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Saturday, March 25, 2023

# Yardsticks to Metric Tensor Fields

I requested myself why completely different scientists perceive the identical factor seemingly in a different way, particularly the idea of a metric tensor. If we ask a topologist, a classical geometer, an algebraist, a differential geometer, and a physicist “What’s a metric?” then we get 5 completely different solutions. I imply it’s all about distances, isn’t it? “Sure” continues to be the reply and all do truly imply the identical factor. It’s their perspective that’s completely different. This text is meant to elucidate how.

The picture reveals medieval requirements of comparability at a church in Regensburg, Germany, Schuh (shoe)Elle (ulna), and Klafter.

## Topology

The topologist has in all probability the best idea. A metric is a property of sure topological areas, metric areas. It signifies that the topological house is provided with a yardstick. A operate that measures distances. The situations are clear:

### Symmetry

We don’t hassle whether or not we measure from proper to left or from left to proper.

### Constructive Definiteness

The gap is rarely destructive, and 0 if and provided that there isn’t any distance between our factors, i.e. it’s the identical level. Sounds foolish, however there are certainly unique topological areas through which this isn’t a triviality. We simply don’t wish to hassle about these pathological examples.

### Triangle Inequality

If we’re going from one level to a different and making a cease at a 3rd level that isn’t in our direct method, then it’s a detour.

Everyone will perceive that this can be a passable description of what our yardstick is meant to do.

## Classical Geometry

Not so the classical geometer! He doesn’t care about distances. His enterprise is comparisons. No, not shorter or longer. His view is how usually one distance suits into one other distance. One might argue that that is nonetheless the idea of a yardstick: how usually does a yard match right into a distance? Nevertheless, the distinction is delicate. The classical geometer doesn’t wish to outline a unit, be it a yard or a meter. He at all times compares any two lengths and tells us the quotient of them, e.g. the golden ratio is outlined by
\$\$
\$\$
The traditional Greeks didn’t solely take care of distances. If we have a look at the trigonometric capabilities, particularly quotients of two lengths once more, we get to the opposite vital measure in classical geometry: angles and therewith triangles, ideally proper ones. We now have to think about instructions, too.

Angles are clearly associated to the multiplication of vectors, and that calls for two different vital situations, the distributive legal guidelines.

##langle vec{a}+vec{b},vec{c}rangle=langle vec{a},vec{c} rangle +langle vec{b},vec{c} rangle## and ##langle vec{a},vec{b}+vec{c} rangle =langle vec{a},vec{b} rangle +langle vec{a},vec{c} rangle##

### Stretches and Compressions

i.e. Scalar Multiples

## alpha langle vec{a},vec{b} rangle =langle alpha vec{ a},vec{b} rangle= langle vec{a},alpha vec{b} rangle##

If we now outline a size by
\$\$
|vec{a}| :=sqrt{langle vec{a},vec{a} rangle}=sqrt{sum_{ok=1}^n a_k^2},
\$\$
we get a metric within the topological sense
\$\$
d(vec{a},vec{b}):=|vec{a}-vec{b}|=sqrt{sum_{ok=1}^n (a_k-b_k)^2}.
\$\$
and an angle
\$\$
cos(sphericalangle(vec{a},vec{b}))=dfrac{langle vec{a},vec{b} rangle}{|vec{a}|cdot |vec{b}|}= dfrac{sum_{ok=1}^na_kb_k}{sqrt{sum_{ok=1}^n a_k^2}sqrt{sum_{ok=1}^n b_k^2}}.
\$\$

