Connections and Curvature · Study tools

InteractiveTu, Ch. 1 (§§1–2): Riemannian Manifolds · Curves

Curves and curvature

Curvature starts with the simplest case: how fast a curve turns per unit of length. For a closed curve in the plane the turning adds up to a whole number of full turns, however you bend it. That pattern, local curvature adding up to a global integer, runs through the whole subject.

The idea

Riemannian metric. An inner product

\langle\ ,\ \rangle_p

on each tangent space

T_pM

, varying smoothly: for smooth vector fields

X,Y

the function

p\mapsto\langle X_p,Y_p\rangle_p

is smooth. Every manifold has one (glue local ones with a partition of unity). It gives lengths of curves,

\ell(\gamma)=\int_a^b|\gamma'(t)|\,dt

.

Regular curve, arc length.

\gamma

is regular if

\gamma'(t)\neq0

for all

t

. Then

s(t)=\int_a^t|\gamma'(u)|\,du

is strictly increasing, and reparametrising by

s

gives a unit-speed curve.

Signed curvature. For a unit-speed plane curve let

T=\gamma'

and let

N

be

T

rotated by

+90^\circ

. Then

T'(s)=\kappa(s)\,N(s)

. For any regular parametrisation

\gamma=(x,y)

,

\kappa = \frac{x'y''-y'x''}{(x'^2+y'^2)^{3/2}} .

$\kappa>0$ means the curve turns left. Near a point with $\kappa\neq0$ , the curve is best approximated by its osculating circle, of radius $1/|\kappa|$ and centre $\gamma+N/\kappa$ .

Turning tangents. Write

T=(\cos\theta,\sin\theta)

with

\theta

continuous. Then

\kappa=d\theta/ds

, so for a closed regular curve of length

\ell

\int_0^\ell\kappa\,ds=\theta(\ell)-\theta(0)=2\pi I,\qquad I\in\Z .

I

is the rotation index. For a simple closed curve,

I=\pm1

(Hopf).

In space. A curve in

\R^3

has no preferred side to turn to, so its curvature

\kappa=|T'|\ge0

carries no sign.

Explore

Start from

Marked point, s / ℓ

Drag the hollow control points. The dashed circle is the osculating circle at the marked point.

\kappa>0

(turning left)

\kappa<0

(turning right), against arc length

s

. The signed area is

\int\kappa\,ds

.

Worked example

Study the curvature of the cubic $\gamma(t)=(t,\,t^3)$ , traversed left to right: its sign, where it is largest, and how much the tangent turns in total.

Compute $\gamma'$ , $\gamma''$ and $\kappa(t)$ .

$\gamma'=(1,3t^2)$ and $\gamma''=(0,6t)$ , so $x'y''-y'x''=6t$ and

$\kappa(t)=\frac{6t}{(1+9t^4)^{3/2}} .$
Where does $\kappa$ change sign, and what does that look like?

$\kappa<0$ for $t<0$ (turning right) and $\kappa>0$ for $t>0$ (turning left). At $t=0$ the curve has an inflection point: curvature zero, and the osculating "circle" is a straight line.
Where is $\kappa$ largest?

Differentiate: $\kappa'(t)=6(1+9t^4)^{-5/2}\big(1-45t^4\big)$ , which vanishes at $t^4=1/45$ , $t\approx0.386$ . There $\kappa=6t/(1.2)^{3/2}\approx1.76$ : the sharpest bend is neither at the inflection nor far out, where the curve straightens.
By how much does the tangent direction turn over the whole curve, $-\infty<t<\infty$ ?

$\int\kappa\,ds=\theta(+\infty)-\theta(-\infty)$ . The tangent $(1,3t^2)$ tends to straight up at both ends, so the total is $\pi/2-\pi/2=0$ : the right turn for $t<0$ exactly cancels the left turn for $t>0$ . No integral was needed.

Check your understanding

You run around a circle of radius 2, first at speed 1, then at speed 5. What is its curvature?

$1/2$ both times
$1/2$ at speed 1 and $25/2$ at speed 5Speed changes the acceleration, but curvature is turning per unit of length, which does not depend on the timing.
$2$ Curvature is the reciprocal of the radius: a big circle turns slowly.

Replacing

t

by

ct

multiplies

x'y''-y'x''

by

c^3

and

|\gamma'|^3

by

c^3

. Curvature belongs to the curve and its direction, not to the parametrisation.

You traverse the same plane curve in the opposite direction. What happens to the signed curvature at a given point?

It changes sign; $|\kappa|$ stays the same.
Nothing changes.Reversing flips $T$ , and $N$ is $T$ rotated by $+90^\circ$ , so $N$ flips too. The curve still bends toward the same side of the plane.
It becomes $1/\kappa$ .Reversal does not change how sharply the curve bends, only which side counts as "left".

A left turn becomes a right turn when you drive the same road the other way. In the Kidney preset, the dent is where

\kappa<0

.

What is the total signed curvature $\int\kappa\,ds$ of a figure eight?

$0$
$4\pi$ The two lobes are traversed in opposite senses, one turning left and one turning right.
$2\pi$ $\pm2\pi$ is Hopf's answer for simple closed curves. A figure eight crosses itself.

The tangent makes one full turn to the left on one lobe and one to the right on the other: rotation index 0. In the Figure eight preset, the blue and gold areas of the

\kappa

graph are equal.

Practice

The logarithmic spiralCurvature against arc length

For $\gamma(t)=(e^t\cos t,\ e^t\sin t)$ , compute $\kappa(t)$ and the arc length $s(t)$ measured from $t=-\infty$ . Show that $\kappa\,s$ is constant.

Hint

$\gamma'=e^t(\cos t-\sin t,\ \sin t+\cos t)$ , so $|\gamma'|=\sqrt2\,e^t$ .

Solution

$\gamma''=e^t(-2\sin t,\ 2\cos t)$ and $x'y''-y'x''=2e^{2t}$ , so $\kappa=2e^{2t}/(2\sqrt2\,e^{3t})=e^{-t}/\sqrt2$ . Also $s(t)=\int_{-\infty}^t\sqrt2\,e^u\,du=\sqrt2\,e^t$ . Hence $\kappa\,s=1$ : the curvature is the reciprocal of the distance travelled along the spiral from its centre.

Constant curvature means a circleA Frenet argument

Show that a unit-speed plane curve with constant curvature $\kappa\equiv c\neq0$ lies on a circle of radius $1/|c|$ .

Hint

First show $N'=-\kappa T$ . Then differentiate the candidate centre $\gamma+N/c$ .

Solution

$N=JT$ where $J$ is rotation by $90^\circ$ , so $N'=JT'=\kappa JN=-\kappa T$ . Then $(\gamma+N/c)'=T-T=0$ : the point $z=\gamma+N/c$ is fixed, and $|\gamma-z|=|N|/|c|=1/|c|$ .

More practice: Tu, Problems 2.2, 2.5 and 17.2.

InteractiveTu, Ch. 1 (§§3–5): Surfaces in Space · Directional Derivatives · The Shape Operator

The shape operator

A surface can bend differently in different directions. The shape operator collects all of that into one symmetric $2\times2$ matrix at each point: its eigenvalues are the principal curvatures, its determinant is the Gaussian curvature, and it is minus the derivative of the Gauss map.

The idea

Directional derivative. In

\R^n

, for a vector field

Y=\sum Y^i\partial_i

and a tangent vector

X_p

,

\ D_{X_p}Y=\sum_i (X_pY^i)\,\partial_i

: differentiate the components.

Shape operator. Let

M\subset\R^3

be a surface with a smooth unit normal field

N

. For

X\in T_pM

set

L(X)=-D_XN

. Since

\langle N,N\rangle=1

,

D_XN\perp N

, so

L:T_pM\to T_pM

. It is self-adjoint,

\langle LX,Y\rangle=\langle X,LY\rangle

, and

\mathrm{II}(X,Y)=\langle LX,Y\rangle

is the second fundamental form.

Principal, Gaussian, mean curvature. The eigenvalues

\kappa_1\ge\kappa_2

of

L

are the principal curvatures, with orthonormal eigenvectors (principal directions).

K=\det L=\kappa_1\kappa_2

and

H=\tfrac12\operatorname{tr}L=\tfrac12(\kappa_1+\kappa_2)

.

Normal curvature (Euler). For a unit

u\in T_pM

at angle

\varphi

from the first principal direction,

\kappa_n(u)=\langle Lu,u\rangle=\kappa_1\cos^2\varphi+\kappa_2\sin^2\varphi .

It is the curvature at

p

of the normal section (the curve cut out by the plane through

p

spanned by

u

and

N

), positive when it bends toward

N

.

Gauss map.

\nu:M\to S^2

,

\nu(p)=N_p

. Since

T_{\nu(p)}S^2=N_p^\perp=T_pM

,

d\nu_p

is a map of

T_pM

to itself, and

d\nu_p=-L_p

. So

K=\det d\nu_p

: the ratio of the area of the Gauss image of a small region to the area of the region, negative when the image is traversed the other way round.

Choice of normal. Replacing

N

by

-N

replaces

L

by

-L

:

\kappa_i

and

H

change sign,

K

does not.

Explore

a b c Direction u

Try:

z=\tfrac12(ax^2+2bxy+cy^2)

near

p=0

, upward normal. Drag to turn.

\kappa_1

direction

\kappa_2

direction normal section along

u

In

T_pM

: unit directions

u

their images

d\nu(u)=-Lu

. Arrowheads show which way each is traversed.

Worked example

For the saddle $z=xy$ with upward normal, compute the shape operator at the origin, the principal data there, and $K$ at every point.

Write down the upward unit normal at a general point $(x,y,xy)$ .

The surface is the level set $z-xy=0$ , with gradient $(-y,-x,1)$ . So $N=\dfrac{(-y,\,-x,\,1)}{\sqrt{1+x^2+y^2}}$ .
Compute $L=-D N$ at the origin on the basis $\partial_x,\partial_y$ of $T_0M$ (the $xy$ -plane).

The square-root factor has zero derivative at the origin, so there $D_{\partial_x}N=(0,-1,0)$ and $D_{\partial_y}N=(-1,0,0)$ . Hence $L\partial_x=\partial_y$ , $L\partial_y=\partial_x$ :

$L=\begin{pmatrix}0&1\\1&0\end{pmatrix},$

which is the Hessian of $xy$ . (For any graph $z=f(x,y)$ with $f(0)=0$ , $\nabla f(0)=0$ , the shape operator at 0 is the Hessian.)
Principal curvatures and directions, $K$ and $H$ at the origin?

Eigenvalues $\pm1$ with eigenvectors $(1,1)/\sqrt2$ and $(1,-1)/\sqrt2$ . So $K=-1$ and $H=0$ . The normal curvature vanishes along the $x$ - and $y$ -axes, which lie on the surface.
What is $K$ at a general point?

For a graph, $K=\dfrac{f_{xx}f_{yy}-f_{xy}^2}{(1+f_x^2+f_y^2)^2}$ , which here gives

$K(x,y)=\frac{-1}{(1+x^2+y^2)^2}.$

Negative everywhere, most negative at the origin. Set $a=c=0$ , $b=1$ in the explorer: the image of the unit circle is traversed clockwise, because $\det d\nu=-1$ .

Check your understanding

You replace the unit normal $N$ by $-N$ . What happens?

$\kappa_1,\kappa_2$ and $H$ change sign; $K$ stays the same.
Everything changes sign, including $K$ . $K=\kappa_1\kappa_2$ is a product of two numbers that both flip.
Nothing changes: curvature is a property of the surface. $L=-DN$ is defined using $N$ , so it flips with $N$ .

L\mapsto-L

, and

\det(-L)=\det L

for a

2\times2

matrix. This is a first hint that

K

is more intrinsic than

H

.

