Riem Geom 11

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    Riemannian Geometry

    it is a draft of Lecture Notes of H.M. Khudaverdian.Manchester, 20-th May 2011

    Contents

    1 Riemannian manifolds 41.1 Manifolds. Tensors. (Recalling) . . . . . . . . . . . . . . . . . 41.2 Riemannian manifoldmanifold equipped with Riemannian

    metric . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.2.1 Pseudoriemannian manifold . . . . . . . . . . . . . . 11

    1.3 Scalar product. Length of tangent vectors and angle betweenvectors. Length of the curve . . . . . . . . . . . . . . . . . . . 11

    1.3.1 Length of the curve . . . . . . . . . . . . . . . . . . . . 121.4 Riemannian structure on the surfaces embedded in Euclidean

    space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151.4.1 Internal and external coordinates of tangent vector . . 151.4.2 Explicit formulae for induced Riemannian metric (First

    Quadratic form) . . . . . . . . . . . . . . . . . . . . . . 171.4.3 Induced Riemannian metrics. Examples. . . . . . . . . 201.4.4 Induced metric on two-sheeted hyperboloid embedded

    in pseudo-Euclidean space. . . . . . . . . . . . . . . . . 271.5 Isometries of Riemanian manifolds. . . . . . . . . . . . . . . . 28

    1.5.1 Examples of local isometries . . . . . . . . . . . . . . . 291.6 Volume element in Riemannian manifold . . . . . . . . . . . . 311.6.1 Volume of parallelepiped . . . . . . . . . . . . . . . . . 311.6.2 Invariance of volume element under changing of coor-

    dinates . . . . . . . . . . . . . . . . . . . . . . . . . . . 331.6.3 Examples of calculating volume element . . . . . . . . 34

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    2 Covariant differentiaion. Connection. Levi Civita Connec-

    tion on Riemannian manifold 362.1 Differentiation of vector field along the vector field.Affine

    connection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 362.1.1 Definition of connection. Christoffel symbols of con-

    nection . . . . . . . . . . . . . . . . . . . . . . . . . . . 372.1.2 Transformation of Christoffel symbols for an arbitrary

    connection . . . . . . . . . . . . . . . . . . . . . . . . . 392.1.3 Canonical flat affine connection . . . . . . . . . . . . . 402.1.4 Global aspects of existence of connection . . . . . . . 43

    2.2 Connection induced on the surfaces . . . . . . . . . . . . . . . 44

    2.2.1 Calculation of induced connection on surfaces in E3

    . . . 452.3 Levi-Civita connection . . . . . . . . . . . . . . . . . . . . . . 472.3.1 Symmetric connection . . . . . . . . . . . . . . . . . . 472.3.2 Levi-Civita connection. Theorem and Explicit formulae 472.3.3 Levi-Civita connection on 2-dimensional Riemannian

    manifold with metric G = adu2 + bdv2. . . . . . . . . . 492.3.4 Example of the sphere again . . . . . . . . . . . . . . . 49

    2.4 Levi-Civita connection = induced connection on surfaces in E3 50

    3 Parallel transport and geodesics 523.1 Parallel transport . . . . . . . . . . . . . . . . . . . . . . . . . 52

    3.1.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . 523.1.2 Parallel transport is a linear map . . . . . . . . . . . 533.1.3 Parallel transport with respect to Levi-Civita connection 54

    3.2 Geodesics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 543.2.1 Definition. Geodesic on Riemannian manifold. . . . . . 543.2.2 Un-parameterised geodesic . . . . . . . . . . . . . . . . 553.2.3 Geodesics on surfaces in E3 . . . . . . . . . . . . . . . 56

    3.3 Geodesics and Lagrangians of free particle on Riemannianmanifold. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 583.3.1 Lagrangian and Euler-Lagrange equations . . . . . . . 58

    3.3.2 Lagrangian of free particle . . . . . . . . . . . . . . . 583.3.3 Equations of geodesics and Euler-Lagrange equations . 593.3.4 Examples of calculations of Christoffel symbols and

    geodesics using Lagrangians. . . . . . . . . . . . . . . . 603.3.5 Variational principe and Euler-Lagrange equations . . 62

    3.4 Geodesics and shortest distance. . . . . . . . . . . . . . . . . . 63

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    3.4.1 Again geodesics for sphere and Lobachevsky plane . . . 65

    4 Surfaces in E3 674.1 Formulation of the main result. Theorem of parallel transport

    over closed curve and Theorema Egregium . . . . . . . . . . . 674.1.1 GauTheorema Egregium . . . . . . . . . . . . . . . . . 69

    4.2 Derivation formulae . . . . . . . . . . . . . . . . . . . . . . . 704.2.1 Gauss condition (structure equations) . . . . . . . . . 72

    4.3 Geometrical meaning of derivation formulae. Weingarten op-erator and second quadratic form in terms of derivation for-mulae. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 734.3.1 Gaussian and mean curvature in terms of derivation

    formulae . . . . . . . . . . . . . . . . . . . . . . . . . . 754.4 Examples of calculations of derivation formulae and curvatures

    for cylinder, cone and sphere . . . . . . . . . . . . . . . . . . 764.5 Proof of the Theorem of parallel transport along closed curve. 80

    5 Curvtature tensor 845.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84

    5.1.1 Properties of curvature tensor . . . . . . . . . . . . . . 855.2 Riemann curvature tensor of Riemannian manifolds. . . . . . . 865.3 Curvature of surfaces in E3.. Theorema Egregium again . . . 87

    5.4 Relation between Gaussian curvature and Riemann curvaturetensor. Straightforward proof of Theorema Egregium . . . . . 885.4.1 Proof of the Proposition (5.25) . . . . . . . . . . . . . 90

    5.5 Gauss Bonnet Theorem . . . . . . . . . . . . . . . . . . . . . . 94

    6 Appendices 976.1 Integrals of motions and geodesics. . . . . . . . . . . . . . . . 97

    6.1.1 Integral of motion for arbitrary Lagrangian L(x, x) . . 976.1.2 Basic examples of Integrals of motion: Generalised

    momentum and Energy . . . . . . . . . . . . . . . . . . 97

    6.1.3 Integrals of motion for geodesics . . . . . . . . . . . . 986.1.4 Using integral of motions to calculate geodesics . . . . 1006.2 Induced metric on surfaces. . . . . . . . . . . . . . . . . . . . 101

    6.2.1 Recalling Weingarten operator . . . . . . . . . . . . . . 1016.2.2 Second quadratic form . . . . . . . . . . . . . . . . . . 1026.2.3 Gaussian and mean curvatures . . . . . . . . . . . . . . 103

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    6.2.4 Examples of calculation of Weingarten operator, Sec-

    ond quadratic forms, curvatures for cylinder, cone ands p h e r e . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 0 3

    1 Riemannian manifolds

    1.1 Manifolds. Tensors. (Recalling)

    I recall briefly basics of manifolds and tensor fields on manifolds.An n-dimensional manifold is a space such that in a vicinity of any point

    one can consider local coordinates

    {x1, . . . , xn

    }(charts). One can consider

    different local coordinates. If both coordinates {x1, . . . , xn}, {y1, . . . , yn} aredefined in a vicinity of the given point then they are related by bijectivetransition functions (functions defined on domains in Rn and taking valuesin Rn).

    x1

    = x1

    (x1, . . . , xn)

    x2

    = x2

    (x1, . . . , xn)

    . . .

    xn1

    = xn1

    (x1, . . . , xn)

    xn

    = xn

    (x1, . . . , xn)

    We say that manifold is differentiable or smooth if transition functions arediffeomorphisms, i.e. they are smooth and rank of Jacobian is equal to k, i.e.

    det

    x1

    x1x1

    x2. . . x

    1

    xn

    x2

    x1x2

    x2. . . x

    2

    xn

    . . .xn

    x1xn

    x2. . . x

    n

    xn

    = 0 (1.1)

    A good example of manifold is an open domain D in n-dimensional vectorspace Rn. Cartesian coordinates on Rn define global coordinates on D. Onthe other hand one can consider an arbitrary local coordinates in different

    domains in Rn.E.g. one can consider polar coordinates {r, } in a domain D = {x, y : y >

    0} of R2 (or in other domain of R2) defined by standard formulae:{x = r cos

    y = r sin ,

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    detxr

    x

    yr

    y = detcos r sin sin r cos = r (1.2)

    or one can consider spherical coordinates {r,,} in a domain D = {x,y,z: x >0, y > 0, z > 0} of R3 (or in other domain of R3) defined by standard for-mulae

    x = r sin cos

    y = r sin sin

    z = r cos

    ,

    ,

    det

    x

    r

    x

    x

    yr

    y

    y

    zr

    z

    z

    = detsin cos r cos cos r sin sin sin sin r cos sin r sin cos cos r sin 0

    = r2 sin (1.3)

    Choosing domain where polar (spherical) coordinates are well-defined wehave to be award that coordinates have to be well-defined and transitionfunctions (1.1) have to be diffeomorphisms.

    Examples of manifolds: Rn, Circle S1, Sphere S2, in general sphere Sn,torus S1 S1, cylinder, cone, . . . .

    We also have to recall briefly what are tensors on manifold.

    Tensors on ManifoldFor every point p on manifold M one can consider tangent vector space

    TpMthe space of vectors tangent to the manifold at the point M.Tangent vector A = Ai

    xi. Under changing of coordinates it transforms

    as follows:

    A = Ai

    xi= Ai

    xm

    xi

    xm= Am

    xm

    Hence

    Ai

    =xi

    xi Ai (1.4)

    Consider also cotangent space TpM (for every point p on manifold M)space of linear functions on tangent vectors, i.e. space of 1-forms whichsometimes are called covectors.:

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    One-form (covector) = idxi transforms as follows

    = mdxm = m

    xm

    xmdxm

    = mdxm .

    Hence

    m =xm

    xmm . (1.5)

    Tensors:

    One can consider contravariant tensors of the rank p

    T = Ti1i2...ip

    xi1

    xi2

    xik

    with components {Ti1i2...ik}.One can consider covariant tensors of the rank q

    S = Sj1j2...jqdxj1 dxj2 . . . d xjq

    with components {Sj1j2...jq}.One can also consider mixed tensors:

    Q = Qi1i2...ipj1j2...jq

    xi1

    xi2

    xik dxj1 dxj2 . . . d xjq

    with components {Qi1i2...ipj1j2...jq}. We call these tensors tensors of the typepq.Tensors of the type

    p0

    are called contravariant tensors of the rank p.

    Tensors of the type

    0q

    are called covariant tensors of the rank q.

    Having in mind (1.4) and (1.5) we come to the rule of transformation for

    tensors of the type

    pq

    :

    Q

    i1i2...ip

    j1j2...jq =

    xi

    1

    xi1xi

    2

    xi2 . . .

    xi

    p

    xipxj1

    xj1

    xj2

    xj2 . . .

    xjq

    xjq Q

    i1i2...ip

    j1j2...jq (1.6)

    E.g. if Sik is a covariant tensor of rank 2 (tensor of the type

    pq

    ) then

    Sik =xi

    xixk

    xkSik . (1.7)

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    If Aik is a tensor of rank 11 (linear operator on TpM) thenAi

    k =xi

    xixk

    xkAik

    Remark Transformations formulae (1.4)(1.7) define vectors, covectorsand in generally any tensor fields in components. E.g. covariant tensor (co-variant tensor field) of the rank 2 can be defined as matrix Sik (matrix valuedfunction Sik(x)) such that under changing of coordinates {x1, x2, . . . , xn} {x1, x2 , . . . , xn} Sik change by the rule (1.7).

