# Questions regarding the projective space

In geometry, we have (kind of) introduced the projective space. Sadly, I have problems understanding some connections and I hope somebody here might help me out, as wikipedia's entry and my professor's notes were of no avail.

Let $K$ be a field. We call $\mathbb{A}^n(K) := K^n$ the affine space over $K$ with dimension $n \in \mathbb{N}_0.$

Question 1: What properties does an affine space have? This is the first time our professor used this notion and I don't see how this "definition" really defines an affine space. German wikipedia tells me it is just some space where we have points, lines and some axioms (only naming the parallel axiom). Which other axioms hold in an affine space?

We say that $x,y \in K^{n+1} \setminus \{0\}$ are equivalent, if $\exists t \in K \setminus \{0\}: y = tx$. This is an equivalence relation. Let $\mathbb{P}^n(K)$ denote the set of equivalence classes. So $\mathbb{P}^n(K)$ is the set of all "terms" $(x_0: \dots : x_n)$ where $x_0, ..., x_n \in K$ and not all $x_0, ..., x_n$ are zero. We see that the map

$$\mathbb{A}^n(K) \to \{(x_0: \dots:x_n) \in \mathbb{P}^n(K) | \; x_0 \neq 0\}, (x_1, \dots, x_n) \mapsto (1:x_1:\dots:x_n)$$

and the complement

$$\mathbb{P}^{n-1}(K) \to \{(0:x_1:\dots:x_n) \in \mathbb{P}^n(K)\}, (x_1 : \dots : x_n) \mapsto (0:x_1:\dots:x_n)$$

are bijective.

Question 2: I can see why these two functions are bijective. But what exactly do we need them for? What is their meaning? And why is the second map called "the complement"?

We have $\mathbb{P}^1(K) = \mathbb{A}^1(K) \sqcup \mathbb{P}^0(K)$, where $\mathbb{P}^1(K)$ is a projective line and $\mathbb{P}^0(K) = \{\infty\}$.

Question 3: How does one obtain this equality? I mean $\mathbb{P}^1(K)$ is a set containing equivalence classes whereas $\mathbb{A}^1(K)$ is a set only containing 1-dimensional points. Also, why is $\mathbb{P}^0(K)=\{\infty\}$?

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Q1. "Properties" is not the word you want; a better word is "structures." Affine space is a setting for doing affine geometry, the study of geometric notions invariant under affine transformations. This does not include lengths or angles (so we aren't doing Euclidean geometry), but does include

• points
• lines
• planes
• hyperplanes

and a notion of intersection of any of these. There are axioms for affine geometry, but I don't think most people think of affine geometry in terms of those axioms; following the Erlangen program it is more natural to think in terms of affine transformations.

Affine space also carries the structure of an affine variety. This means, in particular, that it is equipped with a topology, the Zariski topology, as well as a notion of regular function (function defined by polynomials).

Q2. The map establishes that a subset of projective space can be identified with affine space. You should think of projective space as affine space with the addition of "points at infinity" (roughly speaking corresponding to certain limits that you want to exist in affine space but that don't) and this map formalizes that idea.

Q3. Use the map from Q2 and look at the set of points it doesn't hit.

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I have two follow-up questions: 1. You say that an affine space is some space not including lengths or angles. Also, you mention affine transformations. Wikipedia states that affine transformations maintains parting ratios (if applied to three collinear points). Isn't this a contradiction? How can one obtain maintenance of parting ratios if one does not speak about lengths or angles? 2. (Q3) For $n=1$, the map doesn't hit points in the equivalence class $(0:1)$. Still, to my mind, $\mathbb{P}^0(K) = \{(1)\}$ and I don't see how this could turn into $\{\infty\}$. Can you explain? –  Huy Dec 5 '11 at 12:12
@Huy: I don't know what a parting ratio is. $\infty$ is just a name for the point $(0 : 1)$. –  Qiaochu Yuan Dec 5 '11 at 12:20
On english wikipedia, it is called "ratios of distances". How do I get the zero as a first "coordinate"? Isn't the disjoint union of two sets $X$ and $Y$ defined as $X \sqcup Y = \{(1,x),(2,y)| \; x \in X, y \in Y\}$? –  Huy Dec 5 '11 at 12:29
@Huy: affine transformations preserve proportions, which are not the same as lengths (proportions do not have units and lengths do). As for disjoint unions, you are getting too hung up on set-theoretic stuff that does not matter. $\mathbb{P}^0(K)$ is a point. It doesn't matter what you call that point. –  Qiaochu Yuan Dec 5 '11 at 13:33
I understand the motivation for $\infty$ being a name for the point $(0:1)$ in this case. But can you explain where this point is in $\mathbb{A}^1(K) \sqcup \mathbb{P}^0(K)$? Or maybe the problem lies elsewhere: Can you give me a definition of the disjoint union (which is used here)? –  Huy Dec 5 '11 at 13:40

