nLab
projective space

Contents

Definition

Let TT be an abelian Lawvere theory (one containing the theory of abelian groups). Write 𝔸 1\mathbb{A}^1 for its canonical line object and 𝔾 m\mathbb{G}_m for the corresponding multiplicative group object.

The projective space n\mathbb{P}_n of TT is the quotient

n:=(𝔸 n+1{0})/𝔾 m \mathbb{P}_n := (\mathbb{A}^{n+1} - \{0\})/\mathbb{G}_m

of the (n+1)(n+1)-fold product of the line with itself by the canonical action of 𝔾 m\mathbb{G}_m. Any point (x 0,x 1,,x n)𝔸 n+1{0}(x_0,x_1,\ldots,x_n)\in \mathbb{A}^{n+1} - \{0\} gives homogeneous coordinates for its image under the quotient map. When considered in this fashion, one often writes [x 0:x 1::x n][x_0:x_1:\ldots:x_n]. Homogeneous coordinates were introduced in Möbius 27

More generally, for (X,0)(X,0) a pointed space with (pointed) 𝔾 m\mathbb{G}_m-action, the quotient

(X):=(X{0})/𝔾 m \mathbb{P}(X) := (X-\{0\})/\mathbb{G}_m

is the corresponding projective space.

If instead of forming the quotient one forms the weak quotient/action groupoid, one speaks of the projective stack

^(X)(X{0})//𝔾 m. \hat \mathbb{P}(X) \coloneqq (X-\{0\})//\mathbb{G}_m \,.

Examples

For commutative rings and algebras

For TT the theory of commutative rings or more generally commutative associative algebras over a ring kk, 𝔸 k 1\mathbb{A}_k^1 is the standard affine line over kk. In this case k n\mathbb{P}^n_k is (…) A closed subscheme of k n\mathbb{P}^n_k is a projective scheme?.

Proposition

For RR a commutative kk-algebra, there is a natural isomorphism between

  • \mathbb{Z}-gradings on RR;

  • 𝔾 m\mathbb{G}_m-actions on SpecRSpec R.

The proof is spelled out at affine line.

Real and complex projective space

We discuss how complex projective space for kk the real numbers or the complex numbers equipped with their Euclidean metric topology is a topological manifold and naturally carries the structure of a smooth manifold (prop. 5 below).

For more see at these entries.

Definition

(topological projective space)

Let nn \in \mathbb{N}. Consider the Euclidean space k n+1k^{n+1} equipped with its metric topology, let k n+1{0}k n+1k^{n+1} \setminus \{0\} \subset k^{n+1} be the topological subspace which is the complement of the origin, and consider on its underlying set the equivalence relation which identifies two points if they differ by multiplication with some ckc \in k (necessarily non-zero):

(x 1x 2)(ck(x 2=cx 1)). (\vec x_1 \sim \vec x_2) \;\Leftrightarrow\; \left( \underset{c \in k}{\exists} ( \vec x_2 = c \vec x_1 ) \right) \,.

The equivalence class [x][\vec x] is traditionally denoted

[x 1:x 2::x n+1]. [x_1 : x_2 : \cdots : x_{n+1}] \,.

Then the projective space kP nk P^n is the corresponding quotient topological space

kP n(k n+1{0})/. k P^n \;\coloneqq\; \left(k^{n+1} \setminus \{0\}\right) / \sim \,.
Lemma

(canonical inclusion of projective spaces)

For nn \in \mathbb{N} the function between topological projective spaces from def. 1 given by

kP n kP n+1 [x 1::x n+1] [x 1::x n+1:0] \array{ k P^n &\overset{}{\longrightarrow}& k P^{n+1} \\ [x_1 : \cdots : x_{n+1}] &\mapsto& [ x_1 : \cdots : x_{n+1} : 0] }

is a continuous function.

Proof

There is a commuting square of functions of underlying sets of the form

(x 1,,x n+1) (x 1,,x n+1,0) k n+1 k n+2 kP n kP n+1 [x 1::x n+1] [x 1::x n+1:0], \array{ (x_1, \cdots, x_{n+1}) &\mapsto& (x_1, \cdots, x_{n+1}, 0) \\ k^{n+1} & \overset{}{\longrightarrow} & k^{n+2} \\ \downarrow &\searrow& \downarrow \\ k P^n &\longrightarrow& k P^{n+1} \\ [x_1 : \cdots : x_{n+1}] &\mapsto& [ x_1 : \cdots : x_{n+1} : 0] } \,,

where the two vertical functions are the defining quotient co-projections, which are continuous functions by nature of quotient spaces. The top function is clearly continuous (polynomials are continuous) and hence so is its composite with the right co-projection, inducated by the diagonal arrow in the above diagram.