## Summary Algebra

An algebraist is within the constructions, not a lot in coordinates, so we outline a metric tensor by the properties above fairly than coordinates. A metric tensor for a vector house ##V## is due to this fact merely a component of ##V^*otimes V^*.## Nicely, that is true, brief, and doesn’t assist us in any respect. What we wish to outline is an interior product ##langle , . ,, , ,, . , rangle .## That, at its core, is a symmetric, constructive particular, bilinear mapping
\$\$
g, : ,V instances V longrightarrow mathbb{R}
\$\$
Such a mapping will be written as a matrix ##G,## i.e. ##g(vec{a},vec{b})=vec{a}cdot G cdotvec{b}.## We’ve got thought-about ##g(vec{a},vec{b})=vec{a}cdot operatorname{Id}cdotvec{b}## thus far, the place ##operatorname{Id}## is the id matrix. Nevertheless, there isn’t any motive to limit ourselves to the id matrix. Any constructive particular matrix ##G## will do. That is principally saying ##gin V^*otimes V^*.## In a while, we are going to even drop this situation and solely demand that ##G## is non-degenerate, i.e. that for each vector ##vec{a}neq vec{0}## there’s a vector ##vec{b}## such that ##g(vec{a},vec{b})neq vec{0}.## With ##G=operatorname{Id}## we’ve got a particular instance of a vector house ##V=mathbb{R}^n## that provides the development a that means.

The metric tensor ##g_pin V^*otimes V^*## above an affine level house ##A={p}+V## with translation house ##V## is a mapping that assigns to ##{p}## a symmetric, constructive particular, bilinear mapping ##g_p.## Size, distance and angle for ##vec{a},vec{b}in V ## are actually given as

start{align*}
|vec{a}|_p&=sqrt{g_p(vec{a},vec{a})}
d_p(vec{a},vec{b})&=|vec{a}-vec{b}|_p
cos(sphericalangle_p(vec{a},vec{b}))&=dfrac{g_p(vec{a},vec{b})}{|vec{a}|_pcdot|vec{b}|_p}
finish{align*}
If we’ve got an affine linear monomorphism ##varphi , : ,{p}+U rightarrowtail {varphi(p)}+V## then any metric tensor on ##{varphi(p)}+V## defines a metric tensor on ##{p}+U## by the definition

\$\$(varphi^* g)_p(vec{u}_1,vec{u}_2):=g_{varphi(p)}(varphi_*(vec{u}_1),varphi_*(vec{u}_2)).\$\$

This pullback, or likewise the truth that ##g_pin T(V^*),## is the explanation the algebraist calls the metric tensor contravariant: Given one other monomorphism ##psi , : ,{varphi(p)} +V rightarrowtail {psi(varphi(p))}+W## then ##(psi circvarphi)^*(g_p)=g_{varphi(p)}circ g_{psi(varphi(p))}=(varphi^*circ psi^*)(g_p).## This alteration of order means for mathematicians contravariance. (Consideration: physicists name it covariant for various causes, see under!)

You will need to discover that each one three portions depend upon the purpose ##p.##

## Manifolds

The entire above has been fairly theoretical thus far. Positive, the topologist might clarify our yardstick and the traditional Greeks had been actually glorious in Euclidean geometry. However it isn’t till we meet the differential geometer and the physicist to inform us one thing about the actual world, the non-Euclidean realities, and the truth that the algebraist’s perspective was not fully irrelevant. They deal e.g. with saddles:

If we enhance the decision, rotate the saddle a bit, and at last zoom in, and focus on that small marked sq.

then we will faux that this can be a small flat airplane that represents the tangent house sooner or later throughout the sq.. The smaller we select our sq.. the smaller the error in any calculations as a consequence of curvature. After which we will do the identical with any neighboring sq.. We solely demand that the calculations of any overlapping components need to be the identical, for brief: charts, an atlas, and an interior product on ##T_pM##.

That’s it. We simply have understood the idea of Riemannian manifolds the place its native coordinates throughout the sq. are the parts of foundation tangent vectors of the tangent house at a sure level throughout the sq.. Since there isn’t any globally flat coordinate system on the saddle, we’ve got to patch all of the tiny flat squares, our charts, and bind them to an atlas. That’s what the differential geometer does. Be aware the large distinction between a worldwide view (left) and the native view (proper).

