Smooth Manifolds · Study tools

InteractiveLee, Ch. 1: Smooth Manifolds

Charts and transition maps

A manifold is a space that looks like $\R^n$ up close. Charts are the local pictures. Whether calculus makes sense depends on one thing only: how two overlapping charts translate between each other.

The idea

Topological manifold. A space

M

that is Hausdorff, second countable, and locally Euclidean of dimension

n

: every point has a neighbourhood

U

and a homeomorphism

\varphi: U \to \hat U \subseteq \R^n

onto an open set. The pair

(U,\varphi)

is a chart.

Two charts $(U,\varphi)$ and $(V,\psi)$ overlap on $U\cap V$ . Going from one chart's coordinates to the other's is the transition map

\psi\circ\varphi^{-1} : \varphi(U\cap V) \longrightarrow \psi(U\cap V),

a map between open subsets of $\R^n$ , where we already know what "smooth" means.

Smooth structure. Charts are smoothly compatible if every transition map is smooth with a smooth inverse. A smooth atlas is a collection of compatible charts covering

M

; a maximal one is a smooth structure.

Below, the circle $S^1$ gets two atlases. Drag the point and watch both coordinates and the transition map between them.

Explore

Atlas on the circle

Drag the point p around the circle.

The transition map from the first chart's coordinate to the second's.

Worked example

Stereographic projection from the north pole $N=(0,\dots,0,1)$ of $S^n\subset\R^{n+1}$ sends $x$ to the point where the line from $N$ through $x$ meets $\R^n\times\{0\}$ :

\sigma(x) = \frac{(x^1,\dots,x^n)}{1-x^{n+1}}, \qquad \tilde\sigma(x) = \frac{(x^1,\dots,x^n)}{1+x^{n+1}} \ \text{(from the south pole } S).

Show these two charts form a smooth atlas for $S^n$ .

What is the overlap of the two chart domains, and what does $\sigma$ send it to?

The domains are $S^n\smallsetminus\{N\}$ and $S^n\smallsetminus\{S\}$ , so the overlap is $S^n\smallsetminus\{N,S\}$ . Since $\sigma(S)=0$ , its image under $\sigma$ is $\R^n\smallsetminus\{0\}$ .
Find the inverse $\sigma^{-1}(u)$ for $u\in\R^n$ .

Points on the line from $N$ through $(u,0)$ are $(su,\,1-s)$ . Requiring $s^2|u|^2+(1-s)^2=1$ gives $s = 2/(|u|^2+1)$ , so

$\sigma^{-1}(u) = \frac{\big(2u,\ |u|^2-1\big)}{|u|^2+1}.$
Compute the transition map $\tilde\sigma\circ\sigma^{-1}$ .

With $x=\sigma^{-1}(u)$ we have $1+x^{n+1} = \dfrac{2|u|^2}{|u|^2+1}$ , so

$\tilde\sigma\circ\sigma^{-1}(u) = \frac{2u}{|u|^2+1}\cdot\frac{|u|^2+1}{2|u|^2} = \frac{u}{|u|^2}, \qquad u\neq 0.$
Why does this finish the proof?

$u\mapsto u/|u|^2$ is smooth on $\R^n\smallsetminus\{0\}$ and is its own inverse, so it is a diffeomorphism. By symmetry the same holds for $\sigma\circ\tilde\sigma^{-1}$ . The two charts are smoothly compatible and cover $S^n$ , so they form a smooth atlas. (For $n=1$ this is the map $u\mapsto 1/u$ graphed above.)

Check your understanding

The "line with two origins" is two copies of $\R$ glued along $\R\smallsetminus\{0\}$ . It is locally Euclidean and second countable. What goes wrong?

It is not Hausdorff: the two origins cannot be separated by disjoint open sets.
It is not locally Euclidean at the origins.Each origin has a neighbourhood that is a copy of an open interval, so it is locally Euclidean everywhere.
Nothing; it is a 1-manifold.Look at any two neighbourhoods of the two origins: they always meet.

Every neighbourhood of one origin contains points

x\neq 0

near 0, and so does every neighbourhood of the other, so they always intersect. That is exactly why the Hausdorff condition is part of the definition: it makes limits unique.

On $\R$ , take the charts $\varphi(x)=x$ and $\psi(x)=x^3$ . Are they smoothly compatible?

Yes: both are homeomorphisms and $x^3$ is smooth.Compatibility needs both transition maps to be smooth, not just one.
No: $\varphi\circ\psi^{-1}(y)=y^{1/3}$ is not differentiable at $0$ .
No: $x^3$ is not a smooth function. $x^3$ is smooth. The problem is its inverse.

\psi\circ\varphi^{-1}(x)=x^3

is smooth, but the other direction is the cube root, which has infinite slope at 0. So

\psi

defines a different smooth structure on

\R

.

Continuing: write $\R_\psi$ for $\R$ with the smooth structure defined by $\psi(x)=x^3$ . Is $\R_\psi$ diffeomorphic to the usual $\R$ ?