This implies that the bottom function is continuous by the nature (the universal property) of the quotient space topology.

Proposition

(projective space as quotient space of an nn-sphere)

For nn \in \mathbb{C} there are homeomorphisms

  1. S n/(/2)P nS^{n}/(\mathbb{Z}/2) \simeq \mathbb{R}P^n

    between

    1. the quotient space of the Euclidean n-sphere canonically regarded as a subspace of the Euclidean space n+1\mathbb{R}^{n+1} by the equivalence relation which identifies two points p n+1\vec p \in \mathbb{R}^{n+1} if they differ by multiplication by 1-1

    2. real projective space (def. 1)

  2. S 2n+1/S 1P nS^{2n+1}/S^1 \simeq \mathbb{C}P^{n}

    between

    1. the quotient space of the Euclidean (2n+2)-sphere, canonically regarded as a subspace of the Euclidean space 2n+2 n+1\mathbb{R}^{2n+2} \simeq \mathbb{C}^{n+1} by the equivalence relation which identifies two points p n+1\vec p \in \mathbb{C}^{n+1} if they differ by multiplication by an complex number of unit norm

    2. complex projective space (def. 1).

Proof

It is clear that there is a bijection of underlying sets as claimed: Under the equivalence relation defining projective space, every element x=(x 1,,x n+1)k n+1\vec x = (x_1, \cdots, x_{n+1}) \in k^{n+1} is equivalent to one of unit norm, namely 1|x|x\frac{1}{\vert \vec x\vert} \vec x, hence lying on the unit sphere. Representatives of this form are unique up to further multiplication by elements in k{0}k \setminus \{0\} of unit norm.

It remains to see that this bijection is a homeomorphism. For definiteness of notation, we discuss this for the case k=k = \mathbb{C}, the case k=k = \mathbb{R} works verbatim the same, with the evident substitutions.

So we have a commuting diagram of functions of underlying sets

S 2n+1 n+1{0} q S 2n+1 f q n+1 S 2n+1/S 1 P n \array{ S^{2n+1} &\hookrightarrow& \mathbb{C}^{n+1} \setminus \{0\} \\ {}^{\mathllap{q_{S^{2n+1}}}}\downarrow &\searrow^{\mathrlap{f}}& \downarrow^{\mathrlap{q_{\mathbb{C}^{n+1}}}} \\ S^{2n+1}/S^1 &\longrightarrow& \mathbb{C}P^n }

where the top horizontal and the two vertical functions are continuous, and where the bottom function is is a bijection. Since the diagonal composite is also continuous, the nature of the quotient space topology implies that the bottom function is also continuous. To see that it is a homeomorphism it hence remains to see that it is an open map (by this prop.).

So let US 2n+1/S 1U \subset S^{2n+1}/S^1 be an open set, which means that q S 2n+1 1(U)S 2n+1q_{S^{2n+1}}^{-1}(U) \subset S^{2n+1} is an open set. We need to see that f(q S 2n+1 1(U))P nf(q_{S^{2n+1}}^{-1}(U)) \subset \mathbb{C}P^{n} is open, hence that q n+1 1(f(q S 2n+1 1(U))) n+1q_{\mathbb{C}^{n+1}}^{-1}(f(q_{S^{2n+1}}^{-1}(U))) \subset \mathbb{C}^{n+1} is open. Now by the nature of the Euclidean metric topology, the open subset q S 2n+1 1(U)q_{S^{2n+1}}^{-1}(U) is a union of open balls B x (ϵ)B^\circ_x(\epsilon) in n+1\mathbb{C}^{n+1} intersected with S 2n+1S^{2n+1}. But then q n+1 1(f(B x (ϵ)| S 2n+1))q_{\mathbb{C}^{n+1}}^{-1}(f(B^\circ_x(\epsilon)\vert_{S^{2n+1}})) is their orbit under the multiplicative action by {0}\mathbb{C} \setminus \{0\}, hence is a cylinder B x (ϵ)| S 2n+1×({0})B^\circ_x(\epsilon)\vert_{S^{2n+1}} \times (\mathbb{C} \setminus \{0\}). This is clearly open.