At $p$ , $L=\begin{pmatrix}3&0\\0&-1\end{pmatrix}$ in an orthonormal basis. How many tangent lines through $p$ have normal curvature 0?

Two
NoneThe normal curvature varies continuously from 3 down to −1 as the direction turns, so it passes through 0.
OneSolve $3\cos^2\varphi-\sin^2\varphi=0$ : there are two solutions in $[0,\pi)$ .
Every lineThat would need $L=0$ , a planar point.

\tan^2\varphi=3

, so

\varphi=\pm60^\circ

from the first principal direction. These asymptotic directions exist exactly when

K\le0

.

The Gauss image of a cylinder is one great circle of $S^2$ . What does that say about $K$ ?

$K=0$ everywhere: the image has no area.
$K$ is constant but nonzero, since every point looks the same.Homogeneity forces $K$ to be constant; the area of the image decides which constant.
Nothing: the Gauss map only sees the normal, not the curvature. $d\nu=-L$ , so the Gauss map carries all of the shape operator.

The normal is constant along each ruling, so

d\nu

kills the ruling direction and

\det d\nu=0

. The surface still bends (

H\neq0

); it just does so in one direction only.

Practice

Euler's formulaWhy the principal curvatures are the extremes

Prove $\kappa_n(\varphi)=\kappa_1\cos^2\varphi+\kappa_2\sin^2\varphi$ , and deduce that $\kappa_1$ and $\kappa_2$ are the largest and smallest normal curvatures at $p$ .

Solution

With $Le_i=\kappa_ie_i$ and $u=\cos\varphi\,e_1+\sin\varphi\,e_2$ , $\langle Lu,u\rangle=\kappa_1\cos^2\varphi+\kappa_2\sin^2\varphi$ . This is a weighted average of $\kappa_1$ and $\kappa_2$ with weights summing to 1, so it lies between them and attains each at $\varphi=0$ and $\varphi=\pi/2$ .

The monkey saddleA planar point that isn't flat

For $z=x^3-3xy^2$ , find $L$ at the origin, then the sign of $K$ at every other point.

Hint

Use the graph formula for $K$ from the worked example.

Solution

The Hessian is $\begin{pmatrix}6x&-6y\\-6y&-6x\end{pmatrix}$ , zero at the origin, so $L=0$ there: a planar point, $K=H=0$ . Elsewhere $f_{xx}f_{yy}-f_{xy}^2=-36(x^2+y^2)<0$ , so $K<0$ . Zero curvature at a single point does not make a surface flat.

More practice: Tu, Problems 5.5, 5.6 and 5.8.

InteractiveTu, Ch. 1 (§§6–7): Affine Connections · Vector Bundles

Affine connections

To differentiate a vector field along a surface, differentiate it in space and throw away the part sticking out of the surface. What is left obeys a short list of rules, and those rules, not the ambient space, are what define a connection.

The idea

Affine connection. A map

(X,Y)\mapsto\nabla_XY

on vector fields that is

\R

-bilinear,

C^\infty

-linear in

X

(

\nabla_{fX}Y=f\nabla_XY

), and satisfies the Leibniz rule in

Y

:

\nabla_X(fY)=(Xf)Y+f\nabla_XY

. The directional derivative

D

on

\R^n

is one.

Torsion and curvature.

T(X,Y)=\nabla_XY-\nabla_YX-[X,Y],\qquad R(X,Y)Z=\nabla_X\nabla_YZ-\nabla_Y\nabla_XZ-\nabla_{[X,Y]}Z .

Both are

C^\infty

-linear in every slot, so they are tensors: their value at

p

depends only on the vectors at

p

. For

D

on

\R^n

both vanish.

Fundamental theorem. On a Riemannian manifold there is exactly one connection that is torsion-free and compatible with the metric,

X\langle Y,Z\rangle=\langle\nabla_XY,Z\rangle+\langle Y,\nabla_XZ\rangle

: the Riemannian (Levi-Civita) connection.

Surfaces in $\R^3$ .

\nabla_XY=(D_XY)^{\tan}

is the Riemannian connection of the induced metric, and

D_XY=\nabla_XY+\langle LX,Y\rangle\,N .

Vector bundle. A smooth map

\pi:E\to M

whose fibres

E_p

are

r

-dimensional vector spaces, locally isomorphic to

U\times\R^r

by maps that are linear on fibres. Sections are smooth

s

with

s(p)\in E_p

; a frame on

U

is

r

sections that form a basis at each point. An affine connection is a way to differentiate sections of

TM

.

Watch out.

(\nabla_XY)_p

depends on

X

only through

X_p

, but on

Y

along a curve through

p

with velocity

X_p

.

Explore

Drag to turn the sphere.

V

dV/dt

in space tangential part

\nabla_{\gamma'}V

normal part

Vector field V along the circle

Latitude λ Position t on the circle

Worked example

$M\subset\R^3$ is a surface with the induced metric, and $\nabla_XY=(D_XY)^{\tan}$ for tangent vector fields $X,Y$ (extended arbitrarily off $M$ ). Show that $\nabla$ is the Riemannian connection, and identify what was thrown away.

Why is $\nabla$ an affine connection?

$D$ has the three properties, and taking the tangential part is $C^\infty$ -linear: $(D_{fX}Y)^{\tan}=f(D_XY)^{\tan}$ and $(D_X(fY))^{\tan}=(Xf)Y+f(D_XY)^{\tan}$ , because $Y$ is already tangent. The result doesn't depend on the extension, since $D_XY$ at $p$ only uses $Y$ along a curve in $M$ .
Why is it torsion-free?

In $\R^3$ , $D_XY-D_YX=[X,Y]$ , and the bracket of two fields tangent to $M$ is tangent to $M$ . Taking tangential parts gives $\nabla_XY-\nabla_YX=[X,Y]$ .
Why is it compatible with the metric?

$X\langle Y,Z\rangle=\langle D_XY,Z\rangle+\langle Y,D_XZ\rangle$ in $\R^3$ . Since $Z$ is tangent, $\langle D_XY,Z\rangle=\langle(D_XY)^{\tan},Z\rangle=\langle\nabla_XY,Z\rangle$ , and the same for the other term.
Conclude, and compute the normal part of $D_XY$ .

By uniqueness, $\nabla$ is the Riemannian connection. Differentiating $\langle Y,N\rangle=0$ gives $\langle D_XY,N\rangle=-\langle Y,D_XN\rangle=\langle LX,Y\rangle$ . So $D_XY=\nabla_XY+\mathrm{II}(X,Y)\,N$ : the intrinsic part and the shape operator, cleanly separated. In the explorer the teal arrow is the first term and the red one the second.

Check your understanding

On the unit sphere you walk east along the circle of latitude $60^\circ$ N, carrying the unit east vector $E$ . Which way does $\nabla_{\gamma'}E$ point?

North, toward the pole
South, toward the equatorIn space, $dE/dt$ points toward the Earth's axis. Its tangential part points uphill, toward the pole.
It is zero.That happens only on the equator, the one circle of latitude that is a geodesic.
Straight into the sphereThat is the normal part of $dE/dt$ , which $\nabla$ throws away.

\nabla_{\gamma'}E=\sin\lambda\;U

with

U

the unit north vector, which is

\tfrac{\sqrt3}{2}U

at

60^\circ

. Walking a latitude circle in the northern hemisphere, you are always turning left. Choose "East E" and

\lambda=60^\circ

to check.

$\nabla$ and $\nabla'$ are affine connections on $M$ . What is $A(X,Y)=\nabla_XY-\nabla'_XY$ ?

A tensor: $C^\infty$ -linear in $X$ and in $Y$
Another affine connectionSubtracting kills the $(Xf)Y$ term of the Leibniz rule, so $A(X,fY)=fA(X,Y)$ : no Leibniz rule.
Only $\R$ -bilinear in generalCheck $C^\infty$ -linearity in $Y$ : the derivative terms cancel.

So once you have one connection, every other is that one plus a tensor field of type (1,2). In a chart, the Christoffel symbols are exactly this tensor, comparing

\nabla

with the flat connection of the coordinates.

On $\R^2$ define $\nabla_XY=D_XY+dx(X)\,Y$ . Which is true?

It is an affine connection, but it has torsion and is not compatible with the metric.
It is not a connection: the extra term breaks the Leibniz rule.The extra term is $C^\infty$ -linear in both slots. A connection plus a tensor is a connection.
It is torsion-free, because $D$ is. $T(X,Y)=dx(X)Y-dx(Y)X$ . Try $X=\partial_x$ , $Y=\partial_y$ .

T(\partial_x,\partial_y)=\partial_y\neq0

, and

\langle\nabla_XY,Z\rangle+\langle Y,\nabla_XZ\rangle=X\langle Y,Z\rangle+2\,dx(X)\langle Y,Z\rangle

. Torsion-free and metric are real conditions; together they single out one connection.

Practice

Changing a connection by a tensorWhen does torsion stay zero?

Let $\nabla$ be torsion-free and $\nabla'_XY=\nabla_XY+A(X,Y)$ for a (1,2)-tensor $A$ . Show that $\nabla'$ is torsion-free if and only if $A$ is symmetric.

Solution

$T'(X,Y)=T(X,Y)+A(X,Y)-A(Y,X)=A(X,Y)-A(Y,X)$ . This vanishes for all $X,Y$ exactly when $A$ is symmetric.

Polar coordinatesChristoffel symbols of the flat plane

On $\R^2\smallsetminus\{0\}$ with polar coordinates, compute $D_{\partial_r}\partial_r$ , $D_{\partial_r}\partial_\theta$ and $D_{\partial_\theta}\partial_\theta$ , and read off the Christoffel symbols.

Hint

$\partial_r=(\cos\theta,\sin\theta)$ and $\partial_\theta=(-r\sin\theta,\ r\cos\theta)$ . Differentiate the components.

Solution

$D_{\partial_r}\partial_r=0$ , $D_{\partial_r}\partial_\theta=(-\sin\theta,\cos\theta)=\tfrac1r\partial_\theta$ , $D_{\partial_\theta}\partial_\theta=(-r\cos\theta,-r\sin\theta)=-r\,\partial_r$ . So $\Gamma^\theta_{r\theta}=\Gamma^\theta_{\theta r}=1/r$ , $\Gamma^r_{\theta\theta}=-r$ , and the rest vanish. Nonzero Christoffel symbols, zero curvature.

More practice: Tu, Problems 6.2, 6.5 and 6.6.

InteractiveTu, Ch. 1 (§§8–9): Gauss's Theorema Egregium · Generalizations to Hypersurfaces in ℝⁿ⁺¹

Theorema Egregium

Roll a sheet of paper and its principal curvatures change, but their product stays zero. Gauss's "remarkable theorem" says this is no accident: $K$ can be computed from lengths measured inside the surface, so bending without stretching never changes it.

The idea

Isometry. A diffeomorphism

F:M\to\tilde M

with

\langle F_*X,F_*Y\rangle=\langle X,Y\rangle

. A bending is a family of surfaces that are all isometric to each other: lengths of curves on the surface never change.

Gauss equation. For a surface in

\R^3

with its Riemannian connection,

R(X,Y)Z=\langle LY,Z\rangle\,LX-\langle LX,Z\rangle\,LY .

So for an orthonormal basis

e_1,e_2

of

T_pM

:

\ \langle R(e_1,e_2)e_2,e_1\rangle=\langle Le_1,e_1\rangle\langle Le_2,e_2\rangle-\langle Le_1,e_2\rangle^2=\det L=K.

Theorema Egregium.

R

is built from the Riemannian connection, which the metric alone determines. So

K

is intrinsic, and a local isometry preserves it. In coordinates with

F=0

(metric

E\,du^2+G\,dv^2

):

K=-\frac{1}{2\sqrt{EG}}\left[\Big(\frac{E_v}{\sqrt{EG}}\Big)_v+\Big(\frac{G_u}{\sqrt{EG}}\Big)_u\right].

