Download MA455 Manifolds Exercises III Solutions May 2008 1. Let V and W

MA455 Manifolds Exercises III Solutions May 2008
1. Let V and W be oriented vector spaces of the same dimension, and let E and F be positive
bases for V and W respectively. Show that an isomorphism f : V → W preserves orientation if
E
det[f ]E
F > 0 and reverses orientation if det[f ]F < 0.
Solution: f preserves orientation iff f (E) is a positive basis of W . The change of basis matrix
f (E)
[I]F
is equal to [f ]E
F.
2. (i) Example 3.10(i) claims to orient the preimage M := f −1 (t) of a regular value t of a
smooth function f : Rn+1 → R, by stipulating that a basis v1 , . . . , vn for Tx M is positive if the
basis ∇f (x).v1 , . . . , vn of Rn+1 is positive. Prove that this definition satisfies the compatibility
condition for an orientation, 3.7(iii).
Solution: If v1 , . . . , vn is a tangent frame field for M on U , then ∇f, v1 , . . . , vn is a tangent
frame field for Rn+1 . Hence det(∇f (x), v1 (x), . . . , vn (x)) 6= 0 for all x ∈ U . So by continuity of
the determinant (and the connectivity of U ), det(∇f (x), v1 (x), . . . , vn (x)) > 0 for all x ∈ U or
det(∇f (x), v1 , . . . , vn ) < 0 for all x ∈ U . In the first case, v1 (x), . . . , vn (x) is a positive basis for
Tx M for all x ∈ U , and In the second case, v1 (x), . . . , vn (x) is a negative basis for Tx M for all x ∈ U .
(ii) Ditto for the orientation defined in 3.10(ii).
Solution: Essentially the same as (i).
3.Prove that the antipodal map S 1 → S 1 preserves orientation but that the antipodal map S 2 → S 2
reverses orientation.
Solution: We orient both spheres by using the outward pointing unit normal. Then v is positive
basis for Tx S 1 if x, v is positive basis for R2 . As the antipodal map A is the restriction of the
linear map multiplication by −1 on the ambient space, its derivative is also multiplication by −1.
Multiplying both columns of a 2 × 2 matrix by −1 leaves its determinant unchanged, so
=⇒
v + ive basis for Tx S 1
−x, −v + ive basis for R2
=⇒
=⇒
x, v + ive basis for R2
.
−v + ive basis for T−x S 1
So A : S 1 → S 1 preserves orientation.
On the other hand multiplying all three columns of a 3 × 3 determinant by −1 has the effect of
multipplying its determinant by −1, so
=⇒
v1 , v2 + basis forTx S 2
−x, −v1 , −v2 − basis forR3
=⇒
=⇒
x, v1 , v2 + basis forR3
−v1 , −v2 − basis forT−x S 2
so A : S 2 → S 2 reverses orientation.
5. Let M m and N n be oriented manifolds, and give M × N and N × M the product orientation.
Prove that the map
i : M × N → N × M, i(x, y) = (y, x)
preserves or reverses orientation according to whether mn is even or odd.
Solution: Let v1 , . . . , vm be a positive basis for Tx M and w1 , . . . , wm be a positive basis for Ty N .
Then in the product orientation on M × N ,
(v1 , 0), . . . , (vm , 0), (0, w1 ), . . . , (0, wn )
is a positive basis for T(x,y) M × N . Applying the derivative of the map M × N → N × N , we get
the basis
(0, v1 ), . . . , (0, vm ), (w1 , 0), . . . , (wn , 0).
To transform this to the positive basis
(w1 , 0), . . . , (wn , 0), (0, v1 ), . . . , (0, vm )
requires mn transpositions (try it!). Hence the change of basis matrix has determinant equal to
(−1)mn .
6. Suppose that M is an orientable manifold, and suppose given two, possibly distinct, orientations
of M .
(i) Let U be a connected subset of the domain of a chart. Show that either the two orientations
agree everywhere on U , or they disagree everywhere on U . Hint: find a frame field on U .
