2.3 Examples of Manifolds 21 2.3 Examples of Manifolds We will now discuss some basic examples of manifolds. In each case, the manifold structure is given by a finite atlas; hence the countability property is immediate. We will not spend too much time on verifying the Hausdorff property; while it may be done ‘by hand’, we will later have better ways of doing this. 2.3.1 Spheres The construction of an atlas for the 2-sphere S2, by stereographic projection, also works for the n-sphere Sn = (x0,...,xn) (x0)2 + +(xn)2 = 1 . { | ··· } Let U be the subsets obtained by removing ( 1,0,...,0). Stereographic projection ± ⌥ defines bijections j : U Rn, where j (x0, x1,...,xn)=(u1,...,un) with ± ± ! ± xi ui = . 1 x0 ± For the transition function one finds (writing u =(u1,...,un)) 1 u (j j+− )(u)= . − ◦ u 2 || || We leave it as an exercise to check the details. An equivalent atlas, with 2n+2 charts, + + is given by the subsets U0 ,...,Un ,U0−,...,Un− where + n j n j U = x S x > 0 , U− = x S x < 0 j { 2 | } j { 2 | } n for j = 0,...,n, with j± : U± R the projection to the j-th coordinate plane (in j j ! other words, omitting the j-th component x j): 0 n 0 j 1 j+1 n j±j (x ,...,x )=(x ,...,x − ,x ,...,x ). 2.3.2 Real projective spaces The n-dimensional projective space RPn, is the set of all lines ` Rn+1. It may also ✓ be regarded as a quotient space‡ n n+ RP =(R 1 0 )/ \{ } ⇠ for the equivalence relation x x0 l R 0 : x0 = lx. ⇠ ,9 2 \{ } ‡ See the appendix to this chapter for some background on quotient spaces. 22 2 Manifolds n+1 Indeed, any x R 0 determines a line, while two points x,x0 determine the 2 \{ } same line if and only if they agree up to a non-zero scalar multiple. The equivalence class of x =(x0,...,xn) under this relation is commonly denoted [x]=(x0 : ...: xn). RPn has a standard atlas A = (U ,j ),...,(U ,j ) { 0 0 n n } defined as follows. For j = 0,...,n, let 0 n n j Uj = (x : ...: x ) RP x = 0 { 2 | 6 } be the set for which the j-th coordinate is non-zero, and put 0 j 1 j+1 n n 0 n x x − x x j j : Uj R , (x : ...: x ) ( ,..., , ,..., ). ! 7! x j x j x j x j This is well-defined, since the quotients do not change when all xi are multiplied by a fixed scalar. Put differently, given an element [x] RPn for which the j-th component 2 x j is non-zero, we first rescale the representative x to make the j-th component equal to 1, and then use the remaining components as our coordinates. As an example (with n = 2), 7 2 7 2 j (7:3:2)=j :1: = , . 1 1 3 3 3 3 n From this description, it is immediate that j j is a bijection from Uj onto R , with inverse map 1 1 n 1 j j+1 n j−j (u ,...,u )=(u : ...: u :1:u : ...: u ). n n+1 n Geometrically, viewing RP as the set of lines in R , the subset Uj RP ✓ consists of those lines ` which intersect the affine hyperplane n+1 j Hj = x R x = 1 , { 2 | } and the map j j takes such a line ` to its unique point of intersection ` Hj, followed n j \ by the identification Hj = R (dropping the coordinate x = 1). ⇠ n Let us verify that A is indeed an atlas. Clearly, the domains Uj cover RP , since any element [x] RPn has at least one of its components non-zero. For i = j, the 2 6 intersection U U consists of elements x with the property that both components i \ j xi, x j are non-zero. 1 Exercise 8. Compute the transition maps ji j− , and verify they are smooth. (Hint: You will need to distinguish between the cases and .) i j i j > ◦ j < To complete the proof that this atlas (or the unique maximal atlas containing it) defines a manifold structure, it remains to check the Hausdorff property. This can be done with the help of Lemma 2.2, but we postpone the proof since we will soon have a simple argument in terms of smooth functions. See Proposition 3.1 below. 