Proposition

There is a CW-complex structure on real projective space P n\mathbb{R}P^n (def. 1) for nn \in \mathbb{N}, given by induction, where P n+1\mathbb{R}P^{n+1} arises from P n\mathbb{R}P^n by attaching a single cell of dimension n+1n+1 with attaching map the projection S nP nS^{n} \longrightarrow \mathbb{C}P^n from prop. 1:

S n S n/(/2)P n ι n+1 (po) D n+1 q P n+1. \array{ S^{n} &\longrightarrow& S^{n}/(\mathbb{Z}/2) \simeq \mathbb{R}P^n \\ {}^{\mathllap{\iota_{n+1}}}\downarrow &(po)& \downarrow \\ D^{n+1} &\underset{q}{\longrightarrow}& \mathbb{R}P^{n+1} } \,.

Similarly, there is a CW-complex structure on complex projective space P n\mathbb{C}P^n (def. 1) for nn \in \mathbb{N}, given by induction, where P n+1\mathbb{C}P^{n+1} arises from P n\mathbb{C}P^n by attaching a single cell of dimension 2(n+1)2(n+1) with attaching map the projection S 2n+1P nS^{2n+1} \longrightarrow \mathbb{C}P^n from prop. 1:

S 2n+1 S 2n+1/S 1P n ι 2n+2 (po) D 2n+2 q P n+1. \array{ S^{2n+1} &\longrightarrow& S^{2n+1}/S^1 \simeq \mathbb{C}P^n \\ {}^{\mathllap{\iota_{2n+2}}}\downarrow &(po)& \downarrow \\ D^{2n+2} &\underset{q}{\longrightarrow}& \mathbb{C}P^{n+1} } \,.
Proof

we discuss the case k=k = \mathbb{C}. The case k=k= \mathbb{R} works verbatim the same, with the evident substitutions.

Given homogeneous coordinates (z 0,z 1,,z n,z n+1,z n+2) n+2(z_0 , z_1 , \cdots , z_n , z_{n+1} , z_{n+2}) \in \mathbb{C}^{n+2} for P n+1\mathbb{C}P^{n+1}, let

ϕarg(z n+2) \phi \coloneqq -arg(z_{n+2})

be the phase of z n+2z_{n+2}. Then under the equivalence relation defining P n+1\mathbb{C}P^{n+1} these coordinates represent the same element as

1|z|(e iϕz 0,e iϕz 1,,e iϕz n+1,r), \frac{1}{\vert \vec z\vert}(e^{i \phi} z_0, e^{i \phi}z_1,\cdots, e^{i \phi}z_{n+1}, r) \,,

where

r=|z n+2|[0,1) r = {\vert z_{n+2}\vert}\in [0,1) \subset \mathbb{C}

is the absolute value of z n+2z_{n+2}. Representatives z\vec z' of this form (|z|=1{\vert \vec z' \vert = 1} and z n+2[0,1]z'_{n+2} \in [0,1]) parameterize the 2n+2-disk D 2n+2D^{2n+2} with boundary the (2n+1)(2n+1)-sphere at r=0r = 0.

The resulting function q:D 2n+2P n+1q \colon D^{2n+2} \to \mathbb{C}P^{n+1} is continuous: It may be factored as

q D 2n+2: D 2n+2 AAA n+2{0} q n+2 P n+1 (Re(z 1),Im(z 1),,Re(z n+1),Im(z n+1),r) (z 1,,z n+1,r) [z 1::z n+1:r] \array{ q_{D^{2n+2}} \colon & D^{2n+2} &\overset{\phantom{AAA}}{\hookrightarrow}& \mathbb{C}^{n+2} \setminus \{0\} &\overset{q_{\mathbb{C}^{n+2}}}{\longrightarrow}& \mathbb{C}P^{n+1} \\ & (Re(z_1), Im(z_1), \cdots, Re(z_{n+1}), Im(z_{n+1}), r) &\mapsto& (z_1, \cdots, z_{n+1}, r) &\mapsto& [ z_1 : \cdots : z_{n+1} : r ] }

and here the first map is the embedding of the disk D 2n+2D^{2n+2} as a hemisphere in 2n+1hookrightarow 2n+2 2n+2\mathbb{R}^{2n+1} \hookrightarow \mathbb{R}^{2n+2} \simeq \mathbb{C}^{2n+2}, while the second is the defining quotient space projection. Both of these are continuous, and hence so is their composite.

The only remaining part of the action of {0}\mathbb{C}-\{0\} which fixes the conditions |z|=0{\vert z'\vert} = 0 and z n+2z'_{n+2} is S 1{0}S^1 \subset \mathbb{C} \setminus \{0\} acting on the elements with r={z n+2}=0 r = \{z'_{n+2}\} = 0 by phase shifts on the z 0,,z n+1z_0, \cdots, z_{n+1}. The quotient of this remaining action on D 2(n+1)D^{2(n+1)} identifies its boundary S 2n+1S^{2n+1}-sphere with P n\mathbb{C}P^{n}, by prop. 1.