The other half of the Gauss–Codazzi equations, $(\nabla_XL)Y=(\nabla_YL)X$ (Codazzi–Mainardi), is a condition on how $L$ varies. $H$ , $\kappa_1$ and $\kappa_2$ are not intrinsic.

Hypersurfaces. For

M^n\subset\R^{n+1}

the same definition of

L

works and the Gauss equation holds word for word. The sectional curvature of the plane spanned by orthonormal

X,Y

is

\langle LX,X\rangle\langle LY,Y\rangle-\langle LX,Y\rangle^2

; for two principal directions it is

\kappa_i\kappa_j

.

Explore

Drag to turn. marked point a curve drawn on the surface

Deformation

Amount

Worked example

Mercator's projection, run backwards, is the map $X(u,v)=(\operatorname{sech}v\cos u,\ \operatorname{sech}v\sin u,\ \tanh v)$ . Compute the metric it induces, find $K$ from the metric alone, and explain why no map of the globe can get all distances right.

Check that $X$ lands on the unit sphere. What do $u$ and $v$ measure?

$\operatorname{sech}^2v+\tanh^2v=1$ , so $|X|=1$ . $u$ is longitude; $v$ runs over $\R$ as the latitude runs over $(-\pi/2,\pi/2)$ , with $\tanh v=\sin(\text{latitude})$ .
Compute $E,F,G$ .

$X_u=\operatorname{sech}v\,(-\sin u,\cos u,0)$ , so $E=\operatorname{sech}^2v$ . $X_v=(-\operatorname{sech}v\tanh v\cos u,\ -\operatorname{sech}v\tanh v\sin u,\ \operatorname{sech}^2v)$ , so $G=\operatorname{sech}^2v(\tanh^2v+\operatorname{sech}^2v)=\operatorname{sech}^2v$ , and $F=0$ . The metric is $\operatorname{sech}^2v\,(du^2+dv^2)$ : conformal, which is why Mercator's map preserves angles.
Compute $K$ from the metric only.

For $E=G=\lambda^2$ the formula above becomes $K=-\Delta(\log\lambda)/\lambda^2$ . Here $\log\lambda=\log\operatorname{sech}v$ , with $(\log\operatorname{sech}v)''=-\operatorname{sech}^2v$ . So

$K=\frac{\operatorname{sech}^2v}{\operatorname{sech}^2v}=1,$

as it should be for the unit sphere. Nothing about the embedding was used beyond computing $E,F,G$ .
Why can no map from a region of the sphere to the plane preserve all lengths?

Such a map would be a local isometry, and by the Theorema Egregium it would carry $K=1$ to $K=1$ . But the plane has $K=0$ . Mercator keeps angles and pays with lengths: it scales them by $\cosh v$ , which blows up toward the poles.

Check your understanding

You roll the paper tighter in the explorer. Which quantity at the marked point changes?

$H$ (and $\kappa_1$ ); $K$ stays 0.
$K$ , because the paper is more curved.One principal curvature grows, but the one along the rulings stays 0, so the product stays 0.
Nothing: rolling is an isometry.An isometry preserves intrinsic quantities. $H$ depends on how the surface sits in space.

The principal curvatures are the rolling curvature and

0

, so

K=0

and

H

is half the rolling curvature. The intrinsic number stays put while the extrinsic ones move.

In "Bending a piece of sphere", the principal curvatures at the marked point change but $K$ stays 1. What is going on?

The deformation keeps the metric, so $K=\kappa_1\kappa_2$ is fixed while the individual factors trade off.
The surface is still part of the round sphere, just rotated.A piece of a round sphere would have $\kappa_1=\kappa_2$ . Here they differ: it is a genuinely different surface in space.
It is a numerical coincidence.The extrinsic and intrinsic $K$ readings are computed independently, and they agree for every amount.

Each surface in the family has metric

dv^2+\cos^2v\,du^2

, the metric of the sphere in latitude–longitude coordinates. It is curved in space like a barrel or a spindle, yet anyone living on it measures the same lengths as on the sphere.

A hypersurface in $\R^4$ has principal curvatures $2,3,5$ at $p$ . What is the sectional curvature of the plane spanned by the principal directions for $3$ and $5$ ?

$15$
$30$ That's the product of all three (the Gauss–Kronecker curvature), which belongs to no single plane.
$8$ Sectional curvature comes from the Gauss equation, a product, not a sum.

For orthonormal principal directions

e_i,e_j

the Gauss equation gives

K(e_i,e_j)=\kappa_i\kappa_j

.

Practice

Compact surfaces have positive curvature somewhereA comparison argument

Show that every compact surface $M\subset\R^3$ has a point where $K>0$ . Conclude that no compact surface in $\R^3$ is isometric to a flat torus.

Hint

Look at a point of $M$ farthest from the origin.

Solution

Let $p$ maximise $|x|$ , with $|p|=R$ . Then $M$ lies inside the ball of radius $R$ and touches the sphere $|x|=R$ at $p$ , so $T_pM=T_pS_R$ . Every normal section of $M$ at $p$ stays inside the ball, so it curves toward the centre at least as much as the sphere: $|\kappa_n|\ge1/R$ with the same sign in every direction. So $K=\kappa_1\kappa_2\ge1/R^2>0$ . A flat torus has $K\equiv0$ , and an isometry would preserve $K$ .

Curvature from a conformal metricStereographic coordinates

Compute $K$ for the metric $\dfrac{du^2+dv^2}{(1+(u^2+v^2)/4)^2}$ on $\R^2$ .

Hint

Use $K=-\Delta(\log\lambda)/\lambda^2$ and the radial Laplacian $g''+g'/r$ .

Solution

With $q=r^2/4$ , $g=\log\lambda=-\log(1+q)$ : $g'=-\tfrac{r/2}{1+q}$ and $g''=-\tfrac{(1-q)/2}{(1+q)^2}$ . So $\Delta g=-\tfrac{1}{(1+q)^2}$ and $K=(1+q)^2/(1+q)^2=1$ . This is the round unit sphere seen through stereographic projection.

More practice: Tu, Problems 5.11 and 8.3.

InteractiveTu, Ch. 2 (§§10–12): Connections on a Vector Bundle · Connection, Curvature, and Torsion Forms · The Theorema Egregium Using Forms

Connection and curvature forms

Choose a frame, and a connection becomes a matrix of 1-forms. That matrix depends on the frame and changes whenever the frame turns. Its curvature does not. Here you can twist a frame and see which numbers move.

The idea

Connection on a vector bundle.

\nabla:\mathfrak X(M)\times\Gamma(E)\to\Gamma(E)

,

C^\infty

-linear in

X

, Leibniz in the section:

\nabla_X(fs)=(Xf)s+f\nabla_Xs

. Its curvature is

R(X,Y)s=\nabla_X\nabla_Ys-\nabla_Y\nabla_Xs-\nabla_{[X,Y]}s

. With a fibre metric,

\nabla

is metric if

X\langle s,t\rangle=\langle\nabla_Xs,t\rangle+\langle s,\nabla_Xt\rangle

.

Connection and curvature matrices. For a frame

e_1,\dots,e_r

on

U

,

\ \nabla_Xe_j=\sum_i\omega^i_j(X)\,e_i

and

R(X,Y)e_j=\sum_i\Omega^i_j(X,Y)\,e_i

. So

\omega=[\omega^i_j]

is a matrix of 1-forms and

\Omega=[\Omega^i_j]

a matrix of 2-forms.

Structure equations.

\Omega=d\omega+\omega\wedge\omega

. For

E=TM

with dual coframe

\theta^1,\dots,\theta^n

,

\nabla

is torsion-free if and only if

d\theta^i=-\sum_j\omega^i_j\wedge\theta^j

. In an orthonormal frame of a metric connection,

\omega

and

\Omega

are skew-symmetric.

Surfaces. For an orthonormal frame

e_1,e_2

there is one form

\omega^1_2=-\omega^2_1

, and

d\theta^1=-\omega^1_2\wedge\theta^2,\qquad d\theta^2=\omega^1_2\wedge\theta^1,\qquad \Omega^1_2=d\omega^1_2=K\,\theta^1\wedge\theta^2 .

The first two equations determine

\omega^1_2

from the coframe alone, so

K

is intrinsic: the Theorema Egregium in two lines.

The explorer uses geodesic polar coordinates on a surface of constant curvature $c$ : the metric is $dr^2+f(r)^2d\theta^2$ with $f=\sin(\sqrt c\,r)/\sqrt c$ , $r$ , or $\sinh(\sqrt{-c}\,r)/\sqrt{-c}$ . For $e_1=\partial_r$ , $e_2=\tfrac1f\partial_\theta$ the worked example gives $\omega^1_2=-f'(r)\,d\theta$ . Then the frame is rotated by $\varphi=k\theta$ , which changes the form to $\omega'^1_2=\omega^1_2-d\varphi$ .

Explore

The geodesic disc, drawn in polar coordinates

(r,\theta)

.

e_1'

e_2'

the circle

r=r_0

Curvature c Radius r₀

Frame twist φ = kθ

Worked example

A surface has metric $dr^2+f(r)^2\,d\theta^2$ with $f>0$ . Find $\omega^1_2$ for the frame $e_1=\partial_r$ , $e_2=\tfrac1f\partial_\theta$ , then $K$ .

Write down the coframe and its exterior derivatives.

$\theta^1=dr$ , $\theta^2=f\,d\theta$ . Then $d\theta^1=0$ and $d\theta^2=f'\,dr\wedge d\theta$ .
Solve the first structure equations for $\omega^1_2$ .

From $d\theta^1=-\omega^1_2\wedge f\,d\theta=0$ , $\omega^1_2$ has no $dr$ component, so $\omega^1_2=g\,d\theta$ . Then $\omega^1_2\wedge\theta^1=g\,d\theta\wedge dr=-g\,dr\wedge d\theta$ must equal $f'\,dr\wedge d\theta$ . So

$\omega^1_2=-f'(r)\,d\theta .$
Compute $\Omega^1_2$ and $K$ .

$\Omega^1_2=d\omega^1_2=-f''\,dr\wedge d\theta=-\frac{f''}{f}\,\theta^1\wedge\theta^2,\qquad K=-\frac{f''}{f}.$

For $f=\sin r$ , $K=1$ ; for $f=r$ , $K=0$ ; for $f=\sinh r$ , $K=-1$ .
The flat plane ( $f=r$ ) gives $\omega^1_2=-d\theta\neq0$ . Is anything wrong?

No. $\omega$ records how the frame turns, and $e_r,e_\theta$ turn as you go around. The curvature $d(-d\theta)=0$ is what measures the plane. In the explorer, set $c=0$ and $k=-1$ : the rotated frame is the constant frame $\partial_x,\partial_y$ , and $\omega'^1_2=0$ .

Check your understanding

You rotate an orthonormal frame on a surface by an angle function $\varphi$ . What changes?

$\omega^1_2$ changes by $-d\varphi$ ; $\Omega^1_2$ is unchanged.
Both change: they are computed from the frame. $\Omega^1_2$ changes by $-d(d\varphi)=0$ .
Neither: both are geometric.Twist the frame in the explorer and watch $\omega'^1_2$ move.

\Omega^1_2=K\,\theta^1\wedge\theta^2

, and

\theta^1\wedge\theta^2

is the area form, which a rotation doesn't change. The connection form is frame data; the curvature form is geometry.

For a metric connection and an orthonormal frame, the connection matrix $\omega$ is…

skew-symmetric
symmetricDifferentiate $\langle e_i,e_j\rangle=\delta_{ij}$ : you get $\omega^j_i+\omega^i_j=0$ .
zeroOnly if the frame is parallel, which a curved surface never allows on an open set.