Yes: $F(x)=x^3$ is a diffeomorphism $\R_\psi\to\R$ .
No: different smooth structures are never diffeomorphic.Different structures can still be diffeomorphic by a map other than the identity.
No: the identity map $\R_\psi\to\R$ is not smooth.That's true, but it only rules out one particular map.

In the charts

\psi

and

\mathrm{id}

,

F

reads

F\circ\psi^{-1}(y) = (y^{1/3})^3 = y

, the identity, so

F

is a diffeomorphism. "Different structure" and "different manifold up to diffeomorphism" are different questions.

What is the smallest number of charts in an atlas for $S^1$ ?

1A single chart would make $S^1$ homeomorphic to an open subset of $\R$ . $S^1$ is compact; no nonempty open subset of $\R$ is.
2
3The stereographic atlas above already does it with fewer.
4

One chart is impossible because

S^1

is compact, and two stereographic charts suffice.

In the angle atlas above, the transition map is $\theta\mapsto\theta$ on one part of the overlap and $\theta\mapsto\theta+2\pi$ on the other. Is it smooth?

Yes. The overlap has two components, and on each the map is smooth.
No, it jumps by $2\pi$ .The jump happens between two separate pieces of the domain, never inside one.

Smoothness is a local property. The domain

(-\pi,0)\cup(0,\pi)

is disconnected, and on each piece the map is a translation.

Practice

Real projective lineCharts and a transition map

$\RP^1$ is the set of lines through 0 in $\R^2$ , written $[x:y]$ . Use the charts $u=y/x$ on $\{x\neq0\}$ and $v=x/y$ on $\{y\neq0\}$ . Compute the transition map and conclude the charts form a smooth atlas.

Hint

On the overlap both $x$ and $y$ are nonzero. Express $v$ in terms of $u$ .

Solution

On the overlap $u\neq0$ and $v = x/y = 1/u$ . The transition $u\mapsto 1/u$ on $\R\smallsetminus\{0\}$ is smooth and its own inverse. Notice it is the same transition map as stereographic projection of $S^1$ . That is a hint that $\RP^1$ and $S^1$ are diffeomorphic.

Graphs are manifoldsA single chart

Let $f:U\to\R^k$ be continuous on an open $U\subseteq\R^n$ . Show its graph $\Gamma(f)=\{(x,f(x))\}\subseteq\R^{n+k}$ is a topological $n$ -manifold covered by one chart.

Hint

Try the projection $(x,y)\mapsto x$ .

Solution

$\pi(x,y)=x$ restricted to $\Gamma(f)$ is continuous with continuous inverse $x\mapsto(x,f(x))$ , so it is a homeomorphism onto $U$ . Hausdorff and second countable are inherited from $\R^{n+k}$ . With one chart there are no transition maps, so this chart also defines a smooth structure, even when $f$ is not smooth.

A torus atlasProducts of charts

Using the stereographic charts on $S^1$ , write down a smooth atlas on $T^2=S^1\times S^1$ . How many charts did you use, and what are the transition maps?

Solution

Take all products $\sigma_a\times\sigma_b$ with $a,b\in\{N,S\}$ : four charts. Each transition map acts coordinate by coordinate as either $u\mapsto u$ or $u\mapsto1/u$ , so all of them are smooth.

InteractiveLee, Ch. 3: Tangent Vectors

Tangent vectors and the differential

On a manifold there is no ambient space for arrows to live in, so a tangent vector is defined by what it does: it differentiates functions. A smooth map pushes these vectors forward, and in coordinates that push-forward is just the Jacobian matrix.

The idea

Tangent vector. A derivation at $p$ : a linear map

v:C^\infty(M)\to\R

with

v(fg)=f(p)\,vg+g(p)\,vf

. These form the tangent space

T_pM

, which has dimension

n

and basis

\left.\pd{}{x^1}\right|_p,\dots,\left.\pd{}{x^n}\right|_p

in any chart.

Differential. For smooth

F:M\to N

,

\ dF_p:T_pM\to T_{F(p)}N

is

dF_p(v)f = v(f\circ F).

In coordinates it is the Jacobian:

dF_p\!\left(\left.\pd{}{x^i}\right|_p\right) = \sum_j \pd{F^j}{x^i}(p)\left.\pd{}{y^j}\right|_{F(p)}.

Velocities. If

\gamma

is a curve with

\gamma(0)=p

and

\gamma'(0)=v

, then

(F\circ\gamma)'(0)=dF_p(v)

. This is the chain rule, and it is what you can see below.

Explore

The map F

Domain. Drag the point p and the tip of the vector v.

dF_p(v)

image of the curve

p+tv

Worked example

On $\R^2\smallsetminus\{0\}$ use polar coordinates, $x=r\cos\theta,\ y=r\sin\theta$ . Express the polar coordinate vectors in Cartesian terms, and a Cartesian vector in polar terms.

Write $\pd{}{r}$ in terms of $\pd{}{x}$ and $\pd{}{y}$ .