(ii) Show that if the two orientations agree at x ∈ M , then they agree on a neighbourhood of x, and
if they disagree at x, then they disagree on a neighbourhood of x.
(iii) Prove that if M is connected then the two orientations either agree everywhere or disagree
everywhere. Hint: by (ii), the set of points where they agree is open, and so is the set of points
where they disagree.
(iv) Conclude that on a connected orientable manifold there are just two possible orientations.
Solution: Suppose given two orientations. Each point x in M has a connected neighbourhood V
on which a frame field F is defined. F is either everywhere positive or everywhere negative for
each of the two orientations. So either they agree everywhere on V , or they disagree everywhere
on V . It follows that the set of points in U where they agree is open, and so is the set of points
where they disagree. As U is connected, one or other is all of U . If they differ, each orientation is
obtained by reversing the other. So there are exactly two.
7. Do exercise 3.7:
(i) Show that if M ⊂ Rn is a smooth manifold then for each x0 ∈ M there is neighbourhood U of x0
such that for all x ∈ U , the orthogonal projection π : RN → Tx0 M restricts to give an isomorphism
πx : T x M → T x 0 M .
(ii) Show that the local compatibility condition 3.5(iii) is equivalent to the condition that every
x0 ∈ M has a neighbourhood U such that for all x ∈ U , πx : Tx M → Tx0 M not only is an isomorphism but also preserves orientation.
Solution:(i) Let F = v1 , . . . , vm be a frame field on a neighbourhood of x. Let E = e1 , . . . , em be
a basis for Tx0 M . The matrix
A(x) = π(v1 (x)· · ·· · ·π(vm (x)) E
whose columns are the expressions of the projected vectors π(vi (x)) with respect to the basis E,
varies smoothly with x. This is most obvious if E is orthonormal, for then the entries in A(x) are
just the dot products vi (x) · ej . When x = x0 , A(x) is invertible. Hence there is a neighbourhood
U of x in M on which A continues to be invertible. For x ∈ U , π : Tx M → Tx0 M is an isomorphism.
F (x)
(ii) A(x) = [dx π]E
.
8. (i) Suppose that f : M → N is a map of oriented manifolds of the same dimension, with M
connected. Suppose that dx f is an isomorphism for all x ∈ M . Show that either for all x ∈ M dx f
preserves orientation, or for all x ∈ M dx f reverses orientation.
(ii) Let N be the image of the map f : S 2 → R6 of Example 1.34. Show that N is not orientable
by considering the diagram
S2 A
A
AA
AA
A
f AA
N
/ S2
}
}
}}
}} f
}
}~
where A is the antipodal map.
9. Find an explicit diffeomorphism from some neighbourhood U in B n+1 of the point N = (0, . . . , 0, 1),
to an open set in the half-space H n+1 .
Solution: Let U = B n+1 ∩ {xn+1 > 0} be the upper half-ball. Define φ : U → H n+1 by
s X
x2j .
φ(x1 , . . . , xn , xn+1 ) = x1 , . . . , xn , xn+1 − 1 −
j=1,... ,n
10. Show that if M m is a smooth manifold, and f : M → R has t0 as a regular value, and
f −1 (t0 ) 6= ∅, then f −1 ((−∞, t0 ]) and f −1 ([t0 , ∞)) are m-dimensional manifolds with boundary,
and each has boundary equal to f −1 (t0 ).
Solution: Consider M+ := f −1 ([0, ∞)). The set of points x where f (x) > t0 is already a manifold:
since it’s open in M , you can use the existing charts on M , restricting their domains suitably.
These will give diffeos of open sets in M+ to open sets in Rm ; composed with a suitable embedding
of Rm in H m , such as (x1 , . . . , xm ) 7→ (x1 , . . . , xm−1 , exp(xm )), they become diffeos of open sets in
M+ to open sets in H m .