0=X\langle e_i,e_j\rangle=\omega^j_i(X)+\omega^i_j(X)

. That's why a surface needs only the one form

\omega^1_2

.

In the explorer with $c=0$ and $k=0$ , $\oint_{r=r_0}\omega^1_2=-2\pi$ , but $\iint K\,dA=0$ . Why doesn't Stokes' theorem apply?

The frame $e_r,e_\theta$ is not defined at the centre, so $\omega^1_2$ isn't a form on the whole disc.
Stokes' theorem needs $K\neq0$ .Stokes holds for any smooth form on a compact region with boundary.
Because $\omega^1_2$ is not closed.Here $d\omega^1_2=0$ . The problem is where $\omega^1_2$ is defined.

With

k=-1

the frame extends smoothly over the centre, and then

\oint\omega'^1_2=\iint K\,dA

exactly. For other

k

the difference is

-2\pi(1+k)

,

2\pi

times the number of turns the frame makes around the centre. This is the germ of the Gauss–Bonnet theorem.

Practice

Conformal metricsA formula for K

For the metric $e^{2\phi}(dx^2+dy^2)$ , with $\theta^1=e^\phi dx$ , $\theta^2=e^\phi dy$ , show $\omega^1_2=\phi_y\,dx-\phi_x\,dy$ and $K=-e^{-2\phi}\Delta\phi$ . Apply it to $\phi=-\log y$ on $y>0$ .

Solution

$d\theta^1=-e^\phi\phi_y\,dx\wedge dy$ and $d\theta^2=e^\phi\phi_x\,dx\wedge dy$ . Writing $\omega^1_2=p\,dx+q\,dy$ , $-\omega^1_2\wedge\theta^2=-pe^\phi\,dx\wedge dy$ gives $p=\phi_y$ , and $\omega^1_2\wedge\theta^1=-qe^\phi\,dx\wedge dy$ gives $q=-\phi_x$ . Then $d\omega^1_2=-(\phi_{xx}+\phi_{yy})\,dx\wedge dy=-e^{-2\phi}\Delta\phi\;\theta^1\wedge\theta^2$ . For $\phi=-\log y$ , $\Delta\phi=1/y^2$ and $K=-y^2/y^2=-1$ : the hyperbolic half-plane of the geodesics module.

A curved connection on a trivial bundleCurvature from a matrix of 1-forms

On $\R^2\times\R^2$ let $\nabla s=ds+\omega s$ with $\omega=\begin{pmatrix}0&-x\,dy\\x\,dy&0\end{pmatrix}$ . Compute $\Omega$ . Is there a frame in which $\omega=0$ ?

Solution

$\omega\wedge\omega=0$ because each product contains $dy\wedge dy$ . So $\Omega=d\omega=\begin{pmatrix}0&-dx\wedge dy\\dx\wedge dy&0\end{pmatrix}\neq0$ . Curvature transforms by conjugation under a change of frame, so it is nonzero in every frame, and no frame has $\omega=0$ .

More practice: Tu, Problems 11.1 and 12.5.

AnimatedTu, Ch. 3 (§§13–14): More on Affine Connections · Geodesics

Parallel transport and holonomy

Carry an arrow around a loop on a sphere, never turning it relative to the surface. When you get back it points somewhere else. The angle it has turned through is exactly the area you went around: curvature, measured by a walk.

The idea

Covariant derivative along a curve. For a vector field

V

along

\gamma

,

D_tV

is the unique operator that is

\R

-linear, satisfies

D_t(fV)=f'V+fD_tV

, and equals

\nabla_{\gamma'}\tilde V

whenever

V

extends to a vector field

\tilde V

. In coordinates,

D_tV=\sum_k\Big(\dot V^k+\sum_{i,j}\Gamma^k_{ij}\,\dot\gamma^i V^j\Big)\partial_k .

Parallel transport.

V

is parallel if

D_tV=0

. This is a linear ODE, so each

v\in T_{\gamma(a)}M

extends uniquely to a parallel

V

. The map

P_\gamma:v\mapsto V(b)

is a linear isomorphism

T_{\gamma(a)}M\to T_{\gamma(b)}M

, and an isometry when

\nabla

is metric.

Holonomy on a surface. If a simple closed curve bounds a region

D

homeomorphic to a disc, and is traversed with

D

on its left, then parallel transport around it rotates every tangent vector by the angle

\iint_DK\,dA

(mod

2\pi

).

On a surface in $\R^3$ , $D_tV=(dV/dt)^{\tan}$ , so "parallel" means the derivative in space is normal to the surface. On the unit sphere $K=1$ , so the rotation angle is the enclosed area.

Path dependence. On a curved surface the result depends on the path. There is no way to identify tangent spaces at different points without choosing one.

Watch

Drag the sphere to turn it. transported vector where it started the loop

Loop

Latitude λ Progress around the loop

Worked example

On the unit sphere, transport a vector around the circle of latitude $\lambda$ , $\gamma(t)=(\cos\lambda\cos t,\ \cos\lambda\sin t,\ \sin\lambda)$ , $0\le t\le2\pi$ . Use the orthonormal frame $E$ (east) and $U$ (north) along $\gamma$ .

Compute $D_tE$ and $D_tU$ .

$E=(-\sin t,\cos t,0)$ and $U=(-\sin\lambda\cos t,\ -\sin\lambda\sin t,\ \cos\lambda)$ . Then $dE/dt=(-\cos t,-\sin t,0)=\sin\lambda\,U-\cos\lambda\,\gamma$ , with tangential part $\sin\lambda\,U$ . And $dU/dt=(\sin\lambda\sin t,-\sin\lambda\cos t,0)=-\sin\lambda\,E$ , already tangent. So $D_tE=\sin\lambda\,U$ , $D_tU=-\sin\lambda\,E$ .
Write $V=\cos\alpha\,E+\sin\alpha\,U$ . When is $V$ parallel?

$D_tV=(\alpha'+\sin\lambda)\,(-\sin\alpha\,E+\cos\alpha\,U),$

so $V$ is parallel if and only if $\alpha'=-\sin\lambda$ .
By how much has $V$ turned after one lap?

Relative to $E,U$ , which return to their starting values, $V$ has turned by $\alpha(2\pi)-\alpha(0)=-2\pi\sin\lambda$ .
Compare with the area the circle encloses.

Moving east, the northern cap is on the left. Its area is $2\pi(1-\sin\lambda)$ , and $-2\pi\sin\lambda\equiv2\pi(1-\sin\lambda)\pmod{2\pi}$ , as the theorem predicts with $K=1$ . At the equator the angle is $2\pi\equiv0$ : a geodesic loop around a hemisphere.

Check your understanding

On the unit sphere you transport a vector around the octant triangle, which has three right angles. By how much does it rotate?

$90^\circ$
$270^\circ$ That's the angle sum. The rotation is the enclosed area, which is the excess over $180^\circ$ .
$0^\circ$ , because the sides are geodesics.Along each side the vector keeps a fixed angle with the side, but the side's direction jumps at every corner.

The octant is one eighth of the sphere: area

4\pi/8=\pi/2

. It's the default triangle in the explorer.

In the flat plane you transport a vector around a closed loop. What happens?

It comes back unchanged.
It rotates by $2\pi$ times the winding number.That's what the loop's own tangent vector does. A parallel vector is not the tangent.
It rotates by the enclosed area.Area counts only weighted by $K$ , and $K=0$ in the plane.

In

\R^2

,

D_tV=dV/dt

, so parallel means constant components.

For the Riemannian connection, parallel transport along a curve preserves…

lengths and angles: it is a linear isometry
lengths but not anglesApply compatibility to two parallel fields $V,W$ : $\langle V,W\rangle$ is constant, which controls both.
nothing in particularMetric compatibility gives $\frac{d}{dt}\langle V,W\rangle=\langle D_tV,W\rangle+\langle V,D_tW\rangle$ .

If

D_tV=D_tW=0

then

\frac{d}{dt}\langle V,W\rangle=0

. So holonomy on an oriented surface is a rotation, determined by one angle.

Practice

Around the tip of a coneCurvature concentrated at a point

Cut a sector of angle $\beta$ out of a flat paper disc and tape the edges to make a cone. Transport a vector once around a loop that circles the tip counterclockwise. By how much does it rotate?

Hint

Unroll the cone. Away from the tip it is flat, so a parallel vector is constant in the unrolled picture. What happens when you cross the taped seam?

Solution

The unrolled cone is a sector of angle $2\pi-\beta$ . Crossing the seam identifies one edge with the other by a rotation of $-(2\pi-\beta)$ , so the vector comes back rotated by $\beta-2\pi\equiv\beta$ . The cone is flat except at the tip, where it carries "total curvature" $\beta$ , the angle defect.

Small loops measure KHolonomy as a curvature detector

Use the holonomy theorem to estimate the rotation around a small geodesic circle of radius $\varepsilon$ about $p$ . How could you measure $K(p)$ without leaving the surface?

Solution

$\iint_DK\,dA=K(p)\,\pi\varepsilon^2+o(\varepsilon^2)$ by continuity of $K$ . So $K(p)=\lim_{\varepsilon\to0}(\text{rotation angle})/(\text{area enclosed})$ : walk a small loop, compare the arrow you carried with where it started, and divide by the area.

More practice: Tu, Problems 13.1 and 13.2.

InteractiveTu, Ch. 3 (§§14–16): Geodesics · Exponential Maps · Distance and Volume

Geodesics and the exponential map

A geodesic is a curve that never turns: its velocity is parallel along itself. In the hyperbolic half-plane they are semicircles, a geodesic circle is an off-centre Euclidean circle, and its circumference grows exponentially with its radius.

The idea

Geodesic. A curve with

D_t\gamma'=0

. In coordinates,

\ \ddot\gamma^k+\sum_{i,j}\Gamma^k_{ij}\dot\gamma^i\dot\gamma^j=0.

For the Riemannian connection the speed

|\gamma'|

is constant. For each

p

and

v\in T_pM

there is a unique maximal geodesic

\gamma_v

with

\gamma_v(0)=p

,

\gamma_v'(0)=v

.

Exponential map.

\exp_p(v)=\gamma_v(1)

. Since

\gamma_{tv}(1)=\gamma_v(t)

, it sends the line through

v

onto the geodesic.

d(\exp_p)_0=\mathrm{id}

, so

\exp_p

is a diffeomorphism from a ball around

0

onto a neighbourhood of

p

: normal coordinates.

Distance and volume.

d(p,q)

is the infimum of the lengths of piecewise smooth curves from

p

to

q

; geodesics minimise length locally. In oriented coordinates the volume form is

\sqrt{\det(g_{ij})}\,dx^1\wedge\dots\wedge dx^n

.

The half-plane.

\mathbb H=\{y>0\}

with

g=(dx^2+dy^2)/y^2

has

K=-1

and Christoffel symbols

\Gamma^x_{xy}=\Gamma^x_{yx}=-1/y

,

\Gamma^y_{xx}=1/y

,

\Gamma^y_{yy}=-1/y

, so the geodesic equations are

\ddot x-\frac{2}{y}\dot x\dot y=0,\qquad \ddot y+\frac1y(\dot x^2-\dot y^2)=0 .

The geodesics are vertical half-lines and semicircles centred on the

x

-axis, and

\cosh d(p,q)=1+\dfrac{|p-q|^2}{2\,y_p\,y_q}

.

Explore

Drag p. geodesic, by solving the equations the predicted semicircle geodesic circle of radius

L

Show

Initial direction Hyperbolic length L

Worked example

In the half-plane, compare two paths from $a=(-1,1)$ to $b=(1,1)$ : the horizontal segment and the geodesic.

How long is the horizontal segment?

