Math 531: Topology/Geometry II

2:20-3:40p, Spring 2008

Department of Mathematics

SUNY at Stony Brook

This course will offer an introduction to the elementary theory of smooth manifolds. Topics will include:
vector fields, differential forms, vector bundles, Lie derivatives, Stokes' theorem, and de Rham cohomology.

Instructor: Dr. Corbett Redden

Math Tower 3-114. Phone: 632-8261. email: redden at math dot sunysb dot edu

Office Hours: M 4-5p, Tu 10a-12p, or drop-in, or by appointment.

Grader: Yakov Savelyev

e-mail: yasha at math dot sunysb dot edu

Homework: Working homework problems is the only way to really learn the material. While you are
encouraged to work with others, you must write up all solutions on your own. Homework sets will usually be
collected in class on Thursdays.

Due 2/7: 2. 6, 8, 18

Due 2/14: 2. 26, 28, 32, 33

Due 2/21: §3: 5, 7, 10, 23

Due 2/28: §3: 20, 24, 26, 30

Due 3/6: §4: 1, 2, 6, 8

Due 3/13: §5: 2, 6, 12, 14ab

Recommended by 3/22: Review/Practice Exam (and solutions)

Due 3/28: Midterm Exam (and solutions)

Due 4/3: §6: 5. §7: 2, 18, 20, 27

Due 4/10: §8: 2, 3, 13, 14

Due 4/17: §8: 17, 23, 32

Due 4/24: Short HW

Due 5/1: §11: 2, 3

Due 5/13: Ch 11 #11, 13

Midterm: Take Home midterm due Friday, March 28. Review/Practice HW, Review Solutions, MIDTERM
EXAM. Midterm Solutions

Final Exam: Thursday, May 15. 2:00p.m. Physics P129. Final Exam.

Textbook:: A Comprehensive Introduction to Differential Geometry, Volume 1 (3rd edition), by Michael

Spivak. Publish or Perish Inc., 1999.

Course Grade: Homework: 50%, Midterm: 20%, Final: 30%

Disabilities: If you have a physical, psychological, medical

or learning disability that
may impact your
course work, please contact Disability Support Services, ECC (Educational Communications Center)
Building, room 128, (631)
632-6748. They will determine with you what accommodations are necessary and

appropriate. All information and documentation is confidential.

Students requiring emergency evacuation are
encouraged to discuss their
needs with their professors and Disability Support Services. For
procedures
and
information, go to the following web site:
http://www.www.ehs.stonybrook.edu/fire/disabilities.shtml
 Math 531 Midterm Review

The midterm will be a timed, take-home exam (taken on the honor sys-
tem). You will have 3 hours to complete the exam. You may use your notes
and Spivak, but you may not obtain outside help from anyone. The problems
will be a mix of abstract theorems and concrete calculations. There will be
multiple problems, and you will be able to choose which ones you do (e.g.
work 2 out of 3 problems). Here are a few good problems to practice with:

1. Let f : R2 → R be defined by

f (x, y) = x3 + xy + y 3 + 1.

For which points p = (0, 0), p = ( 31 , 13 ), p = ( −1

3
, −1
3
), is f −1 (f (p)) an embed-
ded submanifold in R2 ?

2. Let M be a compact manifold of dimension n, and let f : M → Rn

be a smooth map. Does f have to have a critical point? In other words,
must there exist a point p ∈ M such that f∗p is not injective?

3. Let S 2 be the 2-sphere. Let UN = S 2 − {N }, US = S 2 − {S} be the

open sets obtained by removing the “North Pole” and the “South Pole,”
respectively. On both UN and US there exist standard stereographic projec-
tions to R2 (dealt with in a previous homework). These coordinate charts
give a trivialization of the tangent bundle over each open set. QUESTION:
Compute the transition functions for the tangent bundle on the overlap. i.e.
if φN , φS are the local trivializations of T S 2 induced by the two stereographic
projections (φN : T UN → UN × R2 , φS : T US → US × R2 ), calculate φS ◦ φ−1 N .