In the neighbourhood of a point where f (x) = t0 , use the local normal form of a submersion:
there is a chart φ on M around x such that f ◦ φ−1 (x1 , . . . , xm ) = xm . The restriction of φ to
M+ gives a diffeomorphism of a neighbourhood of x in M+ to an open set in H m , and sends
M+ ∩ f −1 (t0 ) to ∂H m .
11. Show that if M m is a manifold and N n is a manifold with boundary then M × N is an m + ndimensional manifold with boundary, and ∂(M × N ) = M × ∂N . What if both M and N have
non-empty boundary?
12. Prove Lemma 3.20.
13. Prove that if f : M m → N n and Z n−m ⊂ N with all manifolds oriented, M compact and Z
closed in N , and if g : M 0 → M is a diffeomorphism, then if g preserves orientation everywhere,
(f ◦ g · Z)N = (f · Z)N and if g reverses orientation everywhere, (f ◦ g · Z) = −(f · Z). (This is
used in the proof of 3.23).
Solution: Clearly g determines a bijection (f ◦ g)−1 (Z) → f −1 (Z). Let x0 ∈ (f ◦ g)−1 (Z), let
x = g(x0 ) and y = f (x). Let E 0 be a positive base for Tx0 M 0 , write E = dx0 g(E 0 ), and F a positive
base for Ty Z.
If g preserves orientation, then E is a positive base for Tx M , and sign(dx f (E), F ) = (f · Z)x .
As dx f (E) = dx0 (f ◦ g)(E 0 ), this shows that (f ◦ g · Z)x0 = (f · Z)x .
If g reverses orientation, then E is a negative base for Tx M , and sign(dx f (E), F ) = −(f · Z)x .
This shows that (f ◦ g · Z)x0 = −(f · Z)x .
14. Prove Lemma 3.31: Suppose F : W p → N n is a smooth map, and Z is a submanifold of N of
codimension k. If F tZ and ∂F tZ, then F −1 (Z) is a p − k-dimensional manifold with boundary,
and ∂(F −1 (Z)) = (∂F )−1 (Z). Hint: after composing F with the inverse of a chart on W , you can
replace W by an open set in H p . Use the local normal formm of a submersion.
Solution: As in Exercise 10, for points in F −1 (Z) away from ∂W , there is not much to say - just
use Corollary 1.29. However in the neighbourhood of a point x0 ∈ (∂F )− −1(Z) = F −1 (Z) ∩ ∂W ,
we need a new argument.
Let φ be a chart on W around x0 and write F̃ = F ◦ φ−1 , x̃0 = φ(x0 ). F̃ , defined on an open
set in H m , has a local smooth extension G to a neighbouhhood U of x0 in Rm . Provided U is
chosen small enough, GtZ, and so G−1 (Z) is a smooth manifold. Then F̃ −1 (Z) is the intersection
of G−1 (Z) with H m .
Let f : G−1 (Z) → R be the restriction of the m’th coordinate and is thus f −1 ([0, ∞)). Then
F̃ −1 (Z) = f −1 ([0, ∞)). To show that F̃ −1 (Z) is a smooth manifold with boundary equal to
∂ F̃ −1 (Z), all we need do is apply the result of Exercise 7 - after checking that the hypothesis
applies, of course. That is, we must show that f is a submersion at x0 .
The kernel of dx0 f is the “horizontal” part of Tx̃0 G−1 (Z) - its intersection with {xm = 0}. But
this intersection is equal to (dx̃0 ∂ F̃ )−1 (Ty Z), and has dimension one less than the dimension of
G−1 (Z), by transversality. This shows that f is a submersion at x0 .
15. (i) Check that the sequence
j
i
0 → R2 −→ R5 −→ R3 → 0
is exact, when i and j are the linear maps with matrices


1 3


 2 −1 
0
0
1
0
0


 0 0 
 3 2 0 −7 0 
and


 1 1 
0 1 0 −2 3
0 1
(1)
.
(ii) Let a1 = (1, 0), a2 = (0, 1) be the standard basis of R2 , and give R2 the standard orientation (so
that this basis is positive). Give R4 the standard orientation also. Find three vectors b1 , b2 , b3 ∈ R5
such that i(a1 ), i(a2 ), b1 , b2 , b3 is a positive basis for R5 .