Along $y=1$ the metric is $dx^2$ , so the length is $2$ . (It is not a geodesic: with $\dot y=0$ the second equation would need $\dot x^2/y=0$ .)
Which semicircle joins $a$ and $b$ ?

Its centre is on the $x$ -axis and equidistant from $a$ and $b$ , so it is the origin, and the radius is $\sqrt2$ . Parametrise it by $x=\sqrt2\cos s$ , $y=\sqrt2\sin s$ , $\pi/4\le s\le3\pi/4$ .
Compute its hyperbolic length.

$|\dot\gamma|_g=\sqrt2/(\sqrt2\sin s)=1/\sin s$ , so

$\ell=\int_{\pi/4}^{3\pi/4}\frac{ds}{\sin s}=\Big[\log\tan\tfrac s2\Big]_{\pi/4}^{3\pi/4}=\log\frac{\tan(3\pi/8)}{\tan(\pi/8)}=2\log(1+\sqrt2)\approx1.763 .$

Check with the distance formula: $\cosh d=1+4/2=3$ , and $\operatorname{arccosh}3=\log(3+2\sqrt2)=2\log(1+\sqrt2)$ .
Why does the shortest path bow upward?

Lengths are Euclidean lengths divided by $y$ , so travel is cheaper higher up. The geodesic climbs to $y=\sqrt2$ to save about 12% over the straight segment, even though it is longer in the Euclidean sense ( $\pi/\sqrt2\approx2.22$ ).

Check your understanding

A unit-speed geodesic in the half-plane passes through height $y$ . What is its Euclidean speed there?

$y$
$1$ Unit speed in the metric $g$ means $|\gamma'|_{\text{Eucl}}/y=1$ .
$1/y$ Lengths are divided by $y$ , so near the boundary you cover one hyperbolic unit with only $y$ Euclidean units.

So geodesics slow down in Euclidean terms as they approach the

x

-axis and never reach it in finite time. Compare "Euclidean length" with

L

in the explorer.

Why is $\exp_p$ a diffeomorphism near $0\in T_pM$ ?

Its differential at $0$ is the identity, so the inverse function theorem applies.
Because geodesics from $p$ never meet again.They can: great circles from $p$ all meet at the antipode.
Because it is linear. $\exp_p$ is linear only for a flat metric in linear coordinates.

d(\exp_p)_0(v)=\frac{d}{dt}\exp_p(tv)\big|_{t=0}=\gamma_v'(0)=v

. This is a local statement; how far it extends depends on the geometry.

On the unit sphere, where does $\exp_p$ send every vector of length $\pi$ ?

To the antipode $-p$
Back to $p$ That takes length $2\pi$ , all the way round.
To the equator relative to $p$ That's length $\pi/2$ .

So

\exp_p

collapses a whole circle to one point there: it is a diffeomorphism on the open ball of radius

\pi

and no larger ball. In the half-plane, with

K<0

,

\exp_p

is a diffeomorphism on all of

T_p\mathbb H

.

Practice

Vertical geodesicsSolving the equations

Show that $\gamma(t)=(x_0,\,y_0e^t)$ is a unit-speed geodesic of the half-plane, and find $d\big((0,1),(0,b)\big)$ for $b>1$ .

Solution

$\dot x=\ddot x=0$ , so the first equation holds. $\dot y=\ddot y=y_0e^t=y$ , so $\ddot y-\dot y^2/y=y-y=0$ . The speed is $|\dot y|/y=1$ . From $(0,1)$ to $(0,b)$ takes time $\log b$ ; the distance formula agrees: $\cosh d=1+\frac{(b-1)^2}{2b}=\frac{b+b^{-1}}2$ , so $d=\log b$ .

Area of a hyperbolic discVolume in polar coordinates

In geodesic polar coordinates about any point of $\mathbb H$ the metric is $dr^2+\sinh^2r\,d\theta^2$ . Find the area and circumference of the disc of radius $L$ , and compare with the plane.

Solution

The area form is $\sinh r\,dr\wedge d\theta$ , so the area is $2\pi(\cosh L-1)$ and the circumference $2\pi\sinh L$ . Both grow like $e^L$ , against $\pi L^2$ and $2\pi L$ in the plane. For small $L$ , $2\pi\sinh L\approx2\pi L(1+L^2/6)$ : negative curvature makes circles longer than flat ones.

More practice: Tu, Problems 14.2, 14.3 and 16.5.

InteractiveTu, Ch. 3 (§17): The Gauss–Bonnet Theorem

Gauss–Bonnet

Push a dent into a sphere and the curvature moves around, some of it turning negative. The total never changes: $4\pi$ for anything shaped like a sphere, $0$ for a torus. On a triangulated surface the theorem becomes exact arithmetic with angles, which you can check here.

The idea

Geodesic curvature. For a unit-speed curve on an oriented surface,

D_s\gamma'=\kappa_g\,J\gamma'

, where

J

is rotation by

+90^\circ

in the tangent plane.

\kappa_g

measures turning within the surface;

\gamma

is a geodesic exactly when

\kappa_g=0

.

Gauss–Bonnet for a polygon. Let

D

be a region homeomorphic to a closed disc whose boundary is piecewise smooth, traversed with

D

on the left, with exterior (turning) angles

\varepsilon_1,\dots,\varepsilon_k

at the corners. Then

\iint_DK\,dA+\int_{\partial D}\kappa_g\,ds+\sum_i\varepsilon_i=2\pi .

For a geodesic triangle with interior angles

\alpha_i

:

\ \alpha_1+\alpha_2+\alpha_3=\pi+\iint_DK\,dA

.

Gauss–Bonnet for a surface. For a compact oriented surface

M

,

\iint_MK\,dA=2\pi\chi(M),\qquad \chi=V-E+F

for any triangulation, and

\chi=2-2g

for genus

g

.

Discrete version (Descartes). For a closed surface made of flat triangles, let the angle defect at a vertex be

\delta_v=2\pi-(\text{sum of the angles at }v)

. Then

\sum_v\delta_v=2\pi\chi

exactly. The defect plays the role of

K\,dA

concentrated at the vertex.

Explore

Drag to turn. Colour shows the angle defect per unit area: positive negative.

Surface

Deformation

Triangulation

Worked example

Prove the discrete Gauss–Bonnet theorem: for a closed triangulated surface built from flat triangles, $\sum_v\delta_v=2\pi(V-E+F)$ .

What is the sum of all the angles of all the triangles?

Each flat triangle has angle sum $\pi$ , so the total is $\pi F$ .
How are $E$ and $F$ related on a closed surface?

Each triangle has 3 edges and each edge borders exactly 2 triangles, so $3F=2E$ .
Add up the defects.

Every angle sits at exactly one vertex, so

$\sum_v\delta_v=2\pi V-\pi F=2\pi V-2\pi E+2\pi F=2\pi\chi,$

using $\pi F=2\pi E-2\pi F$ from step 2.
How does this relate to the smooth theorem?

Inscribe finer and finer triangulations in a smooth surface. Near each vertex the defect approximates $\iint K\,dA$ over the region closest to it (a geodesic triangle with angle sum $\pi+\iint K\,dA$ flattened onto a flat one), so $\iint_MK\,dA$ is a limit of sums that all equal $2\pi\chi$ . This is a heuristic; the book's proof uses the polygon formula, which comes from Stokes' theorem applied to $\omega^1_2$ .

Check your understanding

A geodesic triangle on a surface with $K\equiv-1$ has area $0.5$ . What is its angle sum?

$\pi-0.5$
$\pi+0.5$ The angle sum is $\pi+\iint K\,dA$ , and here $K$ is negative.
$\pi$ Only when $\iint K\,dA=0$ .

Hyperbolic triangles are thin: their angles fall short of

\pi

by their area. In particular every hyperbolic triangle has area less than

\pi

.

You push a deep dent into a sphere, creating a region where $K<0$ . What is $\iint K\,dA$ now?

Still $4\pi$
Less than $4\pi$ New positive curvature appears too, around the rim of the dent, and it compensates exactly.
It depends on how deep the dent is. $\iint K\,dA=2\pi\chi$ , and a deformation doesn't change $\chi$ .

In the explorer, the positive and negative defect totals both change as you deform, but their sum stays

2\pi\chi

.

A compact orientable surface has $\iint K\,dA=-8\pi$ . What is its genus?

3
4Solve $2\pi(2-2g)=-8\pi$ .
2Genus 2 gives $\chi=-2$ , so $\iint K\,dA=-4\pi$ .

2-2g=-4

, so

g=3

. The total curvature determines the topology.

Practice

No positively curved torusUsing the total

Can a metric on the torus have $K>0$ everywhere? Can a torus in $\R^3$ have $K\equiv0$ ?

Solution

No to both. $\iint K\,dA=2\pi\cdot0=0$ , so $K$ cannot be positive everywhere. A torus in $\R^3$ is compact, so it has a point with $K>0$ (Theorema Egregium module, first practice problem), and then it must also have points with $K<0$ . Flat tori exist, but only as abstract surfaces or in $\R^4$ .

Spherical polygonsThe polygon formula

Show that a geodesic $n$ -gon on the unit sphere with interior angles $\alpha_1,\dots,\alpha_n$ has area $\sum\alpha_i-(n-2)\pi$ .

Solution

The sides are geodesics, so $\kappa_g=0$ , and the exterior angles are $\varepsilon_i=\pi-\alpha_i$ . Gauss–Bonnet with $K=1$ gives $\text{Area}+n\pi-\sum\alpha_i=2\pi$ .

More practice: Tu, Problems 17.1 and 17.4.

InteractiveTu, Ch. 4 (§§18–21): Tools from Algebra and Topology

Tensors, exterior powers, vector-valued forms

Before characteristic classes, some algebra: tensor products, exterior powers, and forms whose values are vectors or matrices. One fact carries a lot of weight later: on the top exterior power, a linear map acts as multiplication by its determinant.

The idea

Tensor product.

V\otimes W

comes with a bilinear map

(v,w)\mapsto v\otimes w

such that every bilinear

\beta:V\times W\to Z

factors as

\beta(v,w)=\tilde\beta(v\otimes w)

for a unique linear

\tilde\beta

. If

e_i

and

f_j

are bases, the

e_i\otimes f_j

are a basis:

\dim V\otimes W=(\dim V)(\dim W)

.

Exterior power.

\Lambda^kV

does the same for alternating

k

-linear maps. The products

e_{i_1}\wedge\dots\wedge e_{i_k}

with

i_1<\dots<i_k

form a basis, so

\dim\Lambda^kV=\binom nk

. A linear map

A

induces

\Lambda^kA(v_1\wedge\dots\wedge v_k)=Av_1\wedge\dots\wedge Av_k

.

Top degree.

\Lambda^nV

is one-dimensional, and

\Lambda^nA

is multiplication by

\det A

.

Operations on bundles. Natural constructions on vector spaces work fibre by fibre:

E\oplus F

,

E\otimes F

,

\mathrm{Hom}(E,F)

,

E^\vee

,

\Lambda^kE

. For

f:M\to N

and

E\to N

, the pullback

f^*E

has fibre

(f^*E)_p=E_{f(p)}

.

Vector-valued forms. A

k

-form with values in

V

is a section of

\Lambda^kT^*M\otimes V

. For matrix-valued forms,

(\alpha\wedge\beta)^i_j=\sum_k\alpha^i_k\wedge\beta^k_j

. For forms with values in a Lie algebra,

[\alpha,\beta](X,Y)=[\alpha(X),\beta(Y)]-[\alpha(Y),\beta(X)]

for 1-forms; for matrices,

[\alpha,\alpha]=2\,\alpha\wedge\alpha

.