4. Is the Klein bottle orientable? Justify your answer.

5. On R3 with standard coordinates, consider the vector fields

∂ ∂
X=z −y
∂y ∂z
∂ ∂
Y = −z +x
∂x ∂z
∂ ∂
Z=y −x
∂x ∂y

Calculate the Lie bracket on the vector fields X, Y, Z. What sort of structure
does this look like?

1
 Math 531 Midterm Review Solutions

1. Let f : R2 → R be defined by

f (x, y) = x3 + xy + y 3 + 1.

For which points p = (0, 0), p = ( 31 , 13 ), p = ( −1

3
, −1
3
), is f −1 (f (p)) an embed-
ded submanifold in R2 ?
The answer here is quite subtle. The points p = (1/3, 1/3) and p =
(−1/3, −1/3) give submanifolds, while p = (0, 0) does not. However, p =
(−1/3, −1/3) does not follow from the Implicit Function Theorem.
Let us first do a few calculations:

f (x, y) = x3 + xy + y 3 + 1
f (0, 0) = 1
f (1/3, 1/3) = 32/27
f (−1/3, −1/3) = 28/27

h i
∂f ∂f
 
0 0 = Df = ∂x ∂y
 
= 3x + y x + 3y 2
2

The two points where Df has rank 0 (where Df does not have full rank)
occur at (0, 0) and (−1/3, −1/3). Therefore, since f has full rank at all points
M = f −1 (32/37), we know that M is a codimension 1 submanifold of R2 .
However, we do not yet know whether the other two sets are submani-
folds or not. The above process merely guarantees a submanifold structure
in certain situations; it does eliminate the possibility. In general, there is
no standard method for showing that something is not a manifold, though
usually you assume you have a submanifold and try to obtain some sort of
contradiction. We investigate the two unknown situations from above more
carefully.
First, let us consider, f −1 f (−1/3, −1/3). Some sneaky algebra (it helps
to use a computer) shows us that

x3 + xy + y 3 + 1 = 28/27
x3 + xy + y 3 − 1/27 = 0
1
(3x + 3y − 1)(9x2 + 3x − 9xy + 3y + 9y 2 + 1) = 0
27
We know that 3x + 3y − 1 = 0 is a submanifold of R2 . Further investigation
shows that the second factor

9x2 + 3x − 9xy + 3y + 9y 2 + 1 ≥ 0.

1
The above follows from a bit of Calc 3. We can show that this quadratic
function has a critical point at (−1/3, −1/3), f (−1/3, 1/3) = 0, and the
Hessian is positive definite (f is “concave up” in every direction). Therefore,

f −1 f (−1/3, −1/3) = {(x, y)|3x + 3y − 1 = 0} ∪ (−1/3, −1/3)

= {(x, y)|3x + 3y − 1 = 0.}

So, this particular subset is a submanifold, though you cannot simply use
the Implicit Function Theorem.
On the other hand, f −1 f (0, 0) is not a manifold because of the point
(0, 0). Seeing this is a little trickier, and I won’t expect you to do anything
like this on the exam. I’ll try to write up a clean solution to show this is not,
but this is not the important part of the problem. It is, though, good to see
that these things can happen.

2. Let M be a compact manifold of dimension n, and let f : M → Rn be a

smooth map. Then f must have at least one critical point.
Proof. Suppose there are no critical points. Then, by the Inverse Function
Theorem, for every p ∈ M , there exists an open neighborhood U of p such
that f|U is a diffeomorphism. This implies that f is an open map (it takes
open sets to open sets), since these open neighborhoods form a basis for the
open sets in M . Therefore, f (M ) ⊂ Rn is open subset. However, M is
compact, and therefore the image f (M ) ⊂ Rn is also compact. We have now
reached a contradiction, as there are no open compact subsets of Rn (other
than the empty set). The map f must have at least one critical point.
3. Let S 2 be the 2-sphere. Let UN = S 2 − {N }, US = S 2 − {S} be
the open sets obtained by removing the “North Pole” and the “South Pole,”
respectively. On both UN and US there exist standard stereographic projec-
tions to R2 (dealt with in a previous homework). These coordinate charts
give a trivialization of the tangent bundle over each open set. QUESTION:
Compute the transition functions for the tangent bundle on the overlap. i.e.
if φN , φS are the local trivializations of T S 2 induced by the two stereographic
projections, calculate φS ◦ φ−1 N .