One final, however crucial notice: our manifold right here, the saddle, is given by ##M={(x,y,x^2-y^2),|,x,yin mathbb{R}}.## This implies we’ve got two unbiased parameters, ##x## and ##y,## so ##dim M=2,## a floor. And though we’ve got embedded it in our three-dimensional house, we won’t care about this embedding, solely in regards to the factors on ##M.## The one motive for an embedding is, that we will have fancy photos. Individuals on earth don’t acknowledge its curvature until they have a look at the horizon at sea, large grass savannas, or from outer house.

## Physics

Earlier than we take heed to the differential geometer who will inform us one thing about sections and assist clarify the algebraist, allow us to hear what the physicist has to say. Rapidly we discover ourselves in a well-known state of affairs.

The affine house is just ##{p}+V=T_pM,## the tangent house on the level ##p## of the manifold ##M## outlined by the saddle operate ##z=x^2-y^2.## The metric tensor assigns a bilinear kind to ##T_pM.## So, relying on the place we’re, we’ve got size, distance, and angle. However physics is all about measurement. How a lot differs one thing if one thing else occurs? In scientific this implies: We want coordinates! And since we already understood the Riemannian manifold ##M,## they are going to be native coordinates; solely legitimate on ##T_pM## and permitting us an inexpensive measurement: symmetric, bilinear (distribution legal guidelines), and non-degenerate (to permit relativity principle) fairly than constructive definiteness.

We’ve got seen that all of it relies upon (easily) on the situation ##p,## which isn’t any shock since differentiation is an area course of. However we will patch all native views to an atlas. This is the reason we converse of fields as a substitute of areas. A subject is an object obtained by gathering all factors ##pin M.## Due to this fact we’ve got a metric tensor subject, and a tangent vector subject. It’s normally merely referred to as a vector subject, whereas the vector house of tangents is named tangent house.
start{align*}
g=bigsqcup_{pin M}g_p=bigcup_{pin M}{p}instances g_p&quadtext{ metric tensor subject}
TM=bigsqcup_{pin M}T_pM=bigcup_{pin M}{p}instances T_pM&quadtext{ (tangent) vector subject}
finish{align*}

## Coordinates

Let’s think about our saddle ##M={(x,y,x^2-y^2):x,yin mathbb{R}}.## A tangent is the rate vector of a curve ##gamma :[0,1]rightarrow M .## We think about ##gamma^1(t)=(t,p_y,t^2-p_y^2)## and ##gamma^2(t)=(p_x,t,p_x^2-t^2)## by way of a degree ##p=(p_x,p_y,p_z)in M## and get the tangents ##dotgamma^1(t)=(1,0,2t)## and ##dotgamma^2(t)=(0,1,-2t).## They correspond to the vectors fields ##X_1(p)=D_p(x)=(1,0,2p_x)## and ##X_2(p)=D_p(y)=(0,1,-2p_y).##

When it comes to stage units, we’ve got ##f(x,y)=x^2-y^2## and ##(p_x,p_y)in f^{-1}(p_x^2-p_y^2).## Then ##nabla f(p)=(2p_x,-2p_y).## Let additional be ##x,y:[-1,1]rightarrow f^{-1}(p_x^2-p_y^2)## outlined by ##x(t)=(t+p_x,sqrt{(t+p_x)^2-(p_x^2-p_y^2)}),x(0)=(p_x,p_y)## and ##y(t)=(sqrt{(t+p_y)^2+(p_x^2-p_y^2)},t+p_y),y(0)=(p_x,p_y).##
Then
start{align*}
nabla(f(x(0)))cdot dot x(0)&=(2p_x,-2p_y)cdot (1,p_x/p_y)=0
nabla(f(x(0)))cdot dot y(0)&=(2p_x,-2p_y)cdot (p_y/p_x,1)=0
finish{align*}
and the tangent house to the extent set in ##mathbb{R}^2## is ##left{nabla f(p)proper}^perp = operatorname{lin}_mathbb{R}left{(p_y,p_x)proper}##. Thus ##(p_y,p_x,0)=p_yX_1(p)+p_xX_2(p)## is one tangent vector to ##M## at ##p## within the ##z##-plane.

