Download MA455 Manifolds Solutions 1 May 2008 1. (i) Given real numbers a

MA455 Manifolds Solutions 1 May 2008
1. (i) Given real numbers a < b, find a diffeomorphism (a, b) → R.
Solution: For example first map (a, b) to (0, π/2) and then map (0, π/2) diffeomorphically to R using the function
tan.
(ii) Find a diffeomorphism (0, ∞) → R.
Solution: Use log : (0, ∞) → R.
(iii) Let f : R → R be a smooth map, and let graph(f ) be the set {(x, y) ∈ R × R : y = f (x)}. Show that the map
x 7→ (x, f (x)) is a diffeomorphism from R to graph(f ).
Solution: The map x 7→ (x, f (x)) is clearly smooth and has smooth (new) inverse obtained as the restriction to
graph(f ) of the smooth (old) projection (x, y) 7→ x.
(iv) Let C0 be the cylinder S 1 × (0, ∞) ⊂ and let C be the cylinder S 1 × R (you can draw both in R3 ). Find
diffeomorphisms C0 → R2 \ {0} and C → R2 \ {0}.
Solution: C0 = {(x, y, z) ∈ R2 : x2 + y 2 = 1, 0 < z < ∞}. Define a smooth map φ : C0 → R2 \ {(0, 0)} by
φ(x, y, z) = (zx, zy). I leave you to find its inverse. The diffeomorphism of (ii) enables you to define a diffeo
ψ
C0 = (0, ∞) × S 1 −→ R × S 1 = C, and now φ ◦ ψ −1 is a diffeo C → R2 \ {(0, 0)}.
2. (i) Let ∆n be the regular n-simplex
{(t1 , . . ., tn+1 ) ∈ Rn+1 :
X
ti = 1, ti ≥ 0 for all i}.
i
Draw ∆2 .
Solution:
(ii) Let Bn be the set
{(x1 , . . ., xn )} ∈ Rn : 0 ≤ x1 ≤ · · · ≤ xn ≤ 1}.
Find a diffeomorphism Bn → ∆n . Hint: given (x1 , . . ., xn ) ∈ Bn , there is a natural choice of an (n + 1)-tuple of
non-negative numbers with sum equal to 1.
Solution: (x1 , . . ., xn ) 7→ (x1 , x2 − x1 , . . ., xn − xn−1 , 1 − xn ).
3. Show that if U ⊂ Rn is open, f : U → Rn is smooth, and det(dx f ) 6= 0 for all x ∈ U , then f (U ) is open in Rn .
Note that in particular this holds if f : U → f (U ) is a diffeomorphism.
Solution: Use the inverse function theorem.
4. Let X0 be the subset of R2 consisting of the three lines {x1 = 0}, {x2 = 0} and {x1 = x2 }. Let `1 , `2 and `3 be
any three distinct lines through (0, 0) in R2 , and let X = `1 ∪ `2 ∪ `3 . Find a diffeomorphism — indeed, a linear
isomorphism — taking X0 to X.
Solution: Let v1 and v2 be non-zero vectors lying on `1 and `2 respectively. Define a linear isomorphism R2 → R2
by sending (1, 0) to v2 , (0, 1) to v1 and extending linearly. This maps {x1 = 0} to `1 , {x2 = 0} to `2 and {x1 = x2 }
to a third line. A linear isomorphism whose matrix in the basis v1 , v2 is diagonal will re-scale the two axes so that
this line becomes the line `3 .
5. Show that if X ⊂ Y ⊂ Rn are manifolds and i : X → Y is inclusion then for each x ∈ X, dx i : Tx X → Tx Y is
also inclusion. It’s obvious if you use the right definition of derivative.
Solution: Use the definition of dx i as the map sending γ 0 (0) to i ◦ γ 0 (0) (version (i) of definition 1.13).
6. i) Let f : X × Y → X be the projection. Show that d(x,y) f : T(x,y) X × Y → Tx X is also projection.
ii)Fixing any y ∈ Y gives an injection mapping f : X → X×Y , defined by f (x) = (x, y). Show that dx f (x̂) = (x̂, 0).
iii) Let f : X → Y, g : X 0 → Y 0 be smooth maps, and define f × g : X × Y → X 0 × Y 0 by f × g(x, y) = (f (x), g(y)).
Show that d(x,y) (f × g) = dx f × dy g.
Solution: (i) Suppose that X ⊂ RN and Y ⊂ RP . The projection X × Y → X is the restriction to X × Y of the
projection π : RN × RP → RN . So its derivative is the restriction to T(x,y) X × Y = Tx X × Ty Y of the derivative
of π. But π is linear, so d(x,y) π = π.