First, let us set up the following notation:

S 2 = {(x, y, z) ∈ R3 |x2 + y 2 + z 2 = 1}
U = UN , V = US
x y
(u1 , u2 ) = ( , ) = u(x, y, z)
1−z 1−z
x y
(v 1 , v 2 ) = ( , ) = v(x, y, z)
1+z 1+z

2
Here, u and v are the standard coordinates induced from the stereographic
projection, and on overlap U ∩ V , we have functions u ◦ v −1 and v ◦ u−1 given
by the relations
ui
vi =
(u1 )2 + (u2 )2
vi
ui = 1 2
(v ) + (v 2 )2
The two trivializations u, v both give rise to a trivialization of the tangent
bundle over U and V , respectively. Explicitly, this is an equivalence of bun-
dles φU
φ
U × R2 ←−
U
TU
∂ ∂
(u, (a1 , a2 )) 7−→ a1 1
+ a2 2
∂u ∂u
along with the similarly defined equivalence φV . The transition function
φV ◦ φ−1
U is then a bundle equivalence

φV ◦ φ−1 2 2
U : (U ∩ V ) × R → (U ∩ V ) × R .

Since φV ◦ φ−1
U is a bundle map, we can consider this as a smooth map

φV ◦ φ−1
U : U ∩ V → Gl(2, R).

Remembering that the same vector X can be expressed as

∂ ∂
X = ai i
= bi i ,
∂u ∂v
we obtain the standard relationship
∂v j i
bj = a.
∂ui
(Remark: this can be seen directly from the chain rule, which states
∂ ∂ui ∂
= . )
∂v j ∂v j ∂ui
Now, we see that the transition function φV ◦ φ−1
U , which maps

φV ◦φ−1
(a1 , a2 ) 7−→U (b1 , b2 ),

is given by the linear operators (matrices)

 i
−1 ∂v
φV ◦ φU (p) = = D(v ◦ u−1 )p .
∂uj |p

3
A simple calculation now shows us that

( i (ui )2 )δji − 2ui uj

P
∂v i
= P
∂uj ( i (ui )2 )2
 2  
1 −(u1 )2 + (u2 )2 −2u1 u2
= P i 2
i (u )
−2u1 u2 (u1 )2 − (u2 )2
 2  
1 −x2 + y 2 −2xy
=
1+z −2xy x2 − y 2

So, we have the transition function gV U = φV ◦ φ−1U : (U ∩ V ) → Gl(2, R) is

given by
 2    2  
1 −(u1 )2 + (u2 )2 −2u1 u2 1 −x2 + y 2 −2xy
gV U = P i 2 = ,
i (u )
−2u1 u2 (u1 )2 − (u2 )2 1+z −2xy x2 − y 2

in u coordinates and (x, y, z) coordinates, respectively (one may be easier to

use than the other). We can also compute gU V = gV−1U = φU ◦ φ−1 V . This
i
∂u
can be done by performing the above calculation for , or by taking the
∂v j
inverse of the matrix gV U above. Either way, we end up with
 2    2  
1 −(v 1 )2 + (v 2 )2 −2v 1 v 2 1 −x2 + y 2 −2xy
gU V = P i 2 =
i (v )
−2v 1 v 2 (v 1 )2 − (v 2 )2 1−z −2xy x2 − y 2

4. The Klein bottle is not orientable.

Proof. The Klein bottle contains the open Mobius band as a submanifold.
We know that if a manifold M contains a non-orientable submanifold, then
the manifold M is not orientable.
Alternatively, we can think of the Klein bottle M as a quotient of the
cylinder. In other words, (R × S 1 )/Z, where Z acts by the group action

(x, θ) ∼ (x + m, (−1)m θ), m ∈ Z.

