As taught by Prof. Bruce Hughes
Vanderbilt University - Spring Semester 2016
Transcribed by David K. Zhang
The notes below were taken live in-class during Prof. Hughes' MATH 2501 lectures at Vanderbilt University. They are provided here exclusively as a historical record of what topics were covered during each day of class. For readers who aim to learn this material for the first time, they are incomplete in several crucial ways:
Time permitting, I plan to release an expanded and revised version of these notes in the future to address these deficiencies.
Introduction to manifolds. The main new topic of the second semester will be integration of functions . We will generalize the one-dimensional integral
to more exciting higher-dimensional domains called manifolds (possibly with boundary).
Rough idea of manifolds. We call a -dimensional manifold if each point in has a neighborhood that can be described by a -dimensional curvilinear coordinate system.
Definition: is a -dimensional manifold if for each , there exist open sets and and a function such that
.
.
is smooth (i.e., infinitely differentiable).
in injective.
is continuous.
For all , . (Recall that is the dimension of the image of . We call this image the tangent space to the image of at .)
Example: Let be the solution set of . This is the unit paraboloid, i.e., the graph of . Given , let and . Define
Then, we have that
.
.
is smooth (since it has polynomial components).
is injective.
is given by .
has rank 2.
Hence, is a 2-dimensional manifold.
Recall: is a -dimensional manifold if , , such that
.
is smooth.
is continuous.
.
Manifold Characterization Theorem: Let . The following are equivalent:
is a -dimensional manifold in .
, such that
.
.
is smooth.
, .
Remark: This theorem gives an implicit characterization of manifolds as local level sets of smooth locally invertible functions. Its proof requires machinery which will be developed later in the course.
Examples:
The sphere is defined by
Define by , and let . Then , is smooth, and
has rank for all . Hence, by the manifold characterization theorem, is a manifold.
We can also verify that is a manifold directly by definition. (This part omitted. Proceeds straightforwardly by pushing a disk up to a half-sphere.)
Consider the configuration space of all ordered pairs of orthogonal unit vectors in .
Let and define by
Then , is smooth, and
has rank (any smaller and would be a scalar multiple of ).
The Inverse Function Theorem. Let , and suppose is a smooth function with invertible. Then is invertible near . More precisely, there exists such that , is a bijection, and the inverse of is smooth. Moreover, .
Example: Let be defined by . Then near , a local inverse of is given by . In this case, we can take
and for , we have
so indeed .
Using IFT to prove the Manifold Characterization Theorem. We will show that (II) implies (I). (The converse will be proven later.) Recall that we are given a smooth function with for all . We want to show that is a -dimensional manifold.
Let be given. Since , there are free variables in the system . For notational simplicity, assume the free variables are . Then the RREF of is . Define by
Then
Hence, is invertible, and IFT applies, giving us with where has a smooth inverse . This gives the necessary data to show that is a -dimensional manifold. QED
Banach's Contraction Mapping Principle. This will be our main tool for proving the implicit function theorem.
Definition: Let . A function is a contraction mapping if such that and
for all .
Examples:
Let be defined by . Then , so is a contraction mapping with .
Let be defined by . Then , so is not a contraction mapping.
Theorem (CMP): Suppose is nonempty and closed. If is a contraction mapping, then there exists a unique such that . (In other words, has a unique fixed point.)
Examples:
Take from example (1) above. Then the unique fixed point of is .
Every linear transformation has a fixed point . This may or may not be the only fixed point.
Two Facts about Infinite Series.
If , for all , and is bounded, then converges.
If converges, then so does .
Proof of Existence in CMP: Fix any , and let . We claim that converges. To see this, let . Then
This shows that is a positive series bounded above by a geometric series. It follows by fact 1 that converges, and by fact 2 that converges. Since is closed, exists and is a member of .
By homework problem #3, every contraction mapping is continuous. It follows that
as desired. QED
(Proving uniqueness is a homework problem.)
The Operator Norm and Mean Value Theorem. These will be useful tools for the proof of the inverse function theorem.
