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My projective geometry textbook says the following:

Degrees of freedom (dof). It is clear that in order to specify a point two values must be provided, namely its $x$- and $y$-coordinates. In a similar manner a line is specified by two parameters (the two independent ratios $\{a : b : c \}$) and so has two degrees of freedom.

The author has not defined "independent" in "independent ratios". What makes these ratios "independent"? In what sense are they "independent"? For instance, let's say we have $\dfrac{a}{b}$ and $\dfrac{c}{b}$; how are these "independent" when they have $b$ in common?

Thank you.

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Pick any two numbers at all -- call them $p$ and $q$. For example, you might like to choose $p = 3/4$ and $q = 19$. The numbers are independent in the sense that choosing one number imposes no restrictions whatsoever on the choice of the other number.

Now (to take your example) it is always possible to find three numbers $a, b, c$ with $\frac ab = p$ and $\frac cb = q$. For example, with $ p = 3/4$ and $q = 19$ as above, we could choose $a = 3, b = 4$ and $c = 76$. This is not the only possible choice of $a, b, c$, however, we could also choose $a = 6, b = 8, c = 152$, or $a = \frac 32, b = 2, c = 38$. In any of these choices, we would still have $\frac ab = p$ and $\frac cb = q$.

What the excerpt you quoted is trying to say is that regardless of which specific choices of $a, b, c$ we make, we end up with the same line. The triplets $$[3 : 4 : 76]$$ $$[6 : 8 : 152]$$ $$\left[\frac 32 : 2 : 38\right]$$ all determine the same line.

Remember, just because you have chosen the ratio $\frac ab = \frac 34$ does not mean you have chosen $a$ and $b$!

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Maybe this should be seen in the proper context. The projective plane is defined on the set of points in a three dimensional space. It is defined to be the set of lines going trough the origin. Therefore a line (a point of the projective plane) is defined by three coordinates $(a,b,c)$. But any point $(\frac{a}{r},\frac{b}{r},\frac{c}{r})$ defines the same line (with $r \neq 0$ of course). On could use coordinates $(x,y,1)$ by posing $r = c$, but this is not possible if the point lies in the $x,y-$plane, in that case one could chose $r = b$ unless the point lies on the $x-$axis, in which case one choses $r=a$. The case $a=b=c=0$ is evidently excluded since it defines no line through the origin. So locally a point of the projectie plane can be expressed by two coordinates but one cannot use these coordinates on the global projective plane. An analoguous problem exists on the sphere.

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  • $\begingroup$ "but this is not possible if the point lies in the $x, y$-plane"; why is that? Also, I'm still not clear on the meaning of "independence" in your explanation? $\endgroup$ Commented Jun 19, 2019 at 14:22
  • $\begingroup$ Because in that case the vector has coordinates $(a,b,0)$ and the reduction to the form $(x,y,1)$ is impossible because one can not divide by zero. Independence? Where is dat word implied? $\endgroup$ Commented Jun 19, 2019 at 18:19
  • $\begingroup$ But the $x, y$-plane is independent of the third coordinate $z$, so we could have any $z$ in $(x, y, z)$ and there will be an "$x, y$-plane for that $z$". So $z$ can be $0$, as you said, or it can be any other value. Am I misunderstanding something? And when I say "independence", I'm referring to its use in my main post; what makes the ratios "independent"? $\endgroup$ Commented Jun 19, 2019 at 20:22
  • $\begingroup$ I think maybe the original text is getting at the fact that a line in the projective plane is of the form $\{ [x : y : z] \in \mathbb{P}^2 \mid ax + by + cz = 0 \}$ - which is well-defined since scaling $x,y,z$ simultaneously doesn't change whether the equation is satisfied. But there also, scaling $a,b,c$ simultaneously results in the same line (which then results in a duality between the lines in $\mathbb{P}^2$ and the corresponding points $[a : b : c]$ of $\mathbb{P}^2$). $\endgroup$ Commented Jun 19, 2019 at 21:42
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Suppose all of $a,b,c$ are nonzero.

If you want to make particular small changes to $\frac ac$ and $\frac bc$, you can do that in several ways, for example

  • hold $c$ constant and change $a$ and $b$, or
  • hold $a$ constant change $c$ such that $\frac ac$ changes by the amount your want, and then first change $b$ to adjust for how $c$ just changed, and then change $b$ again to make $\frac bc$ move in the amount you want.

or lots of similar combinations.

However, once you have gotten $\frac ab$ and $\frac bc$ to move to the values you chose, there is now only one possible value for $\frac ab$.

In other words we can vary $\frac ac$ and $\frac bc$ independently of each other. but what happens to $\frac ab$ is then dependent on the other two.

This is not specific to $\frac ab$, of course -- you can pick any two of the ratios and vary them independently, but the third one will be determined by what you did to the two you picked.


If one or more of the numbers are zero, you don't have the same freedom in which ratios you consider independent, but it there will always be some two ratios you can vary independently.

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