The quotient map π : R × S 1 → M induces a map on the tangent bundles

π
T (R × S 1 ) →∗ T M,

and for any m ∈ Z, we have the following commutative diagram

g∗
T (R × S 1 ) / T (R × S 1 )
LLL r
LLπL∗
LLL rrrrr
r π∗
L% ry rr
TM

4
Now, we notice that the map

(x, θ) 7→ (x + 1, −θ),

which is the action of 1 ∈ Z, reverses the orientation of R × S 1 . (This is an

easy calculation.) Therefore, by the lemma below, M is not orientable.

Lemma 1. Let X is a connected orientable manifold with map g : X → X,

where g is orientation-reversing. If there exists a local diffeomorphism π :
X → M such that the following diagram commutes
g∗
T X GG / TX
GGπ∗ ww
GG ww
GG wwwπ∗
# w{
TM
then M is not orientable.
In particular, if a discrete group G acts on X and there is some g ∈ G
that reverses the orientation, then X/G is non-orientable.
Proof. First, notice that if M is orientable, and X and M are both oriented,
then π∗ must either preserve orientation at every point, or it must reverse
orientation at every point.
Now, let x ∈ X, p = π(x), and choose an orientation on X. Notice that

π∗|x : Tx X → Tp M

either preserves or reverses orientation. However,

(π∗ ◦ g∗ )|x : Tx X → Tp M

does the opposite of π∗x . By the commutativity of the diagram, π∗|x =

(π∗ ◦ g∗ )|x , and thus M cannot admit an orientation.
5. On R3 with standard coordinates, consider the vector fields
∂ ∂
X=z −y
∂y ∂z
∂ ∂
Y = −z +x
∂x ∂z
∂ ∂
Z=y −x
∂x ∂y

Then,
[X, Y ] = Z, [Y, Z] = X, [Z, X] = Y.

5
This is a straight-forward calculation using the definition of the Lie bracket.
     
∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂
[X, Y ] = z −y −z +x − −z +x z −y
∂y ∂z ∂x ∂z ∂x ∂z ∂y ∂z
∂ ∂
=y −x =Z
∂x ∂y
Notice that, since we know the answer will be vector field, we can ignore
terms with higher-order derivatives. Of course, if you just write everything
out, you will see that these cancel out. Similar calculations show [Z, X] =
Y, [Y, Z] = X.
One might notice that this structure looks a bit like the cross-product
structure on R3 (replacing X, Y, Z with i, j, k). In fact, this can be made ex-
plicit. The 3-dimensional vector space spanned by the vector fields {X, Y, Z},
together with the Lie bracket, form a Lie algebra. Likewise, R3 with the
cross-product structure also forms a Lie algebra. These two Lie algebras are
isomorphic under the map

aX + bY + cZ 7−→ abi + bb
j + cb
k.

The above map is linear and invertible, and the two brackets commute with
each other (check this).
Furthermore, it is not too difficult to show that the flow generated by
aX + bY + cZ is actually a 1-parameter subgroup of SO(3, R). This can be
seen through the work done in homework problem Ch. 5, number 6, as well
as a bit of what we talked about in class concerning matrix groups.

6
 Math 531 Midterm

Please abide by the following rules:

• Time Limit: 3 hours

• You are allowed 1 untimed break

• This test is open notes and open Spivak. You may not consult any additional sources (or
people)

• You are to answer 2 of the first 3 questions and 1 of the last 2 questions.

• Exam due Friday, March 28 (by end of day).

The test is not meant to be confusing. Any extra information contained in the questions is
designed to clarify and not obscure, and the questions are not designed to trick you. Please let me
know if anything is unclear.
 Math 531 Midterm

Solve any 2 of the following 3 problems:

1. In multivariable calculus, given a function f : Rn → R, you define the gradient vector ∇f by
X ∂f ∂
∇f = .
i
∂xi ∂xi

∂f
In other words, it is vector field with component functions .
∂xi
Consider the more general situation where M is a smooth manifold, and f : M → R a smooth
function.
∂f
a. Show that the functions i (where x is a local coordinate system) do not form the component
∂x
functions of a vector field. Show, however, that they do naturally form the component functions of
a 1-form.
b. Suppose we have a tensor g ∈ C ∞ (M, T M ∗ ⊗ T M ∗ ), and further suppose that at each point
x ∈ M , the tensor gx is symmetric and non-degenerate. In other words,

gx (v1 , v2 ) = gx (v2 , v1 ),

and the rewritten

gx : Tx M ∗∗ → Tx M ∗
∂f
is invertible. Then, show that g can be used to change the components ∂xi
into components of a
vector field.