    Remark Einstein summation rulesIn our lectures we always use so called Einstein summation convention. it

    implies that when an index occurs more than once in the same expression inupper and in.. postitions.., the expression is implicitly summed over all pos-sible values for that index. Sometimes it is called dummy indices summationrule.

    1.2 Riemannian manifoldmanifold equipped with Rie-mannian metric

    Definition The Riemannian manifold is a manifold equipped with a Rie-mannian metric.

    The Riemannian metric on the manifold M defines the length of thetangent vectors and the length of the curves.

    Definition Riemannian metric G on n-dimensional manifold Mn definesfor every point p M the scalar product of tangent vectors in the tangentspace TpM smoothly depending on the point p.

    It means that in every coordinate system (x1, . . . , xn) a metric G =gikdx

    idxk is defined by a matrix valued smooth function gik(x) (i = 1, . . . , n; k =1, . . . n) such that for any two vectors

    A = Ai(x)

    x

    i, B = Bi(x)

    x

    i,

    tangent to the manifold M at the point p with coordinates x = (x1, x2, . . . , xn)(A, B TpM) the scalar product is equal to:

    A, BGp

    = G(A, B)p

    = Ai(x)gik(x)Bk(x) =

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    (A1 . . . An

    )g11(x) . . . g1n(x)

    . . . . . . . . .gn1(x) . . . gnn(x)

    B1

    Bn

    (1.8)where

    G(A, B) = G(B, A), i.e. gik(x) = gki(x) (symmetricity condition) G(A, A) > 0 if A = 0, i.e.

    gik(x)uiuk 0, gik(x)uiuk = 0 i ff u1 = = un = 0 (positive-

    definiteness) G(A, B)

    p=x, i.e. gik(x) are smooth functions.

    One can say that Riemannian metric is defined by symmetric covariantsmooth tensor fieldG of the rank 2 which defines scalar product in the tangentspaces TpM smoothly depending on the point p. Components of tensor fieldG in coordinate system are matrix valued functions gik(x):

    G = gik(x)dxi dxk . (1.9)

    The matrix ||gik|| of components of the metric G we also sometimes denoteby G.

    Rule of transformation for entries of matrix gik(x)gik(x)-entries of the matrix ||gik|| are components of tensor field G in a

    given coordinate system.How do these components transform under transformation of coordinates

    {xi} {xi}?

    G = gikdxi dxk = gik

    xi

    xidxi

    xk

    xkdxk

    =

    xi

    xigik

    xk

    xkdxi

    dxk = gikdxi dxk

    Hence

    gik =xi

    xigik

    xk

    xk. (1.10)

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    One can derive transformations formulae also using general formulae (1.7)

    for tensors.Important remark

    gik =

    xi,

    xk

    (1.11)

    Later by some abuse of notations we sometimes omit the sign of tensorproduct and write a metric just as

    G = gik(x)dxidxk .

    Examples

    Rn with canonical coordinates {xi} and with metric

    G = (dx1)2 + (dx2)2 + + (dxn)2

    G = ||gik|| = diag [1, 1, . . . , 1]Recall that this is a basis example of n-dimensional Euclidean space,where scalar product is defined by the formula:

    G(X, Y) = X, Y = gikXiYk = X1Y1 + X2Y2 + + XnYn .

    In the general case if G = ||gik|| is an arbitrary symmetric positive-definite metric then

    G(X, Y) = X, Y = gikXiYk .

    One can show that there exist a new basis {ei} such that in this basis

    G(ei, ek) = ik .

    This basis is called orthonormal basis. (See the Lecture notes in Ge-

    ometry)Scalar product in vector space defines the same scalar product at allthe points. In general case for Riemannian manifold scalar productdepends on the point.

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    R2 with polar coordinates in the domain y > 0 (x = r cos , y =

    r sin ):dx = cos dr r sin d, dy = sin dr + r cos d. In new coordi-nates the Riemannian metric G = dx2 + dy2 will have the followingappearance:

    G = (dx)2+(dy)2 = (cos drr sin d)2+(sin dr+r cos d)2 = dr2+r2(d)2

    We see that for matrix G = ||gik||

    G = gxx gxygyx gyy =

    1 00 1

    in cartesian coordinates

    , G = grr grgr g =

    1 00 r

    in polar coordinates

    CircleInterval [0, 2) in the line 0 x < 2 with Riemannian metric

    G = a2dx2 (1.12)

    Renaming x we come to habitual formula for metric for circle ofthe radius a: x2 + y2 = a2 embedded in the Euclidean space E2:

    G = a2d2{x = a cos

    y = a sin , 0 < 2, (1.13)

    Cylinder surfaceDomain in R2 D = {(x, y) : , 0 x < 2 with Riemannian metric

    G = a2dx2 + dy2 (1.14)

    We see that renaming variables x , y h we come to habitual,familiar formulae for metric in standard polar coordinates for cylinder

    surface of the radius a embedded in the Euclidean space E3

    :

    G = a2d2 + dh2

    x = a cos

    y = a sin

    z = h

    , 0 < 2, < h <

    (1.15)

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    Sphere

    Domain in R2, 0 < x < 2, 0 < y < with metric G = dy2 +sin2 ydx2

    We see that renaming variables x , y h we come to habitual,familiar formulae for metric in standard spherical coordinates for spherex2 + y2 + z2 = a2 of the radius a embedded in the Euclidean space E3:

    G = a2d2+a2 sin2 d2

    {x = a sin cos

    y = a sin sin z = a cos , 0 < 2, < h <

    (1.16)

    1.2.1 Pseudoriemannian manifold

    If we omit the condition of positive-definiteness for Riemannian metric wecome to so called Pseudorimannian metric. Manifodl equipped with pseu-doriemannan metric is called pseudoriemannian manifold. Pseudoriemannianmanifolds appear in applications in the special and general relativity theory.

    Example Consider n+1-dimensional linear space Rn+1 with pseudomet-ric

    (dx0)2 (dx1)2 (dx2)2 (dxn)2in coordinates x0, x1, . . . , xn. In the case n = 3 it is so called Minkovskispace. The coordinate x0 the role of the time: x0 = ct, where c is the value

    of the speed of the light.

    1.3 Scalar product. Length of tangent vectors and an-gle between vectors. Length of the curve

    The Riemannian metric defines scalar product of tangent vectors attachedat the given point. Hence it defines the length of tangent vectors and anglebetween them. If X = Xm

    xm, Y = Ym

    xmare two tangent vectors at the

    given point p of Riemannian manifold with coordinates x1, . . . , xn, then wehave that lengths of these vectors equal to

    |X| = X, X = gik(x)XiXk, |Y| = Y, Y = gik(x)YiYk,(1.17)

    and he angle between these vectors is defined by the relation

    cos =X, Y|X| |Y| =

    gikXiYk

    gik(x)XiXk

    gik(x)YiYk(1.18)

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    Example Let M be 3-dimensional Riemanian manifold. Consider the

    vectors X = 2x + 2y z and Y = x2y2z attached at the point p ofM with local coordinates (x,y,z), where x = y = 1, z = 0. Find the lengthsof these vectors and angle between them if the expression of Riemannianmetric in these coordinates is coordinates dx

    2+dy2+dz2

    (1+x2+y2)2.

    We see that matrix of Riemannian gik = (x,y,z)ik, where (x,y,z) =1

    (1+x2+y2+z2)2ik = 1 i f i = k and ik = 0 i f i = k, i.e. matrix ||gik|| is

    proportional to unity matrix. According to formulae above

    |X| =

    X, X =

    gik(x)XiXk =

    (x,y,z)

    XiXi = 3

    (x,y,z) =

    3 .

    The same answer for|Y

    |. The scalar product between vectors X, Y equal to

    zero:X, Y = (x,y,z)ikXiYk = 0

    Hence these vectors are orthogonal to each other.

    1.3.1 Length of the curve

    Let : xi = xi(t), (i = 1, . . . , n)) (a t b) be a curve on the Riemannianmanifold (M, G).

    At the every point of the curve the velocity vector (tangent vector) isdefined:

    v(t) =

    x1(t)

    xn(t)

    (1.19)

    The length of velocity vector v TxM (vector v is tangent to the manifoldM at the point x) equals to

    |v|x = v, vGx = gikvivkx = gik

    dxi(t)

    dt

    dxk(t)

    dt xx (1.20)The length of the curve is defined by the integral of the length of velocity

    vector:

    L =

    ba

    v, vG

    x(t)

    dt =

    ba

    gik(x(t))xi(t)xk(t)dt (1.21)

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    Bearing in mind that metric (1.9) defines the length we often write metric

    in the following formds2 = gikdx

    idxk (1.22)

    For example consider 2-dimensional Riemannian manifold with metric

    ||gik(u, v)|| =

    g11(u, v) g12(u, v)g21(u, v) g22(u, v)

    .

    Then

    G = ds2 = gikduidvk = g11(u, v)du

    2 + 2g12(u, v)dudv + g22(u, v)dv2

    The length of the curve : u = u(t), v = v(t), where t0 t t1 according to(1.21) is equal to

    L =

    t1t0

    v, v =

    t1t0

    gik(x)xixk = (1.23)

    t1t0

    g11 (u (t) , v (t)) u2t + 2g12 (u (t) , v (t)) utvt + g22 (u (t) , v (t)) v

    2t dt

    (1.24)The length of the curve defined by the formula(1.21) obeys the following

    natural conditions

    It coincides with the usual length in the Euclidean space En (Rn withstandard metric G = (dx1)2 + + (dxn)2 in cartesian coordinates).E.g. for 3-dimensional Euclidean space

    L =

    ba

    gik(x(t))xi(t)xk(t)dt =

    ba

    (x1(t))2 + (x2(t))2 + (x3(t))2dt

    (1.25)

    It does not depend on parameterisation of the curve

    L =

    ba

    gik(x(t))xi(t)xk(t)dt =

    ba

    gik(x())xi()xk()d,

    (1.26)where xi() = xi(t()), a b while a t b.

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    It does not depend on coordinates on Riemannian manifold M

    L =

    ba

    gik(x(t))xi(t)xk(t)dt =

    ba

    gik(x(t))xi

    (t)xk(t)dt

    (1.27)

    It is additive: if a curve = 1+, i.e. : xi(t), a t b, 1 : xi(t), a t c and 2 : xi(t), c t b where a point c belongs to the interval(a, b) then

    L = L1 + L2, i.e.

    b

    a gik(x(t))xi(t)xk(t)dt =ca

    gik(x(t))xi(t)xk(t)dt +

    bc

    gik(x(t))xi(t)xk(t)dt (1.28)

    Conditions (1.25) and (1.28) evidently are obeyed.Condition (1.26) follows from the fact that

    xi() =dx(t())

    d=

    dx(t())

    dt

    dt

    d= xi(t)

    dt

    d.

