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Color is infinite dimensional. Imagine you shine four lasers each with wavelengths of 100, 200, 300, and 400 nanometers. We might see them smushed together into

Color is infinite dimensional. Imagine you shine four lasers each with wavelengths of 100, 200, 300, and 400 nanometers. We might see them smushed together into a combination that's 3 dimensional (unless you're a tetrachromat, a rare person who has extra rods or cones or something), but physically it's just the four frequencies (and phases). You can't combine the first 100, 200, 300nm laser beams to get a 400nm beam, so it's not three dimensional. With N lasers, it's N frequencies, and you can't reproduce them as combinations of fewer frequencies. N goes arbitrarily high. Follow that logic and the dimension of color space is the dimension of the space of frequency distributions, which is infinite dimensional. Likewise with sound, unless you want to describe it as just the amplitude at a single spatial point versus time, in with case sound and color are just 1D (plus time), which isn't a useful model since you now have to account for all spatial points.

sudosysgen 6y ago
Colour can't be described as 1D, as you can have multiple amplitudes of different frequencies at once. You could have an infinite amount of points of differing amplitudes, but this conception breaks down because of the reality of limited bandwidth, and so the best representation is a frequency-space. If you want, here is something you can do. Get a square, inside of that square plot the x axis for frequency, and the y axis for time. For every frequency x, at every time y, choose a luminosity level going from 0 from null amplitude to 1 corresponding to the entire amount of energy in the observable universe. That square will be able to represent perfectly colour over time, what you claim is "infinite dimensional", in a three-dimensional space.
- 6gvONxR4sf7o 6y ago
  Here's a counter argument against what you just said for N dimensions, in almost exactly your words. Get a square, inside of that square, plot the x axis for dimension, and the y axis for time (the x axis is discrete, taking values of 1, 2, 3, ... up to N). For every dimension x, at every time y, choose a luminosity level going from 0 from null amplitude to 1 corresponding to the entire amount of energy in the observable universe. That square will be able to represent perfectly ND position over time, which I claim is N dimensional (plus time), in a three dimensional space. Much more simply, you can represent an Nd point in 2d, with the first dimension indexing dimension, and the second dimension indexing value. The 2D representation isn't the space. It's a single point in the space, which can vary along some number of degrees of freedom (N in this case). If that first dimension of the two (the index dimension) was continuous, and the second still represents value, then we're representing the set of 1D functions, which is infinite dimensional. That's how we represent a frequency distribution.
  - sudosysgen 6y ago
    The issue with your understanding is that in reality, dimensions are not discrete. Dimensions are continuous. As soon as you index a dimension of an n-dimensional space you do not have n-dimensions, but a set of (n-1) -dimensional spaces. You cannot represent an n-dimensional point in 2D; that means representing an n-dimensional point bijectively onto a finite a̶m̶o̶u̶n̶t̶ ̶o̶f̶ ̶2̶D̶ ̶s̶p̶a̶c̶e̶ number of 2D points. You can represent one n-dimensional point using an entire 2-dimensonal space, but that's not what I am talking about. There is no bijection possible from one spatial dimension into two spatial dimensions because all R->R^2 and so on bijections are impossible to construct continuously . A frequency distribution is not the set of 1D functions. A frequency distribution represents one 2D function. Indeed, the amplitude of each possible frequency is in reality the Fourier transform of a 2D, or one argument function. The function f(x) = 2x^2, is a 2D, one argument function, as it describes the relationship between two dimensions. A 1D wave is actually a 2D function. You cannot abstract away time in this situation. There is no such thing as a 1D function; all functions require at least two dimensions. A 1 argument function necessitates a 2D space. For example, s = vt + a/2(t^2) is a relationship between two dimensions, time and displacement.
    - 6gvONxR4sf7o 6y ago
      > You cannot represent an n-dimensional point in 2D Here's a 3d point in 2d with a discrete dimension and a continuous one: (x1, x2, x3) Or if you want it with a continuous horizontal dimension, using dirac deltas makes it trivial, but you wanted it in a finite space, so we can use all sorts of other sets of orthogonal functions. I'll use sinusiods. To represent any 3D point (x1, x2, x3) as a function that takes a finite amount of 2D space, define g(x1, x2, x3) = f, where f is the following function from R to R: f(x) = x1 * sin(x) + x2 * sin(2x) + x3 * sin(3x). So that's representing any point in R^3 in a finite 2D space, with x and y=f(x), just like in your original construction with x=frequency and y=time and z=luminosity=f(frequency, time). For the reverse, to make it a bijection, make use of the fact that the integral from negative pi to pi of 1/pi sin(m x) sin(n x) is 1 if m == n and 0 otherwise. h(f) = ( integral 1/pi sin(x) f(x), integral 1/pi sin(2 x) f(x), integral 1/pi sin(3 x) f(x) ) = (x1, x2, x3) g and h are inverses, so we have a bijection. Crucially, the bijection is between R^3 and a very particular set of functions from R to R. With the set of sin(i x) functions for different i, you can make a bijection between any R^N and a particular set of functions from R to R, so the fact that there's an (x, y=f(x)) representation doesn't matter. And this why your original construction doesn't work. You didn't make a function from the set of frequency distributions to R^3. You made a function from the set of frequency distributions to the set of functions from R^2 to R. The set of functions from R^2 -> R is an infinite dimensional set, which is a very well established fact. See [0] for a bunch of different explanations of this fact. [0] https://math.stackexchange.com/questions/2599827/proof-that-vector-space-ca-b-of-all-functions-from-a-b-to-mathbbr-i https://math.stackexchange.com/questions/2599827/proof-that-...