2. Consider the polar coordinates on (subsets of) R2 defined by

x = r cos θ
y = r sin θ

∂ ∂
Compute and in polar coordinates .
∂x ∂y

3. In an ODE class, you often encounter the following spring equation:

d2 x dx
= −kx − c ,
dt2 dt
where x(t) is the displacement of a mass at time t, and k, c are constants related to the strength of
the spring force and resistance force, respectively.
dx
a. Let y = , and rewrite the above equation in terms of a vector field on M = R2 .
dt
b. Show that flow induced by the vector field in (a) is a 1-parameter subgroup of Gl(2, R).
Solve 1 of the following 2 problems:
4. Implicit differentiation is a useful tool in calculus. Essentially, if we have

f (x, y) = c,

dy
then we solve for by implicitly differentiating and obtaining
dx
dy ∂f /∂x
=− .
dx ∂f /∂y

Prove that this is mathematically rigorous.

5. Suppose M and N are connected, oriented manifolds and f : M → N is a local diffeomorphism.

Show that if f∗p preserves orientation at some p ∈ M , then f∗ preserves orientation at all points
p ∈ M.
 Math 531 Midterm Solutions

1. In multivariable calculus, given a function f : Rn → R, you define the gradient vector ∇f by

X ∂f ∂
∇f = .
i
∂xi ∂xi

∂f
In other words, it is vector field with component functions .
∂xi
Consider the more general situation where M is a smooth manifold, and f : M → R a smooth
function.
∂f
a. Show that the functions i (where x is a local coordinate system) do not form the component
∂x
functions of a vector field. Show, however, that they do naturally form the component functions of
a 1-form.
b. Suppose we have a tensor g ∈ C ∞ (M, T M ∗ ⊗ T M ∗ ), and further suppose that at each point
x ∈ M , the tensor gx is symmetric and non-degenerate. In other words,

gx (v1 , v2 ) = gx (v2 , v1 ),

and the rewritten

gx : Tx M ∗∗ → Tx M ∗
∂f
is invertible. Then, show that g can be used to change the components ∂xi
into components of a
vector field.

Solution:
Notice that we can define df ∈ C ∞ (M, T M ∗ ) in a coordinate-free way. That is, df (X) = X(f ),
where X is a vector field. Then, a simple calculation shows that
∂f i
df = dx
∂xi
∂f
in local coordinates. This implies ∂x i are the components of a (non-trivial) 1-form, and hence not

a vector field.
More explicitly, suppose ai , bi are the component functions of a vector field in x, y coordinates,
respectively. Then these are related by the equation

∂y j
b j = ai .
∂xi
However, we see that
∂f ∂f ∂xi
= .
∂y j ∂xi ∂y j
This does not satisfy the transformation property for components of vector fields, but instead
satisfies the transformation property for differential forms.
In fact, we can see that that

∂f ∂y j
  i 
∂f i ∂x j ∂f
i
dx = j i j
dy = j dy j ,
∂x ∂y ∂x ∂y ∂y
whereas 2
∂xi

∂f ∂ ∂f ∂ ∂f ∂
i i
= j j 6= .
∂x ∂x ∂y ∂y ∂y j ∂y j ∂y j
∂f
Therefore, ∂xi
dxi is a well-defined 1-form on M , but not a well-defined vector field.