Example: We illustrate how the univariate MVT can be used to show that a function is a contraction mapping. Define by . By the mean value theorem, whenever , there exists such that
It follows that
and since is bounded above on by , we have that is a contraction mapping, since
Moreover, if we wanted to, we could numerically solve the equation by taking any and repeatedly taking its cosine.
Definition: Let be a linear transformation. The operator norm of is
Note that this sup exists by the maximum value theorem.
Theorem: For all , we have .
Proof: The result holds trivially for . For nonzero , we have
by definition. Hence,
as desired. QED
Corollary: If is a linear transformation such that , then is a contraction mapping.
Proof: Let . Then for all , we have
QED
Remark: Note that any such has as its unique fixed point.
Example: Let be a linear transformation, and let . Then for all , and . To see this, observe by Cauchy-Schwarz that
This means that is bounded above by . Moreover, if , then
giving a bound below. For the case, we have .
Multivariable Mean Value Theorem: Suppose and is smooth. Let such that . (Here, is the line segment joining to .) If there exists such that , then .
Inverse Function Theorem: Let and suppose has the property that is invertible. Then is smoothly invertible near . More precisely, there exists such that , , and is a bijection. Moreover, its inverse is smooth.
Proof sketch: By translation, we can take without loss of generality. By replacing by , we can take . We will show that there exists such that and is injective on . We will then take and do a little more work to show that is open and is smooth.
Since is smooth, whenever is close to , the entries of will be close to those of . It follows that there exists for which implies and .
We claim that for all , there exists a unique such that . To see this, it suffices to show that defined by a unique fixed point . For this, we wish to apply Banach's CMP, so we must show that is a contraction mapping taking to .
Observe that
for all . Moreover,
Now, by MMVT,
Hence,
This means that for all , we have
so is a contraction mapping. (Extra work to show that is open and is smooth omitted.) EOS
One-Dimensional Integration. Given a continuous function , we define
where is a partition of into subintervals of equal length , and are arbitrary sample points from these intervals.
Remarks:
If on , then defines the area under the graph about .
This limit always exists (assuming is continuous) and is independent of the sample points.
Recall the fundamental theorem of calculus: there exists such that , and moreover .
Double Integrals. Given a rectangle and a bounded function , we define (sometimes written ) as follows: for each product partition of , let
where is the th subrectangle of in . Then we define the upper and lower sums of according to as
Definition: We say that is integrable on iff there exists a unique number such that for every partition of ,
If so, we define and call it the Riemann integral of on .
Example: Let , and let be defined by
We would guess . Consider the partition of into two rectangles across . Then
and
A First Glimpse of Fubini and the Gap Dichotomy. Fubini's theorem allows us to reduce multiple integrals to iterated integrals.
Fubini's Theorem: If and is continuous, then is integrable on , and
Example: Let and . Then
Definition: Let . A partition of is a tuple of one-dimensional partitions, where is a partition of . We define the volume of to be .
{-dimensional integration.} Similarly to the 2D case, we let be a partition of a rectangle . Then for each , we define
We then define the lower and upper sums
Definition: is integrable on iff there exists a unique such that for every partition of ,
When this occurs, we define .
(Example of a non-integrable function omitted. Take the characteristic function of .)
The Gap Dichotomy. For any function , let be the collection of lower sums of over , and let be the collection of upper sums. Then either there is a gap between and , in which case is not integrable, or and touch, and is integrable.
Corollary: If there exists a sequence of partitions of such that
then is integrable over , and is that common limit.
Definition: Let and be partitions of . We say that is a refinement of if every subrectangle is contained in some subrectangle . This means that is obtained from by adding extra endpoints to .
Proof of Gap Dichotomy: Observe that if is a refinement of , then
Furthermore, if and are two partitions, then they have a common refinement . (Simply draw and on top of each other.) Then
This means that for any two partitions and , we have , as desired. QED
Integration over Bounded Sets and Volume. Let be bounded, and let be a bounded function. Then there exists a rectangle , and we can extend to by defining
This allows us to define integrals over non-rectangular bounded regions. Provided the RHS exists, we define
Warning: The RHS might fail to exist even for really “nice” functions . For example, take , and let be the constant function 1. Then
fails to exist. This tells us that “niceness” of is not sufficient to guarantee integrability; we also need “niceness” of the region .