The tangent house with origin in ##p=(2,1,3)## is
\$\$
T_pM=
operatorname{lin}_mathbb{R}left{X_1(p)=left.dfrac{partial }{partial p_x}proper|_p,X_2(p)=left.dfrac{partial }{partial p_y}proper|_pright}=operatorname{lin_mathbb{R}}left{start{pmatrix}14end{pmatrix},start{pmatrix}01-2end{pmatrix} proper}
\$\$
and the vector subject is
\$\$
TM = bigsqcup_{pin M}operatorname{lin}_mathbb{R}left{X_1=dfrac{partial }{partial p_x},X_2=dfrac{partial }{partial p_y}proper}=bigsqcup_{pin M}operatorname{lin}_mathbb{R}left{start{pmatrix}12p_xend{pmatrix},start{pmatrix}01-2p_yend{pmatrix} proper}.
\$\$
Nevertheless, we don’t wish to hassle with the embedding in ##mathbb{R}^3.## The entire thought about manifolds is, that they don’t require an embedding. There isn’t a Euclidean house across the universe, solely the universe itself. Nicely, we begin a bit smaller and see what the saddle will inform us. However who is aware of, possibly the universe is a saddle, say a hyperbolic paraboloid. The metric tensor wants vectors as inputs. Our vectors are the tangent vectors ##{X_1(p),X_2(p)}## which kind a foundation for ##V=T_pM.## The affine house ##A## we’ve got spoken about within the algebra part is just ##A={p}+T_pM.## These areas are all two-dimensional, so ##g_p## may have 4 elements. To be able to illustrate the dependency on location, we outline in keeping with the body ##mathbf{f}={X_1(p),X_2(p)}##
start{align*}
g_p&triangleq G[mathbf{f}]=g_{ij}[mathbf{f}]=start{pmatrix}
cosh^2 p_x +sinh^2 p_y &0&(1+4p_x^2+4p_y^2)^{-1}
finish{pmatrix}.
finish{align*}
If we alter the coordinates in ##T_pM## by
##
U_k=(A_{ij})cdot X_k=sum_{l}a_{lk}X_l={a^l}_kX_l
##
then (altering the body)
start{align*}
g(U_i,U_j)&= g({a^ok}_i X_k,{a^l}_jX_l)={a^ok}_i{a^l}_jg(X_k,X_l)
&={a^ok}_i{a^l}_jg_{kl}[mathbf{f}]=sum_{ok,l}(A_{ki})g_{kl}(A_{lj})[mathbf{f}]=left(A^tau cdot G[mathbf{f}]cdot Aright)_{ij}
finish{align*}
and ##G[Amathbf{f}]=A^tau G[mathbf{f}]A,## i.e. the metric tensor, or fairly its coordinate illustration, is named covariant in physics since its elements rework like the idea below a change of coordinates in every index of the body.

If we alter the coordinate capabilities ##p_x,p_y:Mrightarrow mathbb{R}## to ##q_x,q_y:Mrightarrow mathbb{R}## then
\$\$
dfrac{partial }{partial q_x}=dfrac{partial p_x}{partial q_x}dfrac{partial }{partial p_x}+dfrac{partial p_y}{partial q_x}dfrac{partial }{partial p_y}; , ;dfrac{partial }{partial q_y}=dfrac{partial p_x}{partial q_y}dfrac{partial }{partial p_x}+dfrac{partial p_y}{partial q_y}dfrac{partial }{partial p_y}
\$\$
and with the Jacobi matrix ##Dq(p)##
start{align*}
g_{q(p)}&=gleft(dfrac{partial }{partial q_x},dfrac{partial }{partial q_y}proper)=gleft(dfrac{partial p_x}{partial q_x}dfrac{partial }{partial p_x}+dfrac{partial p_y}{partial q_x}dfrac{partial }{partial p_y}, , ,dfrac{partial p_x}{partial q_y}dfrac{partial }{partial p_x}+dfrac{partial p_y}{partial q_y}dfrac{partial }{partial p_y}proper)
finish{align*}
\$\$
g_{q(p)}=left(left(Dq(p)proper)^{-1}proper)^tau g(p)left(Dq(p)proper)^{-1}
\$\$
The tangent vectors in our instance are already orthogonal and each diagonal entries are in every single place constructive. ##g_p## has due to this fact the signature ##(+,+)## and is a Riemannian metric.

