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I've really tried to desperation before asking here. In a few words, I simply can't think visually about a projective frame. I don't know what to imagine.

Vector spaces

With a basis of a vector space, I usually think about vectors (arrows) in $\mathbb{R^2}$ or $\mathbb{R^3}$ starting from the origin, which serve as building blocks to construct other vectors, a process that can easily be visualized by literally following the arrows tail to tip in order of appearance in their linear combination. This mental image also easily justifies (intuitively) why $n$ vectors are sufficient for a basis, since their independence guarantees that you can move in every direction (there is a bit of every direction at least in one vector).

Affine spaces

Now, for affine spaces the thing becomes a little more subtle, but it's still feasible. In this case, I usually imagine 2D or 3D space ($\mathbb{A^2}$ or $\mathbb{A^3}$), which is pretty much the same thing as before, but now without axes, origin or any kind of grid, AND with a lot of sparse dots which I know are separated by some vectors all belonging to the underlying vector space.

There are two ways I tackle the idea of a reference frame in affine spaces. The first (the most intuitive to me) is just to take a point as origin, and now we return at the previous case where we need to find a basis of a vector space, so I take $n$ more points (independent, of course) and life is good. Of course, this explains why $n+1$ points are required. The second way (which is the only approach that intersects in a way with my attempts of thinking about projective frames) makes use of homogeneous coordinates.

By looking at our affine space from one dimension up (and here I usually think specifically at $\mathbb{A^2}$ which becomes the plane where $z=1$ in $\mathbb{R^3}$), we are able to describe our original affine space entirely as a subset of it's "wrapper" vector space, so that every affine transformation becomes linear and, more importantly, every affine reference frame is associated with a basis. This, again, also explains why we need $n+1$ points.

Projective spaces

Here is the wall. I've exposed myself to every explanation that I've found, and only partly have managed to create a visual and operative intuition for the kind of work projective spaces are meant to be useful in. First of course I've encountered the formal definitions, which are (unexpectedly!) not so distant from the real meaning and motivation of what they describe and model, our eye in the real world.

So first I started to think of $\mathbb{P^1}$ as all the lines through the origin in $\mathbb{R^2}$, then came the "screen" part where I could interpret that y=1 line as my retina, but extended. Same for $\mathbb{P^2}$ and $\mathbb{R^3}$, cool, we only care about directions of lines, not lines themselves, and represent them as points on a one-dimension-lower space. In the case of $\mathbb{P^2}$ and $\mathbb{R^3}$, I like to think about it as stars in the sky. Every star is a direction and to us (naked eye humans) there is no difference between one or another.

Now the question: how do we orient the stars? How do we find a projective frame, visually? I already know that the way to do it is to take $n+2$ points, $n+1$ of them being the "candidate basis" one dimension higher (in our case, 3 points in $\mathbb{P^2}$ which still represent infinite possible triples of vectors in $\mathbb{R^3}$), and the $(n+2)$th (the 4th one in $\mathbb{P^2}$) declaring which of the candidates wins.

But this argument doesn't make sense to me INSIDE the projective space, or, more romantically, among the stars. In contrast to the two cases above, here the $n+2$ points don't seem to "build stuff" if seen inside the main thing they belong to, but only outside of it. What should we do with the 4 points we chose? There are no arrows to follow, no origins to start from, nothing. And visual representations of that aren't really a thing nowadays. Internet is improving (see 3Blue1Brown etc), but that's still not enough.

Also, if we start from a basis (made of $n+1$ vectors) in the wrapper vector space, we ALREADY GET a frame in the projective one! But those $n+1$ vectors correspond to $n+1$ points (stars), so it doesn't seem true that $n+2$ are always needed. This again wasn't true in the vector and affine space. So much confusion!

(P.S. sorry for the long post)

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  • $\begingroup$ @Ataulfo I'll assume a typo ("star" instead of "start"). I'm mainly referring to real (or at most complex) projective spaces (up to dimension 2-3). I'm asking for intuition in these most famous cases, which preserve the original idea that lead to their general definitions $\endgroup$ Commented Aug 24 at 16:08
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    $\begingroup$ Let the set $\mathcal Z=\{1+\sqrt2\mathbb Z,\space2+\sqrt2\mathbb Z,\space 3+\sqrt2\mathbb Z,\space\cdots\}$ this is an infinite set (each element has the form $n+\sqrt2\mathbb Z$ where $n$ is a natural number). What would you call "star" in $\mathcal Z$? Could you construct a projective space defined over a finite field? Do you know why these spaces are called "proyective"? $\endgroup$ Commented Aug 24 at 16:13
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    $\begingroup$ Yes it was a typo. What I wanted to know is your viewpoint on $\mathbb{P^2}$ using the word "star". Could you apply the same perspective on the set I gave you? It seems to me your knowledge on projective spaces is not complet enough. I wanted to explain some ideas helping you but the resulting text was too long. In particular perhaps you don't know the "natural" topology on $\mathbb{P^2}$. My intention was to maybe help you. (Sorry for bad English) $\endgroup$ Commented Aug 24 at 16:27
  • $\begingroup$ Thank you for trying to help. Sure my knowledge is not complete, I'm only familiar with the usual $\mathbb{P(K)}$ where $\mathbb{K}$ is some power of $\mathbb{R}$ or $\mathbb{C}$. When I say "star" I mean a direction (i.e. equivalence class in $\mathbb{P^2}$) in which to look, because stars in our reference frame are exactly that. Ideally there should be a star for direction, even if it's not true (it's just a romantic idea which I like, wasn't intended to actually work). What does $Z$ stand for? Is it a projective space, or a vector space to embed the projective one into? $\endgroup$ Commented Aug 24 at 16:46
  • $\begingroup$ $\mathcal Z$ is just a set which can be "structured" with one can (if $n\in\mathbb Z$, for example, you do have an evident equivalence relation so a quotient set). For finish this comments, I believe that your doubts are because you don't have a clear understanding of the notion of homogeneus coordinates in your projective space. There are many things to learn about these spaces. Try to be quite clear with higher dimensions and after try to construct with finite fields. And also some topology (suite) $\endgroup$ Commented Aug 24 at 19:25

