What is color?

[arfa brane]

I quoted your links, above, specifically doing exactly that. You posted them as examples of analysts setting up vector spaces etc, and I noted that they were doing that with wavelengths, not colors.

A wavelength isn't a vector, sorry. If it was I could add a pair of wavelengths and get a third wavelength. Besides, a wavelength with no defined amplitude isn't meaningful, at least not in electronics.

Once again, the vector space of colors is a space of functions of wavelength, the functions can be added together to give a third function. If all you have is wavelengths, you don't have a vector space.

 

[arfa brane]

Why do you call those colors "errors"?

I don't, I call the discrimination of the two shades of grey, an "error", except it isn't, it's a normal response to the perception of light and dark regions in proximity.

 

[arfa brane]

Once again, iceaura wants to believe that a set of wavelengths is a vector space (with appropriate addition and scalar multiplication). He hasn't done anything except claim (with zero evidence so far) that wavelengths are a vector space, and that all the links I've posted so far say this (but they don't, they say something he seems to be incapable of reading).

I'm still waiting for someone (anyone) to post a definition of wavelength vectors. Maybe I should hold my breath. But, with such a definition I should be able to add together two wavelengths like 233nm and 341nm, so I have a wavelength of 574nm. Right? And then I should be able to scale a wavelength, with . . . scalar multiplication.

Have at it.

 

[arfa brane]

In post 120, there's a quote from a google online book (you can find it here); here's some more from that text,

clearly the author is defining families of spectral functions, over scalar wavelengths. The function space is the color vector space; there is the space of physical stimuli and the space of responses to the stimuli.

Stimulus-response models don't actually need to include human vision, there are plenty of devices which are much more accurate at discrimination than humans are. However, it makes sense to have human vision in some color models. ya think?

 

[iceaura]

Once again, iceaura wants to believe that a set of wavelengths is a vector space (with appropriate addition and scalar multiplication).

No. Please read my posts - I have multiple quotes and links in them, including your links, posted for the explicit purpose of illustrating how the vector spaces and distance functions you require were constructed by the people you claim (correctly) constructed them.

All of them, every single one of them, were constructed from wavelengths in the visible EMR spectrum.

He hasn't done anything except claim (with zero evidence so far) that wavelengths are a vector space,

I have never made that silly claim. Nobody has - wavelengths measure physical entities.
Vector spaces and distance functions normally are constructed, not found; defined, not discovered.

If all you have is wavelengths, you don't have a vector space

But you can construct one. If all you have is colors, you can't do that.

My claim is that it is impossible to construct a Hilbert space over the "color space" of human vision, it is possible to construct one over a "wavelength space" derived from the EMR spectrum, and that the engineers and analysts you linked who constructed the vector spaces you insisted on including here used EMR wavelengths, not colors.

I'm still waiting for someone (anyone) to post a definition of wavelength vectors.

You posted links to the vector space construction yourself.
You even posted a CIE diagram showing a subspace of it: http://www.sciforums.com/threads/what-is-color.162590/page-5#post-3611104

Once again, the vector space of colors is a space of functions of wavelength

You noticed!
And defining colors that way - as wavelengths and combinations of wavelengths - is how they all manage to construct the vector spaces they need, every single one of them.
Because they can't construct a vector space from the colors themselves. They can from the wavelengths of the EMR spectrum. The wavelength approach sacrifices a good share of the human visual "color space", but it's worth it to someone trying to get a computer to display a color image. It even supports a distance function (the EMR spectral wavelengths are ordered, and fixed in singular relationship to one another) - you can get a Hilbert space, if you want one.

 

[arfa brane]

And defining colors that way - as wavelengths and combinations of wavelengths - is how they all manage to construct the vector spaces they need, every single one of them.

But "they" don't define colors as wavelengths or combinations of wavelengths. That's ridiculous. You can't read very well, I assume.
The vector spaces are spaces of functions of wavelengths. The wavelengths of visible light do not form a vector space. Wavelengths of anything don't form a vector space.

This is something you can figure out by looking at a CIE diagram--the space of wavelengths is the curly line, with one dimension so any wavelength corresponds to a point. I have never seen a vector space defined as a continuous set of points (notice how we perceive a finite set of colors in a spectrum).

The basis vectors in the diagram are not wavelengths, they are the red and green LEDS in my display, or your display, or in the ink in a printout of the diagram, etc.

Your claim that colors can't be used as a vector basis is just wrong. I can see this is wrong by looking at a CIE diagram, for one, but there are any number of examples where color is a vector basis. Clearly the CIE diagram is embedded in $$\mathbb R^2 $$, clearly the x and y axes are colors (not labeled with colors), clearly the amplitude of the red basis vector increases left to right.

 

[iceaura]

But "they" don't define colors as wavelengths or combinations of wavelengths. That's ridiculous

They most certainly and explicitly do. It's what you linked to, what you quoted, what I linked to, and what I quoted.
If you don't like the term "combinations", you can always use "functions" - I have no objection. This isn't a math class.

