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2.3 Examples of Manifolds 21 2.3 Examples of Manifolds 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.
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