algebraic-geometryprojective-geometry
(1)a projective but not perspective mapping indicates that they are different
(2)Perspective Transform & Homography Matrix indicates these are the same thing
(3) projective transformation = homography = collineation
it is evident (1) conflicts with (2) and (3). how can pinpoint these ?
Is the camera plane the projective space of the real world?
It depends.
Usually a physical camera has a limited sensor, so the thing where the image appears is too bounded to be an affine or projective plane. You can say “well, simply extend that plane infinitely”. Then it depends on how exactly you define that extension, whether you end up with an affine or a projective space. If you say “well, take any line on it, and for any distance $d\in\mathbb R$ take the point at position $d$ along that line to be a point of my plane as well” then the result is affine. If you say “well, take any pair of lines on it, and include their point of intersection as well” then it's projective.
And for something planar to be “the projective space” of “the real world”, that real world has to be planar, too. So if you have a camera image of things happening on the flat surface spanned by your table top, then yes, you could obtain an image of that surface in your projective camera plane. But if you take a camera image of a 3d scene, you get a projection, which is a reduction of information and usually not what projective geometry describes. Usually, a projective space is used as something more complete than the underlying affine space: you add points (at infinity), you don't remove them (by combining multiple points in the line of projection into a single point on the image plane).
Is the line which is the image of the horizon the distinguished line?
If you have your geometry sketched on a real paper lying on your real (horizontal) table, then the horizon of your real world is the distinguished line. You draw two parallel lines on your table, you look from an angle and you can almost see them converging on the horizon. Of course, that horizon is infinitely far away, so if you were to go there, the fact that the earth is not flat may break the simile.
If you take a camera image of that situation, then your projective camera sensor plane has its own distinguished line at infinity, namely where two parallel lines in the camera sensor plane intersect. If the image of one distinguished line is the other distinguished line, then your camera plane is parallel to the real world table plane, and the resulting transformation would be called affine. Otherwise it's a generic projective transformation.
Whenever we do an Affine transform do we need to look out for a distinguished line?
It depends on how and why you do affine transformations.
If you do an affine transformation via a multiplication of a matrix of the form $$\begin{pmatrix}*&*&*\\*&*&*\\0&0&1\end{pmatrix}$$ then the form of the matrix already guarantees that points with a $z$ coordinate of zero will get mapped to points which again have a $z$ coordinate of zero, i.e. using the $z=0$ line as the distinguished line in both your spaces is already implied by the form of the transformation.
If you do affine transformations as $$\begin{pmatrix}x\\y\end{pmatrix}\mapsto\begin{pmatrix}ax+by+c\\dx+ey+f\end{pmatrix}$$ then you are using the same scheme computation as above, plus you are using non-homogeneous coordinates, so you might not even see what this $z$ coordinate is about since the 2d affine coordinates don't even allow you to express points at infinity.
If, on the other hand, you have some weird choice of coordinate systems where the distinguished line has a more complicated form, and that form perhaps isn't the same for source and target space of your transformation, and you need to ensure that your transformations are indeed affine, then yes, you need to be explicit about the lines at infinity, and verify that they get mapped correctly. One common example would be trilinear coordinates, where $x+y+z=0$ represents the line at infinity. Mapping from carthesian to trilinear coordinates and back is a situation where you want to make sure that the line at infinity gets transformed correctly.
Why does just a distinction of the geometry a line in the perspective plane make the geometry an Affine geometry?
As the other answers by Joseph Malkevitch and by AdiPiratla have already indicated, there are different ways to treat the relation between affine and projective geometry. One is to say that you get projective geometry from affine geometry if you add a point at infinity for every bundle of parallel lines, and a line at infinity made up from all these points. In this sense, a projective space is an affine space with added points.
Reversing that process, you get an affine geometry from a projective geometry by removing one line, and all the points on it. By convention, one uses the line $z=0$ for this, but it doesn't really matter: the projective space does not depend on the choice of coordinates, and removing any line will turn it into an affine space.
But often one wants to exploit the machinery from projective geometry to perform affine operations. Perhaps you want to combine affine and projective transformations, or some such. In such a setup, you can say that as long as you keep track of which line is the line at infinity, you know how to get from there to affine geometry, so you are already doing affine geometry in a different representation. Affine geometry is like projective geometry with one line (the “distinguished line”) labeled “remove this to obtain an affine plane”. In this sense, an affine space is a projective space with additional information.