    Bearing in mind the above formula we have

    ba

    gik(x())xi()xk()d =

    ba

    gik(x(t()))xi(t)xk(t)

    dt

    d

    2d =

    ba

    gik(x(t()))xi(t)xk(t)

    dtd d =

    ba

    gik(x(t))xi(t)xk(t)dt = L

    Condition (1.27) follows from the condition (1.76):

    b

    a gik(x(t))xi(t)xk(t)dt =

    b

    a gik(x(t))xi

    xi

    xk

    xk

    xi(t)xk(t)dt =

    ba

    gik(x(t))

    xi

    xixi(t)

    xk

    xkxk(t)

    dt =

    ba

    gik(x(t))xi(t)xk(t)dt .

    (1.29)

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    1.4 Riemannian structure on the surfaces embedded

    in Euclidean spaceLet M be a surface embedded in Euclidean space. Let G be Riemannianstructure on the manifold M.

    Let X, Y be two vectors tangent to the surface M at a point p M. AnExternal Observer calculate this scalar product viewing these two vectors asvectors in E3 attached at the point p E3 using scalar product in E3. AnInternal Observer will calculate the scalar product viewing these two vectorsas vectors tangent to the surface M using the Riemannian metric G (see theformula (1.39)). Respectively

    If L is a curve in M then an External Observer consider this curve as acurve in E3, calculate the modulus of velocity vector (speed) and the lengthof the curve using Euclidean scalar product of ambient space. An InternalObserver (an ant) will define the modulus of the velocity vector and thelength of the curve using Riemannian metric.

    Definition Let M be a surface embedded in the Euclidean space. Wesay that metric GM on the surface is induced by the Euclidean metric if thescalar product of arbitrary two vectors A, B TpM calculated in terms ofthe metric G equals to Euclidean scalar product of these two vectors:

    A, BGM = A, BGEuclidean (1.30)In other words we say that Riemannian metric on the embedded surface isinduced by the Euclidean structure of the ambient space if External andInternal Observers come to the same results calculating scalar product ofvectors tangent to the surface.

    In this case modulus of velocity vector (speed) and the length of the curveis the same for External and Internal Observer.

    Before going in details of this definition recall the conception of Internaland External Observers when dealing with surfaces in Euclidean space:

    1.4.1 Internal and external coordinates of tangent vector

    Tangent planeLet r = r(u, v) be parameterisation of the surface M embedded in the Eu-clidean space:

    r(u, v) =

    x(u, v)y(u, v)

    z(u, v)

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    Here as always x,y,zare Cartesian coordinates in E3.

    Let p be an arbitrary point on the surface M. Consider the plane formedby the vectors which are adjusted to the point p and tangent to the surfaceM. We call this plane plane tangent to M at the point p and denote it byTpM.

    For a point p M one can consider a basis in the tangent plane TpMadjusted to the parameters u, v. Tangent basis vectors at any point (u, v)are

    ru =r(u, v)

    u=

    x(u,v)

    uy(u,v)

    uz(u,v)

    u

    =

    x(u, v)

    u

    x+

    y(u, v)

    u

    y+

    z(u, v)

    u

    z

    Every vector X TpM can be expanded over this basis:X = Xuru + Xvrv, (1.31)

    where Xu, Xv are coefficients, components of the vector X.Internal Observer views the basis vector ru TpM, as a velocity vector

    for the curve u = u0 + t, v = v0, where (u0, v0) are coordinates of the point p.Respectively the basis vector rv TpM for an Internal Observer, is velocityvector for the curve u = u0, v = v0 + t, where (u0, v0) are coordinates of thepoint p.

    Let r = r(t) be a curve belonging to the surface C, which passes throughthe point p, r(t) = r(u(t), v(t)) and p = r(t0). Then vector

    rt =dr

    dt=

    dr(u(t), v(t))

    dt(1.32)

    belongs to the tangent plane TpM.Note that for the vector (1.32) components Xu, Xv are equal to Xu =

    ut, Xv = vt because

    rt =dr

    dt

    =dr(u(t), v(t))

    dt

    = utru + vtrv (1.33)

    An External Observer describes the vector rt as a vector in E3 attached at

    the point p. The Internal Observer describes this vector as a vector whichhas components (ut, vt) in the basis ru, rv according to the formula (1.33).

    In general consider an arbitrary tangent vector X TpM. Denote Xu =a, Xv = b

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    X = Xuru+Xvrv = aru+brv = a

    xu(u, v)yu(u, v)

    zu(u, v)

    +b

    xv(u, v)xv(u, v)

    xv(u, v)

    =

    axu(u, v) + bxv(u, v)ayu(u, v) + byv(u, v)

    azu(u, v) + bzv(u, v)

    (1.34)The last column in this formula represents the three components of the vectorX in the ambient space.

    The pair (a, b) can be considered as internal coordinates of the tangentvector X. An Internal Observer, Ant living on the surface, deals with thevector X in terms of coordinates (a, b). External observer which contemplatesthe surface embedded in three-dimensional ambient space deals with vector

    X as with vector with three external coordinates (see the last right columnin the formula (1.34).)

    In condensed notation instead denoting coordinates by (u, v) we oftendenote them by u = (u1, u2). Respectively we denote by

    r =dr

    du, ru = r1, rv = r2

    The formula (1.34) for tangent vector field will have the following appear-ance:

    X = X

    r = X1

    r1 + X2

    r2, (X1

    = Xu, X2

    = Xv) (1.35)

    When using condensed notations we usually omit explicit summationsymbols. E.g. we write ur instead

    2i=1 u

    r or u1r1 + u

    2r2One can consider also differentials du = (du1, du2):

    du(r) = : du

    1(r1) = du2(r2) = 1, du

    1(r2) = du2(r1) = 0 (1.36)

    1.4.2 Explicit formulae for induced Riemannian metric (First Quadraticform)

    Now we are ready to write down the explicit formuale for the Riemannianmetric on the surface induced by metric (scalar product) in ambient Eu-clidean space (see the Definition (1.30)).

    Let M: r = r(u, v) be a surface embedded in E3.The formula (1.30) means that scalar products of basic vectors ru =

    u, rv = v has to be the same calculated in the ambient space and on

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    the surface: For example scalar product

    u, v

    M = guv calculated by the

    Internal Observer is the same as a scalar product ru, rvE3 calculated by theExternal Observer, scalar product v, vM = guv calculated by the InternalObserver is the same as a scalar product rv, rvE3 calculated by the ExternalObserver and so on:

    G =

    guu guvgvu gvv

    =

    u, u u, vv, u v, v

    =

    ru, ruE3 ru, rvE3rv, ruE3 rv, rvE3

    (1.37)

    where as usual we denote by , E3 the scalar product in the ambient Eu-clidean space.

    (Here see also the important remark (1.11))

    Remark It is convenient sometimes to denote parameters (u, v) as (u1, u2)or u ( = 1, 2) and to write r = r(u1, u2) or r = r(u) ( = 1, 2) insteadr = r(u, v)

    In these notations:

    GM =

    g11 g12g12 g22

    =

    ru, ruE3 ru, rvE3ru, rvE3 rv, rvE3

    , g = r, r ,

    GM = gdudu = g11du

    2 + 2g12dudv + g22dv2 (1.38)

    where ( , ) is a scalar product in Euclidean space.The formula (1.38) is the formula for induced Riemannian metric on the

    surfaceFirst Quadratic Form.If X, Y are two tangent vectors in the tangent plane TpC then G(X, Y)

    at the point p is equal to scalar product of vectors X, Y:

    (X, Y) = (X1r1 + X2r2, Y

    1r1 + Y2r2) = (1.39)

    X1(r1, r1)Y1 + X1(r1, r2)Y

    2 + X2(r2, r1)Y1 + X2(r2, r2)Y

    2 =

    X(r, r)Y = XgY

    = G(X, Y)

    We can come to this formula just transforming differentials. In carteisan

    coordinates X, Y = X1

    Y

    1

    + X

    2

    Y

    2

    + X

    3

    Y

    3

    , i.e. the Euclidean metric incartesian coordinates is given by

    GE3 = (dx)2 + (dy)2 + (dz)2 . (1.40)

    The condition that Riemannian metric (1.38) is induced by Euclidean scalarproduct means that

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    GE3r=r(u,v)

    =(

    (dx)2 + (dy)2 + (dz)2)

    r=r(u,v)= GM = gdu

    du (1.41)

    i.e. ((dx)2 + (dy)2 + (dz)2)r=r(u,v)

    =

    x(u, v)

    udu +

    x(u, v)

    udu

    2+

    x(u, v)

    udu +

    x(u, v)

    udu

    2+

    x(u, v)

    udu +

    x(u, v)

    udu

    2=

    (x2u + y2u + z

    2u)du

    2 + 2(xuxv + yuyv + zuzv)dudv + (x2v + y

    2v + z

    2v)dv

    2

    We see that

    GM = gdudu = g11du

    2 + 2g12dudv + g22dv2, (1.42)

    where g11 = guu = (x2u + y

    2u + z

    2u) = ru, ruE3, g12 = g21 = guv = gvu =

    (xuxv + yuyv + zuzv) = ru, rvE3, g22 = gvv = (x2v + y2v + z2v) = rv, rvE3 . Wecome to the same formula.

    (See the examples of calculations in the next subsection.)Check explicitly again that length of the tangent vectors and of the curves

    calculating by External observer (i.e. using Euclidean metric (1.40)) is thesame as calculating by Internal Observer, ant (i.e. using the induced Rie-mannian metric (1.38))

    Consider a vector X = Xr = aru + brv tangent to the surface M.Calculate its length by External and Internal Observer.The square of the length |X| of this vector calculated by External observer

    (he calculates using the scalar product in E3) equals to

    |X|2 = X, X = ru + brv, aru + brv = a2ru, ru + 2abru, rv + b2rv, rv(1.43)

    where , is a scalar product in E3.The internal observer will calculate the length using Riemannian metric

    (1.38):

    G(X, X) =(

    a, b) g11 g12

    g21 G22

    a

    b

    = g11a

    2 + 2g12ab + g22b2 (1.44)

    External observer (person living in ambient space E3) calculate the lengthof the tangent vector using formula (1.43). An ant living on the surfacecalculate length of this vector in internal coordinates using formula (1.44).

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    External observer deals with external coordinates of the vector, ant on the

    surface with internal coordinates. They come to the same answer.Let r(t) = r(u(t), v(t)) a t b be a curve on the surface.Velocity of this curve at the point r(u(t), v(t)) is equal to

    v = X = ru + rvwhere = ut, = vt : v =dr(t)dt

    = utru + vtrv .

    The length of the curve is equal to

    L =

    ba

    |v(t)|dt =ba

    v(t), v(t)E3dt =

    ba

    utru + vtrv, utru + vtrvE3dt =

    (1.45)ba

    ru, ruE3u2t + 2ru, rvE3utvt + rv, rvE3v2t d =b

    a

    g11u2t + 2g12utvt + g22v

    2t dt (1.46)

    An external observer will calculate the length of the curve using (1.45).An ant living on the surface calculate length of the curve using (1.46) usingRiemannian metric on the surface:

    ds2 = gikduiduk = g11du

    2 + 2g12dudv + g22dv2 (1.47)

    They will come to the same answer.