(ii) Use Definition 1.13(i).
(iii) Check first that this is true for smooth (old) maps. Then if F and G are smooth (old) extensions of f and g
respectively, F × G is a smooth (old) extension of f ×. So d(x,y) f × g is the restriction to T(x,y) X × Y = Tx X × Ty Y
of d(x,y) F × G, which you have checked is equal to dx F × dy G.
7. (i) Prove that Rk and R` are not diffeomorphic if k 6= `.
Solution: Suppose g : R` → Rk is a smooth inverse to f — that is, g ◦ f = idRk , f ◦ g = idR` . Then for each
x ∈ Rk and y ∈ R` , we have df (x) g ◦ dx f = dx idRk = idRk and dg(y) f ◦ dy g = dx idR` = idR` . Thus both derivatives
are linear isomorphisms, and so k = `.
(ii) Prove that smooth manifolds M, N of dimension k, ` cannot be diffeomorphic if k 6= `.
Solution: Let x ∈ M and y = f (x) ∈ N , and let φ and ψ be charts around x and y respectively. If f is a
diffeomorphism then so is ψ ◦ f ◦ φ−1 . Now the argument of (i) applies, and shows k = `.
8. Prove that if M k ⊂ Rn and N ` ⊂ Rp are smooth manifolds then M × N ⊂ Rn+p is a smooth manifold of
dimension k + `.
Solution: If (x, y) ∈ M × N , let φ : U1 → V1 be a chart on M around x and let φ2 : U2 → V2 be a chart on N
around y. Then U1 ×U2 is a neighbourhood of (x, y) in M ×N and φ1 ×φ2 : U1 ×U2 → V1 ×V2 is a diffeomorphism.
As V1 × V2 is open in Rk × R` = Rm+` , M × N is a manifold of dimension k + `.
9. A Lie group is a manifold G which is also a group, for which the operations of
(
(
p:G×G → G
i:G→G
multiplication:
and inversion:
(g1 , g2 ) 7→ g1 g2
g 7→ g −1
2
are smooth maps. Show that Gl(n, R) := {invertible n × n real matrices under multiplication} ⊂ Matn (R) = Rn
is a Lie group.
2
Solution: Gl(n, R) is an open subset of matrix space, which is naturally identified with Rn . The group operations
are matrix multiplication and matrix inversion. Each entry in the product of two matrices is a sum of products
of entries of the two, and thus is a polynomial. And polynomials are smooth. The entries in A−1 are quotients
of polynomials in the entries of A, with nowhere-vanishing denominator. Hence these too are smooth functions of
the entries of A. So both group operations are smooth (old).
10∗ . Let V be a k-dimensional vector subspace of Rn . Show that V is a smooth manifold diffeomeorphic to Rk ,
and that all linear maps on V are smooth. For each v ∈ V , find a “natural” isomorphism V ' Tv V .
Solution: Choose a basis v1 , . . ., vk for V . The linear map φ sending v ∈ V to its expression as a linear
combination of the vi P
extends to a linear map from Rn to Rk . All linear maps Rn → Rk are smooth. φ has
inverse (α1 , . . ., αk ) 7→ i αi vi . So we can take φ as chart.
The term “natural” is of course rather vague. In my opinion the following construction deserves the term:
given v ∈ V and x ∈ V , let γ(t) = x + tv. Then v = γ 0 (0) ∈ Tx V. Perhaps the exercise should say “Tx V = V ”.
11∗ . i) If M ⊆ N are manifolds of the same dimension, show that M is open in N (Hint: use exercise 3 and the
inverse function theorem.)
ii) Show that S 1 × S 1 is not diffeomorphic to any manifold M contained in R2 . (Hint: S 1 × S 1 is compact.)
Solution: Let i : M → N be inclusion. Then dx i is also inclusion. As the dim M = dim N , dx i is an isomorphism.
The inverse function theorem tells us that there is a neighbourhood U of x in M and a neighbourhood V of i(x)
in N such that i : U → V is a diffeomorphism. Of course U must equal V ; because V is open in N, so is U , and
thus M is open in N .
(ii) If S 1 × S 1 were diffeomorphic to some M ⊂ R2 then by (i), M would be open in R2 . It would also be compact,
and therefore closed in R2 . The only subsets of R2 whoich are both open and closed are ∅ and R2 , neither of which
is diffeomorphic to S 1 × S 1 .