What’s a size in keeping with a metric tensor? The road aspect will be regarded as a line section related to an infinitesimal displacement vector
\$\$
ds^2=sum_{i,j}g_{ij}dx_idx_j=g_{ij}dx^idx^j=g_{ij}dfrac{dx^i}{dt}dfrac{dx^j}{dt}dt^2 =g[mathbf{f}]
\$\$
because it defines the arclength by
\$\$
s=int_a^b sqrtds^2=int_a^bsqrt{left|g_{ij}(gamma(t))left(dfrac{d}{dt}x^i circ gamma(t)proper)left(dfrac{d}{dt}x^j circ gamma(t)proper)proper|},dt.
\$\$
The road aspect fully defines the metric tensor since we thought-about arbitrary curves. That’s the reason it’s usually used as a characterization of the metric tensor fairly than the precise tensor product. In our instance we’ve got (with ##X_1=D_p(x)=dp_x## and ##X_2=D_p(y)=dp_y##)
\$\$
ds^2=g=(cosh^2 p_x +sinh^2 p_y ),dp_x^2+(1+4p_x^2+4p_y^2)^{-1},dp_y^2
\$\$
The notation of the coordinate capabilities as ##p_x,p_y## are supposed to reveal that we’re contemplating actual valued capabilities ##Mto mathbb{R}.## The standard notation is
\$\$
ds^2=g=(cosh^2 x +sinh^2 y ),dx^2+(1+4x^2+4y^2)^{-1},dy^2.
\$\$
The metric tensor at ##(0,0,0)## is given as ##g(0,0,0)=ds^2=dx^2+dy^2## and due to this fact the Euclidean metric. The angle between the 2 foundation vectors is ##cossphericalangleleft( X_1(0,0,0),X_2(0,0,0)proper)=0,## i.e. the idea vectors are orthogonal. At ##p=(2,1,3)## we’ve got a unique metric however nonetheless an orthogonal foundation since ##g(2,1,3)((1,0),(0,1))=0## as a consequence of the truth that ##g[mathbf{f}]## is diagonal and ##[mathbf{f}]## an orthogonal body. However the lengths of the idea vectors are not ##1.##
start{align*}
X_1(2,1,3)&=|(1,0)|_{(2,1,3)}=sqrt{cosh^2(2)+sinh^2(1)} approx 3.94
X_2(2,1,3)&=|(0,1)|_{(2,1,3)}= sqrt{1/21}approx 0.22
finish{align*}

## Differential Geometry

The differential geometer and the physicist share the toolbox, however not essentially the attitude. The principle distinction may be the diploma of abstraction. Emmy Noether spoke of unbiased variables ##p_x,p_y##, dependent capabilities ##q_x(p_x,p_y),q_y(p_x,p_y),## and the applying of a bunch of transformations. Fashionable differential geometers converse of sections and principal fiber bundles, or Euler-Lagrange equations and differential operators. Emmy Noether’s authentic paper [9] is to some lengthen simpler to know than the trendy therapy of her well-known theorem, cp. [11].

The graph of the operate ##f:Blongrightarrow {z}## is ##{(x,y,x^2-y^2)}=B instances {z}=Econg mathbb{R}^3.## ##B## is named the bottom house, right here the ##(x,y)##-plane, and ##E## the overall house, right here the complete ##mathbb{R}^3.## We’ve got a pure projection ##pi:Elongrightarrow B, , ,(x,y,z)mapsto (x,y).## Any clean operate ##sigma :Blongrightarrow E## such that ##pi(sigma (x,y))=(x,y)## is named a bit, and the set of all sections is denoted by ##Gamma(E).## However what does this need to do with differentiation and the saddle? We already launched the (tangent) vector subject, which can also be referred to as tangent bundle or simply vector bundle.