    1.4.3 Induced Riemannian metrics. Examples.

    We consider here examples of calculating induced Riemanian metric on somequadratic surfaces in E3. using calculations for tangent vectors (see (1.38))or explicitly in terms of differentials (see (1.41) and (1.42)).

    First of all consider the general case when a surface M is defined by theequation z F(x, y) = 0. One can consider the following parameterisationof this surface:

    r(u, v) :

    x = u

    y = v

    z = F(u, v)

    (1.48)

    Then

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    ru = 10

    Fu

    rv = 01Fv

    (1.49),

    (ru, ru) = 1 + F2u , (ru, rv) = FuFv, (rv, rv) = 1 + F

    2v

    and induced Riemannina metric (first quadratic form) (1.38) is equal to

    ||g || =

    g11 g12g12 g22

    =

    (ru, ru) (ru, rv)(ru, rv) (rv, rv)

    =

    1 + F2u FuFvFuFv 1 + F

    2v

    (1.50)

    GM = ds2 = (1 + F2u )du2 + 2FuFvdudv + (1 + F2v )dv2 (1.51)and the length of the curve r(t) = r(u(t), v(t)) on C (a t b) can becalculated by the formula:

    L =

    ba

    (1 + F2u )u

    2t + 2FuFvutvt + (1 + Fv)

    2v2t dt

    One can calculate (1.51) explicitly using (1.41):

    GM = (dx2 + dy2 + dz2

    ) x=u,y=v,z=F(u,v)= (du)2 + (dv)2 + (Fudu + Fvdv)

    2 =

    = (1 + F2u )du2 + 2FuFvdudv + (1 + F

    2v )dv

    2 . (1.52)

    CylinderCylinder is given by the equation x2 + y2 = a2. One can consider the

    following parameterisation of this surface:

    r(h, ) :

    x = a cos

    y = a sin

    z = h

    (1.53)

    We have Gcylinder = (dx2 + dy2 + dz2)

    x=a cos,y=a sin,z=h

    =

    = (a sin d)2 + (a cos d)2 + dh2 = a2d2 + dh2 (1.54)

    The same formula in terms of scalar product of tangent vectors:

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    rh =00

    1

    r = a sin a cos 0

    (1.55),

    (rh, rh) = 1, (rh, r) = 0, (r, r) = a2

    and

    ||g|| =

    (ru, ru) (ru, rv)(ru, rv) (rv, rv)

    =

    1 00 a2

    ,

    G = dh2 + a2d2 (1.56)

    and the length of the curve r(t) = r(h(t), (t)) on the cylinder (a

    t

    b)can be calculated by the formula:

    L =

    ba

    h2t + a

    2tdt (1.57)

    ConeCone is given by the equation x2 + y2 k2z2 = 0. One can consider the

    following parameterisation of this surface:

    r(h, ) : x = kh cos

    y = kh sin z = h

    (1.58)

    Calculate induced Riemannian metric:We have

    Gconus =(

    dx2 + dy2 + dz2)

    x=kh cos,y=kh sin,z=h=

    (k cos dh kh sin d)2 + (k sin dh + kh cos d)2 + dh2

    Gconus = k

    2

    h

    2

    d

    2

    + (1 + k

    2

    )dh

    2

    , ||g|| = 1 + k2 0

    0 k2h2 (1.59)The length of the curve r(t) = r(h(t), (t)) on the cone (a t b) can becalculated by the formula:

    L =

    ba

    (1 + k2)h2t + k

    2h22t dt (1.60)

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    Sphere

    Sphere is given by the equation x2 + y2 + z2 = a2. Consider the following(standard ) parameterisation of this surface:

    r(, ) :

    x = a sin cos

    y = a sin sin

    z = a cos

    (1.61)

    Calculate induced Riemannian metric (first quadratic form)

    GS2 = (dx2 + dy2 + dz2) x=a sin cos,y=a sin sin,z=a cos =(a cos cos da sin sin d)2+(a cos sin d+a sin cos d)2+(a sin d)2 =

    a2 cos2 d2 + a2 sin2 d2 + a2 sin2 d2 =

    , = a2d2 + a2 sin2 d2 , ||g|| =

    a2 00 a2 sin2

    (1.62)

    One comes to the same answer calculating scalar product of tangent vec-tors:

    r =

    a cos cos a cos sin

    a sin

    r =

    a sin sin a sin cos

    0

    (1.63)

    ,(r, r) = a

    2, (rh, r) = 0, (r, r) = a2 sin2

    and

    ||g|| =

    (ru, ru) (ru, rv)(ru, rv) (rv, rv)

    =

    a2 00 a2 sin2

    , GS2 = ds

    2 = a2d2 + a2 sin2 d2

    The length of the curve r(t) = r((t), (t)) on the sphere of the radius a

    (a t b) can be calculated by the formula:L =

    ba

    a

    2t + sin

    2 2t dt (1.64)

    We considered all quadratic surfaces except paraboloid z = x2 y2.It can be rewritten as z = xy () and is call sometimes saddle

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    Saddle (paraboloid)

    Saddle is given by the equation z xy = 0.Why paraboloid?Exercise Show that the equation of saddle can be rewritten as z = x2y2(This surface is a ruled surface containing lines...)Consider the following (standard ) parameterisation of this surface:

    r(u, v) :

    x = u

    y = v

    z = uv

    (1.65)

    Calculate induced metric:

    Gsaddle =(

    dx2 + dy2 + dz2)

    x=u cos,y=v sin,z=uv=

    du2 + dv2 + (udv + vdu)2 =

    Gsaddle = (1 + v2)du2 + 2uvdudv + (1 + u2)dv2 . (1.66)

    The length of the curve r(t) = r(u(t), v(t)) on the sphere of the radius a(a t b) can be calculated by the formula:

    L =ba

    a

    (1 + v2)u2t + 2uvutvt + (1 + u2)v2t dt (1.67)

    One-sheeted and two-sheeted hyperboloids.Consider surface given by the equation

    x2 + y2 z2 = c

    If c = 0 it is a cone. We considered it already above.If c > 0 it is one-sheeted hyperboloidconnected surface in E3 If c < 0

    it is two-sheeted hyperboloid a surface with two sheets z > 0 and z < 0 LConsider these cases separately.

    1) One-sheeted hyperboloid: x2 + y2 z2 = a2. It is ruled surface. Itcontains two family of lines

    Exercise Find the lines on two-sheeted hyperboloid

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    One-sheeted hyperboloid is given by the equation x2 + y2

    z2 = a2. it is

    convenient to choose parameterisation:

    r(, ) :

    x = a cosh cos

    y = a cosh sin

    z = a sinh

    (1.68)

    x2 + y2 z2 = a2 cosh2 a2 sinh2 = a2Compare the calculations with calculations for sphere! We changed functionscos, sin on cosh, sinh

    Induced Riemannian metric (first quadratic form).

    GHyperbolI =(

    dx2 + dy2 + dz2)

    x=a cosh cos,y=a cosh sin,z=a sinh =

    (a sinh cos da cosh sin d)2+(a sinh sin d+a cosh cos d)2+(a cosh d)2 =a2 sinh2 d2 + a2 cosh2 d2 + a2 cosh2 d2 =

    , = a2(1+2 sinh2 )d2+a2 cosh2 d2 , ||g|| =

    1 + 2 sinh2 00 cosh2

    (1.69)

    For two-sheeted hyperboloid calculatiosn will be very similar.

    1) Two-sheeted hyperboloid: z2 x2 y2 = a2. It is not ruled surface!it is convenient to choose parameterisation:

    r(, ) :

    x = a sinh cos

    y = a sinh sin

    z = a cosh

    (1.70)

    z2 x2 y2 = a2 cosh a2 sinh2 = a2

    Compare the calculations with calculations for sphere! We changed functionscos, sin on cosh, sinh

    Induced Riemannian metric (first quadratic form).

    GHyperbolI =(

    dx2 + dy2 + dz2)

    x=a sinh cos,y=a sinh sin,z=a cosh =

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    (a cosh cos d

    a sinh sin d)2+(a cosh sin d+a sinh cos d)2+(a sinh d)2 =

    a2 cosh2 d2 + a2 sinh2 d2 + a2 sinh2 d2 =

    , = a2(1+2 sinh2 )d2+a2 sinh2 d2 , ||g|| =

    1 + 2 sinh2 00 sinh2

    (1.71)

    We calculated examples of induced Riemannian structure embedded inEuclidean space almost for all quadratic surfaces.

    Quadratic surface is a surface defined by the equation

    Ax2 + By2 + Cz2 + 2Dxy + 2Exz + 2F yz + ex + f y + dz+ c = 0

    One can see that any quadratic surface by affine transformation can be trans-formed to one of these surfaces

    cylinder (elliptic cylinder) x2 + y2 = 1 hyperbolic cylinder: x2 y2 = 1) parabolic cylinder z = x2

    paraboloid x2 + y2 = z hyperbolic paraboloid x2 y2 = z cone x2 + y2 z2 = 0 sphere x2 + y2 + z2 = 1 one-sheeted hyperboloid two-sheeted hyperboloid

    (We exclude degenerate cases such as point x2 + y2 + z2 = 0, planes, e.t.c.)

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    1.4.4 Induced metric on two-sheeted hyperboloid embedded in

    pseudo-Euclidean space.

    Consider two-sheeted hyperboloid embedded in pseudo-Euclidean space with pseudo-scalar product defined by bilinear form

    X, Ypseudo = X1Y1 + X2Y2 X3Y3 (1.72)The pseudoscalar product is bilinear, symmetric. It is defined by non-degeneratematrix. But it is not positive-definite: The pseudo-length of vectors X =(a cos , a sin , a) equals to zero:

    X = (a cos , a sin , a) X, Xpseudo = 0, (1.73)This is not scalar product. The pseudo-Riemannian metric is:

    Gpseudo = dx2 + dy2 dz2 (1.74)

    it turns out that the following remarkable fact occurs:Proposition The pseudo-Riemannian metric (1.74) in the ambient3-dimensional

    pseudo-Euclidean space induces Riemannian metric on two-sheeted hyperboloid

    x2 + y2 z2 = 1.Show it. repeat the calculations above for two-sheeted hyperboloid changing in

    the ambient space Riemannian metric G = dx2 + dy2 + dz2 on pseudo-Riemanniandx2 + dy2 dz2:

    Using (1.70) and (1.74) we come now to

    G =(

    dx2 + dy2 dz2) x=a sinh cos,y=a sinh sin,z=a cosh

    =

    (a cosh cos da sinh sin d)2+(a cosh sin d+a sinh cos d)2(a sinh d)2 =a2 cosh2 d2 + a2 sinh2 d2 a2 sinh2 d2

    , GL = a2d2 + a2 sinh2 d2 , ||g || =

    1 00 sinh2

    (1.75)

    The two-sheeted hyperboloid equipped with this metric is called hyperbolic orLobachevsky plane.

    Now express Riemannian metric in stereographic coordinates.Calculations are very similar to the case of stereographic coordinates of 2-

    sphere x2 + y2 + z2 = 1. (See homework 1). Centre of projection (0, 0, 1): Forstereographic coordinates u, v we have u

    x= y

    v= 11+z . We come to{

    u = x1+zv = y1+z

    ,

    x = 2u1u2v2

    y = 2v1u2v2

    z = u2+v2+1

    1u2v2

    (4)

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    The image of upper-sheet is an open disc u2 + v2 = 1 since u2 + v2 = x2+y2

    (1+z)2=

    z21(1+z)2 = z1z+1 . Since for upper sheet z > 1 then 0 z1z+1 < 1.