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Pamfilos' notes on the projective plane has a nice diagram on page 3 explaining the four points in the projective basis for a projective plane. Of the four points $\{A,B,C, D\}$, any three points, say $\{A,B,C\}$ constitute a basis for $\mathbb R^3$. Given the origin $O$, we have three directions $OA, OB, OC$ but the directions are not associated with a length. In other words, we have a direction $OA$, but we do not have a unit vector in that direction. What the fourth or "unit" point $D$ does is to provide a unit vector in each of the three directions $OA, OB, OC$. The unit vector is obtained by projecting $OD$ on these three lines. More visually, Pamfilos draws a parallelepiped such that (a) one vertex is at $O$, (b) the sides are in the directions $OA, OB, OC$, and (c) $OD$ is the grand diagonal of this parallelepiped. The sides of this parallelepiped $a', b', c'$ allow any point $x$ to be expressed using homogeneous coordinates as $x = ua′ + vb′ + wc′$. The coordinates $u,v,w$ are homogeneous in the sense that $ku,kv,kw$ with $k \ne 0$ are equally valid coordinates. The above construction gives the coordinates of the four points $\{A,B,C, D\}$ as $\{(1, 0, 0), (0, 1, 0), (0, 0, 1), (1, 1, 1)\}$. The last point $D$ is the sum of the other three points as is to be expected from the grand diagonal interpretation of this point.

The construction in higher dimensions is similar.

Response to OP's comment on meaning of unit

I like to think of unit as the physicist's notion of unit of measurement. $\overrightarrow{OA}, \overrightarrow{OB}, \overrightarrow{OC}$ are just directions (infinite rays) and have no definite length at all. So we could choose a line segment $OA$ to be a light year, $OB$ to be a kilometer and $OC$ to be a millimeter. Homogeneous coordinates mean that we could live with any of these three units, but we want the same consistent choice everywhere. So we introduce $OD =OA + OB + OC$ (vector addition in the parallelepiped or projection of $OD$ on the three directions) and that enforces a common unit. If we choose $OD$ to be a lightyear long, then the projections $OA, OB, OC$ will also have to be very long to add up to $OD$. If we choose $OD$ to be a millimeter, then the projections $OA, OB, OC$ will also have to be very small to add up to $OD$. Choice of different units for $OD$ is just a different choice of homogeneous coordinates.

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  • $\begingroup$ As I've mentioned in the post, this does make sense only when seen outside the projective space (in this case, $\mathbb{A^2}$). Is there a general way of thinking of a projective frame DIRECTLY inside of it? (I suspect that there is because for the projective line the three points can be used as a base for cross-ratio with the fourth) $\endgroup$ Commented Aug 24 at 14:36
  • $\begingroup$ I think you can see it algebraically inside the space, and you can see it visually from outside the space. It is like asking how do we see that the surface of a balloon is curved. A 2-D creature on the balloon can see it algebraically by computing the intrinsic curvature, and a 3-D creature can see it visually. Can a 2-D creature see the curvature visually? I think not. I thought that with affine spaces, you were willing to go to an extra dimension to see the need for $n+1$ points. Once you see that, the grand diagonal of the parallellipiped gives $n+2$ points in projective space. $\endgroup$ Commented Aug 25 at 9:29
  • $\begingroup$ I did consider the extra dimension in affine spaces in my post, but that was second to the internal view of an affine frame (which makes sense to the 2-D creature). This doesn't SEEM the case for projective spaces, BUT then why it does for at least the projective line? I mean, in $\mathbb{P^1}$ one can say that three points are needed so that each fourth point will have a unique, invarianti cross-ratio with them. This does not require homogeneous coordinates and doesn't seem, at least to me, "seeing algebraically" $\endgroup$ Commented Aug 25 at 11:05
  • $\begingroup$ In the affine plane, there was no natural zero (origin), so you added it without compunction. In the projective plane there is no natural one (unit) so you need to add that as well. And I think the cross ratio is algebra just as $d=a'+b'+c'$ in Pamfilos is algebra. $\endgroup$ Commented Aug 25 at 13:23
  • $\begingroup$ that's a much better way of saying that to me. But how does this unit work? Usually when I think of unit in algebra, it's something that I can invert and multiply other elements by. Why is it called unit? $\endgroup$ Commented Aug 27 at 15:17

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