First, I think it may be a mistake to think about "translation", "rotation" and "scale", which is one particular decomposition of the affine group -- perhaps it's better to think about what transformations can be effected by affine maps.
For affine maps: We can move any collection of three noncollinear points to any other collection of three points (which must be noncollinear if we want the map to be invertible).
For projective maps: we can move any collection of four points (no three collinear) to any collection of four points. (Although to make complete sense of this, points and lines at infinity must be included.)
Similar characterizations for smaller groups:
Translation: we can move any point to any other.
Rotation: we can move any line through the origin to any other line through the origin
T + R: we can move any point-line pair to any other point-line pair, where a "point line pair" means a line L and a point P that lies on L.
I'll let you work out descriptions of the transformative power of things like "all scales and rotations", etc.
ADDITIONAL REMARKS
Although a homography has 9 entries, there are really only 8 free parameters, in the sense that two matrices that differ by a multiplicative (nonzero) constant represent the same homography. So we might as well simplify a bit by dividing through by h9 to get a matrix whose lower right entry is a 1. (That'll miss out on describing matrices whose lower-right entry is 0, but this is a small set, and once you understand the others, this last set won't give you any problems.
Such a matrix can now be factored into \begin{align} \begin{bmatrix} h_1 & h_2 & h_3 \\ h_4 & h_5 & h_6 \\ h_7 & h_8 & 1 \end{bmatrix} & = \begin{bmatrix} h_1- h_3 h_7& h_2 - h_3 h_8& h_3 \\ h_4 - h_6 h_7 & h_5 - h_6 h_8 & h_6 \\ 0 & 0 & 1 \end{bmatrix} \cdot \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ h_7 & h_8 & 1 \end{bmatrix} \end{align} i.e., your transformation becomes a combination of an affine transform (on the left), albeit one slightly different from the one you "see" in the top 6 matrix entries of your original matrix, and an transform whose only interesting entries are in the bottom row. So since you understand affine xforms already, let's look at the rightmost matrix, which I'll rewrite $$ \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ u & v & 1 \end{bmatrix} $$ to avoid having to type subscripts. Note that if $(u, v) = (0, 0)$, then this is an affine transformation and you know about this, so from here on, we'll assume that $u$ and $v$ are not both zero.
What does this to a point $(x, y)$ of the plane? Well, we write $(x,y)$ as a column vector by appending a "1", so we get $$ \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ u & v & 1 \end{bmatrix} \begin{bmatrix} x\\ y \\ 1 \end{bmatrix} = \begin{bmatrix} x\\ y \\ ux + vy + 1 \end{bmatrix} $$ which, when "rehomogenized" (i.e., when divided by its last coordinate to make the last coordinate by "1"), becomes $$ \begin{bmatrix} x/ (ux + vy + 1)\\ y/ (ux + vy + 1) \\ 1 \end{bmatrix}. $$ In short, we get the transformation $$ (x,y) \mapsto (\frac{x}{ux + vy + 1}, \frac{y}{ux + vy + 1}). $$ What does that "look like"? Well, it sends the line where $ux + vy = -1$ to infinity. It takes the line where $ux + vy + 1 = 1$ to itself (i.e., it fixes every point on that line). But as for the details...let's simplify a little.
To be more explicit: you can think of this as transforming the plane by fixing the $x$-axis, and transforming each line through $(0, -1)$ into a vertical line. If the line $L$ passes through $(0, -1)$ and $(a, 0)$, then the transformed line will pass through $(a, 0)$ and be vertical. People in computer graphics sometimes call this the "unhinging" transformation, thinking of two diagonal lines through $(0, -1)$ as forming a "hinge", while after transformation, they become parallel vertical lines.
Best Answer
In projective geometry a projective transformation is a product of perspective transformations. A perspective transformation is a projective transformation, but a projective transformation is not necessarily a perspective transformation.
A projective transform is a homography is a collineation.
In general, the transformation between four corresponding pairs of points is a projective transform.
The blog post (2) gets it wrong. The OpenCV's
getPerspectiveTransformfunction seems to be incorrectly named. It should be calledgetProjectiveTransform, I suppose, but presumably nobody in that community objects.So it's actually (2) that conflicts with (1) and (3), and I'd venture that's because (1) and (3) are math while (2) is computer vision software, where terminology may differ. It could be that in computer vision the most common use of a projective transform is to remove or add perspective.