    G = (dx2 + dy2 dz2)x=x(u,v),y=y(u,v),z=z(u,v)

    =

    d

    2u

    1 u2 v22

    +

    d

    2v

    1 u2 v22

    d

    u2 + v2 + 1

    1 u2 v22

    =

    (Compare with calculations for sphere x2 + y2 + z2 = 1). We have G =

    2du1 u2 v2 +2u(2udu + 2vdv)

    (1 u2

    v2

    )2

    2

    + 2dv1 u2 v2 +2v(2udu + 2vdv)

    (1 u2

    v2

    )2

    2

    2udu + 2vdv

    1 u2 v2 +(u2 + v2 + 1)(2udu + 2vdv)

    (1 u2 v2)22

    =4(du)2 + 4(dv)2

    (1 + u2 + v2)2

    (To perform these calculations it is convenient to denote by s = 1 u2 v2).

    1.5 Isometries of Riemanian manifolds.

    Let (M1, G(1)), (M2, G(2)) be two Riemannian manifoldsmanifolds equippedwith Riemannian metric G1 and G2 respectively.

    Definition

    We say that these Riemannian manifolds are isometric if there exists adiffeomorphism F (one-one smooth map) which preserves the distances. Thismeans the following:

    We say that these Riemannian manifolds are isometric if there exists adiffeomorphism F (one-one smooth map) such that

    FG(2) = G(1) ,

    which means the following:Let p1 be an arbitrary point on manifold M1 and p2 M2 be its image:F(p1) =

    p2. Let {xi} be coordinates in a vicinity of a point p1 M1 and {ya} becoordinates in a vicinity of a point p2 M2. Let Riemannian metrics G1 onM1 has local expression G(1) = g(1)ik(x)dx

    idxk in coordinates {xi} and re-spectively Riemannian metrics G(2) has local expression G2 = g(2)ab(y)dy

    adyb

    in coordinates {yi} on M2. Then

    g(1)ik(x)dxidxk = g(2)ab(y)dy

    adyb = g(2)ab(y(x))ya(x)

    xidxi

    y b(x)

    xkdxk (1.76)

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    i.e.

    g(1)ik(x) = ya

    (x)xi

    g(2)ab(y(x)) yb

    (x)xk

    . (1.77)

    where yi = yi(x) is local expression for diffeomorphism F.

    Definition We say that two Riemannian manifolds (M1, G(1)), (M2, G(2))are locally isometric if the following conditions hold:

    For arbitrary point p1 on the manifold M1 there exists a point p2 on themanifold M2 such that there exist coordinates {xi} in a vicinity of a pointp1 M1 and coordinates {ya} in a vicinity of a point p2 M2 such that localexpression for metric G(1) on M1 in coordinates {xi} and local expression formetric G(2) on M2 in coordinates

    {ya

    }are related via formuale (1.76), (1.29).

    Locally isometric Riemannian manifolds have not to be diffeomorphic.E.g. Euclidean plane is locally isometric to cylinder, but they are not diffeo-morphic.

    1.5.1 Examples of local isometries

    Consider examples.

    Example 1 Cylinder-ConePlane,Riemannian metric on cylinder is Gcylinder = a

    2d2 + dh2 and on the coneGconus = k

    2h2d2 + (1 + k2)dh2 (see formulae (1.54) and (1.59)).To show that cylinder is isometric to Euclidean plan we have to find

    new local coordinates u, v on the cylinder such that in these coordinates themetric on the cylinder equals to du2 + dv2. If we put u = a, v = h then

    du2 + dv2 = d(a)2 + dh2 = a2d2 + dh2 = Gcylinder . (1.78)

    Thus we prove the local isometry. Of course the coordinate u = is notglobal coordinate on the surface of cylinder. It is evident that cylinder andplane are not globally isometricsince there are no diffeomorphism of cylinderon the plane. (They are different non-homeomorphic topological spaces.)

    Now show that cone is locally isometric to the plane.This means that we have to find local coordinates u, v on the cone such

    that in these coordinates induced metric G|c on cone would have the appear-ance G|c = du2 + dv2.

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    First of all calculate the metric on cone in natural coordinates h, where

    r(h, ) :

    x = kh cos

    y = kh sin

    z = h

    .

    (x2 + y2 k2z2 = k2h2 cos2 + k2h2 sin2 k2h2 = k2h2 k2h2 = 0.Calculate metric Gc on the cone in coordinates h, induced with the

    Euclidean metric G = dx2 + dy2 + dz2:

    Gc =

    (dx2 + dy2 + dz2

    ) x=kh cos,y=kh sin,z=h= (k cos dh kh sin d)2+

    (k sin dh + kh cos d)2 + dh2 = (k2 + 1)dh2 + k2h2d2 .

    In analogy with polar coordinates try to find new local coordinates u, v such

    that

    {u = h cos

    v = h sin , where , are parameters. We come to du2 + dv2 =

    ( cos dh h sin d)2+( sin dh + h cos d)2 = 2dh2+22h2d2.Comparing with the metric on the cone Gc = (1 + k

    2)dh2 + k2h2d2 we seethat if we put = k and = k

    1+k2then du2 + dv2 = 2dh2 + 22h2d2 =

    (1 + k2

    )dh2

    + k2

    h2

    d2

    .Thus in new local coordinates{u =

    k2 + 1h cos k

    k2+1

    v =

    k2 + 1h sin kk2+1

    induced metric on the cone becomes G|c = du2 + dv2, i.e. cone locally isisometric to the Euclidean plane

    Of course these coordinates are local. Cone and plane are not homeo-morphic, thus they are not globally isometric.

    One can show also that2) Plane with metric 4R2(dx2+dy2)

    (1+x2+y2)2is isometric to the sphere with radius

    R(a) = ...3) Disc with metric du

    2+dv2

    (1u2v2)2 is isometric to half plane with metricdx2+dy2

    4y2.

    (see exercises in Homeworks and Coursework.)

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    1.6 Volume element in Riemannian manifold

    The volume element in n-dimensional Riemannian manifold with metric G =gikdx

    idxk is defined by the formuladet gik dx

    1dx2 . . . d xn (1.79)

    If D is a domain in the n-dimensional Riemannian manifold with metricG = gikdx

    i then its volume is equal to to the integral of volume element overthis domain.

    V(D) =

    D

    det gik dx

    1dx2 . . . d xn (1.80)

    Remark Students who know the concept of exterior forms can read thevolume element as det gik dx

    1 dx2 dxn (1.81)

    Note that in the case of n = 1 volume is just the length, in the case ifn = 2 it is area.

    1.6.1 Volume of parallelepiped

    Note that the formula (1.79) gives the volume ofn-dimensional parallelepiped.Show this. Let En be Euclidean vector space with orthonormal basis

    {ei

    }.

    Let vi be an arbitrary basis in this vector space (vectors vi in general have notunit length and are not orthogonal to each other). Consider n-parallelepipedspanned by vectors {vi}:

    vi : r = tivi, 0 ti 1.

    The volume of this parallelepiped equals to

    V ol(vi) = det ||ami || , (1.82)where A = ||ami || is transition matrix, vi = emami . On the other hand

    r = xiei = tmvm, hence xi = aimtm, where vm = eiaim.

    Let G = (dx1)2 + + (dxn)2 = gikdtidtk be usual Euclidean metric in newcoordinates ti. The

    G = (dx1)2 + + (dxn)2 = dxiikdxk = dtixi

    tiik

    xk

    tkdtk.

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    Since xi

    ti= ai

    i then

    gik =i

    ai

    i ai

    k det g = (det A)2, det g = det A.

    and according to the formula (1.80)

    V ol(vi) =

    0ti1

    det gdt1dt2 . . . d tn = det A .

    We come to (1.82).

    Perform these calculations in detail for 3-dimensional case.E.g. if Euclidean space is 3-dimensional then the parallelepiped spanned

    by basis vectors {a, b, c}a,b,c = t

    1a + t2b + t3c , 0 t1, t2, t3 1.Volume of parallelepiped equals to

    V ola,b,c = det

    ax bx cxay by cy

    az bz cz

    , (1.83)

    and

    xyz =

    ax bx cxay by cyaz bz cz

    t1

    t2

    t3 .

    The Riemannian metric in new coordinates (t1, t2, t3) equals to

    dx2 + dy2 + dz2 =(

    dt1 dt2 dt3)ax ay azbx by bz

    cx cy cz

    ax bx cxay by cy

    az bz cz

    dt1dt2

    dt3

    ,

    i.e. in coordinates ti Riemannian metric G = gikdtidtk where

    g11 g12 g13g21 g22 g23

    g31 g32 g33=

    ax ay azbx by bz

    cx cy czax bx cxay by cy

    az bz czi.e.

    det gik = det

    ax bx cxay by cy

    az bz cz

    = V ola,b,c . (1.84)

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    1.6.2 Invariance of volume element under changing of coordinates

    Prove that volume element is invariant under coordinate transformations, i.e. ify1, . . . , yn are new coordinates: x1 = x1(y1, . . . , yn), x2 = x2(y1, . . . , yn)...,

    xi = xi(yp), i = 1, . . . , n , p = 1, . . . , n

    and gpq(y) matrix of the metric in new coordinates:

    gpq(y) =xi

    ypgik(x(y))

    xk

    yq. (1.85)

    Then

    det gik(x) dx

    1dx2 . . . d xn =

    det gpq(y) dy

    1dy2 . . . d yn (1.86)

    This follows from (1.85). Namelydet gik(y) dy

    1dy2 . . . d yn =

    det

    xi

    ypgik(x(y))

    xk

    yq

    dy1dy2 . . . d yn

    Using the fact that det(ABC) = det A det B det C and det

    xi

    yp

    = det

    xk

    yq

    1

    we see that from the formula above follows:det gik(y) dy

    1dy2 . . . d yn =

    det

    xi

    ypgik(x(y))

    xk

    yq

    dy1dy2 . . . d yn =

    detxiyp2det gik(x(y))dy1dy2 . . . d yn =

    det gik(x(y)) det

    xi

    yp

    dy1dy2 . . . d yn = (1.87)

    Now note that

    det

    xi

    yp

    dy1dy2 . . . d yn = dx1 . . . d xn

    according to the formula for changing coordinates in n-dimensional integral 2.Hence

    det gik(x(y)) det xi

    yp dy1dy2 . . . d yn = det gik(x(y))dx1dx2 . . . d xn (1.88)1determinant of matrix does not change if we change the matrix on the adjoint, i.e.

    change columns on rows.2Determinant of the matrix

    xi

    yp

    of changing of coordinates is called sometimes Ja-

    cobian. Here we consider the case if Jacobian is positive. If Jacobian is negative thenformulae above remain valid just the symbol of modulus appears.

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    Thus we come to (1.86).