The chain rule, which is the change-of-coordinates formula for tangent vectors, gives

$\pd{}{r} = \pd{x}{r}\pd{}{x}+\pd{y}{r}\pd{}{y} = \cos\theta\,\pd{}{x}+\sin\theta\,\pd{}{y} = \frac{x\,\partial_x + y\,\partial_y}{r}.$
Now do $\pd{}{\theta}$ .

$\pd{}{\theta} = -r\sin\theta\,\pd{}{x}+r\cos\theta\,\pd{}{y} = -y\,\pd{}{x}+x\,\pd{}{y}.$

This is the rotation field. Note it grows with $r$ : the coordinate vector $\partial_\theta$ is not a unit vector.
Sanity check: apply both to $f=x^2+y^2$ .

$\partial_r f = (x\cdot 2x + y\cdot 2y)/r = 2r$ and $\partial_\theta f = -y\cdot 2x + x\cdot 2y = 0$ . That matches $f=r^2$ , which doesn't depend on $\theta$ .
Write $v = 3\,\partial_x$ at $p=(1,1)$ in the polar basis.

Go the other way with the inverse Jacobian: $\partial_x = \pd{r}{x}\partial_r+\pd{\theta}{x}\partial_\theta = \tfrac{x}{r}\partial_r - \tfrac{y}{r^2}\partial_\theta$ . At $(1,1)$ , $r=\sqrt2$ , so

$v = \frac{3}{\sqrt2}\,\partial_r - \frac32\,\partial_\theta.$

Check: $\partial_r=(\partial_x+\partial_y)/\sqrt2$ and $\partial_\theta=-\partial_x+\partial_y$ there, and $\tfrac{3}{\sqrt2}\cdot\tfrac{1}{\sqrt2}(1,1)-\tfrac32(-1,1)=(3,0)$ .

Check your understanding

Let $v\in T_pM$ be a derivation and $c$ a constant function. What is $v(c)$ ?

$0$
$c$ Try the product rule on $1\cdot1$ .
It depends on $v$ .The product rule pins it down for every derivation.

v(1)=v(1\cdot1)=1\cdot v(1)+1\cdot v(1)=2v(1)

, so

v(1)=0

, and by linearity

v(c)=c\,v(1)=0

.

For smooth $F:M\to N$ , $v\in T_pM$ and $f\in C^\infty(N)$ , what is $dF_p(v)\,f$ ?

$v(f\circ F)$
$F(vf)$ $f$ is a function on $N$ , so $v$ can't act on it directly.
$(vf)\circ F$ $v$ only differentiates functions on $M$ .

Pull

f

back to

M

by composing with

F

, then let

v

differentiate it. Everything else in the theory of differentials follows from this formula.

Polar coordinates $F(r,\theta)=(r\cos\theta,r\sin\theta)$ : what is the rank of $dF$ at a point with $r=0$ ?

2Set $r=0$ in the Jacobian. One column vanishes.
1
0The $\partial_r$ column is $(\cos\theta,\sin\theta)$ , which is never zero.

The Jacobian has columns

(\cos\theta,\sin\theta)

and

(-r\sin\theta,r\cos\theta)

. At

r=0

the second is zero. Moving in

\theta

doesn't move the image point at all, as you can check above.

$M$ is an $n$ -manifold. What is the dimension of its tangent bundle $TM$ ?

$n$
$2n$
$n^2$

A point of

TM

is a base point (

n

coordinates) together with a tangent vector there (

n

components).

Practice

A paraboloid mapComputing a differential

Let $F(u,v)=(u,\,v,\,u^2+v^2)$ . Compute $dF_{(1,2)}(\partial_u)$ and $dF_{(1,2)}(\partial_v)$ , and describe the image of $dF_{(1,2)}$ .

Solution

The Jacobian columns are $(1,0,2u)$ and $(0,1,2v)$ . At $(1,2)$ : $dF(\partial_u)=\partial_x+2\partial_z$ and $dF(\partial_v)=\partial_y+4\partial_z$ . The image is the plane spanned by these, which is the tangent plane to the paraboloid at $(1,2,5)$ . $F$ is an immersion.

Derivations are localBump functions

Show that if $f=g$ on a neighbourhood of $p$ , then $vf=vg$ for every $v\in T_pM$ .

Hint

Take a bump function $\psi$ that is $1$ near $p$ and supported where $f=g$ . Look at $\psi\,(f-g)$ .

Solution

Choose $\psi$ supported in the neighbourhood where $f=g$ , with $\psi(p)=1$ . Then $\psi(f-g)\equiv0$ , so

0 = v(\psi(f-g)) = \psi(p)\,v(f-g) + (f-g)(p)\,v\psi = v(f-g).

This is what lets derivations be computed in a chart.

The coordinate vectors are independentUsing coordinate functions

In a chart $(x^1,\dots,x^n)$ , show that $\left.\pd{}{x^1}\right|_p,\dots,\left.\pd{}{x^n}\right|_p$ are linearly independent.

Solution

If $\sum_i a^i\partial_i|_p=0$ , apply it to the coordinate function $x^j$ (extended to a global function by a bump function). This gives $\sum_i a^i\delta_i^j=a^j=0$ for each $j$ .