A Second Glimpse of Fubini. Suppose we have two functions for which for all . Let , and let be the region bounded by the graphs of and . Then by Fubini's theorem,
Example: Find where is the region in bounded by and . (Solution omitted.)
Definition: Let be bounded. The -dimensional volume of is
if the RHS exists. (Some discussion omitted.)
(Discussion of the definition of -dimensional volume omitted.)
Theorem: iff for every , there exists a partition of such that
Proof (): Let be the characteristic function of . Given , the gap dichotomy implies the existence of a partition such that
as desired.
(): We first show that is integrable. Let be given. By hypothesis, there exists a partition for which
It follows that
It follows that the upper and lower sums of get arbitrarily close; hence, exists. Moreover, , so , as desired. QED
Corollary: If , then does not contain any -dimensional rectangle.
Theorem: Let be a rectangle and let be continuous. Then exists.
Proof: Let be given. By the gap dichotomy, it suffices to find a partition of such that
Because is compact, is uniformly continuous on , and there exists such that
for all . Choose such that every subrectangle has side lengths strictly less than . This guarantees that the diameter of is less than , so that
as desired. QED
Fubini's Theorem. Let be integrable in . We want to show that
Notation: For all , define by
for all in . Assume further that for all , is integrable on , and define by
This is the inner integral in the statement of Fubini's theorem.
Theorem: is integrable on , and
Proof: Let be a partition of . We can write where and are partitions of and , respectively. We will show that
This will prove the theorem by the gap dichotomy. Indeed, we are given that no gap exists between and . The preceding inequality will show that no gap exists between and . Let
Then every is of the form for some , and we can write
Now note that for all , we have
It follows that for all and , we have
Now multiply both sides by and sum over to obtain
This shows that
as desired. The inequality for the upper sums follows similarly. QED
Introduction to Change of Variables. We will often find that a multiple integral can be broken into iterated integrals more naturally in a non-Cartesian coordinate system. For example, the area of a circular sector is easier to calculate in polar coordinates.
Polar Coordinates . These coordinates are defined by
Let be the coordinate transformation function
We usually restrict . Note that
Area Distortion. Let be a domain in the -plane and let . In general, we will find
because the coordinate transformation does not preserve area. We need to multiply the RHS by to compensate for this.
(Example omitted.)
Cylindrical Coordinates in . Let be defined by
Then .
(Example omitted.)
Discussion of test review omitted.
Spherical Coordinates. This is the coordinate system defined by , where
(Note that is the azimuthal angle, and is the polar angle.) The determinant of this transformation is
(Example omitted: volume of sphere.)
Definition: An determinant is an alternating multilinear function with the normalization property .
Theorem: For every , there exists a unique determinant.
Proof: The theorem is clear in the case: for a scalar , we must have .
To demonstrate existence for higher , we can inductively define as follows: for , let be the matrix obtained by deleting the th row and th column of . Let be the cofactors of . Then we define
The function defined in this way (so-called “cofactor expansion”) is an determinant. (We omit details of the verification here.)
The proof of uniqueness requires the following 12-step program: Let be an determinant.
If has two distinct equal columns, then .
If is obtained from by multiplying one of its columns by , then .
If is a column of , then .
If is obtained from by adding a scalar multiple of column to column , then .
\setcounter{enumi}{4}
Using steps 1-4, we can easily compute determinants of elementary matrices.
If with elementary, then performs an elementary column operation on .
If is elementary, then .
If , then , since we can express one of the columns of as a linear combination of the others.
. Indeed, if , then both sides are zero, and if , we can write as a product of elementary matrices and repeatedly apply step 8.
If is invertible, then and . This follows by considering .
The following are equivalent:
.
is invertible.
. Indeed, the equivalence of (b) and (c) was shown last semester. Step 10 shows (b) implies (a), and the contrapositive of step 8 shows (a) implies (c).