sudosysgen 6y ago

Colour can't be described as 1D, as you can have multiple amplitudes of different frequencies at once. You could have an infinite amount of points of differing amplitudes, but this conception breaks down because of the reality of limited bandwidth, and so the best representation is a frequency-space. If you want, here is something you can do. Get a square, inside of that square plot the x axis for frequency, and the y axis for time. For every frequency x, at every time y, choose a luminosity level going from 0 from null amplitude to 1 corresponding to the entire amount of energy in the observable universe. That square will be able to represent perfectly colour over time, what you claim is "infinite dimensional", in a three-dimensional space.

6gvONxR4sf7o 6y ago
Here's a counter argument against what you just said for N dimensions, in almost exactly your words. Get a square, inside of that square, plot the x axis for dimension, and the y axis for time (the x axis is discrete, taking values of 1, 2, 3, ... up to N). For every dimension x, at every time y, choose a luminosity level going from 0 from null amplitude to 1 corresponding to the entire amount of energy in the observable universe. That square will be able to represent perfectly ND position over time, which I claim is N dimensional (plus time), in a three dimensional space. Much more simply, you can represent an Nd point in 2d, with the first dimension indexing dimension, and the second dimension indexing value. The 2D representation isn't the space. It's a single point in the space, which can vary along some number of degrees of freedom (N in this case). If that first dimension of the two (the index dimension) was continuous, and the second still represents value, then we're representing the set of 1D functions, which is infinite dimensional. That's how we represent a frequency distribution.
- sudosysgen 6y ago
  The issue with your understanding is that in reality, dimensions are not discrete. Dimensions are continuous. As soon as you index a dimension of an n-dimensional space you do not have n-dimensions, but a set of (n-1) -dimensional spaces. You cannot represent an n-dimensional point in 2D; that means representing an n-dimensional point bijectively onto a finite a̶m̶o̶u̶n̶t̶ ̶o̶f̶ ̶2̶D̶ ̶s̶p̶a̶c̶e̶ number of 2D points. You can represent one n-dimensional point using an entire 2-dimensonal space, but that's not what I am talking about. There is no bijection possible from one spatial dimension into two spatial dimensions because all R->R^2 and so on bijections are impossible to construct continuously . A frequency distribution is not the set of 1D functions. A frequency distribution represents one 2D function. Indeed, the amplitude of each possible frequency is in reality the Fourier transform of a 2D, or one argument function. The function f(x) = 2x^2, is a 2D, one argument function, as it describes the relationship between two dimensions. A 1D wave is actually a 2D function. You cannot abstract away time in this situation. There is no such thing as a 1D function; all functions require at least two dimensions. A 1 argument function necessitates a 2D space. For example, s = vt + a/2(t^2) is a relationship between two dimensions, time and displacement.
  - 6gvONxR4sf7o 6y ago
    > You cannot represent an n-dimensional point in 2D Here's a 3d point in 2d with a discrete dimension and a continuous one: (x1, x2, x3) Or if you want it with a continuous horizontal dimension, using dirac deltas makes it trivial, but you wanted it in a finite space, so we can use all sorts of other sets of orthogonal functions. I'll use sinusiods. To represent any 3D point (x1, x2, x3) as a function that takes a finite amount of 2D space, define g(x1, x2, x3) = f, where f is the following function from R to R: f(x) = x1 * sin(x) + x2 * sin(2x) + x3 * sin(3x). So that's representing any point in R^3 in a finite 2D space, with x and y=f(x), just like in your original construction with x=frequency and y=time and z=luminosity=f(frequency, time). For the reverse, to make it a bijection, make use of the fact that the integral from negative pi to pi of 1/pi sin(m x) sin(n x) is 1 if m == n and 0 otherwise. h(f) = ( integral 1/pi sin(x) f(x), integral 1/pi sin(2 x) f(x), integral 1/pi sin(3 x) f(x) ) = (x1, x2, x3) g and h are inverses, so we have a bijection. Crucially, the bijection is between R^3 and a very particular set of functions from R to R. With the set of sin(i x) functions for different i, you can make a bijection between any R^N and a particular set of functions from R to R, so the fact that there's an (x, y=f(x)) representation doesn't matter. And this why your original construction doesn't work. You didn't make a function from the set of frequency distributions to R^3. You made a function from the set of frequency distributions to the set of functions from R^2 to R. The set of functions from R^2 -> R is an infinite dimensional set, which is a very well established fact. See [0] for a bunch of different explanations of this fact. [0] https://math.stackexchange.com/questions/2599827/proof-that-vector-space-ca-b-of-all-functions-from-a-b-to-mathbbr-i https://math.stackexchange.com/questions/2599827/proof-that-...