Home Physics Yardsticks to Metric Tensor Fields

### 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 area is provided with a yardstick. A perform that measures distances. The circumstances are clear:

### Symmetry

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

### Constructive Definiteness

The gap is rarely damaging, 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 wherein this isn’t a triviality. We simply don’t need to trouble 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’s not in our direct means, then it’s a detour.

Everyone will perceive that it is 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 typically one distance suits into one other distance. One might argue that that is nonetheless the idea of a yardstick: how typically does a yard match right into a distance? Nevertheless, the distinction is refined. The classical geometer doesn’t need to outline a unit, be it a yard or a meter. He all the time 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 cope with distances. If we take a look at the trigonometric capabilities, specifically 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 circumstances, 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_{okay=1}^n a_k^2},
\$\$
we get a metric within the topological sense
\$\$
d(vec{a},vec{b}):=|vec{a}-vec{b}|=sqrt{sum_{okay=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_{okay=1}^na_kb_k}{sqrt{sum_{okay=1}^n a_k^2}sqrt{sum_{okay=1}^n b_k^2}}.
\$\$

## Summary Algebra

An algebraist is within the buildings, not a lot in coordinates, so we outline a metric tensor by the properties above somewhat than coordinates. A metric tensor for a vector area ##V## is due to this fact merely a component of ##V^*otimes V^*.## Effectively, that is true, quick, and doesn’t assist us in any respect. What we need to outline is an interior product ##langle , . ,, , ,, . , rangle .## That, at its core, is a symmetric, optimistic particular, bilinear mapping
\$\$
g, : ,V instances V longrightarrow mathbb{R}
\$\$
Such a mapping may be written as a matrix ##G,## i.e. ##g(vec{a},vec{b})=vec{a}cdot G cdotvec{b}.## We now have thought of ##g(vec{a},vec{b})=vec{a}cdot operatorname{Id}cdotvec{b}## up to now, the place ##operatorname{Id}## is the identification matrix. Nevertheless, there isn’t any motive to limit ourselves to the identification matrix. Any optimistic particular matrix ##G## will do. That is mainly saying ##gin V^*otimes V^*.## Afterward, we’ll 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 a particular instance of a vector area ##V=mathbb{R}^n## that offers the development a that means.

The metric tensor ##g_pin V^*otimes V^*## above an affine level area ##A={p}+V## with translation area ##V## is a mapping that assigns to ##{p}## a symmetric, optimistic 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 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 variation of order means for mathematicians contravariance. (Consideration: physicists name it covariant for various causes, see beneath!)

It is very important discover that each one three portions rely upon the purpose ##p.##

## Manifolds

The entire above has been somewhat theoretical up to now. Certain, 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 fake that it is a small flat airplane that represents the tangent area sooner or later inside 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 elements should 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 inside the sq. are the parts of foundation tangent vectors of the tangent area at a sure level inside the sq.. Since there isn’t any globally flat coordinate system on the saddle, we’ve to patch all of the tiny flat squares, our charts, and bind them to an atlas. That’s what the differential geometer does. Word the massive 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 two impartial parameters, ##x## and ##y,## so ##dim M=2,## a floor. And though we’ve embedded it in our three-dimensional area, we is not going to 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. Folks on earth don’t acknowledge its curvature except they take a look at the horizon at sea, massive grass savannas, or from outer area.

## Physics

Earlier than we hearken 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. Swiftly we discover ourselves in a well-known scenario.

The affine area is just ##{p}+V=T_pM,## the tangent area on the level ##p## of the manifold ##M## outlined by the saddle perform ##z=x^2-y^2.## The metric tensor assigns a bilinear type to ##T_pM.## So, relying on the place we’re, we’ve 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’d like 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) somewhat than optimistic definiteness.