Let and be two determinants. Then for all , . Indeed, if , then both sides are zero by step 8, and if , then we write as a product of elementary matrices and apply steps 5 and 7.
Theorem: If , then .
Proof: If , then since .
If , write as a product of elementary matrices. It is easy to show by cases that . Hence, by the product rule. QED
Theorem: If has an -dimensional volume and is a linear transformation, then has -dimensional volume, and
Proposition: Let be a rectangle in and be a linear transformation. Then
Proof: If is not invertible, then is a linear subspace of of dimension . Any bounded subset of thus has volume .
If is elementary of type I or II, then takes rectangles to rectangles, and the theorem is established with simple casework. (We omit details here.)
If is elementary of type III, then has one nonzero off-diagonal entry. Say performs the elementary row operation , so that the nonzero entry occurs at location . Then the region is bounded by
and a simple application of Fubini's theorem finishes the proof. QED
Proposition: Let be a linear transformation, and suppose has -dimensional volume. Then , provided is either singular or elementary.
Proof: Let for some rectangle , so that . For any partition of , we have
It follows, by the preceding proposition, that
A similar argument for lower sums shows that
Now, either , in which case , , in which case
QED
Theorem: Let be a linear transformation, and suppose has -dimensional volume. Then .
Proof: If is singular, then both sides are zero by the preceding proposition. If is invertible, write as a product of elementary matrices and repeatedly apply the preceding proposition using the product rule for determinants. QED
The Linear Change of Variables Theorem: Let be a linear transformation, , and be bounded. Then
assuming either integral exists.
Proof sketch (assuming both integrals exist): For simplicity of notation, let
and assume is a rectangle. Let be a partition of . Then
A similar argument for lower sums shows that is between every upper and lower sum of , so the two are equal. QED
The General Change of Variables Theorem: For smooth,
{Parallelepipeds and -dimensional volume.} Let be linearly independent vectors in .
Definition: The -dimensional parallelepiped spanned by is
For , we define the unit -cube .
Let . Then . If , we define
which is motivated by our Linear Change of Variables Theorem.
{Volume with .} If is a linear subspace of and is bounded, we would like to have a reasonable definition of . We will take the following approach:
Find a linear isometry , that is, a linear transformation such that is an isometry.
Define .
Prove that this definition is independent of the choice of .
Theorem: If has rank , then
Note that for , this reduces to , as before.
Recall: is orthogonal if , that is, if the columns of form an orthonormal basis.
Theorem: If is a -dimensional linear subspace, then has an orthonormal basis.
Proof: Build up the basis incrementally. Take any unit vector , take , and so on. QED
Thus, we can always find with , , and .
Proposition: If satisfies (i.e., the columns of are orthonormal), then preserves dot products of vectors in . That is, for all , we have .
Proof: Since , we can write and for some . It follows that
as desired. QED
Corollary: If satisfies , then is an isometric embedding of in .
Definition: If is a subset of a -dimensional linear subspace of , then we define , where is an isometric embedding of in . (By the preceding proposition, such an embedding always exists by taking the columns of to be an orthonormal basis of .)
The following theorem shows that is well-defined.
Theorem: Let have rank . Let be bounded. Then
Proof: Let be a matrix whose columns form an orthonormal basis for . By definition,
where we note that is square. Thus, it suffices to show that
Observe that
We claim that . To see this, let be arbitrary. Then there exists such that . It follows that
Since this holds for arbitrary , we have , as desired. This completes the proof. QED
Corollary: If has rank , then .
Proof: We have already seen that is nonnegative, so it suffices to show that is nonzero. Suppose . Then
where the last implication follows from the rank-nullity theorem. Thus, is invertible, and . QED
Corollary: If has rank , then
Integration of Scalar-Valued Functions on Manifolds. Let be a bounded subset of an open set in , and let be smooth. Then is a -parameterized region in . We would like to construct a reasonable definition for the integral of over .
Design Criteria:
If is a constant function and is injective, our definition should give the -dimensional volume of .
Our definition should be consistent with previous theorems. In particular,
If is a linear transformation and , we should get .