Now, suppose we have the tensor g. Such a tensor g is a non-degenerate inner product on
each tangent space and is called a metric (if g is positive-definite, then it is a Riemannian metric).
(Remark: we don’t need the symmetric property here, but it makes indices a little easier to deal
with.)
If g ∈ C ∞ (M, T M ∗ ⊗ T M ∗ ), then we can equivalently consider

g ∈ C ∞ (M, Hom(T M, T M ∗ ))

(using the generic natural equivalences Hom(V, W ) ∼

= V ∗ ⊗ W and V ∗∗ ∼
= V ). The non-degeneracy
of g implies that we have

g −1 ∈ C ∞ (M, Hom(T M ∗ , T M )) ∼
= C ∞ (M, T M ⊗ T M ).

Therefore, we can construct ∇f = g −1 (df ) by the following:

d g −1
C ∞ (M ) −→ C ∞ (M, T M ∗ ) −→ C ∞ (M, T M )
f 7−→ df 7−→ g −1 (df ) = ∇f

For fun, let’s write this in local coordinates. We know that

g = gij dxi ⊗ dxj .

Since g is non-degenerate, the matrix gij is invertible. Hence, we also have the tensor

∂ ∂
g −1 = g ji ⊗ .
∂xi ∂xj
We now see that we can use g −1 to transform the above df into a vector field. We define
 
−1 kj ∂ ∂ ∂f i
∇f = g (df ) = g ⊗ k dx
∂xj ∂x ∂xi
∂ ∂f
= g kj j δki i
∂x ∂x
∂f ∂
= g ij i j
∂x ∂x
It is simple to check that ∇f is now a well-defined vector field. Suppose that
∂ ∂
g −1 = g βα .
∂y α ∂y β
Then, we see that
α
∂y β ∂f ∂xk
   l 
αβ ∂f ∂ ij ∂y ∂x ∂
g = g
∂y α ∂y β ∂xi ∂xj ∂xk ∂y α ∂y β ∂xl
∂f ∂
= g ij δik δjl k l
∂x ∂x
ij ∂f ∂
=g
∂xi ∂xj
Therefore, we see that ∇f is a well-defined vector field.
The above demonstrates the general principle that if some quantity is written in local coordi-
nates, each index should appear “up” and “down” an equal number of times.
∂f
In summary, we see that we can use the tensor g to transform the functions ∂x i into components

of a vector field. In the situation where M = Rn with standard coordinate x, and gij = δij is the
usual Euclidean metric, we see that
∂f ∂ ∂f ∂ ∂f ∂
∇f = g ij i j
= δ ij i j = .
∂x ∂x ∂x ∂x ∂xi ∂xi

2. Consider the polar coordinates on (subsets of) R2 defined by

x = r cos θ
y = r sin θ
∂ ∂
Compute and in polar coordinates .
∂x ∂y
Solution:
We know that
∂ ∂r ∂ ∂θ ∂
= +
∂x ∂x ∂r ∂x ∂θ
∂ ∂r ∂ ∂θ ∂
= +
∂y ∂y ∂r ∂y ∂θ

Letting u and p denote the Euclidean and polar coordinate systems, respectively, we know that
D(p ◦ u−1 ) = D((u ◦ p−1 )−1 ) = [D(u ◦ p−1 )]−1 ].
As easy calculation shows us that
∂x ∂x
   
−1 ∂r ∂θ cos θ −r sin θ
D(u ◦ p ) = ∂y ∂y = .
∂r ∂θ
sin θ r cos θ
Therefore,
 ∂r ∂r   −1
−1 ∂x ∂y cos θ −r sin θ
D(p ◦ u ) = ∂θ ∂θ =
∂x ∂y
sin θ r cos θ
 
cos θ sin θ
= − sin θ cos θ
r r
Therefore, we see that
∂ ∂ 1 ∂
= cos θ − sin θ
∂x ∂r r ∂θ
∂ ∂ 1 ∂
= sin θ + cos θ
∂y ∂r r ∂θ

3. In an ODE class, you often encounter the following spring equation:

d2 x dx
2
= −kx − c ,
dt dt
where x(t) is the displacement of a mass at time t, and k, c are constants related to the strength of
the spring force and resistance force, respectively.
dx
a. Let y = , and rewrite the above equation in terms of a vector field on M = R2 .
dt
b. Show that flow induced by the vector field in (a) is a 1-parameter subgroup of Gl(2, R).