\$\$
E=TM=bigsqcup_{pin M}T_pM=bigcup_{pin M}{p}instances T_pM
\$\$
the place the tangent areas are the bottom house isomorphic to ##mathbb{R}^2## and ##sigma(p_x,p_y)=(p_x,p_y,p_x^2-p_y^2)## are the sections. The preimage ##pi^{-1}(p_x,p_y)={p}instances T_pM## is named fiber (over ##p=(p_x,p_y)##). Bundle as a result of we think about units of factors, i.e. bundles (units) of fibers. That is additionally the explanation why we use the notation with ##sqcup_p.## To be exact, we should always have launched the cotangent bundle as a substitute as a result of our metric tensor is outlined as
\$\$
Gamma((TMotimes TM)^*ni g , , ,g_p in T_p^*Motimes T_p^*M = (T_pMotimes T_pM)^*
\$\$
If we’ve got a bunch working on the fibers then we converse of principal bundles. Emmy Noether’s group of clean transformations is the group that operates on a precept bundle, and its smoothness qualifies it as a Lie group.

## Levi-Civita Connection

A connection is a directional by-product when it comes to vector fields comparable with the gradient. The Levi-Civita connection is torsion-free and preserves the Riemannian metric tensor. It is usually referred to as the covariant by-product alongside a curve
\$\$
nabla_{dotgamma(0)} sigma(p) =(sigma circ gamma )'(0)=lim_{t to 0}dfrac{sigma (gamma (t))-sigma (p)}{t},
\$\$
or Riemannian connection. Its existence and uniqueness are typically referred to as the basic theorem of Riemmanian geometry. An (affine) connection is formally outlined as a bilinear map
\$\$
nabla, : ,Gamma(TM)timesGamma(TM) longrightarrow Gamma(TM), , ,(X,Y)longmapsto nabla_XY
\$\$
that’s ##C^infty (M)##-linear within the first argument, ##mathbb{R}##-linear within the second, and obeys the Leibniz rule ##nabla_X(hsigma )=X(h)cdot sigma +hcdot nabla_X(sigma).## The Levi-Civita connection is an affine connection that’s torsion-free, i.e. ##nabla_XY-nabla_YX=[X,Y]## with the Lie bracket of vector fields, and preserves the metric, i.e.
\$\$
Z(g(X,Y))=g(nabla_Z X,Y)+g(X,nabla_Z Y))textual content{ or brief }nabla_Z g(X,Y)=0.
\$\$
The distinction to a Lie by-product is, {that a} Lie by-product is ##mathbb{R}##-linear however doesn’t need to be ##C^infty (M)##-linear. We additionally point out the Koszul components
start{align*}
g(nabla_{X}Y,Z)&={tfrac{1}{2}}{Large {}X{bigl (}g(Y,Z){bigr )}+Y{bigl (}g(Z,X){bigr )}-Z{bigl (}g(X,Y){bigr )}&phantom{=}+g([X,Y],Z)-g([Y,Z],X)-g([X,Z],Y){Large }}.
finish{align*}
If we write with Christoffel symbols ##Gamma_{ij}^ok##
\$\$
nabla_jpartial_k =Gamma_{jk}^lpartial_l
\$\$
then \$\$
nabla_XY=X^jleft(partial_j(Y^l)+Y^kGamma_{jk}^lpartial_lright) \$\$
and ##nabla## is suitable with the metric if and provided that
\$\$partial_{i}{bigl (}g(partial_{j},partial_{ok}){bigr )}=g(nabla_{i}partial_{j},partial_{ok})+g(partial_{j},nabla_{i}partial_{ok})=g(Gamma_{ij}^{l}partial_{l},partial_{ok})+g(partial_{j},Gamma_{ik}^{l}partial_{l}),\$\$

i.e. ##partial_{i}g_{jk}=Gamma_{ij}^{l}g_{lk}+Gamma_{ik}^{l}g_{jl}.## It’s torsion-free if

