Is the projection matrix in OpenGL really a "projection matrix"?

Question

A projection matrix projects a vector from a higher dimensional space onto a subspace. I would have expected the projection matrix in OpenGL to project a point in R³ onto a 2 dimensional plane. This seems to be supported by a lot of literature on the internet. Many sites imply that the projection matrix projects the 3D world onto a plane and this is what is drawn. However I get the feeling that most of these explanations are skipping several steps. Many of them seem to contradict each other so I'd like some clarification of the conclusions I have drawn from my own analysis.

Can someone please confirm (or correct if wrong) that:

1. The projection transformation in OpenGL is not actually a projection matrix, but rather transforms a point into clip space (which is still part of the R³ domain) and the actual projection onto a 2D plane happens later as a fixed function of the pipeline.
2. The projection matrix doesn't apply the perspective divide; however it does need to set the w coordinate so that when the perspective divide happens later (as a fixed function of the pipeline) points are correctly placed either inside or outside of NDC.
3. Clip space is a box between (-1,+1) on the x,y axis, and (n,f) on the z-axis while NDC is a box betwen (-1,+1) on all axis.

I analysed the following projection matrix to come to the above conclusions:

[ 2n/(r-l)     0     (r+l)/(r-b)      0     ]
[    0     2n/(t-b)  (t+b)/(t-b)      0     ]
[    0         0    -(f+n)/(f-n) -2fn/(f-n) ]
[    0         0         -1           0     ]

From that analysis I concluded that any point that is within the frustum will be within the clip boundaries along the x,y axis; it may be outside the boundaries along the z axis, however once the perspective divide happens (with w now being the old -z) the point will fully inside clip space.

From this I have also concluded that for a point to be visible after the MVP transformation it's x,y and z/w coordinates must be between +/-1, and that the perspective divide and actual projection happen after the vertex shader.

If applicable answers specific to modern OpenGL (3.3 core or later) only please.

Answer 1

1. The projection matrix in OpenGL transforms points into clip space. But this is already a projection. The only thing that has to be done after the matrix multiplication is the perspective divide.

2. True

3. Clip space is the space from [-w to w] on each axis, since the only operation that happens between clip space and NDC is the perspective divide. NDC is from [-1 to 1] on each axis.

Additional notes:

• Mathematically, an OpenGL projection matrix maps a 4D space (P^4) into another 4D space (clip space). This can easily be seen by the form of a matrix (4x4 matrix maps 4D -> 4D). With the perspective divide the 4D clip space is truncated by homogenization into the 3D NDC (R^3) space.
• A point is visible after the projection, when it's x,y,z coordinates are between [-w, w]. The reason why clipping happens before the perspective divide is, that NDC is not necessarily a cubic space (it is one in OpenGL, but in DirectX, e.g., NDC is x,y in [-1,1], z in [0,1])

• A geometric projection in general is defined as a mapping p from one space (O) into another one (T). this would be written as

  O --p--> T

  In some cases such a mapping can be described by a tranformation matrix in euclidean space (a parallel projection, for example, would work), but in a lot of cases this is not possible (especially in cases where parallel lines in O are not parallel any more in T). This is why projective spaces are required.

I better stop here now since it gets more and more complicated from the mathematical point of view, but if you want to dig more into this topic, I suggest the following articles:

Wikipedia Projective Space
Wikipedia Projective Geometry
Video about projection in general (this, and the next one)