Interactive galleryLee, Ch. 4: Submersions, Immersions, and Embeddings

Immersions and embeddings

"Smooth", "immersion", "injective immersion" and "embedding" are four different conditions, and the gap between each pair is easiest to see in a picture. Here is a gallery of curves, one for each gap.

The idea

Rank. The rank of

F

at

p

is the rank of

dF_p

.

F

is an immersion if

dF_p

is injective everywhere (rank

=\dim M

), and a submersion if it is surjective everywhere (rank

=\dim N

).

Embedding. An immersion that is also a homeomorphism onto its image with the subspace topology.

Rank theorem. If

F

has constant rank

r

near

p

, there are charts in which

F(x^1,\dots,x^m) = (x^1,\dots,x^r,0,\dots,0)

.

Watch out. A smooth curve can have a corner (when its velocity vanishes), and an injective immersion can still fail to be an embedding.

Explore

velocity

\gamma'(t)

the part traced so far

Curve

Parameter t

Worked example

Show that $\beta(t)=(\sin2t,\ \sin t)$ for $t\in(-\pi,\pi)$ (the figure eight in the gallery) is an injective immersion but not an embedding.

Show $\beta$ is an immersion.

$\beta'(t)=(2\cos2t,\ \cos t)$ . For this to vanish we need $\cos t=0$ , so $t=\pm\pi/2$ . But then $\cos2t=\cos(\pm\pi)=-1\neq0$ . So $\beta'(t)\neq0$ everywhere, and for a curve that means rank 1.
Show $\beta$ is injective.

Suppose $\beta(t)=\beta(s)$ . If $\sin t=\sin s\neq0$ , then $\sin2t=2\sin t\cos t$ forces $\cos t=\cos s$ , and sine and cosine together determine the angle, so $t=s$ . If $\sin t=0$ , the only such $t$ in $(-\pi,\pi)$ is $0$ .
Find a sequence that shows $\beta^{-1}$ (defined on the image) is not continuous.

Take $t_k\to\pi$ . Then $\beta(t_k)\to(0,0)=\beta(0)$ , but $t_k\not\to0$ . So $\beta$ is not a homeomorphism onto its image.

Another way to see it: the image is the whole figure eight, which is compact, while $(-\pi,\pi)$ is not. A homeomorphism would preserve compactness.
Is the image an embedded submanifold of $\R^2$ at all?

No. Removing the crossing point from a small neighbourhood of it in the image leaves four pieces, but removing a point from an interval leaves two. So no neighbourhood of the crossing is homeomorphic to an interval.

Check your understanding

$\gamma(t)=(t^3,t^3)$ . Its image is the line $y=x$ , a perfectly good embedded submanifold. Is $\gamma$ an immersion?

Yes, because its image is a smooth line.Being an immersion is a property of the map, not of its image.
No, because $\gamma'(0)=0$ .

\gamma'(t)=(3t^2,3t^2)

vanishes at

t=0

. The map pauses there, even though the image looks perfectly smooth. Try it in the gallery and watch the velocity arrow shrink.

If $F:M^m\to N^n$ is an immersion, which must hold?

$m\le n$
$m\ge n$ That's the condition for a submersion.
$m=n$

An injective linear map

T_pM\to T_{F(p)}N

can only exist if

\dim T_pM\le\dim T_{F(p)}N

.

$M$ is compact and $F:M\to N$ is an injective immersion. Is $F$ an embedding?

Yes, always.
Not necessarily: the figure eight is a counterexample.The figure eight is parametrised by $(-\pi,\pi)$ , which is not compact.

A continuous bijection from a compact space onto a Hausdorff space is a homeomorphism. Closed subsets of

M

are compact, their images are compact, hence closed.

The line of irrational slope on the torus $T^2$ is an injective immersion of $\R$ whose image is dense. Is its image an embedded submanifold?

Yes, it's the image of an injective immersion.That makes it an immersed submanifold. Embedded is stronger.
No. Every small open set of the torus meets it in infinitely many separate segments.

An embedded submanifold looks like a single slice

\R^k\times\{0\}

in some chart around each of its points. Here every neighbourhood contains infinitely many parallel pieces of the line.

The cusp $\gamma(t)=(t^2,t^3)$ . Which is true?

It is not smooth, because the image has a corner.Both components are polynomials, so the map is smooth.
It is smooth and injective, but not an immersion at $t=0$ .
It is an immersion that is not injective.Different $t$ give different values of $t^3$ , so it is injective.

\gamma'(0)=(0,0)

. A smooth map can have a sharp image exactly where its derivative vanishes.

Practice

The circle, twiceImmersion vs. embedding

Show $F(t)=(\cos t,\sin t)$ , $F:\R\to\R^2$ , is an immersion but not injective. Then show the inclusion $S^1\hookrightarrow\R^2$ is an embedding.

Solution

$|F'(t)|=1$ , so $F$ is an immersion, and $F(t+2\pi)=F(t)$ , so it is not injective. The inclusion of $S^1$ is an injective immersion of a compact manifold, so it is an embedding (see the third concept check).

