along the semi-rays through the origin).

(5) My intuition is that for a semi-linearly closed and topologically closed subset $A\subseteq{\bf R}^n$, there exists a linear isomorphism taking it to ${\bf R}^n_{\ge 0}$ if and only if and only if there is no point $x\neq 0$ with $x,-x\in A$.

If someone can confirm my thoughts, they are encouraged to put them in an answer!

  • $\begingroup$ Why are there three votes on close? Too many questions? Should I strip away some context and ask the questions one by one in purely linear algebra related terms? $\endgroup$
    – fweth
    Jul 23, 2016 at 17:35
  • $\begingroup$ I think this is a very interesting question. What references on the subject would you recommend? $\endgroup$ Jul 23, 2016 at 17:37
  • $\begingroup$ Thank you! I was reading mainly those two documents. $\endgroup$
    – fweth
    Jul 23, 2016 at 17:44
  • $\begingroup$ @fweth two persons have voted to close because they think what you're asking is unclear $\endgroup$ Jul 23, 2016 at 17:46
  • $\begingroup$ It is unclear that deriving $T$ from $T'$ has a specific mathematical meaning. While the context is interesting and (in my opinion) helpful, the Question would be easier to respond to if the requirements on $T$ were framed separately from the application. This might be simply a matter of reformatting and reorganizing your existing text. $\endgroup$
    – hardmath
    Jul 23, 2016 at 18:05


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