I have noticed weird thing when working with HSL - unlike RGB, it has some blind spots where certain value just does not matter. I'm sure we were taught about this when I had Linear algebra lectures about Linear space transformations.
Say RGB(r,g,b) and HSL(h,s,l) are both vectors. Now every value of RGB produces different color. It's a space of a cube and every point is different:
HSL is not this case. For $s=0$ or $l \in \{0,1\}$ the value of $h$ does not matter. And specifically in case $l \in \{0,1\}$ nothing matters. In other words some HSL vectors project into same RGB vector. There was a term in the linear algebra for this situation.
There is a 3D plot for HSL too:
But in fact, you can also reference values that are out of this cone - they are just not included because they are the same colors.
I was mostly thinking about this because it occurred to me to have an idea that maybe 3 coordinates are too much for color system that uses hue. Could there be one angle for color and one for both saturation and lightness without excluding any colors?
Could you explain me how come that three coordinates can contain different amounts of information?
Both RGB and HSL denote colours as coordinates in a 3D space. That the correspondence is not strictly one-to-one can be understood by thinking of the the analogy (pointed out in comments) of the two common ways of specifying points in the 2D space: cartesian vs polar (and, indeed, if we think of the RGB space as cartesian coordinates, both HSL and HSV should be thought as cylindrical coordinates). Both cartesian and polar coordinates contains the same amount of information, but in the polar coordinates, for the case $r=0$ the value of the angle $\phi$ does not matter - so, in that sense you could argue that the cartesian coordinate $(x=0,y=0)$ maps to a full set of polar coordinates $(r=0,\phi)$. I don't think that this fact would imply that one coordinate system has "more information" than the other.
Perhaps you are thinking in terms of quantized representations, such that, for example, we have 8 bits for each "channel" - in this discrete space, RGB would allow for $256^3$ different colours, while if we do the same for HSL, we'd get fewer colours. That's true, but it's rather irrelevant.