Where a submersion failsCritical points

Where does $f(x,y)=x^2-y^2$ fail to be a submersion $\R^2\to\R$ ? What does the rank theorem say near any other point?

Solution

$df=2x\,dx-2y\,dy$ is zero only at the origin. Everywhere else the rank is 1, so near those points there are coordinates in which $f(u,v)=u$ : the level sets near that point are parallel lines.

InteractiveLee, Ch. 5: Submanifolds

Regular level sets

Most manifolds you meet in practice are carved out by equations: spheres, matrix groups, constraint surfaces. The regular level set theorem says when an equation defines a manifold. You can watch it work, and fail, as you move the level.

The idea

Regular value.

c\in N

is a regular value of

F:M\to N

if

dF_p

is surjective at every

p\in F^{-1}(c)

. Otherwise

c

is a critical value.

Regular level set theorem. If

c

is a regular value of

F:M^m\to N^n

, then

F^{-1}(c)

is a properly embedded submanifold of dimension

m-n

, and

T_p F^{-1}(c)=\ker dF_p

.

Only one direction. If

c

is critical the theorem says nothing. The level set may still be a manifold.

For $f:\R^2\to\R$ , "regular" just means $\nabla f\neq0$ at every point of the curve $f=c$ . Then the curve is smooth, and its tangent line is perpendicular to $\nabla f$ .

Explore

the level set

f=c

\nabla f

along it critical points

Function

Level c

Worked example

Show that the orthogonal group $O(n)=\{A: A^{\mathsf T}A=I\}$ is an embedded submanifold of the space $M(n,\R)\cong\R^{n^2}$ of all $n\times n$ matrices, and find its dimension.

Pick the defining map. What should its target be?

$\Phi(A)=A^{\mathsf T}A$ . Its values are always symmetric, so take the target to be the space of symmetric matrices $\mathrm{Sym}(n)$ , of dimension $n(n+1)/2$ . (Using all of $M(n,\R)$ as the target would make every value critical.)
Compute $d\Phi_A(B)$ .

Differentiate along $A+sB$ :

$d\Phi_A(B)=\frac{d}{ds}\Big|_{s=0}(A+sB)^{\mathsf T}(A+sB) = B^{\mathsf T}A + A^{\mathsf T}B.$
Show $d\Phi_A$ is onto $\mathrm{Sym}(n)$ whenever $A\in O(n)$ .

Given symmetric $C$ , try $B=\tfrac12AC$ . Then

$d\Phi_A(B) = \tfrac12 C^{\mathsf T}A^{\mathsf T}A + \tfrac12 A^{\mathsf T}AC = \tfrac12C+\tfrac12C = C,$

using $A^{\mathsf T}A=I$ and $C^{\mathsf T}=C$ . So $I$ is a regular value.
Conclude: what are the dimension and the tangent space at $I$ ?

$\dim O(n) = n^2-\frac{n(n+1)}2 = \frac{n(n-1)}2, \qquad T_IO(n)=\ker d\Phi_I = \{B: B^{\mathsf T}+B=0\}.$

The tangent space at the identity is the skew-symmetric matrices. This is the Lie algebra $\mathfrak{o}(n)$ , which comes back in Chapter 8.

Check your understanding

For $f(x,y)=x^2-y^2$ , is $0$ a regular value?

Yes, because $f$ takes the value 0.Every value in the image could be regular or critical. What matters is $\nabla f$ on the level set.
No: the origin lies on $f=0$ and $\nabla f(0,0)=0$ .

The level set is two crossing lines. Pick "saddle" in the explorer and move

c

through 0 to watch the two branches pinch together.

$g(x,y)=(x^2+y^2-1)^2$ . The level set $g=0$ is the unit circle, yet $0$ is a critical value. What follows?

The unit circle is not a submanifold.The theorem only goes one way: a critical value proves nothing.
The theorem doesn't apply to this $g$ . The circle is still a submanifold, as another defining function shows.
The theorem is wrong here.

\nabla g = 4(x^2+y^2-1)(x,y)

vanishes on the whole circle. But

h=x^2+y^2

has 1 as a regular value with the same level set. Being a submanifold is a property of the set; regularity is a property of the function you chose. The "trap" option in the explorer shows this one.

$F:\R^5\to\R^2$ is smooth and $c$ is a regular value. What is the dimension of $F^{-1}(c)$ ?

2
3
5
7

Codimension equals the dimension of the target:

5-2=3

.

At $p\in S^n=\{|x|^2=1\}$ , what is $T_pS^n$ as a subspace of $\R^{n+1}$ ?

$\{v: v\cdot p=0\}$
$\mathrm{span}(p)$ That's the normal line, the direction of the gradient.
All of $\R^{n+1}$

T_pS^n=\ker d(|x|^2)_p=\{v: 2p\cdot v=0\}

. It is the tangent space you'd expect from calculus.

$SL(2,\R)=\det^{-1}(1)$ . What is its dimension?

1
2
3
4

1

is a regular value of

\det:\R^4\to\R

(see the practice problem below), so

\dim=4-1=3

.