We now have seen that all of it relies upon (easily) on the placement ##p,## which isn’t any shock since differentiation is an area course of. However we will patch all native views to an atlas. Because of this we converse of fields as an alternative of areas. A discipline is an object obtained by gathering all factors ##pin M.## Due to this fact we’ve a metric tensor discipline, and a tangent vector discipline. It’s often merely referred to as a vector discipline, whereas the vector area of tangents is named tangent area.
start{align*}
g=bigsqcup_{pin M}g_p=bigcup_{pin M}{p}instances g_p&quadtext{ metric tensor discipline}
TM=bigsqcup_{pin M}T_pM=bigcup_{pin M}{p}instances T_pM&quadtext{ (tangent) vector discipline}
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)## via some extent ##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 degree units, we’ve ##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 area 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 area 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 discipline 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 need to trouble with the embedding in ##mathbb{R}^3.## The entire concept about manifolds is, that they don’t require an embedding. There isn’t any Euclidean area across the universe, solely the universe itself. Effectively, we begin a bit smaller and see what the saddle will inform us. However who is aware of, perhaps 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 type a foundation for ##V=T_pM.## The affine area ##A## we’ve spoken about within the algebra part is just ##A={p}+T_pM.## These areas are all two-dimensional, so ##g_p## can have 4 parts. As a way to illustrate the dependency on location, we outline in line 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 modify 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^okay}_i X_k,{a^l}_jX_l)={a^okay}_i{a^l}_jg(X_k,X_l)
&={a^okay}_i{a^l}_jg_{kl}[mathbf{f}]=sum_{okay,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 somewhat its coordinate illustration, is named covariant in physics since its parts rework like the idea below a change of coordinates in every index of the body.

If we modify 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 optimistic. ##g_p## has due to this fact the signature ##(+,+)## and is a Riemannian metric.

What’s a size in line with a metric tensor? The road component may be considered 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 sqrt=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 component fully defines the metric tensor since we thought of arbitrary curves. That’s the reason it’s typically used as a characterization of the metric tensor somewhat than the precise tensor product. In our instance we’ve (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 exhibit 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 a special 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 now 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 is perhaps the diploma of abstraction. Emmy Noether spoke of impartial variables ##p_x,p_y##, dependent capabilities ##q_x(p_x,p_y),q_y(p_x,p_y),## and the appliance of a bunch of transformations. Trendy 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 prolong simpler to know than the trendy therapy of her well-known theorem, cp. [11].

The graph of the perform ##f:Blongrightarrow {z}## is ##{(x,y,x^2-y^2)}=B instances {z}=Econg mathbb{R}^3.## ##B## is named the bottom area, right here the ##(x,y)##-plane, and ##E## the entire area, right here the complete ##mathbb{R}^3.## We now have a pure projection ##pi:Elongrightarrow B, , ,(x,y,z)mapsto (x,y).## Any easy perform ##sigma :Blongrightarrow E## such that ##pi(sigma (x,y))=(x,y)## is named a piece, and the set of all sections is denoted by ##Gamma(E).## However what does this should do with differentiation and the saddle? We already launched the (tangent) vector discipline, 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 area 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 must always have launched the cotangent bundle as an alternative 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 a bunch working on the fibers then we converse of principal bundles. Emmy Noether’s group of easy 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 spinoff by way of vector fields comparable with the gradient. The Levi-Civita connection is torsion-free and preserves the Riemannian metric tensor. It’s also referred to as the covariant spinoff 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 generally 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 quick }nabla_Z g(X,Y)=0.
\$\$
The distinction to a Lie spinoff is, {that a} Lie spinoff is ##mathbb{R}##-linear however doesn’t should be ##C^infty (M)##-linear. We additionally point out the Koszul components
start{align*}
g(nabla_{X}Y,Z)&={tfrac{1}{2}}{Massive {}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){Massive }}.
finish{align*}
If we write with Christoffel symbols ##Gamma_{ij}^okay##
\$\$
nabla_jpartial_k =Gamma_{jk}^lpartial_l
\$\$
then \$\$
nabla_XY=X^jleft(partial_j(Y^l)+Y^kGamma_{jk}^lpartial_lright) \$\$
and ##nabla## is appropriate with the metric if and provided that
\$\$partial_{i}{bigl (}g(partial_{j},partial_{okay}){bigr )}=g(nabla_{i}partial_{j},partial_{okay})+g(partial_{j},nabla_{i}partial_{okay})=g(Gamma_{ij}^{l}partial_{l},partial_{okay})+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 examine by direct calculation that
\$\$Gamma_{jk}^{l}={tfrac{1}{2}}g^{lr}left(partial_{okay}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 easy curve on the manifold ##M.## A piece ##sigma in Gamma(TM)## alongside ##gamma ## is named parallel in line 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 route of ##dotgamma(t).## A vector discipline 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 discipline ##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 perform
\$\$ 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 unusual linear differential equation methods. We now have 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 area to the opposite.

We now have 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 ##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 remedy
\$\$
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 (easy) part of ##Gamma(TM).## The parallel transport is just given by preserving the parts 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}.
\$\$