tangent bundle in nLab

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Differential geometry

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Definition

(tangency relation on differentiable curves)

Let XX be a differentiable manifold of dimension nn and let x∈Xx \in X be a point. On the set of smooth functions of the form

γ:ℝ 1⟶X \gamma \;\colon\; \mathbb{R}^1 \longrightarrow X

such that

γ(0)=x \gamma(0) = x

define the relations

(γ 1∼γ 2)≔∃ℝ n→ϕchartU i⊂XU i⊃{x}(ddt(ϕ −1∘γ 1)(0)=ddt(ϕ −1∘γ 2)(0)) (\gamma_1 \sim \gamma_2) \coloneqq \underset{ { { \mathbb{R}^n \underoverset{}{\phi \, \text{chart}}{\to} U_i \subset X } } \atop { U_i \supset \{x\} } }{ \exists } \left( \frac{d}{d t}(\phi^{-1}\circ \gamma_1)(0) = \frac{d}{d t}(\phi^{-1}\circ \gamma_2)(0) \right)

and

(γ 1∼′γ 2)≔∀ℝ n→ϕchartU i⊂XU i⊃{x}(ddt(ϕ −1∘γ 1)(0)=ddt(ϕ −1∘γ 2)(0)) (\gamma_1 \sim' \gamma_2) \coloneqq \underset{ { { \mathbb{R}^n \underoverset{}{\phi \, \text{chart}}{\to} U_i \subset X } } \atop { U_i \supset \{x\} } }{ \forall } \left( \frac{d}{d t}(\phi^{-1}\circ \gamma_1)(0) = \frac{d}{d t}(\phi^{-1}\circ \gamma_2)(0) \right)

saying that two such functions are related precisely if either there exists a chart around xx such that (or else for all charts around xx it is true that) the first derivative of the two functions regarded via the given chart as functions ℝ 1→ℝ n\mathbb{R}^1 \to \mathbb{R}^n, coincide at t=0t = 0 (with tt denoting the canonical coordinate function on ℝ\mathbb{R}).

Lemma

(tangency is equivalence relation)

The two relations in def. are equivalence relations and they coincide.

Proof

First to see that they coincide, we need to show that if the derivatives in question coincide in one chart ℝ n→≃ϕ iU i⊂X\mathbb{R}^n \underoverset{\simeq}{\phi_i}{\to} U_i \subset X, that then they coincide also in any other chart ℝ n→≃ϕ jU j⊂X\mathbb{R}^n \underoverset{\simeq}{\phi_j}{\to} U_j \subset X.

For brevity, write

U ij≔U i∩U j U_{i j} \coloneqq U_i \cap U_j

for the intersection of the two charts.

First of all, since the derivative may be computed in any open neighbourhood around t=0t = 0, and since the differentiable functions γ i\gamma_i are in particular continuous functions, we may restrict to the open neighbourhood

V≔γ 1 −1(U ij)∩γ 2 −1(U ij)⊂ℝ V \coloneqq \gamma_1^{-1}( U_{i j} ) \cap \gamma_2^{-1}(U_{i j}) \;\subset\; \mathbb{R}

of 0∈ℝ0 \in \mathbb{R} and consider the derivatives of the functions

γ n i≔(ϕ i −1| U ij∘γ n| V):V⟶ϕ i −1(U ij)⊂ℝ n \gamma_n^i \;\coloneqq\; (\phi_i^{-1}\vert_{U_{i j}} \circ \gamma_n\vert_{V}) \;\colon\; V \longrightarrow \phi_i^{-1}(U_{i j}) \subset \mathbb{R}^n

and

γ n j≔(ϕ j −1| U ij∘γ n| V):V⟶ϕ j −1(U ij)⊂ℝ n. \gamma_n^j \;\coloneqq\; (\phi_j^{-1}\vert_{U_{i j}} \circ \gamma_n \vert _{V}) \;\colon\; V \longrightarrow \phi_j^{-1}(U_{i j}) \subset \mathbb{R}^n \,.