Solution: (i) Exactness can be checked by first checking that j ◦ i = 0, and then that dim ker j =
dim im i.
(ii) The two columns of the matrix of i are i(a1), i(a2 ). The three additional vectors (0, 0, 1, 0, 0), (0, 0, 0, 1, 0)
and (0, 0, 0, 0, 1), together with i(a1 ) and i(a2 ), make up a basis for R5 . Its sign is that of the determinant
1 1 0 0 0 2 −1 0 0 0 0 0 1 0 0 1 1 0 1 0 0 1 0 0 1 which is negative. So we get a positive basis by interchanging,say, the last two. Thus, we take
b1 = (0, 0, 1, 0, 0), b2 = (0, 0, 0, 0, 1), b3 = (0, 0, 0, 1, 0).
(iii) Let R3 also have the standard orientation. What is the sign of the short exact sequence (1)?
4
Solution: It is the sign of the basis j(b1 ), j(b2 ), j(b3 ) of R3 . That is, the sign of the determinant
of the last three columns of the matrix of j, with the last two columns interchanged. Since
1 0 0 0 0 −7 > 0
0 3 −2 the s.e.s. is positive.
(iv) Now orient R2 by taking the standard basis to be positive, and orient R3 by taking as positive
the basis c1 = (1, −1, 2), c2 = (−1, −1, 3), c3 = (0, 2, 5). Orient R5 (by exhibiting a positive basis)
so that the sequence is positive.
Solution: The new orientation of R3 is opposite to the old, since
1 −1 0 −1 −1 2 < 0,
2
3 5 so since the s.e.s. was positive with respect to the old orientation of R3 , to restore positivity now we
reverse the orientation of R5 . For example we could take as positive basis i(a1 ), i(a2 ), (0, 0, 1, 0, 0), (0, 0, 0, 1, 0), (0, 0, 0
Another correct procedure would be to find b̃1 , b̃2 , b̃3 in R5 such that j(b̃i ) = ci and take as
positive basis for R5 the vectors i(a1 ), i(a2 ), b̃1 , b̃2 , b̃3 .
(v) (v) Now orient R5 and R3 by taking as positive the bases b1 = (0, 1, 0, 0, 0), b2 = (0, 0, 1, 0, 0), b3 =
(0, 0, 0, −1, 0), b4 = (1, 1, 1, 1, 0), b5 = (1, 2, 3, 4, 5) and c1 = (1, −2, 2), c2 = (2, 0, 0), c3 = (0, 1, 2) respectively. As explained in the lecture notes, there is a unique orientation on R2 making the
sequence (1) exact. Is it the standard orientation or the opposite orientation?
17. Suppose that A, B and C are oriented vector spaces, and that 0 → A → B → C → 0 is a
positive short exact sequence. Let f : A → A0 , g : B → B 0 and h : C → C 0 be isomorphisms,
and suppose A0 , B 0 and C 0 are all oriented (though we do not suppose that f , g and h necessarily
preserve orientation). Then the procedure of the previous exercise gives short exact sequences
0 → A0 → B → C → 0,
0 → A → B0 → C → 0
and
0 → A → B → C 0 → 0.
Give conditions on f, g and h for these to be positive. When is the exact sequence
0 → A0 → B 0 → C 0 → 0,
constructed by using the same procedure three times, positive?
Solution: The four sequences can be seen in the diagrams
0
→
&
A
l
A0
0
→
A
→
%
→
B
B
→
→
&
C
→
0
0
→
A
→
&
B
l
B0
→
%
C
→
0
C
l
C0
→
%
0
0
→
&
A
l
A0
→
B
l
B0
→
C
l
C0
→
%
0
5
→
→
The sign of 0 → A0 → B → C → 0 is the same as the sign of 0 → A → B → C → 0 if A → A0
preserves orientation, and opposite otherwise. Similarly for the next two cases. The last is the
product of three replacements, and has the same sign as 0 → A → B → C → 0 iff an even number
of the isomorphisms A → A0 , B → B 0 and C → C 0 preserve orientation.