Tensor criterion. A map

\Gamma(E_1)\times\dots\times\Gamma(E_k)\to\Gamma(F)

comes from a bundle map, so that its value at

p

depends only on the arguments at

p

, if and only if it is

C^\infty

-linear in each argument.

Explore

The linear map A

Drag the tips of v and w. The shaded parallelogram is

v\wedge w

: blue if positively oriented, gold if not.

Drag the columns

Ae_1

,

Ae_2

(hollow dots) to change

A

. The shaded region is

Av\wedge Aw

.

Worked example

Let $\omega=A\,dx+B\,dy$ be a $\mathfrak{gl}(2,\R)$ -valued 1-form on $\R^2$ with constant matrices $A,B$ . Compute $\omega\wedge\omega$ , and the curvature of the connection $d+\omega$ on the trivial bundle $\R^2\times\R^2$ .

Express $\omega\wedge\omega$ in terms of $A$ and $B$ .

Matrix entries multiply in order, and $dx\wedge dx=dy\wedge dy=0$ :

$\omega\wedge\omega=AB\,dx\wedge dy+BA\,dy\wedge dx=[A,B]\,dx\wedge dy .$
Take $A=\begin{pmatrix}0&1\\0&0\end{pmatrix}$ , $B=\begin{pmatrix}0&0\\1&0\end{pmatrix}$ .

$AB=\begin{pmatrix}1&0\\0&0\end{pmatrix}$ , $BA=\begin{pmatrix}0&0\\0&1\end{pmatrix}$ , so $\omega\wedge\omega=\begin{pmatrix}1&0\\0&-1\end{pmatrix}dx\wedge dy\neq0$ .
What is the curvature $\Omega=d\omega+\omega\wedge\omega$ ?

The coefficients are constant, so $d\omega=0$ and $\Omega=\operatorname{diag}(1,-1)\,dx\wedge dy$ . A connection with constant coefficients can be curved, because moving in the $x$ and $y$ directions does not commute. For an ordinary 1-form, $\alpha\wedge\alpha=0$ always; for matrices it is a commutator.
Check $[\omega,\omega]=2\,\omega\wedge\omega$ here.

$[\omega,\omega](\partial_x,\partial_y)=[A,B]-[B,A]=2[A,B]$ , and $(\omega\wedge\omega)(\partial_x,\partial_y)=[A,B]$ .

Check your understanding

What are $\dim(\R^4\otimes\R^4)$ and $\dim\Lambda^2\R^4$ ?

16 and 6
8 and 6Dimensions of tensor products multiply: $4\cdot4$ .
16 and 12 $e_j\wedge e_i=-e_i\wedge e_j$ and $e_i\wedge e_i=0$ , so only pairs $i<j$ count.

\binom42=6

:

e_1\wedge e_2,\ e_1\wedge e_3,\ e_1\wedge e_4,\ e_2\wedge e_3,\ e_2\wedge e_4,\ e_3\wedge e_4

.

Is every element of $\Lambda^2\R^4$ of the form $v\wedge w$ ?

No: $e_1\wedge e_2+e_3\wedge e_4$ is not.
Yes, since such products span $\Lambda^2\R^4$ .They span it, but a sum of two of them needn't be a single one.

If

\xi=v\wedge w

then

\xi\wedge\xi=0

. But

(e_1\wedge e_2+e_3\wedge e_4)^{\wedge2}=2\,e_1\wedge e_2\wedge e_3\wedge e_4\neq0

. (This quadratic expression is the Pfaffian in disguise; see the module on Euler classes.)

In the explorer, $v\wedge w=2\,e_1\wedge e_2$ and $\det A=-3$ . What is $Av\wedge Aw$ ?

$-6\,e_1\wedge e_2$
$6\,e_1\wedge e_2$ A negative determinant reverses orientation, and the sign survives.
$-3\,e_1\wedge e_2$ $\Lambda^2A$ multiplies by $\det A$ ; it doesn't replace the value.

Av\wedge Aw=\Lambda^2A(v\wedge w)=\det A\cdot v\wedge w

, whatever

v

and

w

are. Drag them and the ratio in the explorer doesn't move.

Practice

The top exterior powerWhere the determinant comes from

Show that $Ae_1\wedge\dots\wedge Ae_n=\det(A)\,e_1\wedge\dots\wedge e_n$ .

Solution

Write $Ae_j=\sum_ia_{ij}e_i$ and expand by multilinearity. Terms with a repeated $e_i$ vanish, and the rest are indexed by permutations: $\sum_\sigma a_{\sigma(1)1}\cdots a_{\sigma(n)n}\,e_{\sigma(1)}\wedge\dots\wedge e_{\sigma(n)}$ . Reordering gives the sign $\operatorname{sgn}\sigma$ , and $\sum_\sigma\operatorname{sgn}\sigma\,a_{\sigma(1)1}\cdots a_{\sigma(n)n}=\det A$ .

Abelian connection matricesWhy surfaces have

\Omega=d\omega

Show that $\omega\wedge\omega=0$ for every $\mathfrak{so}(2)$ -valued 1-form $\omega=\begin{pmatrix}0&-\alpha\\\alpha&0\end{pmatrix}$ .

Solution

$\omega\wedge\omega=\begin{pmatrix}-\alpha\wedge\alpha&0\\0&-\alpha\wedge\alpha\end{pmatrix}=0$ . So for an orthonormal frame on a surface, $\Omega=d\omega$ , which is why $\Omega^1_2=d\omega^1_2$ . The same holds for 1×1 matrices (line bundles). Both Lie algebras are abelian.

More practice: Tu, Problems 18.4, 20.2 and 21.4.

Worked examplesTu, Ch. 5 (§22): Connections and Curvature Again

Change of frame and the Bianchi identities

Connection matrices live on one frame at a time. To get global objects you need to know how they change when the frame does. Curvature transforms cleanly, by conjugation; the connection picks up an extra term. That difference is why characteristic classes are built from $\Omega$ .

The idea

Change of frame. If

e'=ea

on

U

, where

e=[e_1\ \cdots\ e_r]

is a row of sections and

a:U\to GL(r,\R)

, then

\omega'=a^{-1}\omega a+a^{-1}da,\qquad \Omega'=a^{-1}\Omega a .

Bianchi identities. Second:

d\Omega=\Omega\wedge\omega-\omega\wedge\Omega

. First, for a torsion-free connection on

TM

with dual coframe

\theta

:

\Omega\wedge\theta=0

, equivalently

R(X,Y)Z+R(Y,Z)X+R(Z,X)Y=0

.

Symmetries. For the Riemannian connection,

\langle R(X,Y)Z,W\rangle

is skew in

X,Y

, skew in

Z,W

, and equal to

\langle R(Z,W)X,Y\rangle

.

Ricci and scalar curvature.

\operatorname{Ric}(Y,Z)=\operatorname{tr}\big(X\mapsto R(X,Y)Z\big)

, a symmetric 2-tensor; the scalar curvature is its trace,

\sum_i\operatorname{Ric}(e_i,e_i)

for an orthonormal frame.

Worked example: the transformation laws

Let $e'=ea$ . Derive $\omega'$ and $\Omega'$ , then the second Bianchi identity.

Write $\nabla e=e\,\omega$ and apply $\nabla$ to $e'=ea$ .

By the Leibniz rule entry by entry, $\nabla(ea)=(\nabla e)a+e\,da=e(\omega a+da)$ . Substituting $e=e'a^{-1}$ , $\nabla e'=e'\big(a^{-1}\omega a+a^{-1}da\big)$ . So $\omega'=a^{-1}\omega a+a^{-1}da$ .
Compute $d\omega'$ , using $d(a^{-1})=-a^{-1}(da)\,a^{-1}$ .

$d\omega'=-a^{-1}da\,a^{-1}\wedge\omega a+a^{-1}d\omega\,a-a^{-1}\omega\wedge da-a^{-1}da\,a^{-1}\wedge da .$

The third sign comes from moving $d$ past the 1-form $\omega$ .
Compute $\omega'\wedge\omega'$ and add.

$\omega'\wedge\omega'=a^{-1}\omega\wedge\omega\,a+a^{-1}\omega\wedge da+a^{-1}da\,a^{-1}\wedge\omega a+a^{-1}da\,a^{-1}\wedge da .$

Adding, every term containing $da$ cancels: $\Omega'=a^{-1}(d\omega+\omega\wedge\omega)a=a^{-1}\Omega a$ .
Apply $d$ to $\Omega=d\omega+\omega\wedge\omega$ .

$d\Omega=d\omega\wedge\omega-\omega\wedge d\omega$ . Substitute $d\omega=\Omega-\omega\wedge\omega$ : the cubic terms cancel and

$d\Omega=\Omega\wedge\omega-\omega\wedge\Omega .$

Worked example: Ricci curvature of a sphere

On the unit sphere $S^n$ , $R(X,Y)Z=\langle Y,Z\rangle X-\langle X,Z\rangle Y$ . Find the sectional, Ricci and scalar curvatures.

Sectional curvature of the plane spanned by orthonormal $X,Y$ ?

$\langle R(X,Y)Y,X\rangle=\langle Y,Y\rangle\langle X,X\rangle-\langle X,Y\rangle^2=1$ , for every plane.
Compute $\operatorname{Ric}(Y,Z)$ with an orthonormal basis $e_1,\dots,e_n$ .

$\operatorname{Ric}(Y,Z)=\sum_i\langle R(e_i,Y)Z,e_i\rangle=\sum_i\big(\langle Y,Z\rangle-\langle e_i,Z\rangle\langle Y,e_i\rangle\big)=(n-1)\langle Y,Z\rangle .$
Scalar curvature?

$\sum_j\operatorname{Ric}(e_j,e_j)=n(n-1)$ . For $S^2$ this is $2=2K$ , and for $S^1$ it is $0$ : a curve has no intrinsic curvature.

Check your understanding

Under a change of frame $e'=ea$ , which statement is right?

$\Omega'=a^{-1}\Omega a$ , but $\omega'=a^{-1}\omega a+a^{-1}da$ .
Both transform by conjugation.The Leibniz rule produces an extra $a^{-1}da$ in $\omega'$ .
$\Omega'=\Omega$ .That holds for rank 1, where matrices commute. In general $\Omega$ is conjugated.

So

\Omega

is an

\mathrm{End}(E)

-valued 2-form, a global object, while

\omega

is not. Any polynomial

P

with

P(a^{-1}Xa)=P(X)

gives a global form

P(\Omega)

.

For a line bundle (rank 1), what does the second Bianchi identity say?

$d\Omega=0$
$\Omega=0$ Line bundles can be curved; see the Chern–Weil module.
$d\omega=0$ For rank 1, $d\omega=\Omega$ , which needn't vanish.

For

1\times1

matrices,

\Omega\wedge\omega=\omega\wedge\Omega

(a 2-form and a 1-form commute), so the right side vanishes. With

\Omega=d\omega

this is just

d^2=0

.

What is the scalar curvature of a round sphere of radius $a$ in $\R^3$ ?

$2/a^2$
$1/a^2$ That's $K$ . In dimension 2 the scalar curvature is $2K$ .
$2/a$ Curvature scales like 1/length².

Scaling the metric by

a^2

divides every curvature by

a^2

, so

n(n-1)

becomes

n(n-1)/a^2

.

Practice

The trace of the curvature is closedA first characteristic form

Show that $\operatorname{tr}\Omega$ is a well-defined closed 2-form on $M$ , independent of the frame.

Solution

$\operatorname{tr}(a^{-1}\Omega a)=\operatorname{tr}\Omega$ , so the local forms agree on overlaps. By Bianchi, $d\operatorname{tr}\Omega=\operatorname{tr}(\Omega\wedge\omega)-\operatorname{tr}(\omega\wedge\Omega)$ , and $\operatorname{tr}(\Omega\wedge\omega)=\sum_{i,k}\Omega^i_k\wedge\omega^k_i=\sum_{i,k}\omega^k_i\wedge\Omega^i_k=\operatorname{tr}(\omega\wedge\Omega)$ , since even-degree forms commute with everything.