12∗ . If G is a Lie group, the tangent space Te G is known as the Lie algebra of G. Show that the following are
Lie groups, and determine their Lie algebras: i) Sl(n, R) = {A ∈ Gl(n, R)|det(A) = 1} ii) 0(k, n − k) = {A ∈
Gl(n, R)|At Ik,n−k A = Ik,n−k }, where Ik,n−k is the n × n matrix with k 1’s followed by (n − k) -1’s down the
diagonal, and 0’s elsewhere.
Solution: (i) By Exercise 11 below, for each invertible matrix A, dA det is surjective. In particular 1 is a regular
value of det, so Sl(n, R) is at least a manifold. Since the group operations are the restriction of the smooth (old)
group operations in those of the ambient manifold Gl(n, R), they are smooth (new) and so Sl(n, R) is a Lie group.
Its tangent space at I is equal to the kernel of dI det. The formula given in the solution to Exercise 11 shows that
this is equal to the set of all matrices with trace 0.
(ii) Let f : Matn×n (R) → Symn (R) be the map sending A to At Ik,n−k A. Then O(k, n − k) = f −1 (Ik,n−k ), and it
is necessary to show that Ik,n−k is a regular value. This is easily done by a small modification of the argument in
1.28(ii) in the Lecture Notes, as is the calculation of Te O(k, n − k).
13∗ . i) Find the derivative of the map det : M (n, R) → R, where M (n, R) is the vector space of n × n matrices
with real entries. Don’t attempt to differentiate a complicated polynomial in the entries; instead, use the fact that
the determinant is linear in each row of the matrix.
ii) Prove that the set of all 2 × 2 matrices of rank 1 is a three-dimensional submanifold of R4 = space of 2 × 2
matrices. Hint: prove that the determinant function is a submersion at every point of the manifold of non-zero
2 × 2 matrices.
Solution: (i) Represent a each matrix as an n-tuple of columns: A = (a1 , . . ., an ), B = (b1 , . . ., bn ), etc. Then
dA det(B) = lim
h →
det(A + hB) − det A
0
h
n det(a + hb , . . ., a + hb ) − det(a + hb , . . ., a
1
1
n
n
1
1
n−1 + hbn−1 , an )
h → 0
h
det(a1 + hb1 , . . ., an−1 + hbn−1 , an ) − det(a1 + hb1 , αn−2 + hbn−2 , an−1 , an )
+
+ ···
h
det(a1 + hb1 , a2 , . . ., an ) − det(a1 , . . ., an ) o
+
h
Each summand is the difference of the determinants of two matrices differing only in one column. By linearity in
the i-th column,
= lim
det(a1 , . . ., a0i , . . ., an ) − det(a1 , . . ., a00i , . . ., an ) = det(a1 , . . ., a0i − a00i , . . ., an )
and so dA det(B) =
lim
h →
n det(a + hb , . . ., a
det(a1 + hb1 , . . ., ai−1 + hbi−1 , hbi , ai+1 , . . ., an )
1
1
n−1 + hbn−1 , hbn )
+ ··· +
0
h
h
+· · · +
det(hb1 , a2 , . . ., an ) o
h
n
det(a1 + hb1 , . . ., an−1 + hbn−1 , bn ) + · · · + det(a1 + hb1 , . . ., ai−1 + hbi−1 , bi , ai+1 , . . ., an )
→ 0
o
+· · · + det(b1 , a2 , . . ., an )
o
= det(a1 , . . ., an−1 , bn ) + · · · + det(a1 , . . ., ai−1 , bi , ai+1 , . . ., an ) + · · · + det(b1 , a2 , . . ., an )
= lim
h
= det(a1 , . . ., an−1 , bn ) + · · · + det(a1 , . . ., ai−1 , bi , ai+1 , . . ., an ) + · · · + det(b1 , a2 , . . ., an ).
(ii) If A has rank 1 then A 6= 0. Suppose a1 6= 0. Then there is some λ ∈ R such that a2 = λa1 . Using the formula
from (i), it is now easy to find B such that dA det(B) 6= 0.
14.(i) Let i : Gl(n, R) → Gl(n, R) be the inversion map, i(A) = A−1 . Find dA i. Hint: Proposition 1.17 of the
lecture notes gives the answer when A is the identity matrix In . For a general A, apply the chain rule to the
commutative diagram
i
(Gl(n, R), In ) −→
(Gl(n, R), In )
`A ↓
↓ rA−1
(Gl(n, R), A)
i
−→
(Gl(n, R), A−1 )
in which `A is left-multiplication by A, `A (B) = AB, and rA−1 is right-multiplication by A−1 , rA−1 (C) = CA−1 ;
note that both `A and rA−1 are the restriction to Gl(n, R) of linear maps on the ambient space of all matrices
2
(= Rn ).
