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The parametric equations of a projective line passing through $P = \left[X_1:Y_1:Z_1\right]$ and $Q = \left[X_2:Y_2:Z_2\right]$ are :

$\left[X:Y:Z\right] = \alpha\left[X_1:Y_1:Z_1\right] + \beta\left[X_2:Y_2:Z_2\right]$, $\alpha,\beta \in \mathbb{R}$?.

Is it an abuse of notation?

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  • $\begingroup$ You should take $\beta=1-\alpha.$ $\endgroup$ – Jean Marie Feb 1 '17 at 17:19
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    $\begingroup$ At present; this parameterization allows $\alpha=\beta=0$. But $[0:0:0]$ isn't in $\Bbb P^2$. $\endgroup$ – Semiclassical Feb 1 '17 at 17:21
  • $\begingroup$ Because $\beta = 1 - \alpha$? if consider $(\alpha,\beta) \in \mathbb{R^2}$ \ $\lbrace 0 \rbrace $? $\endgroup$ – Orested Feb 1 '17 at 17:34
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Your equation is perfectly valid (with the caveat spotted by @Semiclassical).

The parametric equation of a line defined by 2 points in projective space is usually derived from a reasoning on the rank of matrix:

$$\left(\begin{array}{l l l} X_1 & X_2 & X \\ Y_1 & Y_2 & Y \\ Z_1 & Z_2 & Z \end{array}\right) $$

By definition of a projective line, this rank has to be 2; which is equivalent to the fact that the last column can be expressed as linear combination of the first two columns:

$$\tag{1}M:=\left(\begin{array}{l l l} X \\ Y \\ Z \end{array}\right)=\alpha_1\left(\begin{array}{l l l} X_1 \\ Y_1 \\ Z_1 \end{array}\right)+\alpha_2\left(\begin{array}{l l l} X_2 \\ Y_2 \\ Z_2 \end{array}\right)$$

Edit (following a remark by @MvG pointing an error of mine in a first draft):

Two cases can occur:

  • either $\alpha_1+\alpha_2 \neq 0$, and point $M$ is said "an ordinary point". In this case, we can further normalize this linear combination (considering that we can take taking $\alpha_1+\alpha_2=1$), due do the fact that projective coordinates are defined up to a multiplicative factor. Thus (1) implies:

$$M:=\left(\begin{array}{l l l} X \\ Y \\ Z \end{array}\right) \sim k\alpha_1\left(\begin{array}{l l l} X_1 \\ Y_1 \\ Z_1 \end{array}\right)+k\alpha_2\left(\begin{array}{l l l} X_2 \\ Y_2 \\ Z_2 \end{array}\right)$$

where $k$ is any nonzero real number, for example $1/(\alpha_1+\alpha_2)$, and symbol $\sim$ means "proportional to" (= with equivalence relationship "defined up to a non zero multiplicative factor"). We check immediately that the new coefficients $\alpha'_1:=\alpha_1/(\alpha_1+\alpha_2),\alpha'_2:=\alpha_2/(\alpha_1+\alpha_2)$ sum up to $1$.

Remark: this way of writing things is important because, in this way, as we are in affine geometry, the bridge is made with the barycentrical approach.

  • or $\alpha_1+\alpha_2=0$. In this case, (1) becomes:

$$M:=\left(\begin{array}{l l l} X \\ Y \\ Z \end{array}\right) \sim \left(\begin{array}{l l l} X_1 \\ Y_1 \\ Z_1 \end{array}\right)-\left(\begin{array}{l l l} X_2 \\ Y_2 \\ Z_2 \end{array}\right)$$

i.e., is equivalent to a vector. Point $M$ is said "at infinity".

Remark: A key point in the understanding of projective geometry is the bijective correspondence between vectors (up to a sign change) and points at infinity.

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  • $\begingroup$ I consider the part about further normalization to be confusing. If I understand you correctly, you seem to imply that one could always choose $\alpha$ and $\beta$ such that $\alpha+\beta=1$, but that is not the case for $\alpha=-\beta$. Therefore if you make that assumption you loose one point, turning the projective line into an affine one. $\endgroup$ – MvG Feb 6 '17 at 10:58
  • $\begingroup$ @MvG You are right. I had my head elsewhere. I have modified all the second part of my answer. $\endgroup$ – Jean Marie Feb 6 '17 at 16:19
  • $\begingroup$ Better now. In my opinion this choice of “infinity” is pretty arbitrary. Some people would consider $[\alpha:\beta]=[Z_2:-Z_1]$ as infinity, as it leads to a zero $Z$ coordinate for the result, corresponding to the line at infinity in the standard $Z=1$ embedding. Having you introduce a different concept of infinity might be confusing, particularly since it depends on representatives. I also would consider the homogeneous coordinates to be vectors, so I can't subscribe to your statement that only points at infinity correspond to vectors. $\endgroup$ – MvG Feb 6 '17 at 17:09
  • $\begingroup$ @MvG I understand your point. As the OP has not said that he/she works with the standard "embedding" (Z≠0) I preferred this presentation which is a way one can "projectify" affine plane defined with barycentrical coordinates. I agree that it depends on representatives; but when you are a beginner in this domain... sticking to coordinates is not bad. $\endgroup$ – Jean Marie Feb 6 '17 at 18:19

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