Ricci curvature of a surfaceEverything is K

On a surface, show $R(X,Y)Z=K\big(\langle Y,Z\rangle X-\langle X,Z\rangle Y\big)$ , $\operatorname{Ric}=K\,g$ and scalar curvature $2K$ .

Hint

Use the symmetries: on a 2-dimensional space, how many independent values can $\langle R(e_i,e_j)e_k,e_l\rangle$ have?

Solution

By the skew-symmetries, $\langle R(e_i,e_j)e_k,e_l\rangle$ is zero unless $\{i,j\}=\{k,l\}=\{1,2\}$ , and all those values are $\pm\langle R(e_1,e_2)e_2,e_1\rangle=\pm K$ . The right-hand side has the same symmetries and the same value on $(e_1,e_2,e_2,e_1)$ , so they agree. Then the sphere computation with $n=2$ , scaled by $K$ , gives $\operatorname{Ric}=Kg$ and scalar curvature $2K$ .

More practice: Tu, Problems 22.1 and 22.3.

InteractiveTu, Ch. 5 (§23): Characteristic Classes

Chern–Weil: classes from curvature

Put the curvature matrix into a polynomial that doesn't care about the frame, and you get a closed differential form. Change the connection and the form changes, but only by an exact form, so its integral stays put. Here that integral is a whole number, and you can push the curvature around without moving it.

The idea

Invariant polynomial. A polynomial

P

on

\mathfrak{gl}(r)

with

P(a^{-1}Xa)=P(X)

for all invertible

a

. Examples:

\operatorname{tr}(X^k)

, and the coefficients

f_k

in

\det(I+tX)=\sum_kf_k(X)\,t^k

(so

f_1=\operatorname{tr}

,

f_r=\det

).

Chern–Weil. For a connection on

E

with curvature

\Omega

:
(i)

P(\Omega)

is a well-defined global form, since

\Omega'=a^{-1}\Omega a

;
(ii) it is closed, by the second Bianchi identity;
(iii) its de Rham class does not depend on the connection: for two connections,

P(\Omega_1)-P(\Omega_0)

is exact (join them by

\nabla_t=(1-t)\nabla_0+t\nabla_1

and integrate in

t

).
So

P\mapsto[P(\Omega)]

is an algebra homomorphism into

H^*(M)

, natural under pullback.

First Chern class of a line bundle. A complex line bundle with a Hermitian metric and a metric connection has, in a unit frame,

\omega=-i\,a

with

a

a real 1-form; then

\Omega=-i\,da

, and the first Chern form is

c_1=\tfrac{i}{2\pi}\Omega=\tfrac1{2\pi}da

.

The example below. On $S^2$ (polar angle $\theta$ , azimuth $\varphi$ ) take $a_N=A(\theta)\,d\varphi$ on $S^2\smallsetminus\{S\}$ and $a_S=a_N-n\,d\varphi$ on $S^2\smallsetminus\{N\}$ , with $A(0)=0$ and $A(\pi)=n$ so that each is smooth at its pole. They describe one connection in two unit frames related by $e^{in\varphi}$ , which is a function on the overlap only when $n$ is an integer. The explorer starts from $A=\tfrac n2(1-\cos\theta)$ and adds a bump $\beta h(\theta)$ with $h(0)=h(\pi)=0$ .

Explore

n Bump strength β Bump position θ

Density of

c_1

on

S^2

: positive negative. Drag to turn.

Against

\theta

from N to S:

4\pi\times

density (equals

n

if spread evenly)

\int c_1

over the cap above

\theta

Worked example

For the connection above, compute $\int_{S^2}c_1$ without knowing $A$ in between the poles.

Split the sphere. On which region is each of $a_N$ , $a_S$ defined?

Use the closed northern hemisphere $D_N$ , where $a_N$ is smooth, and the closed southern hemisphere $D_S$ , where $a_S$ is smooth. On each, $c_1=\tfrac1{2\pi}da_N$ or $\tfrac1{2\pi}da_S$ .
Apply Stokes on each hemisphere. How is the equator oriented each time?

With the outward normal, the boundary of $D_N$ is the equator with $\varphi$ increasing; the boundary of $D_S$ is the equator with $\varphi$ decreasing. So

$2\pi\int_{S^2}c_1=\oint_{\varphi\uparrow}a_N-\oint_{\varphi\uparrow}a_S=\oint_{\varphi\uparrow}(a_N-a_S).$
Finish.

$a_N-a_S=n\,d\varphi$ , so $\int_{S^2}c_1=\frac1{2\pi}\oint n\,d\varphi=n$ . Only the gluing entered: the interior of each hemisphere, and therefore the bump, played no part.
Why must $n$ be an integer?

The two unit frames differ by the function $e^{in\varphi}$ on the overlap, and that is single-valued around the equator only if $e^{2\pi in}=1$ . So for a genuine bundle, $\int c_1\in\Z$ . Line bundles over $S^2$ are classified by this integer; $n=\pm1$ gives the tautological bundle over $\C P^1=S^2$ and its dual.

Check your understanding

Which of these is an invariant polynomial on $\mathfrak{gl}(2)$ ?

$\operatorname{tr}(X^2)$
$X_{11}$ , the top-left entryConjugating by a permutation matrix swaps $X_{11}$ and $X_{22}$ .
The sum of all four entriesConjugating by $\operatorname{diag}(1,2)$ changes the off-diagonal entries, so the sum changes.

\operatorname{tr}\big((a^{-1}Xa)^2\big)=\operatorname{tr}(a^{-1}X^2a)=\operatorname{tr}(X^2)

. Invariant polynomials only see the eigenvalues.

Two connections on the same line bundle over $S^2$ have different curvature forms. Their first Chern forms…

differ by an exact form, so they have the same integral
are equalMove the bump in the explorer: the density changes pointwise.
can have different integralsThe integral is determined by the gluing alone, as the worked example shows.

The connection forms differ by a global real 1-form

\eta

(here

\beta h(\theta)\,d\varphi

), so the Chern forms differ by

\tfrac1{2\pi}d\eta

, whose integral over

S^2

is 0.

Why is $P(\Omega)$ a closed form?

The second Bianchi identity $d\Omega=\Omega\wedge\omega-\omega\wedge\Omega$ , plus invariance of $P$
Because $\Omega=d\omega$ is exact.That's true only in rank 1, and even then $\omega$ is local, so it shows only local exactness.
Every 2-form on a surface is closed, and this generalises.Top-degree forms are closed for degree reasons, but $P(\Omega)$ usually isn't of top degree.

Writing

P

as a symmetric multilinear form,

dP(\Omega)=\sum P(\Omega,\dots,d\Omega,\dots,\Omega)

. Bianchi replaces each

d\Omega

by

\Omega\wedge\omega-\omega\wedge\Omega

, and the sum then vanishes by the infinitesimal form of invariance,

\frac{d}{dt}P(e^{-tA}Xe^{tA})=0

.

Practice

Invariant polynomials in rank 2Trace and determinant

Show that $\det(I+tX)=1+t\operatorname{tr}X+t^2\det X$ for $2\times2$ matrices, and deduce that the coefficients are invariant.

Solution

$\det\begin{pmatrix}1+tx_{11}&tx_{12}\\tx_{21}&1+tx_{22}\end{pmatrix}=1+t(x_{11}+x_{22})+t^2(x_{11}x_{22}-x_{12}x_{21})$ . Since $\det(I+ta^{-1}Xa)=\det\big(a^{-1}(I+tX)a\big)=\det(I+tX)$ for all $t$ , each coefficient is invariant.

The tangent bundle of the sphere is not trivialCurvature as an obstruction

Use connection forms to show that $TS^2$ has no global frame.

Hint

A global frame could be made orthonormal. What would $\omega^1_2$ be then?

Solution

Gram–Schmidt turns a global frame into a global orthonormal frame, whose $\omega^1_2$ is a global 1-form with $d\omega^1_2=K\,dA$ . By Stokes on the closed surface $S^2$ , $\iint K\,dA=\int_{S^2}d\omega^1_2=0$ . For the round metric $\iint K\,dA=4\pi$ . Contradiction.

More practice: Tu, Problems 23.1 and 23.2.

InteractiveTu, Ch. 5 (§§24–26): Pontrjagin Classes · The Euler Class and Chern Classes · Some Applications

Pontrjagin, Euler and Chern classes

For a real bundle with a metric, the curvature matrix is skew-symmetric. That kills half the Chern–Weil classes and adds a new one: the Pfaffian, a square root of the determinant that needs an orientation. Its integral over a manifold is the Euler characteristic.

The idea

Vanishing. For a metric connection on a real bundle,

\Omega

is skew in an orthonormal frame, and

f_k(\Omega)=0

for odd

k

:

\det(I+tX)=\det(I+tX)^{\mathsf T}=\det(I-tX)

when

X^{\mathsf T}=-X

.

Pontrjagin classes.

p_k(E)=\big[f_{2k}(\Omega/2\pi)\big]\in H^{4k}(M)

(Tu writes

f_{2k}\big(\tfrac{i}{2\pi}\Omega\big)

, which differs from this by

(-1)^k

; the convention here is Milnor–Stasheff's, the one under which

p_1=e^2

for an oriented plane bundle and the signature formula below hold). Whitney product formula:

p(E\oplus F)=p(E)\,p(F)

in de Rham cohomology.

Pfaffian. For a skew

2k\times2k

matrix,

\operatorname{Pf}(X)

is a polynomial with

\operatorname{Pf}(X)^2=\det X

and

\operatorname{Pf}(a^{\mathsf T}Xa)=\det(a)\operatorname{Pf}(X)

. In sizes 2 and 4:

\operatorname{Pf}\begin{pmatrix}0&x\\-x&0\end{pmatrix}=x,\qquad \operatorname{Pf}(X)=x_{12}x_{34}-x_{13}x_{24}+x_{14}x_{23}.

Euler class. For an oriented rank-

2k

bundle with a metric connection, use oriented orthonormal frames; changes of frame then lie in

SO(2k)

, so

\operatorname{Pf}(\Omega/2\pi)

is a global closed form, and

e(E)

is its class. Reversing the orientation changes its sign. For a surface,

\operatorname{Pf}(\Omega/2\pi)=\Omega^1_2/2\pi=K\,dA/2\pi

.

Generalised Gauss–Bonnet (Chern). For a compact oriented Riemannian manifold of dimension

2k

,

\displaystyle\int_M\operatorname{Pf}(\Omega/2\pi)=\chi(M)

.

Chern classes. For a complex bundle with a Hermitian metric connection,

c(E)=\big[\det\big(I+\tfrac{i}{2\pi}\Omega\big)\big]=1+c_1+c_2+\cdots

,

c_k\in H^{2k}(M)

.

Integrals of products of these classes of $TM$ over $M$ , the characteristic numbers, are diffeomorphism invariants. Pontrjagin numbers vanish when $M$ is the boundary of a compact oriented manifold, which makes them cobordism invariants; Hirzebruch's signature theorem expresses the signature of a 4-manifold as $\tfrac13\int_Mp_1$ .

Explore

The entries of a curvature matrix are even-degree forms, so they commute, and polynomial identities can be checked with numbers. Top:

\Omega

in frame

e

and in frame

e'=ea

. Bottom: coefficients of

\det(I+t\,\Omega)

.

Entries of Ω above the diagonal

Worked example

Verify $\int e(TM)=\chi(M)=0$ directly for the torus $X(u,v)=\big((R+r\cos v)\cos u,\ (R+r\cos v)\sin u,\ r\sin v\big)$ , $0<r<R$ .