Solution:
dx
First, let y = dt
. The 2nd-order differential equation now becomes a system of first-order equations
(
dx
dt
=y
dy
dt
= −kx − cy

Remembering that we are solving for functions (x(t), y(t)), we think about solutions to the above
equation as paths γ = (γ x , γ y ) : (−, ) → R2 . The equation is then equivalent to
( x
dγ ∂ ∂
dt ∂x
= y ∂x
y
dγ ∂ ∂
dt ∂y
= (−kx − cy) ∂y

In other words, we want

∂ dγ dγ x ∂ dγ y ∂ ∂ ∂
γ∗ ( )= = + =y + (−kx − cy) = Xγ(t) .
∂t dt dt ∂x dt ∂y ∂x ∂y
Solutions to the system of ODEs are equivalent to integral curves to the vector field
∂ ∂
X=y + (−kx − cy) .
∂x ∂y
The above equation is system of linear equations with constant coefficients. In homework prob-
lem number...., we saw that if this is rewritten as
dγ
= Aγ,
dt
where A has constant coefficients, then the integral curve to X with initial point γ(0) = γ0 is given
by
γ(t) = eAt γ0 .
Therefore, we have that (in standard components)
 0    
0 x (t) 0 1 x(t)
γ (t) = 0 =
y (t) −k −c y(t)

Therefore, the integral curves, with γ(0) = (x0 , y0 ), will be of the form
     
x(t) 0 1 x0
γ(t) = = exp t .
y(t) −k −c y0

The flow φt : R2 → R2 is then given by

φt = eAt ∈ Gl(2, R).

Furthermore, φ is a homomorphism R → Gl(2, R) (where R is the additive group), showing that

the flow is a 1-parameter subgroup of Gl(2, R). Equivalently, we can explicitly see that

φ0 = e0A = I, φt ◦ φs = eA(s+t) .
Solve 1 of the following 2 problems:
4. Implicit differentiation is a useful tool in calculus. Essentially, if we have

f (x, y) = c,

dy
then we solve for by implicitly differentiating and obtaining
dx
dy ∂f /∂x
=− .
dx ∂f /∂y

Prove that this is mathematically rigorous.

dy
Solution: Suppose we wish to calculate dx at a point (x0 , y0 ), and assume that ∂f
∂y
6= 0 at (x0 , y0 ).
Then, by the implicit function theorem, f (x, y) = c implicitly defines y as a function of x. In other
words, in a neighborhood of (x0 , y0 ), there exists g(x) (=y(x))such that

f (x, g(x)) = c.

Hence, we calculate
d d
|x0 (f (x, g(x))) = |x (c)
dx dx 0
dx dg
D1 (f (x0 , g(x0 ))) |x + D2 (f (x0 , g(x0 ))) |x0 = 0
dx 0 dx
Then, using the shorthand of y(x) = g(x), we obtain the standard expression

∂f ∂f dy
|(x0 ,y0 ) + |(x0 ,y0 ) |(x0 ,y0 ) = 0
∂x ∂y dx
∂f /∂x dy dg
− |(x0 ,g(x0 ) = |(x0 ,y0 ) = |x
∂f /∂y dx dx 0

dy ∂f /∂x
=− .
dx ∂f /∂y
∂f ∂f
Notice that since wish to divide by ∂y
, we would have wanted to require ∂y
to be non-zero.

5. Suppose M and N are connected, oriented manifolds and f : M → N is a local diffeomorphism.

Show that if f∗p preserves orientation at some p ∈ M , then f∗ preserves orientation at all points
p ∈ M.