Solution: The commutativity of the diagram says:
i ◦ `A = rA−1 ◦ i.
(1)
Evaluating both sides on a matrix B, this is simply the familiar fact that
(AB)−1 = B −1 A−1 .
Applying the chain rule to (1) and taking derivatives at In on both sides we get
dA i ◦ dIn `A = dIn rA−1 ◦ dIn i.
(2)
Now the two maps `A and rA−1 are both the restrictions to Gl(n, R) of maps which are linear on the ambient
vector space (of all n × n matrices). Hence the derivative (at any point) of `A is just `A itself, and similarly the
derivative of rA−1 is rA−1 itself. So (2) becomes
dA i ◦ `A = rA−1 ◦ dIn i = −rA−1
(the last equality because we know from 1.17 (i) in the Lecture Notes that dIn i is just multiplication by −1). As
`A is a linear isomorphism with inverse `A−1 , we conclude
dA i = −rA−1 ◦ `A−1 ,
or, in other words, for any element B ∈ TA Gl(n, R) = Matn×n (R), we have
dA i(B) = −A−1 BA−1 .
15∗ . (Stack of records theorem). Suppose that M and N are smooth manifolds of the same dimension, with M
compact, f : M → N is a smooth map, and that y is a regular value of f . Show that f −1 (y) is finite, and that
there exists a neighbourhood V of y in Y such that f −1 (V ) is a disjoint union of open sets of X, each of which is
mapped diffeomorphically to V by f . Does either statement still hold if we drop the requirement that X be compact?
Solution: By the inverse function theorem, for each x ∈ f −1 (y), there exists an open neighbourhood Ux of x on
which f is a diffeomorphism. The union of the Ux cover the compact set f −1 (y), so there is a finite subcover. As
each Ux only contains one preimage of y, there can be only finitely many of the x’s. Number them x1 , . . ., xn , and
write Ui in place of Uxi . Shrink the Ui until Ui ∩ Uj = ∅ if i 6= j. For each i, Vi := f (Ui ) is an open neighbourhood
of y. Let V = ∩i Vi , and let Wi = f −1 (V ) ∩ Ui . Then f −1 (V ) = W1 ∪ · · · ∪ Wn , the Wi are mutually disjoint, and
f : Wi → V is a diffeomorphism for each i.
If M is not compact, f −1 (y) may not be finite - consider, for example, the exponential map f : R → S 1 f (x) = e2πix .
For each x ∈ f −1 (y) there is are neighbourhoods Ux and Vx of x and y such that fx : Ux → Vx is a diffeo, as in the
proof above, but now the intersection of all of the Vx will not necessaarily be open. In the case of the exponential
map f : R → Y , the stack of records theorem still holds, but it’s easy to modify the domain so that it fails. Take
y = 1, for example. Then f −1 (y) = Z ⊂ R. Now remove from the domain the points n + 1/n for n ∈ Z r {0}, and
write M denote R with all these points removed. The map exp : M → S 1 is still a submersion, so at each point
n ∈ f −1 (1) there is a neighbourhood Un of x and a neighbourhoood Vn of 1 such that f : Un → Vn is a diffeo.
Since n + 1/n ∈
/ Un , exp(1 + 1/n ∈
/ Vn . We have exp(1 + 1/n) = exp(1/n), and the sequence exp(1/n) tends to 1,
so ∩n Vn is not a neighbourhood of 1.
16. With the hypotheses of the previous exercise, suppose also that f is a local diffeomorphism and that Y is
connected. Show that f is onto, and indeed that any 2 points in Y have the same number of preimages in X.
Solution: “f is a local diffeomorphism” means that the derivative of f is everywhere an isomorphism, so the
conclusion of the previous exercise holds for all y ∈ Y . It follows that the number of preimages is locally constant:
if y has n preimages, so does every point y 0 ∈ V (notation as in previous
exercise). For each n, the set Vn of points
S
with n preimages is therefore open. Suppose Vn 6= ∅. Let Vn0 = m6=n Vm . As a union of open sets, Vn0 also is
open, and now if it is not empty then Y = Vn ∪ Vn0 is the union of two disjoint non-empty subsets, contradicting
its connectedness.
17. Prove that the following are not manifolds: i) The union of the two coordinate axes in R2 ; ii) the double cone
{(x, y, z) ∈ R3 |x2 + y 2 = z 2 }. iii) The single cone {(x, y, z) ∈ R3 |x2 + y 2 = z 2 , z ≥ 0}
Solution: (i) Many arguments are possible here. For example, removal of the point 0 breaks the set into four connected components, something that cannot happen to any manifold. An alternative argument is that if M 1 ⊂ R2
is any manifold and x ∈ M then by the inverse function theorem 1.16, in some neighbourhood of x in M , either
the projection to the x1 axis, or the projection to the x2 axis is a local diffeomorphism, and in particular is 1-1.