2.3 Examples of Manifolds 23 Remark 2.4. In low dimensions, we have that RP0 is just a point, while RP1 is a circle. n+1 Remark 2.5. Geometrically, Ui consists of all lines in R meeting the affine hyper- plane Hi, hence its complement consists of all lines that are parallel to Hi, i.e., the i n 1 lines in the coordinate subspace defined by x = 0. The set of such lines is RP − . In n n 1 other words, the complement of Ui in RP is identified with RP − . Thus, as sets, RPn is a disjoint union n n n 1 RP = R RP − , t n n 1 where R is identified (by the coordinate map ji) with the open subset Un, and RP − with its complement. Inductively, we obtain a decomposition n n n 1 0 RP = R R − R R , t t···t t where R0 = 0 . At this stage, it is simply a decomposition into subsets; later it will { } be recognized as a decomposition into submanifolds. Exercise 9. Find an identification of the space of rotations in R3 with the 3- 3 samedimensional rotation.) projective space RP . || || − p determined by . Note that for , the vectors and determine the v v v v = 6 || || rotation, if 0, take the rotation by an angle around the oriented axis v v = 2 R (Hint: Associate to any a rotation, as follows: If 0, take the trivial v v = 3 2.3.3 Complex projective spaces In a similar fashion to the real projective space, one can define a complex projective space CPn as the set of complex 1-dimensional subspaces of Cn+1. If we identify C with R2, thus Cn+1 with R2n+2, we have n n+ CP =(C 1 0 )/ \{ } ⇠ where the equivalence relation is z z if and only if there exists a complex l with ⇠ 0 z0 = lz. (Note that the scalar l is then unique, and is non-zero.) Alternatively, letting S2n+1 Cn+1 = R2n+2 be the ‘unit sphere’ consisting of complex vectors of length ✓ z = 1, we have || || n n+ CP = S2 1/ , ⇠ where z z if and only if there exists a complex number l with z = lz. (Note that 0 ⇠ 0 the scalar l is then unique, and has absolute value 1.) One defines charts (Uj,j j) similarly to those for the real projective space: 0 n j n 2n Uj = (z : ...: z ) z = 0 , j j : Uj C = R , | 6 ! 24 2 Manifolds z0 z j 1 z j+1 zn j (z0 : ...: zn)= ,..., − , ,..., . j z j z j z j z j ⇣ ⌘ The transition maps between charts are given by similar formulas as for RPn (just replace x with z); they are smooth maps between open subsets of Cn = R2n. Thus n n CP is a smooth manifold of dimension 2n§. As with RP there is a decomposition n n n 1 0 CP = C C − C C . t t···t t 2.3.4 Grassmannians The set Gr(k,n) of all k-dimensional subspaces of Rn is called the Grassmannian of k-planes in Rn. (Named after Hermann Grassmann (1809-1877).) n 1 As a special case, Gr(1,n)=RP − . We will show that for general k, the Grassmannian is a manifold of dimension dim(Gr(k,n)) = k(n k). − An atlas for Gr(k,n) may be constructed as follows. The idea is to present linear k n k k subspaces of dimension k as graphs of linear maps from R to R − . Here R is viewed as the coordinate subspace corresponding to a choice of k components from 1 n n n k x =(x ,...,x ) R , and R − the coordinate subspace for the remaining coordi- 2 nates. To make it precise, we introduce some notation. For any subset I 1,...,n of the set of indices, let ✓{ } I0 = 1,...,n I { }\ be its complement. Let RI Rn be the coordinate subspace ✓ I n i R = x R x = 0 for all i I0 . { 2 | 2 } I I I If I has cardinality I = k, then R Gr(k,n). Note that R 0 =(R )?. Let | | 2 I0 UI = E Gr(k,n) E R = 0 . { 2 | \ { }} § The transition maps are not only smooth but even holomorphic, making CPn into an exam- ple of a complex manifold (of complex dimension n). 2.3 Examples of Manifolds 25 I I Each E UI is described as the graph of a unique linear map AI : R R 0 , 2 ! that is, I E = y + AI(y) y R . { | 2 } Exercise 10. Let E be as above. Show that there is a unique linear map AI : I I I R R 0 such that E = y + AI(y) y R . ! { | 2 } This gives a bijection I I0 jI : UI Hom(R ,R ), E jI(E)=AI, ! 7! where Hom(F,F0) denotes the space of linear maps from a vector space F to a vector I I k(n k) I I space F0. Note Hom(R ,R 0 ) = R − , because the bases of R and R 0 identify the ⇠ k(n k) space of linear maps with (n k) k-matrices, which in turn is just R − by listing − ⇥ the matrix entries.