Solution: First, f is a local diffeomorphism implies that f∗ is an isomorphism at every point. An

isomorphism of vector spaces gives a bijection between the set of orientations on each vector space.
Hence, it makes sense to talk about f∗ preserving or switching orientation at each point.
The space of orientations at each point p ∈ M is a discrete space. In fact, the space of orientations
naturally forms a double cover, which we shall call Or(M ). That is, there is a topological space
Or(M ) with a free Z/2 action such that Or(M )/Z/2 = M . If M, N are oriented, then the covers
Or(M ) and Or(N ) are disconnected. The choice of orientations M and N give us a choice of
connected components in Or(M ), Or(N ). We therefore have a continuous map

f∗ : Or(M ) → Or(N ).

Because of continuity, f∗ cannot map a connected component to two different connected components.
Therefore, f∗ will preserve orientation at all points.
Alternatively, we can let x0 ∈ M be an arbitrary point. Assume f∗ : Tx0 M → Tf (x0 ) N preserves
orientation. Then, let x ∈ M be any other point. Since M is connected (and connectedness is
equivalent to path connectedness for manifolds), we can choose some path

γ : [0, 1] → M
0 7→ x0
1 7→ x

connecting x0 and x. The tangent spaces of M and N are continuously oriented, and f is a
smooth map. This implies that if f∗p preserves orientation, then f∗ preserves orientation in some
neighborhood of p. We then take open cover containing the image of γ and see that fx must preserve
orientation.
 HW Due Thursday 4/24

1. Let G be a Lie group. A form ω ∈ Ωk (G) is (left) invariant if

L∗g ω = ω ∀g ∈ G.

(Above, Lg is left-multiplication by g.)

a. Show that the restriction of
d d d
0 → C ∞ (G) → Ω1 (G) → · · · → Ωn (G) → 0

to the subspaces of left-invariant forms is also a cochain complex. In other

words, show that the exterior derivative of an invariant form is also invariant.
b. If g is the Lie algebra of left-invariant vector fields, show that the
cochain complex of left-invariant forms is the finite-dimensional complex
d d d d
0 → R → g∗ → Λ2 g∗ → · · · → Λn g∗ → 0

Use the invariant description of d to give a description of d above using only

the Lie algebra structure.
c. Consider the cohomology H ∗ (g, d) of the finite-dimensional complex
from part (b) (this is called Lie algebra cohomology). Prove that

H 1 (g) ∼
= (g/[g, g])∗ .

What is H 1 (gl(n, R), where gl(n, R) denotes the Lie algebra of all n × n real
matrices?

1
 Math 531 Final
May 15, 2008
Solve any 2 of the following 3 problems:

1. Show directly that RP n is a smooth manifold by giving local coordinate patches and calculating
the transition functions (Hint: use homogeneous coordinates).

2. Let M = R3 , and consider the 2-dimensional sub-bundle of T R3 spanned by the vector fields
∂ ∂ ∂ ∂ ∂ ∂
+ + f (x, y) and + + g(x, y) .
∂x ∂y ∂z ∂x ∂y ∂z
For what functions f, g will this be an integrable distribution?

3. Let M = R2 . Apply Stokes Theorem to integrals of 1-forms (in local coordinates) to obtain
two standard results from multi-variable calculus: The Fundamental Theorem of Line Integrals and
Green’s Theorem. (You don’t need to remember what these theorems are to answer this question.)

Solve any 3 of the following 4 problems: (In the following, H k (M ) denotes de Rham coho-
mology.)

4. Let M be a connected manifold. Show that π1 (M ) = 0 implies that H 1 (M ) = 0. (Hint: Show

that the integral of a closed 1-form over a closed 1-manifold is 0.)

5. Let M n be a compact, connected, oriented n-manifold. Let Dn be a closed disc around a point
x0 ∈ M . Calculate the de Rham cohomology of M n − Dn .

6. Let M, N be connected, compact, oriented n-manifolds. Prove that if H k (N ) 6= 0 and H k (M ) = 0

for some k 6= 0, then any map f : M → N must have degree 0. (Hint: Use Poincare duality.)

7. Let M be a compact, connected, oriented, n-manifold. What is the Euler class of the vector
bundle Λn T M ∗ . Specifically, in what cohomology group does it live, and is it non-zero?

Solve the next problem:

8. Make up a good exam question and answer it.