If and , we should get .
Based on these criteria, we can rule out a few possible candidates:
The definition should NOT be , since when , this gives , which vanishes when .
The definition should NOT be , since when and is a linear transformation, we want to get , not .
Definition: The integral of on with respect to -dimensional volume is
Remark: We often write for , for , and or for .
(Omitted discussion of physical interpretation of this integral as measuring total mass for a given density distribution.)
_{(Missed proof that is well-defined. Show that for any two matrices whose columns form an orthonormal basis of . Let such that . Then
as desired.)}_
Recall our new integral
If is injective on , then
Note that in the special case , we obtain the usual formula for the arc length of a parametric curve:
(Examples omitted: arc length of conical helix and surface area of unit 2-sphere.)
(Missed definition of -forms.)
Main Example: Let where are integers between and . we call a multi-index. Define as follows: write
Then we define .
Notation: We typically give this -form the name . We call an elementary -form.
(Basic examples of computations omitted.)
Fact: If , then
Geometric interpretation. Let and . Then the columns of span a -dimensional parallelepiped in . Project to the -dimensional coordinate subspace to obtain a -dimensional parallelepiped in . Then .
Definition: is the set of all -forms on . (This notation is not universal; other authors may call this or , where stands for “alternating.”) We also define a -form to be a number, so that .
Note: if .
Definition: If is strictly increasing, then is called a basic -form.
Theorem:
is a vector space (under the standard addition and scalar multiplication of real-valued functions).
If , then .
For , the basic -forms on form a basis for .
{-forms and differential -forms.} Recall that is the vector space of all -forms on , which are multilinear alternating functions
There are basic -forms on . Thus, is -dimensional.
Proposition: .
(Proof omitted. Elementary combinatorics.)
Recall that a -form on is a multilinear alternating function . We denote by the vector space of all -forms on . If is an increasing multi-index (i.e., if ), then we define the basic -form by , where is the matrix formed by taking rows of .
Theorem: If , then the basic -forms on form a basis for . Hence, .
Special Case: If , then , with . For each , there exists such that . In particular, .
Special Case: If , then the 1-forms are precisely the linear transformations . Thus, can be viewed as the space of row vectors of length , which clearly has dimension .
Remark: We call the dual basis for .
General case: Given , we can write , where the sum is taken over strictly increasing multi-indices , and
It is not difficult to show that this expansion holds for all -forms . We will omit the details here.
Definition: If and , then their wedge product is the -form defined by
for basic forms and extending linearly for other forms. (Example omitted.)
Review for Exam II. Discussion omitted.
Missed this lecture.
The Exterior Derivative, Vector Fields, Work. Let . Recall that denotes the vector space of all differential -forms on , where iff for each increasing multi-index there exists a smooth function such that
We can evaluate at a point to obtain an ordinary -form
We also define set of all smooth functions .
The exterior derivative is a map . For , we define by
Note that . The object on the left is the exterior derivative of a smooth function; the object on the right is a basic 1-form.
For , we define the exterior derivative by declaring
and extending linearly.
(Example of differentiating a 1-form omitted.)
Missed this lecture.
Let be an open subset of .
Definition: is conservative iff there exists such that . In this case, we say that is a potential function for . Similarly, is exact iff there exists such that .
Fundamental Theorem of Line Integrals: Let be a conservative vector field on with potential function , and let be a smooth curve in parameterized by . Then
Here, is the work form of , which is obtained from by replacing unit vectors by unit 1-forms .
Proof: Straightforward chasing through definitions. QED
Example: Consider the “toilet bowl” field
defined on . Observe that is not conservative (since the integral of over the unit circle is nonzero), but .
Missed this lecture.
Stokes' Theorem: Let be a compact oriented -manifold with boundary. Suppose and . Then
Definition: An orientation of is represented by an ordered basis \linebreak of . This orientation is the positive orientation iff
Otherwise, it is the negative orientation.
Examples:
The orientation represented by the basis of is positive if and negative if .
A basis of represents the positive (resp. negative) orientation if is reached from by traveling in a counterclockwise (resp. clockwise) direction.