Practice

The special linear groupA regular value of the determinant

Show $d(\det)_A(B)=\det(A)\,\mathrm{tr}(A^{-1}B)$ for invertible $A$ , and conclude that $SL(n,\R)$ is an embedded submanifold of dimension $n^2-1$ .

Hint

Write $\det(A+sB)=\det A\,\det(I+sA^{-1}B)$ and expand $\det(I+sX)$ to first order in $s$ .

Solution

$\det(I+sX)=1+s\,\mathrm{tr}X+O(s^2)$ , which gives the formula. On $SL(n,\R)$ , taking $B=cA/n$ gives $d(\det)_A(B)=c$ , so the differential is onto $\R$ and $1$ is a regular value. The dimension is $n^2-1$ .

Critical values of a cubicMatches the "cubic" option in the explorer

Find the critical points and critical values of $f(x,y)=y^2-x^3+x$ . For which $c$ is $f^{-1}(c)$ guaranteed to be a smooth curve?

Solution

$\nabla f=(1-3x^2,\ 2y)=0$ at $(\pm1/\sqrt3,\,0)$ , with values $f=\mp\tfrac{2}{3\sqrt3}\approx\mp0.385$ . Every other $c$ is a regular value. At $c\approx0.385$ the curve has a self-crossing (a node). At $c\approx-0.385$ it has an isolated point next to a smooth branch.

Quadric surfacesVarying the level

For which $c$ is $\{x^2+y^2-z^2=c\}$ a regular level set? Describe the surfaces for $c<0$ , $c=0$ , $c>0$ .

Solution

$\nabla=2(x,y,-z)$ vanishes only at the origin, which lies on the level $c=0$ . So every $c\neq0$ is regular. $c>0$ : a one-sheeted hyperboloid. $c<0$ : a two-sheeted hyperboloid. $c=0$ : a double cone, which fails to be a manifold at its tip.

Animated lessonLee, Ch. 8: Vector Fields · Ch. 9: Integral Curves and Flows

Flows and the Lie bracket

The Lie bracket $[X,Y]$ is easy to define algebraically and hard to picture. This short animation shows what it measures: follow $X$ , then $Y$ , then go back along $X$ and then $Y$ , and see how far you are from where you started.

The idea

Flow. An integral curve of

X

satisfies

\gamma'(t)=X_{\gamma(t)}

. The flow

\theta_t

moves every point along its integral curve for time

t

.

Lie bracket.

[X,Y]f = X(Yf)-Y(Xf)

. Second derivatives cancel, so this is again a vector field. In coordinates:

[X,Y] = \sum_j \big(X Y^j - Y X^j\big)\pd{}{x^j}.

What it measures. With

\theta

the flow of

X

and

\psi

the flow of

Y

,

\psi_{-t}\circ\theta_{-t}\circ\psi_{t}\circ\theta_{t}(p) = p + t^2\,[X,Y]_p + O(t^3).

So $[X,Y]=0$ exactly when the flows commute.

Watch

Vector fields

Time on each leg, t 0.80

X

Y

gap

t^2[X,Y]_p

· drag p

Two vector fields: $X$ in blue and $Y$ in gold. Pick a start point $p$ .
Flow along $X$ for time $t$ .
Then along $Y$ for time $t$ .
Then backwards along $X$ for time $t$ .
Then backwards along $Y$ . If the flows commuted, we'd be back at $p$ .
The loop doesn't close. The gap (red) is close to $t^2[X,Y]_p$ (green), and the match improves as $t$ shrinks.

Worked example

On $\R^2$ , let $X=-y\,\partial_x+x\,\partial_y$ (rotation) and $Y=x\,\partial_x+y\,\partial_y$ (scaling). Compute $[X,Y]$ , interpret the answer using the flows, then compute $[X,\partial_x]$ .

Compute $XY^j$ and $YX^j$ for each component.

$Y^1=x,\ Y^2=y$ , so $XY^1=X(x)=-y$ and $XY^2=X(y)=x$ .

$X^1=-y,\ X^2=x$ , so $YX^1=Y(-y)=-y$ and $YX^2=Y(x)=x$ .
Assemble $[X,Y]$ .

$[X,Y] = (-y-(-y))\,\partial_x + (x-x)\,\partial_y = 0.$
Explain the answer using the flows.

The flow of $X$ is rotation by angle $t$ , and the flow of $Y$ is scaling by $e^{s}$ . Rotating and scaling commute, $R_t(e^{s}p)=e^{s}R_tp$ , so the four-leg loop always closes. Choose "rotation & scaling" above and check that the gap is zero for every $t$ .
Now replace $Y$ with $\partial_x$ . Compute $[X,\partial_x]$ .

The components of $\partial_x$ are constant, so the $XY^j$ terms vanish. And $\partial_x X^1=\partial_x(-y)=0$ , $\partial_x X^2=\partial_x(x)=1$ . So

$[X,\partial_x] = -\partial_y.$

In the animation ("rotation & translation"), the gap points straight down with length about $t^2$ , at every starting point.