Since you're working over an arbitrary field, I am going to assume that you want to learn algebraic geometry.

1. I would avoid axioms for now. At this stage it seems to me that the reason one writes $\mathbf{A}^n$ instead of $K^n$ is to remind that the topology on, for example, $\mathbf{A}^n(\mathbf{C})$ is the Zariski topology and not the old metric topology. However, later on $\mathbf{A}^n$ will be the set of prime ideals of the polynomial ring $K[x_1, \ldots, x_n]$, which has far more points.

2. Projective varieties offer us a lot of theoretical advantages, and yet they are seemingly difficult to work with: points are equivalence classes, it's harder to write down functions, etc. But in geometry it's often enough to work locally on arbitrarily small open subsets, and your first map shows that you can then work with a subset of affine space, which might be notationally and psychologically easier. Think of charts and local coordinates on a manifold.

The map also shows that you can take an affine variety and shove it into projective space, where it will enjoy certain benefits—it's best to do examples. Take the zero set of $y = x^2$ in $\mathbf{A}^2$; if we send this through your map $\mathbf{A}^2 \to \mathbf{P}^2$ and take the Zariski closure, thinking of the coordinates on $\mathbf{P}^2$ as being $(W, X, Y)$, we get the zero set of $YW = X^2$ [I'm not explaining why this is the right transformation, but note the need for a homogeneous polynomial if I want to make sense of "the zeros of the polynomial in $\mathbf{P}^n$"]. The points $(a, b)$ which satisfy the equation in $\mathbf{A}^2$ correspond under your map to points $(1, a, b)$ in $\mathbf{P}^2$. But there's one new point $(0, 0, 1)$ which keeps track of how the curve becomes extremely vertical at infinity. That's an improvement!

The second map is called a "complement" because it hits exactly those points that the first map does not.

3. I think Qiaochu has answered this (and your other questions) well.

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I can't follow your example. For $y = x^2$ in $\mathbf{A}^2$, we have points of the form $(x,x^2)$ (is this what you mean by the condition?). Sending it through the first map in my question, we get equivalence classes of the form $(1:x:x^2)$. Is this correct? If not, why not? If yes, how is this related to your statements "taking the Zariski closure, we get ... and there's a new point $(0,1,0)$ ..."? (I don't understand that part of your answer) –  Huy Dec 5 '11 at 12:20
@Huy Two things I could have written better: (1) I was thinking of the version of your map that sends $(a, b) \mapsto (a, b, 1)$, because I wanted to think of $x = X/Z$, $y = Y/Z$. There a lot of copies of $\mathbf{A}^2$ inside of $\mathbf{P}^2$, and I should have used yours. My mistake. (2) I'm glossing over how the closure is taken, mostly because I don't want to rewrite the first two sections of Hartshorne! I'll try to patch this up later. I would recommend reading those two sections, though. –  Dylan Moreland Dec 5 '11 at 13:46
[Sorry about the confusion. I probably dashed this off too quickly. Are you learning algebraic geometry, by the way, or have I seriously misjudged?] –  Dylan Moreland Dec 5 '11 at 13:57
Regarding your second comment, I am attending a lecture called "geometry" and we have just started to talk about projective space. We will only have four more lectures and I don't know whether this would suffice to learn about algebraic geometry. –  Huy Dec 5 '11 at 20:39
@Huy Interesting. I really only understand these things (to the small degree that I understand them) in the algebraic setting, so I may be of little help to you. I made my assumption because usually a geometer will specialize to $K = \mathbf R$ or $\mathbf C$. Anyway, let me know if I can say more about this. –  Dylan Moreland Dec 5 '11 at 20:42