The basis vectors in the diagram are not wavelengths,

Neither are they colors. They are not perceived or measured physical entities at all - they are mathematical abstractions.
The construction of a vector space over the wavelengths of the EMR spectrum is described in your links, as is the reasoning behind assigning particular color names to points or regions of it - regardless of how the corresponding physical entities (wavelengths, or photons if you insist) are perceived in various circumstances. (The variety of perceived colors in the optical "illusions" posted above all have exactly the same color name assigned to them in the RGB system for example. That's because they are generated by exactly the same combination of wavelengths, which defines their color).

Your claim that colors can't be used as a vector basis is just wrong.

Am I mistaken in my observation that perceived colors are neither uniquely fixed in relation to each other nor ordered ?

but there are any number of examples where color is a vector basis.

We would need only one, to make the point.

The basis vectors in the diagram are not wavelengths, they are the red and green LEDS in my display, or your display, or in the ink in a printout of the diagram, etc.

LEDs are not vectors, inks are not vectors, etc.

btw: The basis vectors of the vector spaces constructed over "ink space", "oil paint space", etc, and used to manage color printing or painting or dye mixing and so forth - if any - are quite different from the ones used to manage the generation of colors by emitted light from screens and the like. The vector spaces involved often have four or more dimensions, rather than the three of most emitted light system, for starters. I linked to an example, from book printing, above.

That may be relevant if this discussion ever returns to the OP topic.

 

[arfa brane]

A color is a spectral density or a function of wavelength. According to one author the space P of physical stimuli in the visible range is a semigroup (under addition). He embeds this in a larger space, initially a group (under addition), then defines scalar multiplication to define a vector space in which P is a subspace.

He defines equivalence classes on P, where two stimuli with different spectral densities are perceived to be the same color. This equivalence relation is transitive, for one. Clearly the vector space is defined over a field of wavelengths.

If, in the CIE diagram you can't see that the color red increases in intensity (smoothly or perceivably so), I don't know what to do about it. But the color red, not a mathematical abstraction but an actual physical stimulus, is the basis along the x axis, likewise the color green is the basis along the y axis.

I'm now officially sick to death of this thread. If you don't understand what those links (or any publications about color science) are telling you about colors, I'm not interested. You keep telling yourself whatever it is you keep telling yourself.

Bye

 

[exchemist]

 

[arfa brane]

Am I mistaken in my observation that perceived colors are neither uniquely fixed in relation to each other nor ordered ?

Am I mistaken in my observation, with a magnifying eyepiece, that colors are uniquely fixed in relation to each other, and are ordered in a LED based screen?
In electronics, voltages are in a vector space. Voltages are what make the LEDs "light up". The emitted light you see is generated by voltages.

The design and construction of the display I'm looking has been determined by (a model of) human visual perception and an 'ideal' observer.

What a show, huh? How good's your model?

 

[QuarkHead]

You guys seem to be chasing wild geese.

There is no requirement for a vector space to be ordered. Some are, of course, but that is purely accidental.

Neither is there any requirement that the basis be unique - any basis will do so long as its elements are independent.

The problem comes with the identity, the inverse and the scaling

 

[arfa brane]

To quote one of the online sources I have access to:

Theorem 2.2
The semigroup $$\mathcal P$$ of physical light rays can be embedded in a unique, minimal commutative group $$\mathcal A$$ such that every element of $$\mathcal P$$ has an inverse in $$\mathcal A$$.

[in the case of your display screen] The field of scalars is the field of (discrete) voltages, which determine the intensity (irradiance or what have you, something to do with Watts and an area, meh).

Because the relation between the forward bias voltage (we need to exclude negative bias) and the intensity is not linear, there's a display controller making all the adjustments, so we can perceive say, a linear increase in intensity of some color from the set of colors (also necessarily finite).

So how hard is it to show that a set of physical light rays (where negative rays aren't physically meaningful), is a semigroup under some binary operation? Can I fix a pair of real numbers, call them wavelengths, and show that neither is an affine combination of the other? That just doesn't sound like a hard exam question.

And lastly, LEDs are not vectors, except when the light they emit has a fixed color (which is why they're called light emitting diodes, as I suppose you might call an induction coil a magnetic emitting . . . coil). You don't need an observer for this to be electronically true. So the vector is the emitted color with a voltage dependent intensity (i.e. magnitude), and a fixed wavelength.

The wavelength becomes (abstractly) an angle, when you give the vector space a Euclidean metric. That is, a color-difference metric, abstractly this is embedded in human color discrimination, the space of optical illusions.

Oh shit, why did I just post all that at a site where nobody studies a damn thing?

 

[iceaura]

Am I mistaken in my observation, with a magnifying eyepiece, that colors are uniquely fixed in relation to each other, and are ordered in a LED based screen?

Yes.
You are mistaking an imposed order of wavelengths, mappable to colors only by excluding much if not most actual color perception, for an ordering of the colors themselves.

A color is a spectral density or a function of wavelength.

No, it isn't. I posted counterexamples, above.