\$\$
nabla_{i}partial_{j}-nabla_{j}partial_{i}=(Gamma_{jk}^{l}-Gamma_{kj}^{l})partial_{l}=[partial_{i},partial_{j}]=0,\$\$

i.e. ##Gamma_{jk}^{l}=Gamma_{kj}^{l}.## We are able to test by direct calculation that
\$\$Gamma_{jk}^{l}={tfrac{1}{2}}g^{lr}left(partial_{ok}g_{rj}+partial_{j}g_{rk}-partial_{r}g_{jk}proper)\$\$

the place ##g^{lr}## stands for the inverse metric ##(g_{lr})^{-1}.##

Let ##gamma(t):[-1,1]longrightarrow M## be a clean curve on the manifold ##M.## A piece ##sigma in Gamma(TM)## alongside ##gamma ## is named parallel in keeping with ##nabla## if
\$\$nabla_{dotgamma(t)}sigma (gamma(t))=0 textual content{ for all } t.\$\$
This implies in case of tangent vectors of tangent bundles of a manifold that each one tangent vectors are fixed with respect to an infinitesimal displacement from ##gamma(t)## within the path of ##dotgamma(t).## A vector subject is named parallel whether it is parallel with respect to each curve within the manifold. Let ##t_0,t_1in [-1,1].## Then there’s a distinctive parallel vector subject ##Y:Mrightarrow T_{gamma(t)}M## alongside ##gamma ## for each ##X_pin T_{gamma(t_0)}M## such that ##X_p=Y_{gamma (t_0)}.## The operate
\$\$ P_{gamma(t_{0}),gamma(t_{1})}colon T_{gamma(t_{0})}Mto T_{gamma(t_{1})}M,quad X_pmapsto Y_{gamma(t_{1})} \$\$
is named parallel transport. The existence and uniqueness observe from the worldwide model of the Picard-Lindelöf theorem for odd linear differential equation techniques. We’ve got completely different interior merchandise at ##t_0## and ##t_1,## and thus completely different metrics. A parallel transport determines what occurs alongside a path from one tangent house to the opposite.

We’ve got commuting two-dimensional vector fields in our instance of the two-dimensional saddle ##M## as a result of the partial derivatives commute. This implies we’ve got ##nabla_{X_1}X_2=nabla_{X_2}X_1.## Moreover, the metric is diagonal, i.e. solely stretches and compressions alongside the coordinate capabilities. Let
\$\$
%gamma: [-1,1] rightarrow M, , ,gamma(t)=(2t+1,t+1/2,3(t+1/2)^2)
gamma : [-1,1] rightarrow M, , ,gamma(t)=(2t+1,t+1/2)
\$\$
with ##p=gamma(-1/2)=(0,0)## and ##q=gamma(1/2)=(2,1).## We have to clear up
\$\$
nabla_{(2,1)}sigma(2t+1,t+1/2)= lim_{t’ to t}dfrac{sigma(2t’+1,t’+1/2)-sigma(2t+1,t+1/2)}{t’} =0
\$\$
which is clearly true for a (clean) part of ##Gamma(TM).## The parallel transport is just given by preserving the elements of the vector fields. ##X_p=alpha X_1(p)+beta X_2(p)## maps to ##Y_{q}=alpha X_1(q)+beta X_2(q).## This implies within the three dimensions of the embedding
\$\$
start{pmatrix}alpha beta finish{pmatrix}longmapsto start{pmatrix}2t+1t+1/23(t+1/2)^2end{pmatrix}+start{pmatrix}alpha beta 4alpha t +2alpha -2beta t-beta finish{pmatrix}.
\$\$