This shows that the above square is a pushout diagram of underlying sets.

By the nature of colimits in Top (this prop.) it remains to see that the topology on P n+1\mathbb{C}P^{n+1} is the final topology induced by the functions D 2n+2P n+1D^{2n+2} \to \mathbb{C}P^{n+1} and P nP n+1\mathbb{C}P^n \to \mathbb{C}P^{n+1}, hence that a subset of P n+1\mathbb{C}P^{n+1} is open precisely if its pre-images under these two functions are open.

We saw above that q D 2n+2q_{D^{2n+2}} is continuous. Moreover, also the function i n:P nP n+1i_n \colon \mathbb{C}P^n \to \mathbb{C}P^{n+1} is continuous (by this lemma).

This shows that if a subset of P n+1\mathbb{C}P^{n+1} is open, then its pre-images under these functions are open. It remains to see that if SP n+1S \subset \mathbb{C}P^{n+1} is a subset with q S 2n+2 1(S)D 2n+2q_{S^{2n+2}}^{-1}(S) \subset D^{2n+2} open and i n 1(S)P ni_n^{-1}(S) \subset \mathbb{C}P^n open, then SP n+1S \subset \mathbb{C}P^{n+1} is open.

Notice that q n+2 1(S)q_{\mathbb{C}^{n+2}}^{-1}(S) contains with every point also its orbit under the action of {0}\mathbb{C} \setminus \{0\}, and that every open subset of D 2n+2D^{2n+2} is a unions of open balls. By the above factorization of q D 2n+2q_{D^{2n+2}} this means that if q D 2n+2 1(S)q_{D^{2n+2}}^{-1}(S) is open, then q n+2 1(S)q_{\mathbb{C}^{n+2}}^{-1}(S) is a union of open cyclinders, hence is open. By the nature of the quotient topology, this means that SP nS \subset \mathbb{C}P^n is open.

Definition

(standard open cover of topological projective space)

For nn \in \mathbb{N} the standard open cover of the projective space kP nk P^n (def. 1) is

{U ikP n} i{1,,n+1} \left\{ U_i \subset k P^n \right\}_{i \in \{1, \cdots, n+1\}}

with

U i{[x 1::x n+1]kP n|x i0}. U_i \coloneqq \left\{ [x_1 : \cdots : x_{n+1}] \in k P^n \;\vert\; x_i \neq 0 \right\} \,.

To see that this is an open cover:

  1. This is a cover because with the orgin removed in k n{0}k^n \setminus \{0\} at every point [x 1::x n+1][x_1: \cdots : x_{n+1}] at least one of the x ix_i has to be non-vanishing.

  2. These subsets are open in the quotient topology kP n=(k n{0})/k P^n = (k^n \setminus \{0\})/\sim, since their pre-image under the quotient co-projection k n+1{0}kP nk^{n+1} \setminus \{0\} \to k P^n coincides with the pre-image (pr iι) 1(k{0})(pr_i\circ\iota)^{-1}( k \setminus \{0\} ) under the projection onto the iith coordinate in the product topological space k n+1=i{1,,n}kk^{n+1} = \underset{i \in \{1,\cdots, n\}}{\prod} k (where we write k n{0}ιk npr ikk^n \setminus \{0\} \overset{\iota}{\hookrightarrow} k^n \overset{pr_i}{\to} k).

Proposition

(n-sphere projecting to real projective space is covering space projection)

For nn \in \mathbb{N}, the continuous function p:S nP np \;\colon\; S^n \to \mathbb{R}P^n from prop. 1 is a covering space projection.

Proof

We need to produce an open cover {U iP n} iI\{U_i \subset \mathbb{R}P^n\}_{i \in I} such that the restrictions of the projection to this cover are homeomorphic over the base to a product topological space

U i×Disc( 2) S n| U i U i. \array{ U_i \times Disc(\mathbb{Z}_2) && \overset{\simeq}{\longrightarrow} && S^n|_{U_i} \\ & \searrow && \swarrow \\ && U_i } \,.

Consider the standard open cover from def. 2. Hence i{1,,n+1}i \in \{1, \cdots, n+1\} and U iU_i consists of those lines through the origin in n+1\mathbb{R}^{n+1} which do not lie in the subspace defined by x i=0x_i = 0. The intersection of this subspace with the unit sphere S n n+1S^n \subset \mathbb{R}^{n+1} is an equator of the nn-sphere, and so the complement of this equator is the disjoint union of the two open hemispheres D i ±S nD_i^\pm \subset S^n. Hence

S n| U i D i +D i . \array{ S^n\vert_{U_i} & \simeq D_i^+ \sqcup D_i^- } \,.