But the color red, not a mathematical abstraction but an actual physical stimulus, is the basis along the x axis, likewise the color green is the basis along the y axis.

Colors are not physical stimuli. Colors are mental representations of physical stimuli (second order - representations of representations).

So how hard is it to show that a set of physical light rays (where negative rays aren't physically meaningful), is a semigroup under some binary operation?

Not hard at all. We agreed on that many pages ago.

The wavelength becomes (abstractly) an angle, when you give the vector space a Euclidean metric. That is, a color-difference metric, abstractly this is embedded in human color discrimination,

You cannot construct a Euclidean color-difference metric over the space of human-perceived colors.
- - - - -

There is no requirement for a vector space to be ordered.

There is a requirement that it support addition and scalar (field) multiplication , and that these operations be associative and commutative and appropriately distributive etc.

Also, a Hilbert space must support a distance function.

 

[arfa brane]

You cannot construct a Euclidean color-difference metric over the space of human-perceived colors.

Yes you can. Choose a pair of colors and define an inner product on them which is zero. It doesn't matter what the colors are (as long as they are distinct) or what their order is.
In other words, first define a pair of linearly independent colors as a basis.

 

[iceaura]

Choose a pair of colors and define an inner product on them which is zero.

No inner product can be defined over the space of human color perception.

In other words, first define a pair of linearly independent colors as a basis

You can't.
For counterexamples to any attempts, see various posts and links above.

There is a reason all your links and engineering approaches are based on wavelengths, rather than the colors themselves.

 

[arfa brane]

No inner product can be defined over the space of human color perception.

So what would you say the CIE has managed to define? Do you mean the idea of a standard color basis isn't a rational one? It's all some kind of illusion?

There is a reason all your links and engineering approaches are based on wavelengths, rather than the colors themselves.

And there's a reason that engineers map wavelengths to colors, and vice-versa. Do you have any idea why there's a map? Could it possibly be useful or are the engineers misguided--building a color display is hit and miss (because, as you insist, colors can't be embedded in a Euclidean space, although nobody knows why or why anyone might want to do this).

What does a wavelength have to do with say, yellow colored ink in a printer?

My final question, and I would really like someone to explain their answer here, is why do I get > 6 million hits when I google "euclidean color space"? You understand that a Euclidean space has an inner product by definition? Perhaps a lot of the links are repeated, perhaps they're all misleading and ill-informed? Perhaps engineers are pulling the fast one?

Please explain.

 

[iceaura]

A wavelength isn't a vector, sorry.

Nobody said it was.

So what would you say the CIE has managed to define?

An inner product over a restricted set of wavelengths.

And there's a reason that engineers map wavelengths to colors, and vice-versa.

Yep.
Because they can handle wavelengths much more easily than colors. There are no optical wavelength illusions, for example. They can even construct a vector space over a suitably restricted and organized set of wavelengths, a great convenience in digital processing and manipulation, which is impossible with colors.

What does a wavelength have to do with say, yellow colored ink in a printer?

It's the basis for how the color "yellow" is represented in the printer's software - software which among other things determines how, when, and where the printer puts the appropriate ink on the paper.

Please explain.

Once engineers have mapped colors to wavelengths the use of the color name is an easy and convenient shorthand. It only misleads people who have lost track of the mapping - say, the kinds of people who mistake a line on a topographical map for an actual height, and become confused when informed that the map is in fact flat.

 

[arfa brane]

Yep.
Because they can handle wavelengths much more easily than colors.

In the sRGB color standard, there are only three wavelengths. Any color is a combination of three intensity values, where the intensity of each color is a number from 0 to 255 (note, they don't mention the intensity of a wavelength). So each tricolor pixel has a triple of numbers; this means you can embed every triple in Euclidean 3-space. Euclidean spaces are vector spaces.

 

[James R]

arfa brane:

Surely, after all this time, you're not still confused about this?

The word "color" in things like "sRGB color standard" is not "color" in the sense that iceaura has been using that word. It is an assignment of color names to a particular combination of wavelengths.

If you take one of your sRGB "colors", then under different viewing conditions, different people will potentially perceive different colors.

What this means is that your sRGB system does not reliably produce colors. It does reliably produce light that is a specified combination of wavelengths. You can represent particular combinations of wavelengths as vectors in an RGB vector space if you like, but you're not doing anything with colors when you do that.

 

[Neddy Bate]

For color models like RGB, the intensity values can only be zero or positive, and they have a maximum value of essentially 100%. If we make a 3 dimensional Cartesian coordinate system, the RGB color space would be analogous to a cube with vertices at (x,y,z) = (0,0,0) (1,0,0) (0,1,0) (0,0,1) (1,1,0) (0,1,1) (1,0,1) and (1,1,1).

This cube is just a subset of a vector space, and not technically a vector space itself, because the inverses of the vectors (i.e. the ones with negative coordinates) are not present in the space. Also, the maximum vectors, when multiplied by a scalar, produce vectors which have magnitudes greater than 100% which again are not present in the space.