Find an orthonormal coframe.

The metric is $(R+r\cos v)^2du^2+r^2dv^2$ , so $\theta^1=(R+r\cos v)\,du$ , $\theta^2=r\,dv$ .
Find $\omega^1_2$ .

$d\theta^1=r\sin v\,du\wedge dv$ and $d\theta^2=0$ . From $d\theta^2=\omega^1_2\wedge\theta^1=0$ , $\omega^1_2=g\,du$ ; then $d\theta^1=-\omega^1_2\wedge\theta^2=-gr\,du\wedge dv$ gives $g=-\sin v$ . So $\omega^1_2=-\sin v\,du$ .
Compute the Euler form and $K$ .

$\Omega^1_2=d\omega^1_2=\cos v\,du\wedge dv$ . Since $\theta^1\wedge\theta^2=r(R+r\cos v)\,du\wedge dv$ ,

$K=\frac{\cos v}{r(R+r\cos v)},\qquad \operatorname{Pf}(\Omega/2\pi)=\frac{\cos v}{2\pi}\,du\wedge dv .$
Integrate.

$\displaystyle\frac1{2\pi}\int_0^{2\pi}\!\!\int_0^{2\pi}\cos v\,du\,dv=0=\chi(T^2)$ . The outer half ( $|v|<\pi/2$ , $K>0$ ) cancels the inner half exactly. The Torus option in the Gauss–Bonnet module shows the same cancellation on a triangulation.

Check your understanding

Why is the Euler class defined only for oriented bundles?

$\operatorname{Pf}(a^{\mathsf T}Xa)=\det(a)\operatorname{Pf}(X)$ , so an orientation-reversing frame change flips its sign.
Because the Pfaffian needs complex numbers.The Pfaffian is a polynomial with integer coefficients in the real entries.
Because $\Omega$ is skew only in oriented frames.Every orthonormal frame makes $\Omega$ skew. Orientation decides the sign of $\operatorname{Pf}$ .

Press "Reflect e₁" in the explorer:

\det

and

f_2

are unchanged,

\operatorname{Pf}

changes sign. Restricting to oriented frames makes the sign well defined.

For a metric connection on a real bundle, why does $f_1(\Omega)=\operatorname{tr}\Omega$ vanish?

$\Omega$ is skew-symmetric in an orthonormal frame, so its diagonal is zero.
By the Bianchi identity.Bianchi makes $f_1(\Omega)$ closed, not zero.
Because every real bundle is trivial.The Möbius band and $TS^2$ are not.

All odd

f_k

vanish because

\det(I+tX)=\det(I-tX)

for skew

X

. That's why real bundles have classes only in degrees

4k

(Pontrjagin), plus the Euler class.

$\chi(S^4)=2$ . For an arbitrary Riemannian metric on $S^4$ , what is $\int_{S^4}\operatorname{Pf}(\Omega/2\pi)$ ?

$2$
$0$ , because $S^4$ bounds a ball.Bounding kills Pontrjagin numbers, not the Euler characteristic: $S^4$ bounds and still has $\chi=2$ .
It depends on the metric.Chern–Weil: the class doesn't depend on the connection, so neither does the integral.

Generalised Gauss–Bonnet. The integrand changes with the metric; the integral cannot.

Practice

Block-diagonal curvatureWhitney for the Euler class

Let $X=\begin{pmatrix}0&a&0&0\\-a&0&0&0\\0&0&0&b\\0&0&-b&0\end{pmatrix}$ . Compute $\operatorname{Pf}X$ , $\det X$ and $f_2(X)$ . What does this say about $e(E_1\oplus E_2)$ for oriented plane bundles?

Solution

$\operatorname{Pf}X=ab$ , $\det X=a^2b^2$ , $f_2(X)=a^2+b^2$ . A direct sum of connections has block-diagonal curvature, so $e(E_1\oplus E_2)=e(E_1)\,e(E_2)$ and $p_1(E_1\oplus E_2)=p_1(E_1)+p_1(E_2)$ , and for a plane bundle $p_1=e^2$ . Check in the explorer: set every entry except $x_{12}$ and $x_{34}$ to 0.

Odd classes of real bundlesA one-line proof

Show that $\det(I+tX)$ is an even function of $t$ when $X$ is skew-symmetric.

Solution

$\det(I+tX)=\det\big((I+tX)^{\mathsf T}\big)=\det(I+tX^{\mathsf T})=\det(I-tX)$ .

More practice: Tu, Problems 25.2 and 25.3.

Draw it yourselfTu, Ch. 6 (§§27–32): Principal Bundles and Characteristic Classes

Principal bundles and horizontal lifts

A principal bundle puts a copy of a group over each point, and a connection says which directions are horizontal. Lift a loop horizontally and you may come back to a different point of the same fibre. The gap is the holonomy, and for a circle bundle it is the curvature you enclosed. Draw a loop and see.

The idea

Principal bundle. A map

\pi:P\to M

with a free right action of a Lie group

G

on

P

whose orbits are the fibres, locally isomorphic to

U\times G

compatibly with the action. Example: the frame bundle

\mathrm{Fr}(E)

, whose fibre over

p

is the set of ordered bases of

E_p

, with

G=GL(r,\R)

acting by

e\mapsto ea

.

Vertical vectors.

V_u=\ker d\pi_u

, spanned by the fundamental vector fields

A^\#_u=\frac{d}{dt}\big|_0\,u\cdot\exp(tA)

,

A\in\mathfrak g

.

Connection. A smooth right-invariant choice of horizontal subspaces

H_u

with

T_uP=V_u\oplus H_u

. Equivalently, a

\mathfrak g

-valued 1-form

\omega

with

\omega(A^\#)=A

and

R_g^*\omega=\mathrm{Ad}(g^{-1})\,\omega

; then

H=\ker\omega

.

Horizontal lifts and holonomy. A curve

\gamma

in

M

and a point

u

over

\gamma(0)

have a unique horizontal lift

\tilde\gamma

starting at

u

. If

\gamma

is a loop,

\tilde\gamma(1)=u\cdot g

for some

g\in G

, the holonomy. For a connection on

\mathrm{Fr}(E)

, lifting is parallel transport of frames.

Curvature.

\Omega=d\omega+\tfrac12[\omega,\omega]

. For horizontal

X,Y

,

\ \Omega(X,Y)=-\omega([X,Y])

: curvature measures how far the horizontal distribution is from being integrable. Pulled back by a local section,

\omega

and

\Omega

become the connection and curvature matrices of the earlier modules.

Chern–Weil on $P$ . An

\mathrm{Ad}

-invariant polynomial

f

on

\mathfrak g

gives a form

f(\Omega)

on

P

that descends to a closed form on

M

, with class independent of

\omega

.

The example below. $P=\R^2\times S^1$ with fibre coordinate $\psi$ , $G=U(1)$ , $\mathfrak g=\R$ , and $\omega=d\psi+\pi^*a$ for a 1-form $a$ on the plane. Horizontal means $d\psi=-a(\gamma')$ , so the lift of a loop rises by $\Delta\psi=-\oint_\gamma a=-\iint da$ , each region counted as many times as the loop winds around it. Here $\Omega=\pi^*da$ .

Explore

Curvature da

Or try a ready-made loop

The base

M=\R^2

, shaded by

da

. Draw a loop; it closes itself when you let go.

P

, with the fibre direction

\psi

drawn upward. horizontal lift holonomy gap. Drag to turn.

Worked example

On $P=\R^2\times S^1$ with $\omega=d\psi+a$ , $a=a_x\,dx+a_y\,dy$ , find the horizontal lifts of $\partial_x$ and $\partial_y$ , their bracket, and what it says about small loops.

Find the horizontal lift $X^h$ of $\partial_x$ .

Try $X^h=\partial_x+c\,\partial_\psi$ . Then $\omega(X^h)=c+a_x=0$ , so $X^h=\partial_x-a_x\partial_\psi$ . Likewise $Y^h=\partial_y-a_y\partial_\psi$ .
Compute $[X^h,Y^h]$ .

The coefficients depend only on $x,y$ , so $[X^h,Y^h]=-(\partial_xa_y)\partial_\psi+(\partial_ya_x)\partial_\psi=-da(\partial_x,\partial_y)\,\partial_\psi$ . The bracket of two horizontal fields is vertical.
Relate this to the curvature.

$\mathfrak g=\R$ is abelian, so $[\omega,\omega]=0$ and $\Omega=d\omega=\pi^*da$ . Then $\omega([X^h,Y^h])=-da(\partial_x,\partial_y)=-\Omega(X^h,Y^h)$ . By Frobenius, the horizontal distribution is integrable exactly where $da=0$ .
For uniform curvature $da=B\,dx\wedge dy$ , what happens around a small square of side $\varepsilon$ ?

Following $X^h,Y^h,-X^h,-Y^h$ for time $\varepsilon$ each gives a gap of about $\varepsilon^2[X^h,Y^h]=-B\varepsilon^2\partial_\psi$ , and exactly $-\oint a=-B\varepsilon^2$ by Stokes. This is the four-leg loop of the Lie bracket animation on the Smooth Manifolds page, one level up.

Check your understanding

With the flux tube, a loop goes once around the tube but stays where $da\approx0$ . What is the holonomy?

About $-\Phi$ (mod $2\pi$ ), set by the flux enclosed, not by the curvature along the loop
Zero, because there is no curvature on the loopStokes uses the curvature inside the loop, not on it.
It depends on the shape of the loop.Every loop that winds once around the tube encloses the same flux.

\Delta\psi=-\iint da

over the inside, which is

-\Phi

for any loop winding once around the tube. This is the geometry behind the Aharonov–Bohm effect.

What is a point of the frame bundle $\mathrm{Fr}(TM)$ over $p$ ?

An ordered basis of $T_pM$
A tangent vector at $p$ That's a point of $TM$ , a vector bundle, not a principal bundle.
A matrix in $GL(n,\R)$ Only after choosing a reference frame. The fibre is a copy of $GL(n,\R)$ with no preferred identity element.

GL(n,\R)

acts freely and transitively on the frames at

p

, so each fibre looks like the group but without a chosen origin. A local frame field is exactly a local section.

When is the horizontal distribution of a connection on $P$ integrable?

Exactly when $\Omega=0$
Always: it is a distribution.A distribution is integrable only if it is closed under the bracket (Frobenius).
When $G$ is abelian.In the explorer $G=U(1)$ is abelian, and lifts of loops still fail to close.

The vertical part of

[X,Y]

for horizontal

X,Y

is determined by

-\Omega(X,Y)

. A flat connection has integrable

H

, and then holonomy depends only on the homotopy class of the loop.

Practice

Fundamental vector fields are verticalUsing the free action

Show that $d\pi(A^\#)=0$ , and that $A^\#_u\neq0$ for every $u$ when $A\neq0$ .

Solution

$\pi(u\cdot\exp tA)=\pi(u)$ is constant, so $d\pi(A^\#_u)=0$ . For $c(t)=u\cdot\exp(tA)$ , $c'(t)=d(R_{\exp tA})(A^\#_u)$ . If $A^\#_u=0$ then $c$ is constant, so $u\cdot\exp(tA)=u$ and, by freeness, $\exp(tA)=e$ for all $t$ , which forces $A=0$ .

When does a lift close up?Uniform curvature

With $da=B\,dx\wedge dy$ , lift a circle of radius $\rho$ traversed once counterclockwise. For which $\rho$ does the lift return to its starting point in $P=\R^2\times S^1$ ?

Solution

$\Delta\psi=-B\pi\rho^2$ , and $\psi$ is an angle, so the lift closes exactly when $B\pi\rho^2\in2\pi\Z$ , that is $\rho^2\in(2/B)\,\Z$ .

More practice: Tu, Problems 27.1, 28.1 and 30.3.