18. Suppose f : M → N , Z is a submanifold of N , and f tZ, with f −1 (Z) 6= 0. Explain how to
construct the short exact sequence
0 → Tx f −1 (Z) → Tx M → Ty N/Ty Z → 0
(used in the definition of the preimage orientation 3.26(iii)).
Solution: Begin with
dx f
Tx M −→ Tf (x) N.
The map
Tx M → Tf (x) N/Tf (x) Z
in the s.e.s is the composite
Tx M
dx f
−→
Tf (x) N
↓
Tf (x) N/Tf (x) Z
where the vertical map is projection to the quotient. The formula in 1.44 in the Lecture Notes,
defining transversality, shows that this composite is surjective. Moreover, 1.45 shows that Tx f −1 (Z)
is its kernel. This completes the proof of exactness.
19. Suppose f : M m → N n is smooth and Z n−m ⊂ N , with M compact, Z closed in N , and all
manifolds oriented. Let f˜ : M → M × N be the map x 7→ (x, f (x)), whose image, of course, is the
graph of f .
(i) Find a natural bijection f −1 (Z) → f˜−1 (M × Z), and show that f t Z if and only if f˜ t M × Z.
(ii) Let v1 , . . . , vm be a positive basis for Tx M . Write down the image of this basis under dx f˜.
(iii) Show that if we give M × Z and M × N the product orientations, then
(f˜ · M × Z)M ×N if m is even
(f · Z)N =
−(f˜ · M × Z)M ×N if m is odd
Solution:(i) The bijection f −1 (Z) → f˜−1 (M × Z) is given by x ↔ (x, f (x)).
(ii) The image of v1 , . . . , vm is v1 , dx f (v1 ) , . . . , vm , dx f (vm ) .
(iii) Let z1 , . . . , zn−m be a positive basis for Tf (x) Z. Then assuming f (x) ∈ Z, (v1 , 0), . . . , (vm , 0), (0, z1 ), . . . , (0, zn−m
is a positive basis for T(x,f (x) M × Z. We have
(f˜ · M × Z)x = sign of v1 , dx f (v1 ) , . . . , vm , dx f (vm ) , (v1 , 0), . . . , (vm , 0), (0, z1 ), . . . , (0, zn−m )
as basis of T(x,f (x) M × N . This basis is equivalent to
0, dx f (v1 ) , . . . , 0, dx f (vm ) , (v1 , 0), . . . , (vm , 0), (0, z1 ), . . . , (0, zn−m )
2
and (−1)m transpositions convert it to
(v1 , 0), . . . , (vm , 0), 0, dx f (v1 ) , . . . , 0, dx f (vm ) , (0, z1 ), . . . , (0, zn−m ).
6
By definition of the product orientation on M × N , this basis is positive iff the basis
dx f (v1 ), . . . , dx f (vm ), z1 , . . . , zn−m
of Tx N is positive, i.e. iff f, Z x = +1. Hence
2
(f · Z)x = (−1)m f˜ · M × Z
x
20. Suppose that f : M m → N m is a smooth map, with M compact, and M and N oriented.
For a fixed y ∈ N , let iy : M → M × N be the map iy (x) = (x, y). Show that y is a regular value
of f if and only if iy t graph(f ). What is the precise relation between deg(f ) and (iy ·graph(f ))M ×N ?
Solution: This can be solved by a method very like the last exercise — good practice if you haven’t
already done it. There is a slick way, viewing it as a special case of the last exercise:
2
2
deg(f ) = f · {y} N = (−1)m f˜ · (M × {y}) M ×N = (−1)m graph(f ) · (M × {y}) M ×N
2
2
= (−1)mn (−1)m (M × {y}) · graph(f ) M ×N = (−1)mn (−1)m (iy · graph(f ))M ×N .
7