Check your understanding

$X=\partial_x$ , $Y=x\,\partial_y$ . What is $[X,Y]$ ?

$\partial_y$
$-\partial_y$ Check the order: $XY^j-YX^j$ .
$0$ $Y$ depends on $x$ , which is the direction $X$ moves in.
$x\,\partial_x$

XY^2=\partial_x(x)=1

and every other term is 0. This is the first pair in the animation. Move

p

around and the gap stays the same, because the bracket is constant.

Which identity is false in general?

$[X,Y]=-[Y,X]$ That one is true (antisymmetry).
$[fX,Y]=f[X,Y]$
$[X,[Y,Z]]+[Y,[Z,X]]+[Z,[X,Y]]=0$ That's the Jacobi identity, which always holds.

The correct rule is

[fX,Y]=f[X,Y]-(Yf)X

. The bracket is not

C^\infty

-linear, so it is not a tensor. It involves derivatives of the coefficients.

On $\R$ , what is the integral curve of $X=x^2\,\partial_x$ starting at $x=1$ ?

$x(t)=e^t$ That solves $x'=x$ , not $x'=x^2$ .
$x(t)=1/(1-t)$ , defined only for $t<1$
$x(t)=1+t^2$

Solving

x'=x^2

gives

x(t)=1/(1-t)

, which runs off to infinity in finite time. So

X

is not complete: its flow is not defined for all

t

.

Which vector fields are guaranteed to be complete?

Compactly supported ones (so every vector field on a compact manifold)
Bounded ones on any manifold $\partial_x$ on the open interval $(0,1)$ is bounded but runs off the end.
Polynomial ones on $\R^n$ The previous question is a counterexample.

Outside the support the flow is stationary, and on the compact support there is a uniform time for which all integral curves exist. That uniform time lets you extend every curve forever.

Practice

A hyperbolic flowSolving for a flow

Find the flow of $X=x\,\partial_x-y\,\partial_y$ on $\R^2$ and sketch its integral curves.

Solution

$x'=x,\ y'=-y$ gives $\theta_t(x,y)=(e^tx,\ e^{-t}y)$ . The integral curves are the hyperbolas $xy=\text{const}$ , the four half-axes, and the fixed point at the origin.

The bracket with functionsA product rule

Show $[fX,gY]=fg[X,Y]+f(Xg)Y-g(Yf)X$ .

Solution

Apply both sides to $h$ :

\begin{aligned} fX(gYh)-gY(fXh) &= f(Xg)(Yh)+fg\,XYh\\ &\quad -g(Yf)(Xh)-gf\,YXh. \end{aligned}

The two second-order terms combine to $fg[X,Y]h$ .

A shear flowFlows that don't commute

Find the flows of $\partial_x$ and $x\,\partial_y$ . Compute the four-leg loop from $(0,0)$ exactly and compare it with $t^2[X,Y]$ .

Solution

$\theta_t(x,y)=(x+t,y)$ and $\psi_t(x,y)=(x,y+tx)$ . From $(0,0)$ : $(t,0)\to(t,t^2)\to(0,t^2)\to(0,t^2)$ . The gap is exactly $t^2\partial_y=t^2[X,Y]$ , with no $O(t^3)$ error, because these fields are so simple.

Draw it yourselfLee, Ch. 11 · 14 · 16 · 17

Forms, Stokes and de Rham

Integrate a 1-form around a loop you draw. For an exact form the answer is always zero. For the area form it is the enclosed area (Stokes). For the angle form $d\theta$ it counts how often you wound around the hole, which is the first de Rham cohomology of the punctured plane.

The idea

Line integral. For a 1-form

\omega=P\,dx+Q\,dy

and a curve

\gamma

,

\displaystyle\int_\gamma\omega=\int_a^b\omega_{\gamma(t)}(\gamma'(t))\,dt=\int_a^b \gamma^*\omega.

It doesn't depend on how the curve is parametrised, only on its direction.

Exterior derivative.

d(P\,dx+Q\,dy)=\big(\partial_xQ-\partial_yP\big)\,dx\wedge dy,

and

d\circ d=0

.

\omega

is closed if

d\omega=0

and exact if

\omega=df

.

Stokes.

\displaystyle\int_M d\omega=\int_{\partial M}\omega

. For exact

\omega=df

and a closed loop,

\int_\gamma df = 0

.

de Rham.

H^1_{dR}(M)=\{\text{closed}\}/\{\text{exact}\}

. For the punctured plane it is one-dimensional, spanned by the class of

d\theta

, and

\frac1{2\pi}\int_\gamma d\theta

is the winding number.

Explore

Draw a loop with your mouse or finger. It closes itself when you let go. Arrows show the vector field dual to

\omega

.

The 1-form ω

Or try a ready-made loop

Worked example

On $\R^2\smallsetminus\{0\}$ let $\displaystyle\omega=\frac{-y\,dx+x\,dy}{x^2+y^2}$ . Show that $\omega$ is closed but not exact.

Compute $d\omega$ .