    1.6.3 Examples of calculating volume element

    Consider first very simple example: Volume element of plane in cartesiancoordinates, metric g = dx2 + dy2. Volume element is equal to

    det gdxdy =

    det

    1 00 1

    dxdy = dxdy

    Volume of the domain D is equal to

    V(D) =D

    det gdxdy =

    D

    dxdy

    If we go to polar coordinates:

    x = r cos , y = r sin (1.89)

    Then we have for metric:G = dr2 + r2d2

    because

    dx2 + dy2 = (dr cos r sin d)2 + (dr sin + r cos d)2 = dr2 + r2d2(1.90)

    Volume element in polar coordinates is equal to

    det gdrd =

    det

    1 00 r2

    drd = drd.

    Lobachesvky plane.In coordinates x, y (y > 0) metric G = dx

    2+dy2

    y2, the corresponding matrix

    G = 1/y2 00 1/y2 . Volume element is equal to det gdxdy = dxdyy2 .Sphere in stereographic coordinates Consider the two dimensional plane

    with Riemannian metrics

    G =4R2(du2 + dv2)

    (1 + u2 + v2)2(1.91)

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    (It is isometric to the sphere of the radius R without North pole in stere-

    ographic coordinates (see the Homeworks.))Calculate its volume element and volume. It is easy to see that:

    G =

    4R2

    (1+u2+v2)20

    0 4R2

    (1+u2+v2)2

    det g =

    16R4

    (1 + u2 + v2)4(1.92)

    and volume element is equal to

    det gdudv = 4R2dudv

    (1+u2+v2)2

    One can calculate volume in coordinates u, v but it is better to considervolume form in polar coordinates u = r cos , v = r sin . Then it is easy

    to see that according to (1.90) we have for the metric G = R2(du2+dv2)

    (1+u2+v2)2=

    R2(dr2+r2d2)(1+r2)2

    and volume form is equal to det gdrd = 4R2rdrd(1+r2)2Now calculation of integral becomes easy:

    V =

    4R2rdrd

    (1 + r2)2= 8R2

    0

    rdr

    (1 + r2)2= 4R2

    0

    du

    (1 + u)2= 4R2 .

    Segment of the sphere.Consider sphere of the radius a in Euclidean space with standard Riema-

    nian metrica2d2 + a2 sin2 d2

    This metric is nothing but first quadratic form on the sphere (see (1.4.3)).The volume element is

    det gdd =

    det

    a2 00 a2 sin

    dd = a2 sin dd

    Now calculate the volume of the segment of the sphere between two parallelplanes, i.e. domain restricted by parallels 1 0: Denote by h be theheight of this segment. One can see that

    h = a cos 0 a cos 1 = a(cos 0 a cos 1)There is remarkable formula which express the area of segment via the heighth:

    V =

    10

    (a2 sin

    )dd =

    10

    20

    (a2 sin

    )d

    d =

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    0

    12a2 sin d = 2a2(cos

    0 cos

    1) = 2a(a cos

    0 acos

    1) = 2ah

    (1.93)E.g. for all the sphere h = 2a. We come to S = 4a2. It is remarkableformula: area of the segment is a polynomial function of radius of the sphereand height (Compare with formula for length of the arc of the circle)

    2 Covariant differentiaion. Connection. LeviCivita Connection on Riemannian mani-

    fold2.1 Differentiation of vector field along the vector field.

    Affine connection

    How to differentiate vector fields on a (smooth )manifold M?Recall the differentiation of functions on a (smooth )manifold M.Let X = Xi(x)ei(x) =

    xi

    be a vector field on M. Recall that vectorfield 3 X = Xiei defines at the every point x0 an infinitesimal curve: x

    i(t) =xi0 + tX

    i (More exactly the equivalence class [(t)]X of curves xi(t) = xi0 +

    tXi + . . . ).

    Let f be an arbitrary (smooth) function on M and X = Xi xi . Thenderivative of function f along vector field X = Xi

    xiis equal to

    Xf = Xf = Xi fxi

    The geometrical meaning of this definition is following: If X is a velocityvector of the curve xi(t) at the point xi0 = x

    i(t) at the time t = 0 thenthe value of the derivative Xf at the point xi0 = xi(0) is equal just to thederivative by t of the function f(xi(t)) at the time t = 0:

    if X

    i

    (x)x0=x(0) = dxi(t)

    dt t=0, then Xfxi=xi(0) = ddtf(xi (t)) t=0(2.1)

    Remark In the course of Geometry and Differentiable Manifolds theoperator of taking derivation of function along the vector field was denoted

    3here like always we suppose by default the summation over repeated indices. E.g.X =Xiei is nothing but X =

    ni=1 X

    iei

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    by Xf. In this course we prefer to denote it by

    Xf to have the uniform

    notation for both operators of taking derivation of functions and vector fieldsalong the vector field.

    One can see that the operation X on the space C(M) (space of smoothfunctions on the manifold) satisfies the following conditions:

    X (f + g) = aXf+ bXg where , R (linearity over numbers)

    hX+gY(f) = hX(f) + gY(f) (linearity over the space of functions)

    X(f g) = f

    X(g) + g

    X(f) (Leibnitz rule)

    (2.2)

    Remark One can prove that these properties characterize vector fields:operatoron smooth functions obeying the conditions above is a vector field. (You willhave a detailed analysis of this statement in the course of Differentiable Man-ifolds.)

    How to define differentiation of vector fields along vector fields.The formula (2.1) cannot be generalised straightforwardly because vec-

    tors at the point x0 and x0 + tX are vectors from different vector spaces.(We cannot substract the vector from one vector space from the vector fromthe another vector space, because apriori we cannot compare vectors fromdifferent vector space. One have to define an operation of transport of vec-tors from the space Tx0M to the point Tx0+tXM defining the transport fromthe point Tx0M to the point Tx0+tXM).

    Try to define the operation on vector fields such that conditions (2.2)above be satisfied.

    2.1.1 Definition of connection. Christoffel symbols of connection

    Definition Affine connection on M is the operation which assigns to everyvector field X a linear map, (but not necessarily C(M)-linear map!) (i.e. amap which is linear over numbers not necessarily over functions) X on thespace O(M) of vector fields:

    X (Y + Z) = XY + XZ, for every , R (2.3)

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    (Compare the first condition in (2.2)).

    which satisfies the following conditions:

    for arbitrary (smooth) functions f, g on M

    fX+gY (Z) = fX (Z) + gY (Z) (C(M)-linearity) (2.4)

    (compare with second condition in (2.2))

    for arbitrary function f

    X (fY) = (

    Xf) Y + f

    X (Y) (Leibnitz rule) (2.5)

    Recall that Xf is just usual derivative of a function f along vectorfield: Xf = Xf.(Compare with Leibnitz rule in (2.2)).

    The operationXY is called covariant derivative of vector fieldY alongthe vector field X.

    Write down explicit formulae in a given local coordinates {xi} (i =1, 2, . . . , n) on manifold M.

    Let

    X = Xiei = Xi xi

    Y = Yiei = Yi xi

    The basis vector fields xi

    we denote sometimes by i sometimes by eiUsing properties above one can see that

    XY = XiiYkk = Xi(i (Ykk)) , where i = i (2.6)

    Then according to (2.4)

    i(

    Ykk

    )= i

    (Yk

    )k + Y

    kik

    Decompose the vector field ik over the basis i:ik = mikm (2.7)

    and

    i(

    Ykk)

    =Yk(x)

    xik + Y

    kmikm, (2.8)

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    XY = X

    i Ym(x)

    xim + X

    iYkmikm, (2.9)

    In components

    (XY)m = Xi

    Ym(x)

    xi+ Ykmik

    (2.10)

    Coefficients {mik} are called Christoffel symbols in coordinates {xi}. Thesecoefficients define covariant derivativeconnection.

    If operation of taking covariant derivative is given we say that the con-nection is given on the manifold. Later it will be explained why we us theword connection

    We see from the formula above that to define covariant derivative of vectorfields, connection, we have to define Christoffel symbols in local coordinates.

    2.1.2 Transformation of Christoffel symbols for an arbitrary con-nection

    Let be a connection on manifold M. Let {ikm} be Christoffel symbolsof this connection in given local coordinates {xi}. Then according (2.7) and(2.8) we have

    XY = X

    m Yi

    xm

    xi

    + XmimkYk

    xi,

    and in particularlyimki = mk

    Use this relation to calculate Christoffel symbols in new coordinates xi

    i

    mki = mkWe have that m =

    xm = x

    m

    xm

    xm

    = xm

    xm m. Hence due to properties

    (2.4), (2.5) we have

    i

    m

    k

    i

    = m k

    =

    m xkxk k = xk

    xkmk +

    xm xkxk k =

    xk

    xk

    xm

    xmm

    k +2xk

    xmxkk =

    xk

    xkxm

    xmmk +

    2xk

    xmxkk

    xk

    xkxm

    xmimki +

    2xk

    xmxkk =

    xk

    xkxm

    xmimk

    xi

    xii +

    2xk

    xmxkxi

    xki

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    Comparing the first and the last term in this formula we come to the trans-

    formation law:If{ikm} are Christoffel symbols of the connection in local coordinates

    {xi} and {ikm} are Christoffel symbols of this connection in new localcoordinates {xi} then

    i

    km =xk

    xkxm

    xmxi

    xiikm +

    2xr

    xkxmxi

    xr(2.11)

    Remark Christoffel symbols do not transform as tensor. If the secondterm is equal to zero, i.e. transformation of coordinates are linear (see theProposition on flat connections) then the transformation rule above is the the

    same as a transformation rule for tensors of the type1

    2

    (see the formula

    (1.6)). In general case this is not true. Christoffel symbols does not trans-form as tensor under arbitrary non-linear coordinate transformation: see thesecond term in the formula above.

    2.1.3 Canonical flat affine connection

    It follows from the properties of connection that it is suffice to define con-nection at vector fields which form basis at the every point using (2.7), i.e.

    to define Christoffel symbols of this connection.Example Consider n-dimensional Euclidean space En with cartesian co-

    ordinates {x1, . . . , xn}.Define connection such that all Christoffel symbols are equal to zero in

    these cartesian coordinates {xi}.eiek = mikem = 0, mik = 0 (2.12)

    Does this mean that Christoffel symbols are equal to zero in an arbitrarycartesian coordinates if they equal to zero in given cartesian coordinates?

    Does this mean that Christoffel symbols of this connection equal to zero

    in arbitrary coordinates system?To answer these questions note that the relations (2.12) mean that

    XY = Xm Yi

    xm

    xi(2.13)

    in coordinates {xi}

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    Consider an arbitrary new coordinates xi

    = xi

    (x1, . . . , xn). Recall the

    transformation rule for an arbitrary vector field (see subsection 1.1)

    R = Rm

    xm= Rm

    xm

    xm

    xm, i.e.Rm

    =xm

    xmRm , and , Rm =

    xm

    xmRm

    .