Moreover, each line in n+1\mathbb{R}^{n+1} which corresponds to an element in U iU_i intersects D i +D^+_i as well as D i D^-_i exactly once. In particular therefore the 2\mathbb{Z}_2-action on S nS^n restricts over U iU_i to the interchange of these two hemispheres, and hence prop. 1 gives the required homeomorphism as above.

Proposition

(standard open cover is atlas)

The charts of the standard open cover of def. 2 are homeomorphic to Euclidean space k nk^n.

Proof

If x i0x_i \neq 0 then

[x 1::x i::x n+1]=[x 1x i::1:x n+1x i] [x_1 : \cdots : x_i : \cdots : x_{n+1}] = \left[ \frac{x_1}{x_i} : \cdots : 1 : \cdots \frac{x_{n+1}}{x_i} \right]

and the representatives of the form on the right are unique.

This means that

n ϕ i U i (x 1,,x i1,x i+1,,x n+1) [x 1::1::x n+1] \array{ \mathbb{R}^n &\overset{\phi_i}{\longrightarrow}& U_i \\ (x_1, \cdots, x_{i-1}, x_{i+1}, \cdots, x_{n+1}) &\mapsto& [x_1: \cdots: 1: \cdots : x_n+1] }

is a bijection of sets.

To see that this is a continuous function, notice that it is the composite

n+1{x i=0} ϕ^ i n ϕ i U i \array{ && \mathrlap{\mathbb{R}^{n+1} \setminus \{x_i = 0\}} \\ & {}^{\mathllap{\hat \phi_i}}\nearrow & \downarrow \\ \mathbb{R}^n & \underset{\phi_i}{\longrightarrow} & U_i }

of the function

n ϕ^ i n+1{x i=0} (x 1,,x i1,x i+1,,x n+1) (x 1,,1,,x n+1) \array{ \mathbb{R}^n &\overset{\hat \phi_i}{\longrightarrow}& \mathbb{R}^{n+1} \setminus \{x_i = 0\} \\ (x_1, \cdots, x_{i-1}, x_{i+1}, \cdots, x_{n+1}) &\mapsto& (x_1, \cdots, 1, \cdots ,x_n+1) }

with the quotient projection. Now ϕ^ i\hat \phi_i is a polynomial function and since polynomials are continuous, and since the projection to a quotient topological space is continuous, and since composites of continuous functions are continuous, it follows that ϕ i\phi_i is continuous.

It remains to see that also the inverse function ϕ i 1\phi_i^{-1} is continuous. Since

n+1{x i=0} U i ϕ i 1 n (x 1,,x n+1) (x 1x i,,x i1x i,x i+1x i,,x n+1x i) \array{ \mathbb{R}^{n+1} \setminus \{x_i = 0\} &\overset{}{\longrightarrow}& U_i &\overset{\phi_i^{-1}}{\longrightarrow}& \mathbb{R}^n \\ (x_1, \cdots, x_{n+1}) && \mapsto && ( \frac{x_1}{x_i}, \cdots, \frac{x_{i-1}}{x_i}, \frac{x_{i+1}}{x_i}, \cdots , \frac{x_{n+1}}{x_i}) }

is a rational function, and since rational functions are continuous, it follows, by nature of the quotient topology, that ϕ i\phi_i takes open subsets to open subsets, hence that ϕ i 1\phi_i^{-1} is continuous.

Proposition

(real/complex projective space is smooth manifold)

For k{,}k \in \{\mathbb{R}, \mathbb{C}\} the topological projective space kP nk P^n (def. 1) is a topological manifold.

Equipped with the standard open cover of def. 2 regarded as an atlas by prop. 4, it is a differentiable manifold, in fact a smooth manifold.

Proof

By prop. 4 kP nk P^n is a locally Euclidean space. Moreover, kP nkP^n admits the structure of a CW-complex (by prop. 2) and therefore it is a paracompact Hausdorff space since CW-complexes are paracompact Hausdorff spaces. This means that it is a topological manifold.

It remains to see that the gluing functions of this atlas are differentiable functions and in fact smooth functions. But by prop. 4 they are even rational functions.

References

An introduction to projective spaces over the theory of ordinary commutative rings is in

  • Miles Reid, Graded rings and varieties in weighted projective space (pdf)

  • Aurelio Carboni, Marco Grandis , Categories of projective spaces , JPAA 110 (1996) pp.241-258.

Revised on August 9, 2017 13:21:06 by Urs Schreiber (131.220.184.222)