With $P=-y/r^2$ and $Q=x/r^2$ :

$\partial_xQ = \frac{y^2-x^2}{(x^2+y^2)^2},\qquad \partial_yP = \frac{-(x^2+y^2)+2y^2}{(x^2+y^2)^2}=\frac{y^2-x^2}{(x^2+y^2)^2}.$

They are equal, so $d\omega=0$ .
Pull $\omega$ back along the unit circle $\gamma(t)=(\cos t,\sin t)$ and integrate.

$\gamma^*\omega = \big(-\sin t\cdot(-\sin t)+\cos t\cdot\cos t\big)\,dt = dt, \qquad \int_{S^1}\omega = \int_0^{2\pi}dt = 2\pi.$
Why does this show $\omega$ is not exact?

If $\omega=df$ for a smooth $f$ on $\R^2\smallsetminus\{0\}$ , then $\int_\gamma df=f(\gamma(2\pi))-f(\gamma(0))=0$ for the closed curve $\gamma$ . But the integral is $2\pi$ .
Yet $\omega$ "is" $d\theta$ . Where does $\theta$ live?

On the plane minus any ray from the origin there is a smooth angle function $\theta$ with $d\theta=\omega$ . So $\omega$ is locally exact, as the Poincaré lemma promises for every closed form. No single $\theta$ works on the whole punctured plane. This obstruction is the nonzero class $[\omega]\in H^1_{dR}(\R^2\smallsetminus\{0\})$ .

Check your understanding

What is $d(x\,dy)$ ?

$dx\wedge dy$
$dy\wedge dx$ $d(x\,dy)=dx\wedge dy$ , and $dy\wedge dx$ is its negative.
$0$ $x\,dy$ is not closed; only exact forms are guaranteed to be.
$dx+dy$

d(x\,dy)=dx\wedge dy

. By Stokes,

\oint_\gamma x\,dy

is the area inside

\gamma

when it runs counterclockwise. Choose

x\,dy

in the explorer to see it.

A loop winds twice counterclockwise around the origin. What is $\oint d\theta$ ?

$2\pi$
$4\pi$
$0$ That's what an exact form would give, and $d\theta$ isn't exact.

Each full turn contributes

2\pi

. Use the "twice around" loop in the explorer.

$\omega$ is a closed 1-form on all of $\R^2$ . Must it be exact?

Yes
Not necessarily: $d\theta$ is a counterexample. $d\theta$ is not defined at the origin, so it is not a form on all of $\R^2$ .

\R^2

is star-shaped, so the Poincaré lemma gives

H^1_{dR}(\R^2)=0

: every closed form is exact.

For a smooth map $F$ and a form $\omega$ , how do $d$ and pullback interact?

$F^*(d\omega)=d(F^*\omega)$
$F^*(d\omega)=-d(F^*\omega)$
Only when $F$ is a diffeomorphism

d

commutes with pullback by any smooth map. This is why closed forms pull back to closed forms and exact forms to exact forms, so pullback is defined on cohomology.

On the plane with two punctures $a$ and $b$ , what is $\dim H^1_{dR}$ ?

1
2
0

The angle forms

d\theta_a

and

d\theta_b

around each puncture are independent: integrals around small loops tell them apart. Mayer–Vietoris confirms the dimension is exactly 2. Try the two-puncture form in the explorer.

Practice

Area of an ellipse by StokesGreen's theorem as a special case

Compute $\oint_E (x\,dy-y\,dx)$ around the ellipse $x^2/a^2+y^2/b^2=1$ , counterclockwise, in two ways.

Solution

$d(x\,dy-y\,dx)=2\,dx\wedge dy$ , so by Stokes the integral is twice the area: $2\pi ab$ . Directly, with $x=a\cos t,\ y=b\sin t$ , the pullback is $ab\,dt$ , and integrating over $[0,2\pi]$ gives $2\pi ab$ .

An exterior derivative in ℝ³Computing d

Compute $d\omega$ for $\omega = xy\,dz + e^x\,dy$ , and check that $d(d\omega)=0$ .

Solution

$d\omega = y\,dx\wedge dz + x\,dy\wedge dz + e^x\,dx\wedge dy$ . Then $d(d\omega)=dy\wedge dx\wedge dz+dx\wedge dy\wedge dz+0=0$ , since $dy\wedge dx=-dx\wedge dy$ .

Detecting exactnessOne number decides it

Show that a closed 1-form $\eta$ on $\R^2\smallsetminus\{0\}$ is exact if and only if $\oint_{S^1}\eta=0$ .

Hint

Consider $\eta-\frac{c}{2\pi}d\theta$ with $c=\oint_{S^1}\eta$ . Show its integral over every closed curve vanishes.

Solution

Exact forms integrate to 0 over closed curves, which gives one direction. Conversely, every closed curve in the punctured plane is homotopic to a multiple of the circle, and integrals of closed forms are homotopy invariant. So if $\oint_{S^1}\eta=0$ , every loop integral vanishes, and $f(x)=\int_{x_0}^x\eta$ along any path is well defined with $df=\eta$ .