    Hence we have from (2.13) that

    XY = Xm Yi

    xm

    xi= Xm

    xm(

    Yi)

    xi= Xm

    xm

    xm

    xm

    xi

    xiYi

    xi=

    Xm

    xm

    xi

    xi

    Yi

    xi

    = Xm

    xm Yi

    xi

    xi

    xi+Xm

    2xi

    xm

    xi Yi

    xi

    =

    Xm Yi

    xm

    xi+ Xm

    2xi

    xmxiYi

    xi an additional term

    = (2.14)

    We see that an additional term equals to zero for arbitrary vector fields X, Yif and only if the relations between new and old coordinates are linear:

    2xi

    xmxi= 0, i.e. xi = bi + aikx

    k (2.15)

    Comparing formulae (2.15) and (2.13) we come to simple but very importantProposition Let all Christoffel symbols of a given connection be equal to

    zero in a given coordinate system {xi}. Then all Christoffel symbols of thisconnection are equal to zero in an arbitrary coordinate system {xi} such thatthe relations between new and old coordinates are linear:

    xi

    = bi + aikxk (2.16)

    If transformation to new coordinate system is not linear, i.e. 2xi

    xmxi

    = 0then Christoffel symbols of this connection in general are not equal to zero innew coordinate system

    {xi

    }.

    Definition We call connection flat if there exists coordinate systemsuch that all Christoffel symbols of this connection are equal to zero in agiven coordinate system.

    In particular connection (2.12) has zero Christoffel symbols in arbitrarycartesian coordinates.

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    Corollary Connection has zero Christoffel symbols in arbitrary Cartesian

    coordinates if it has zero Christoffel symbols in a given Cartesian coordinates.Hence the following definition is correct:

    Definition A connection on En which Christoffel symbols vanish in carte-sian coordinates is called canonical flat connection.

    Remark Canonical flat connection in Euclidean space is uniquely defined,sincce cartesian coordinates are defined globally. On the other hand on arbitrary

    manifold one can define flat connection locally just choosing any arbitrary local

    coordinates and define locally flat connectionby condition that Christoffel symbols

    vanish in these local coordinates. This does not mean that one can define flat

    connection globally. We will study this question after learning transformation law

    for Christoffel symbols.

    Remark One can see that flat connection is symmetric connection.

    Example Consider a connection (2.12) in E2. It is a flat connection.Calculate Christoffel symbols of this connection in polar coordinates{

    x = r cos

    y = y sin

    {r =

    x2 + y2

    = arctan yx

    (2.17)

    Write down Jacobians of transformationsmatrices of partial derivatives:

    xr yrx y

    =

    cos sin

    r sin r cos

    ,

    rx xry y

    =

    xx2+y2 yx2+y2y

    x2+y2x

    x2+y2

    (2.18)

    According (2.11) and since Chrsitoffel symbols are equal to zero in cartesiancoordinates (x, y) we have

    i

    km =2xr

    xkxmxi

    xr, (2.19)

    where (x1

    , x2

    ) = (x, y) and (x1

    , x2

    ) = (r, ). Now using (2.18) we have

    rrr =2x

    rr

    r

    x+

    2y

    rr

    r

    y= 0

    rr = rr =

    2x

    r

    r

    x+

    2y

    r

    r

    y= sin cos + sin cos = 0 .

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    r =2x

    r

    x+

    2y

    r

    r

    y=

    x

    x

    r y

    y

    r=

    r .

    rr =2x

    rr

    x+

    2y

    rr

    y= 0 .

    r = r =

    2x

    r

    x+

    2y

    r

    y= sin y

    r2+ cos

    x

    r2=

    1

    r

    =2x

    x+

    2y

    y= xx

    r2 y y

    r2= 0 . (2.20)

    Hence we have that the covariant derivative (2.13) in polar coordinates hasthe following appearance

    rr = rrrr + rr = 0 , , r = rrr + r =r

    r = rrr + r =r

    , = rr + = rr (2.21)Remark Later when we study geodesics we will learn a very quick methodto calculate Christoffel symbols.

    2.1.4 Global aspects of existence of connection

    We defined connection as an operation on vector fields obeying the special axioms(see the subsubsection 2.1.1). Then we showed that in a given coordinates con-nection is defined by Christoffel symbols. On the other hand we know that ingeneral coordinates on manifold are not defined globally. (We had not this troublein Euclidean space where there are globally defined cartesian coordinates.)

    How to define connection globally using local coordinates? Does there exist at least one globally defined connection? Does there exist globally defined flat connection?

    These questions are not naive questions. Answer on first and second questionsis Yes. It sounds bizzare but answer on the first question is not Yes 4

    Global definition of connection

    4Topology of the manifold can be an obstruction to existence of global flat connection.E.g. it does not exist on sphere Sn if n > 1.

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    The formula (2.11) defines the transformation for Christoffer symbols if we go

    from one coordinates to another.Let {(xi), U} be an atlas of charts on the manifold M.If connection is defined on the manifold M then it defines in any chart (local

    coordinates) (xi) Christoffer symbols which we denote by ()ikm. If (x

    i), (x

    i

    ())

    are different local coordinates in a vicinity of a given point then according to (2.11)

    ()i

    km =xk()

    xk

    ()

    xm()

    xm

    ()

    xi

    ()

    xi() ()

    ()imk +

    2xk()

    xm

    ()xk

    ()

    xi

    ()

    xk()

    (2.22)

    Definition Let {(xi), U} be an atlas of charts on the manifold MWe say that the collection of Christoffel symbols

    {

    ()ikm

    }defines globally a

    connection on the manifold M in this atlas if for every two local coordinates(xi()), (x

    i()) from this atlas the transformation rules (2.22) are obeyed.

    Using partition of unity one can prove the existence of global connection con-structing it in explicit way. Let {(xi), U} ( = 1, 2, . . . , N ) be a finite atlas onthe manifold M and let {} be a partition of unity adjusted to this atlas. Denoteby ()ikm local connection defined in domain U such that its components in these

    coordinates are equal to zero. Denote by()()

    ikm Christoffel symbols of this local

    connection in coordinates (xi()) (()()

    ikm = 0). Now one can define globally theconnection by the formula:

    ()ikm(x) =

    (x)()()

    ikm(x) =

    (x) xi()

    xi

    ()

    2

    xi

    ()(x)xk

    ()xm

    ()

    . (2.23)

    This connection in general is not flat connection5

    2.2 Connection induced on the surfaces

    Let M be a manifold (surface) embedded in Euclidean space6. Canonical flatconnection on EN induces the connection on surface in the following way.

    Let X, Y be tangent vector fields to the surface M and can.flat a canonicalflat connection in EN. In general

    Z = can.flatX Y is not tangent to manifold M (2.24)5See for detail the text: Global affine connection on manifold in my homepage:

    www.maths.mancheser.ac.uk/khudian in subdirectory Etudes/Geometry6We know that every n-dimensional manifodl can be embedded in 2n + 1-dimensional

    Euclidean space

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    Consider its decomposition on two vector fields:

    Z = Ztangent + Z, can.flatX , Y =(can.flatX Y)tangent + (can.flatX Y) , (2.25)

    where Z is a component of vector which is orthogonal to the surface Mand Z|| is a component which is tangent to the surface. Define an inducedconnection M on the surface M by the following formula

    M : MX Y : =(can.flatX Y)tangent (2.26)

    Remark One can imply this construction for an arbitrary connection inEN.

    2.2.1 Calculation of induced connection on surfaces in E3.

    Let r = r(u, v) be a surface in E3. Let can.flat be a flat connection in E3.Then

    M : MX Y : =(can.flatX Y)|| = can.flatX Y n(can.flatX Y, n), (2.27)

    where n is normal unit vector field to M. Consider a special exampleExample (Induced connection on sphere) Consider a sphere of the radius

    R in E3:

    r(, ) :

    x = R sin cos y = R sin sin

    z = R cos

    then

    r =

    R cos cos R cos sin

    R sin

    , r =

    R sin sin R sin cos

    0

    , n =

    sin cos sin sin

    cos

    ,

    where r =r(,)

    , r =

    r(,)

    are basic tangent vectors and n is normal unit

    vector.Calculate an induced connection on the sphere.First calculate .

    =

    r

    tangent

    = (r)tangent .

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    On the other hand one can see that r = Rsin cos Rsin sin R cos

    = Rn isproportional to normal vector, i.e. (r)tangent = 0. We come to

    = (r)tangent = 0 = = 0 . (2.28)Now calculate and .

    =

    r

    tangent

    = (r)tangent , =

    r

    tangent

    = (r)tangent

    We have

    = = (r)tangent =R cos sin R cos cos

    0

    tangent

    .

    We see that the vector r is orthogonal to n:

    r, n = R cos sin sin cos + R cos cos sin sin = 0.Hence

    = = (r)tangent = r =R cos sin R cos cos

    0

    = cotan r .We come to

    = = cotan = = 0, = = cotan (2.29)Finally calculate

    = (r)tangent =R sin cos R sin sin

    0

    tangent

    Projecting on the tangent vectors to the sphere (see (2.27)) we have

    = (r)tangent = r nn, r =

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    R sin cos R sin sin 0

    sin cos

    sin sin cos

    (R sin cos sin cos R sin sin sin sin ) = sin cos

    R cos cos R cos sin

    R sin

    = sin cos r,

    i.e.

    = sin cos r = sin cos , = = 0 . (2.30)

    2.3 Levi-Civita connection

    2.3.1 Symmetric connection

    Definition. We say that connection is symmetric if its Christoffel symbolsikm are symmetric with respect to lower indices

    ikm = imk (2.31)

    The canonical flat connection and induced connections considered above aresymmetric connections.

    Invariant definition of symmetric connection

    A connection is symmetric if for an arbitrary vector fields X, YXY YX [X, Y] = 0 (2.32)

    If we apply this definition to basic fields k, m which commute: [k, m] = 0 wecome to the condition

    km mk = imki ikmi = 0and this is the condition (2.31).

    2.3.2 Levi-Civita connection. Theorem and Explicit formulae

    Let (M, G) be a Riemannian manifold.Definition. TheoremA symmetric connection is called Levi-Civita connection if it is com-

    patible with metric, i.e. if it preserves the scalar product:

    XY, Z = XY, Z + Y, XZ (2.33)

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    for arbitrary vector fields X, Y, Z.

    There exists unique levi-Civita connection on the Riemannian manifold.In local coordinates Christoffel symbols of Levi-Civita connection are given

    by the following formulae:

    imk =1

    2gij

    gjmxk

    +gjkxm

    gmkxj

    . (2.34)

    where G = gikdxidxk is Riemannian metric in local coordinates and ||gik|| is

    the matrix inverse to the matrix ||gik||.ProofSuppose that this connection exists and imk are its Christoffel symbols. Con-

    sider vector fields X = m, Y = i and Z = k in (2.33). We have that

    mgik = rmir, k + i, rmkr = rmigrk + girrmk . (2.35)

    for arbitrary indices m,i,k.Denote by mik =

    rmigrk we come to

    mgik = mik + mki, i.e.

    Now using the symmetricity mik = imk since kmi =

    kim we have

    mik = mgik

    mki = mgik

    kmi = mgik

    (kgmi

    kim) =

    mgik kgmi + kim = mgik kgmi + ikm = mgik kgmi + (igkm imk) =mgik kgmi + igkm mik .

    Hence

    mik =1

    2(mgik + igmk kgmi) kim =

    1

    2gkr (mgir + igmr rgmi) (2.36)

    We see that if this connection exists then it is given by the formula(2.34).On the other hand one can see that (2.34) obeys the condition (2.35). We

    prove the uniqueness and existence.

    since ik = mikm.Consider examples.