In , positive/negative orientation of a basis corresponds to right/left-handedness.
Fact: represents the positive orientation iff the matrix
can be column reduced to such that the number of column exchanges and scalar multiplications of columns by negative scalars is even.
Fancier way: Choose a nonzero . Then the bases and represent the same orientation iff and \linebreak have the same sign.
Definition: Let be a -dimensional manifold in . An orientation form for is a differential form defined on some neighborhood of such that is non-vanishing on . This means such that .
Fact: This is true iff for every basis of .
Example: Consider . Define by
This an orientation form on , since is always orthogonal to the plane . This shows that is orientable.
Alternatively, recall that the flux form of a vector field is defined by
so our orientation form is actually the flux form of the identity vector field on .
Orientation of hypersurfaces. Let be a -dimensional manifold embedded in , where .
Definition: A vector field defined on a neighborhood of is a normal field on iff for all , is orthogonal to . It is nowhere zero on iff for all .
Theorem: is orientable iff admits a nowhere zero normal field.
(Some example computations omitted.)
(Example computations with Stokes's theorem. Omitted.)
Stokes's Theorem: If is a compact oriented -manifold in and , then
where has the induced orientation compatible with .
What does compatible mean? Suppose and are orientations for and .
Definition: and are compatible iff , bases of , has the same sign as , where is an outward-pointing tangent vector to at . Here, is a boundary chart mapping to a coordinate neighborhood of in such that .
Example: is a -manifold with boundary in .
Theorem: If is the standard orientation of and is the standard orientation of , then and are compatible.
Proof: Immediate. QED
(Some discussion of partitions of unity omitted. To integrate a form over a manifold, pick a coordinate cover and a partition of unity subordinate to it, and integrate the form as usual in coordinates. Sum up the results.)
Proof of Stokes's Theorem: It suffices to prove Stokes's theorem for compact rectangles in . Indeed, by taking a partition of unity on a manifold with boundary, we reduce integration of a form to integration over coordinate charts and boundary charts.
Given where outside a compact rectangle , write
where the hat denotes omission as usual, and is zero on the boundaries of , except for the boundary of the upper half-space . (This is the boundary that matters in Stokes's theorem.) Then
and it follows that
On the other hand,
where denotes the determinant of omitting the th row. QED
Change of Basis. The goal of our next unit will be to represent a linear transformation with respect to a (possibly more convenient) basis, other than the standard basis .
Set-up. Let be an ordered basis for a vector space . This means that any can be uniquely written as a linear combination of members of the basis:
The scalars are called the coordinates of with respect to the basis . The map sending each vector to its (ordered) tuple of coordinates is clearly a linear isomorphism between and .
Definition: Let be a linear transformation. The {matrix representation of with respect to } is
Example: Let be the set of polynomials of degree , with ordered basis . The derivative operator is a linear transformation with matrix representation
Theorem: The diagram
V \rar{T} \dar[swap]{\CB} & V \dar{\CB}
\R^n \rar{[T]_\B} & \R^n
commutes.
Proof: It suffices to check commutativity on a basis. Observe that
clearly commutes by our definition of . QED
Example: Let be a -dimensional linear subspace of , and let be the orthogonal projection operator from onto . Let
be an ordered basis of , where is a basis for , and is a basis for . Then clearly,
This means that
How can we use this to figure out the matrix representation with respect to the standard basis ?
Change-of-Coordinates Matrix. Suppose we have two ordered bases for . The change-of-coordinates matrix between and is defined by . This is equivalent to defining so that the diagram
commutes. We write , and say that changes -coordinates to -coordinates.
{How to find ?} Say . Then using the preceding diagram,
so we conclude that
Example: Let be the plane defined by , and let . Find .
Solution: First, find bases for and . We will use the basis
and the matrix representation of with respect to is
Let , where
(Note that we define , not , because we know the coordinates of with respect to the standard basis, not the other way around.) Then
and using the change-of-basis formula, we have
Eigenvalues and Eigenvectors. Recall from last lecture that if we have two ordered bases and