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2.3 Examples of 21 2.3 Examples of Manifolds

We will now discuss some basic examples of manifolds. In each case, the 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

The construction of an atlas for the 2- S2, by stereographic , 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). ± ⌥ defines 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 (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

The n-dimensional projective RPn, is the of all lines ` Rn+1. It may also ✓ be regarded as a quotient space‡

n n+ RP =(R 1 0 )/ \{ } ⇠ for the

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 , while two points x,x0 determine the 2 \{ } same line if and only if they agree up to a non-zero scalar multiple. The 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 from Uj onto R , with inverse 1 1 n 1 j j+1 n jj (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 n+1 j Hj = x R x = 1 , { 2 | } and the map j j takes such a line ` to its unique of ` 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 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 , we have that RP0 is just a point, while RP1 is a .

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 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 .

Exercise 9. Find an identification of the space of in R3 with the 3-

3

samedimensional .) 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 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 , one can define a 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 ‘’ 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 l with z = lz. (Note that 0 ⇠ 0 the scalar l is then unique, and has 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 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

The set Gr(k,n) of all k-dimensional subspaces of Rn is called the of k-planes in Rn. (Named after (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 (of complex dimension n). 2.3 Examples of Manifolds 25

I I Each E UI is described as the graph of a unique 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 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. In terms of A , the subspace E U is the range of the injective I 2 I linear map 1 I I I n : R R R 0 = R (2.1) AI ! ⇠ ✓ ◆ where we write elements of Rn as column vectors. To check that the charts are compatible, suppose E U U , and let A and A 2 I \ J I J be the linear maps describing E in the two charts. We have to show that the map

1 I I0 J J0 jJ j : Hom(R ,R ) Hom(R ,R ), AI = jI(E) AJ = jJ(E) I ! 7! is smooth. By assumption, E is described as the range of (2.1) and also as the range of a similar map for J. Here we are using the identifications RI RI0 = Rn and ⇠ RJ RJ0 = Rn. It is convenient to describe everything in terms of RJ RJ0 . Let ⇠ ab I I J J : R R 0 R R 0 cd ! ✓ ◆ be the matrix corresponding to the identification RI RI0 Rn followed by the in- ! verse of RJ RJ0 Rn. For example, c is the inclusion RI Rn as the correspond- ! ! J ing coordinate subspace, followed by projection to the coordinate subspace R 0 . ¶ We then get the condition that the injective linear maps

¶ Put differently, the matrix is the permutation matrix ‘renumbering’ the coordinates of Rn. 26 2 Manifolds

ab 1 I J J 1 J J J : R R R 0 , : R R R 0 cd AI ! AJ ! ✓ ◆✓ ◆ ✓ ◆ have the same range. In other words, there is an S : RI RJ such that ! ab 1 1 = S cd A A ✓ ◆✓ I ◆ ✓ J ◆ as maps RI RJ RJ0 . We obtain ! a + bA S I = c + dA A S ✓ I ◆ ✓ J ◆ Using the first row of this to eliminate the second row of this equation, we obtain the formula 1 AJ =(c + dAI)(a + bAI) .

The dependence of the right hand side on the matrix entries of AI is smooth, by Cramer’s formula for the inverse matrix. It follows that the collection of all jI : k(n k) UI R defines on Gr(k,n) the structure of a manifold of dimension k(n k). ! The number of charts of this atlas equals the number of subsets I 1,...,n of n ✓{ } cardinality k, that is, it is equal to k . (The Hausdorff property may be checked n in a similar fashion to RP . Alternatively, given distinct E1,E2 Gr(k,n), choose a 2 subspace F Gr(k,n) such that F has zero intersection with both E ,E . (Such a 2 ? 1 2 subspace always exists.) One can then define a chart (U,j), where U is the set of subspaces E transverse to F?, and j realizes any such map as the graph of a linear map F F . Thus j : U Hom(F,F ). As above, we can check that this is ! ? ! ? compatible with all the charts (UI,jI). Since both E1,E2 are in this chart U, we are done by Lemma 2.2.)

Exercise 11. Prove the parenthetical remark above: If E ,E Gr(k,n) are dis- 1 2 2 tinct, show that there exists F Gr(k,n) such that F E = F E = 0 . 2 ? \ 1 ? \ 2 { }

n 1 Remark 2.6. As already mentioned, Gr(1,n)=RP . One can check that our sys- n 1 tem of charts in this case is the standard atlas for RP .

Exercise 12. This is a preparation for the following remark. Recall that a linear map P : Rn Rn is an orthogonal projection onto some subspace E Rn ! ✓ if P(x)=x for x E and P(x)=0 for x E . Show that a matrix 2 2 ? P Mat (n) is the matrix of an orthogonal projection if and only if it has the 2 R properties P> = P and PP = P.

What is the matrix of the orthogonal projection onto E?? 2.3 Examples of Manifolds 27

Remark 2.7. For any k-dimensional subspace E Rn, let P E : Rn Rn be the linear ✓ ! map given by orthogonal projection onto E, and let P Mat (n) be its matrix. By E 2 R the exercise, PE> = PE , PE PE = PE ,

Conversely, any square matrix P with the properties P> = P, PP = P with rank(P)= k is the orthogonal projection onto a subspace Px x Rn Rn. This identifies the { | 2 }✓ Grassmannian Gr(k,n) with the set of orthogonal projections of rank k. In summary, we have an inclusion

n2 Gr(k,n) , Mat (n) = R , E PE . ! R ⇠ 7! By construction, this inclusion take values in the subspace Sym (n) n(n+1)/2 of R ⇠= R symmetric n n-matrices. ⇥ Remark 2.8. For all k, there is an identification Gr(k,n) = Gr(n k,n) (taking a k- ⇠ dimensional subspace to the orthogonal subspace).

Remark 2.9. Similar to RP2 = S2/ , the quotient modulo antipodal identification, ⇠ one can also consider M =(S2 S2)/ ⇥ ⇠ the quotient space by the equivalence relation

(x,x0) ( x, x0). ⇠ It turns out that this manifold M is the same as Gr(2,4), where ‘the same’ is meant in the sense that there is a bijection of sets identifying the atlases. Note that this is 3 the first genuinely new manifold, since Gr(1,4)=RP , and Gr(3,4) ⇠= Gr(1,4) by the previous remark. Question: What about the other Gr(k,n) with n 4? 

2.3.5 Complex Grassmannians

Similar to the case of projective spaces, one can also consider the complex Grass- n mannian GrC(k,n) of complex k-dimensional subspaces of C . It is a manifold of dimension 2k(n k), which can also be regarded as a complex manifold of complex dimension k(n k).