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    2.3.3 Levi-Civita connection on 2-dimensional Riemannian mani-

    fold with metric G = adu2 + bdv2.

    Example Consider 2-dimensional manifold with Riemannian metrics

    G = a(u, v)du2 + b(u, v)dv2, G =

    g11 g12g21 g22

    =

    a(u, v) 0

    0 b(u, v)

    Calculate Christoffel symbols of Levi Civita connection.Using (2.36) we see that:

    111 =12

    (1g11 + 1g11 1g11) = 121g11 = 12au

    211 = 121 =12

    (1g12 + 2g11 1g12) = 122g11 = 12av

    221 =12

    (2g12 + 2g12 1g22) = 121g22 = 12bu

    112 =12

    (1g12 + 1g12 2g11) = 122g11 = 12av

    122 = 212 =12

    (2g21 + 1g22 2g21) = 121g22 = 12bu

    222 =12

    (2g22 + 2g22 2g22) = 122g22 = 12bv

    (2.37)

    To calculate ikm = girkmr note that for the metric a(u, v)du2 + b(u, v)dv2

    G1 =

    g11 g12

    g21 g22

    =

    1

    a(u,v)0

    0 1b(u,v)

    Hence

    111 = g11111 =

    au2a

    , 121 = 112 = g

    11121 =av2a

    , 122 = g11221 =

    bu2a

    211 = g22112 =

    av2b

    , 221 = 212 = g

    22122 =bu2b

    , 222 = g22222 =

    bv2b

    (2.38)

    2.3.4 Example of the sphere again

    Calculate Levi-Civita connection on the sphere.On the sphere first quadratic form (Riemannian metric) G = R2d2 +

    R2 sin2 d2. Hence we use calculations from the previous example with

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    a(, ) = R2, b(, ) = R2 sin2 (u = , v = ). Note that a = a = b = 0.

    Hence only non-trivial components of will be:

    =b2a

    = sin2

    2,

    =

    R2 sin22

    , (2.39)

    = =

    b2b

    =cos

    sin

    =

    R2 sin2

    2

    (2.40)

    All other components are equal to zero:

    = =

    =

    =

    = 0

    Remark Note that Christoffel symbols of Levi-Civita connection on thesphere coincide with Christoffel symbols of induced connection calculated inthe subsection Connection induced on surfaces. later we will understandthe geometrical meaning of this fact.

    2.4 Levi-Civita connection = induced connection onsurfaces in E3

    Christoffel symbols of canonical flat connection vanish in Cartesian coordi-nates. On the other hand standard Riemannian metric on Euclidean space

    has the appearance: G = dxiikdxk. The matrix gik has constant entries.Hence according to Levi-Civita formula (2.34) the Christoffel symbols ofLevi-Civita connection vanish also. Hence we proved very important fact:

    Canonical flat connection of Euclidean space is the Levi-Civita connectionof the standard metric on Euclidean space.

    Now we show that Levi-Civita connection on surfaces in Euclidean spacecoincides with the connection induced on the surfaces by canonical flat con-nection.

    We perform our analysis for surfaces in E3.Let M: r = r(u, v) be a surface in E3. Let G be induced Riemannian

    metric on M and LeviCivita connection of this metric.We know that the induced connection (M) is defined in the following

    way: for arbitrary vector fields X, Y tangent to the surface M, MX Y equalsto the projection on the tangent space of the vector field can.flatX Y:

    MX Y =(can.flatX Y)tangent ,

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    where

    can.flat is canonical flat connection in E3 (its Christoffel symbols

    vanish in cartesian coordinates). We denote by Atangent a projection ofthe vector A attached at the point of the surface on the tangent space:A = A n(A, n) , (n is normal unit vector field to the surface.)

    Theorem Induced connection on the surface r = r(u, v) in E3 coincideswith Levi-Civita connection of Riemannian metric induced by the canonicalmetric on Euclidean space E3.

    Proof

    Let M be induced connection on a surface M in E3 given by equationsr = r(u, v). Considering this connection on the basic vectors rh, rv we seethat it is symmetric connection. Indeed

    Muv = (ruv)tangent = (rvu)tangent = Mvu . uuv = uvu, vuv = vvu .

    Prove that this connection preserves scalar product on M. For arbitrarytangent vector fields X, Y, Z we have

    XY, ZE3 = can.flatX Y, ZE3 + Y, can.flatX ZE3 .

    since canonical flat connection in E3 preserves Euclidean metric in E3 (itis evident in Cartesian coordinates). Now project the equation above on

    the surface M. If A is an arbitrary vector attached to the surface andAtangent is its projection on the tangent space to the surface, then for ev-ery tangent vector B scalar product A, BE3 equals to the scalar productAtangent, BE3 = Atangent, BM since vector A Atangent is orthogonal tothe surface. Hence we deduce from (2) that XY, ZM =

    (can.flatX Y)tangent , ZE3+Y, (can. flatX Z)tangentE3 = MX Y, ZM+Y, MX ZM .We see that induced connection is symmetric connection which preservesthe induced metric. Hence due to Levi-Civita Theorem it is unique and isexpressed as in the formula (2.34).

    Remark One can easy to reformulate and prove more general statement: Let M bea submanifold in Riemannian manifold (E, G). Then Levi-Civita connection of the metric

    induced on this submanifold coincides with the connection induced on the manifold by

    Levi-Civita connection of the metric G.

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    3 Parallel transport and geodesics

    3.1 Parallel transport

    3.1.1 Definition

    Let M be a manifold equipped with affine connection .Definition Let C: x(t) with coordinates xi = xi(t), t0 t t1 be a

    curve on the manifold M. Let X = X(t0) be an arbitrary tangent vectorattached at the initial point x0 (with coordinates x

    i(t0)) of the curve C, i.e.X(t0) Tx0M is a vector tangent to the manifold M at the point x0 withcoordinates xi(t0). (The vector X is not necessarily tangent to the curve C)

    We say that X(t), t0 t t1 is a parallel transport of the vector X(t0) Tx0M along the curve C: x

    i = xi(t), t0 t t1 if

    For an arbitrary t, t0 t t, vector X = X(t), (X(t)|t=t0 = X(t0)) isa vector attached at the point x(t) of the curve C, i.e. X(t) is a vectortangent to the manifold M at the point x(t) of the curve C.

    The covariant derivative of X(t) along the curve C equals to zero:

    Xdt

    = vX = 0 . (3.1)In components: if Xm(t) are components of the vector field X(t) andvm(t) are components of the velocity vector v of the curve C ,

    X(t) = Xm(t)

    xm|x(t) , v = dx(t)

    dt=

    dxi

    dt

    xm|x(t)

    then the condition (3.1) can be rewritten as

    dXi(t)

    dt+ vk(t)ikm(x

    i(t))Xm(t) 0 . (3.2)

    Remark We say sometimes that X(t) is covariantly constant along thecurve C If X(t) is parallel transport of the vector X along the curve C. Ifwe consider Euclidean space with canonical flat connection then in Cartesiancoordinates Christoffel symbols vanish and parallel transport is nothing butdXdt

    = vX = 0, X(t) is constant vector.

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    Remark Compare this definition of parallel transport with the defini-

    tion which we consider in the course of Introduction to Geometry wherewe consider parallel transport of the vector along the curve on the surfaceembedded in E3 and define parallel transport by the condition, that onlyorthogonal component of the vector changes during parallel transport, i.e.dX(t)dt

    is a vector orthogonal to the surface (see the Exercise in the Homework7).

    3.1.2 Parallel transport is a linear map

    Consider two different points x0, x1 on the manifold M with connection .Let C be a curve x(t) joining these points. The parallel transport (3.1) X(t)defines the map between tangent vectors at the point x0 and tangent vectorsat the point x1. This map depends on the curve C. Parallel transport alongdifferent curves joining the same points is in general different (if we are notin Euclidean space).

    On the other hand parallel transport is a linear map of tangent spaceswhich does not depend on the parameterisation of the curve joining thesepoints.

    Proposition Let X(t), t0 t t1 be a parallel transport of the vectorX(t0) Tx0M along the curve C: x = x(t), t0 t t1, joining the pointsx

    0= x(t

    0) and x

    1= x(t

    1). Then the map

    C: Tx0M X(t0) X(t1) Tx1M (3.3)is a linear map from the vector space Tx0M to the vector space Tx1M whichdoes not depend on the parametersiation of the curve C.

    The fact that the map (3.3) does not depend on the parameterisationfollows from the differential equation (3.2) also.

    Indeed let t = t(), 0 1, t(0) = t0, t(1) = 1 be anotherparameterisation of the curve C. Then multiplying the equation (3.2) ondtd

    and using the fact that velocity v() = tv(t) we come to differential

    equation:

    dXi(t())

    d+ vk(t())ikm(x

    i(t()))Xm(t()) 0 . (3.4)

    The functions X(t()) with the same initial conditions are the solutionsof this equation.

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    The fact that it is a linear map follows immediately from the fact that

    differential equations (3.2) are linear. E.g. let vector fields X(t), Y(t) becovariantly constant along the curve C. Then since linearity X(t) + Y(t) isa solution too.

    3.1.3 Parallel transport with respect to Levi-Civita connection

    We mostly consider parallel transport on Riemannian manifold. If (M, G) isRiemannian manifold we mostly consider parallel transport with respect toconnection which is Levi-Civita connection of the Riemannian metric G

    Proposition The length of the vector preserves during parallel transportwith respect to Levi-Civita connection.

    Proof follows immediately from the definition (3.1) of aprallel traspotrand the definition (2.33) of Levi-Civita connection

    d

    dtX(t), X(t) = vX(t), X(t) = 2 vX(t), X(t) = 2 0, X(t) = 0 .

    (3.5)Hence |X(t)| is constant.

    Exercise Show that scalar product preserves during parallel transport.

    3.2 Geodesics

    3.2.1 Definition. Geodesic on Riemannian manifold.

    Let M be manifold equipped with connection .Definition A parameterised curve C: xi = xi(t) is called geodesic if

    velocity vector v(t) : vi(t) = dxi(t)dt

    is covariantly constant along this curve,i.e. it remains parallel along the curve:

    vv = vdt

    =dvi(t)

    dt+ vk(t)ikm(x(t))v

    m(t) = 0, i.e. (3.6)

    d2xi(t)

    dt2 +dxk(t)

    dt ikm(x(t))

    dxm(t)

    dt = 0 .

    These are linear second order differential equations. One can prove thatthis equations have solution and it is unique7 for an arbitrary initial data(xi(t0) = x

    i0, x

    i(t0) = xi0. )

    7this is true under additional technical conditions which we do not discuss here

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    In other words the curve C: x(t) is a geodesic if parallel transport of

    velocity vector to along the curve is a velocity vector at any point of thecurve.

    Geodesics defined with Levi-Civita connection on the Riemannian mani-fold is called geodesic on Riemannian manifold. We mostly consider geodesicson Riemannian manifolds.

    Since velocity vector of the geodesics on Riemannian manifold at anypoint is a parallel transport with the Levi-Civita connection, hence due toProposition (3.5) the length of the velocity vector remains constant:

    Proposition If C: x(t) is a geodesics on Riemannian manifold then thelength of velocity vector is preserved along the geodesic.

    Proof Since the connection is Levi-Civita connection then it preserves