But then by definition of the differentiable atlas, there is the differentiable gluing function

α≔ϕ i −1(U ij)⟶≃ϕ iU ij⟶≃ϕ j −1ϕ j −1(U ij) \alpha \;\coloneqq\; \phi_i^{-1}(U_{i j}) \underoverset{\simeq}{\phi_i}{\longrightarrow} U_{i j} \underoverset{\simeq}{\phi_j^{-1}}{\longrightarrow} \phi_j^{-1}(U_{i j})

such that

γ n j=α∘γ n i \gamma_n^j = \alpha \circ \gamma_n^i

for n∈{1,2}n \in \{1,2\}. The chain rule now relates the derivatives of these functions as

ddtγ n j=(Dα)∘(ddtγ n i). \frac{d}{d t} \gamma_n^j \;=\; (D \alpha) \circ \left(\frac{d}{d t} \gamma_n^i \right) \,.

Since α\alpha is a diffeomorphism and since derivatives of diffeomorphisms are linear isomorphisms, this says that the derivative of γ n j\gamma_n^j is related to that of γ n i\gamma_n^i by a linear isomorphism , and hence

(ddtγ 1 i=ddtγ 2 i)⇔(ddtγ 1 j=ddtγ 2 j). \left( \frac{d}{d t} \gamma_1^i = \frac{d}{d t} \gamma_2^i \right) \;\Leftrightarrow\; \left( \frac{d}{d t} \gamma_1^j = \frac{d}{d t} \gamma_2^j \right) \,.

Finally, that either relation is an equivalence relation is immediate.

Definition

(tangent vector)

Let XX be a differentiable manifold and x∈Xx \in X a point. Then a tangent vector on XX at xx is an equivalence class of the the tangency equivalence relation (def. , lemma ).

The set of all tangent vectors at x∈Xx \in X is denoted T xXT_x X.

Lemma

(real vector space structure on tangent vectors)

For XX a differentiable manifold of dimension nn and x∈Xx \in X any point, let ℝ n→≃ϕU⊂X\mathbb{R}^n \underoverset{\simeq}{\phi}{\to} U \subset X be a chart with x∈U⊂Xx \in U \subset X.

Then there is induced a bijection of sets

ℝ n⟶≃T xX \mathbb{R}^n \overset{\simeq}{\longrightarrow} T_x X

from the nn-dimensional Cartesian space to the set of tangent vectors at xx (def. ) given by sending v→∈ℝ n\vec v \in \mathbb{R}^n to the equivalence class of the following differentiable curve:

γ v→ ϕ: ℝ 1 ⟶ϕ −1(x)+(−)⋅v→ ℝ n ⟶≃ϕ U i⊂X t ↦AAA ϕ −1(x)+tv→ ↦AAA ϕ(ϕ −1(x)+tv→). \array{ \gamma^\phi_{\vec v} \colon & \mathbb{R}^1 &\overset{ \phi^{-1}(x) + (-)\cdot \vec v }{\longrightarrow}& \mathbb{R}^n &\underoverset{\simeq}{\phi}{\longrightarrow}& U_i \subset X \\ & t &\overset{\phantom{AAA}}{\mapsto}& \phi^{-1}(x) + t \vec v &\overset{\phantom{AAA}}{\mapsto}& \phi(\phi^{-1}(x) + t \vec v) } \,.

For ℝ n⟶≃ϕ′U′⊂X\mathbb{R}^n \underoverset{\simeq}{\phi'}{\longrightarrow} U' \subset X another chart with x∈U′⊂Xx \in U' \subset X, then the linear isomorphism relating these two identifications is the derivative

d((ϕ′) −1∘ϕ) ϕ −1(x)∈GL(n,ℝ) d \left((\phi')^{-1} \circ \phi \right)_{ \phi^{-1}(x) } \in GL(n,\mathbb{R})

of the gluing function of the two charts at the point xx:

ℝ n ⟶d((ϕ′) −1∘ϕ) ϕ −1(x) ℝ n ≃↘ ↙ ≃ T xX. \array{ \mathbb{R}^n && \overset{ d \left((\phi')^{-1} \circ \phi\right)_{\phi^{-1}(x)}}{\longrightarrow} && \mathbb{R}^n \\ & {}_{\mathllap{\simeq}}\searrow && \swarrow_{\mathrlap{\simeq}} \\ && T_x X } \,.