Neither is 1-1 in this case.
(ii) The double cone is disconnected by removing the vertex; no 2-manifold is disconnected by removing a single
point. An argument like the second one in (i) also can be used here.
(iii) Many planes through (0, 0, 0) meet the single cone in only one point. By contrast, if M is a smooth surface
in R3 and x ∈ M then any plane through x distinct from the affine plane x + Tx M is transverse to M at x, and
therefore meets M in a curve in the neighbourhood of x.
18∗ . Prove that if X ⊆ Rn is a (smooth) (n − 1)-dimensional manifold and x ∈ X then for any (n − 1)-dimensional
vector subspace V of Rn with V 6= Tx X, the intersection of the affine subspace x + V with X is smooth and of
dimension n − 2 in some neighbourhood of x.
Solution: If V 6= Tx X then V + Tx X = Rn , since Tx X is a hyperplane and V is not contained in it. Hence
x + V is transverse to X at x, and, therefore, in some neighbourhood of x, (x + V ) ∩ X is an (n − 2)-dimensional
manifold.
Section C
19∗ . The tangent bundle of a manifold M ⊆ Rn is the set {(x, x̂) ∈ M × Rn |x̂ ∈ Tx M }.
(i) Show that T M is a manifold.
(ii) Show that T S 1 is diffeomorphic to S 1 × R.
(iii) Show that if V ⊂ Rn is a vector subspace then T V is naturally diffeomorphic to V × V (see exercise 7).
Solution: (iii) For each x ∈ V there is an iso ix : V → Tx V . described above in Exercise 10. The map V ×V → T V
sending (x, v) to (x, ix (v)) is a diffeo.
20. Prove that the set of m × n matrices of rank r is a manifold (in Matm×n (R) = Rmn ) and find its dimension.
Hint: Show first that the set of matrices of rank ≥ r is open in Matm×n (R) (recall that rank A ≥ r⇔ A has an
r × r submatrix with non-zero determinant). Next, suppose rank(A) ≥ r, and after permuting rows and columns,
that A has the form
B C
(3)
D E
where B is nonsingular of size r × r. Postmultiply by the nonsingular matrix
I −B −1 C
0
I
to prove that rank(A) = r if and only if E − DB −1P
C = 0.
Solution: Denote the set of matrices of rank r by r . The map f ,
B C
7→ E − DB −1 C,
D E
is defined
P on the open set U of matrices with top left r × r corner invertible. It is a submersion. Hence its zero
locus, r ∩ U , is a smooth
P manifold of codimension (m − r)(n − r) (the dimension of the target space of f ).
Now every point of r has some r ×r minor invertible. On the open set U 0 of matrices for which this particular
minor is non-zero, it is possible to
a map to
of (m − r) × (n − r) matrices,Pof exactly the same
P define
P the space
0
0
nature as f , whose zero locus is r ∩U . Hence r ∩U is also smooth. As every point of r lies in some such
open set, the proof is complete.
21∗ . Let f be the embedding S 1 → S 1 ×S 1 given by eiθ 7−→ (e2iθ , e3iθ ), and let g be the diffeomorphism S 1 ×S 1 → T 2
described in the lecture. Sketch the image of g ◦ f . What is it?
Solution It’s a trefoil knot.
22. Is it possible to find an atlas for T ⊂ R3 consisting of 2 charts?
Solution: There is an atlas mapping each of two open sets in T to an annulus in R2 . The domains of the two
charts are described most easily in S 1 × S 1 :
U1 = {(x1 , yy , x2 , y2 ) ∈ S 1 × S 1 : y1 > −1/2}
U2 = {(x1 , yy , x2 , y2 ) ∈ S 1 × S 1 : y1 < 1/2}.
23. Prove that if U ⊂ Rk is open and f : U → Rp is not smooth then graph(f ) ⊂ Rk ×Rp is not a (smooth) manifold.
24. RP2 is the identification space obtained from S 2 by identifying antipodal points. (Recall that if X is a
topological space and ∼ is an equivalence relation on X, then the identification topology on the quotient space
X/ ∼ is defined by declaring a set in X/ ∼ to be open if and only if its preimage in X is open.) Show that the
f
mapping S 2 → R6 given by f (x, y, z) = (x2 , xy, y 2 , xz, yz, z 2 ) gives rise to a homeomorphism from RP2 to a smooth
manifold X ⊆ R6 .