This is also called the transition function between the two local identifications of the tangent space.

If {ℝ n→≃ϕ iU i⊂X} i∈I\left\{ \mathbb{R}^n \underoverset{\simeq}{\phi_i}{\to} U_i \subset X \right\}_{i \in I} is an atlas of the differentiable manifold XX, then the transition functions

{g ij≔d(ϕ j −1∘ϕ i) ϕ i −1(−):U i∩U j⟶GL(n,ℝ)} i,j∈I \left\{ g_{i j} \coloneqq d( \phi_j^{-1} \circ \phi_i )_{\phi_i^{-1}(-)} \colon U_i \cap U_j \longrightarrow GL(n,\mathbb{R}) \right\}_{i,j \in I}

defined this way satisfy the following Cech cocycle conditions for all i,j∈Ii,j \in I, x∈U i∩U jx \in U_i \cap U_j

1. g ii(x)=id ℝ ng_{i i}(x) = id_{\mathbb{R}^n};

2. g jk∘g ij(x)=g ik(x)g_{j k}\circ g_{i j}(x) = g_{i k}(x).

Proof

The bijectivity of the map is immediate from the fact that the first derivative of ϕ −1∘γ v→ ϕ\phi^{-1}\circ \gamma^\phi_{\vec v} at ϕ −1(x)\phi^{-1}(x) is v→\vec v.

The formula for the transition function now follows with the chain rule:

d((ϕ′) −1∘ϕ(ϕ −1(x)+(−)v→)) 0=d((ϕ′) −1∘ϕ) ϕ −1(x)∘d(ϕ −1(x)+(−)v→) 0⏟=(−)v→. d \left( (\phi')^{-1} \circ \phi( \phi^{-1}(x) + (-) \vec v ) \right)_0 = d \left( (\phi')^{-1} \circ \phi \right)_{\phi^{-1}(x)} \circ \underset{ =(-)\vec v }{\underbrace{ d ( \phi^{-1}(x) +(-)\vec v )_0 }} \,.

Similarly the Cech cocycle condition follows by the chain rule:

g jk∘g ij(x) =d(ϕ k −1∘ϕ j) ϕ j −1(x)∘d(ϕ j −1∘ϕ i) ϕ i −1(x) =d(ϕ k −1∘ϕ j∘ϕ j −1∘ϕ i) ϕ i −1(x) =d(ϕ k −1∘ϕ i) ϕ i −1(x) =g ik(x). \begin{aligned} g_{j k} \circ g_{i j}(x) & = d( \phi_k^{-1} \circ \phi_j )_{\phi_j^{-1}(x)} \circ d( \phi_j^{-1} \circ \phi_i )_{\phi_i^{-1}(x)} \\ & = d( \phi_k^{-1} \circ \phi_j \circ \phi_j^{-1} \circ \phi_i )_{\phi_i^{-1}(x)} \\ & = d( \phi_k^{-1} \circ \phi_i )_{\phi_i^{-1}(x)} \\ &= g_{i k}(x) \end{aligned} \,.

Example

(tangent bundle of Euclidean space)

If X=ℝ nX = \mathbb{R}^n is itself a Euclidean space, then for any two points x,y∈Xx,y \in X the tangent spaces T xXT_x X and T yXT_y X (def. ) are canonically identified with each other:

Using the vector space (or just affine space) structure of X=ℝ nX = \mathbb{R}^n we may send every smooth function γ:ℝ→X\gamma \colon \mathbb{R} \to X to the smooth function

γ′:t↦γ(t)+(x−y). \gamma' \colon t \mapsto \gamma(t) + (x-y) \,.

This gives a linear bijection

ϕ x,y:T xX⟶≃T yX \phi_{x,y} \colon T_x X \overset{\simeq}{\longrightarrow} T_y X

and these linear bijections are compatible, in that for x,y,z∈ℝ nx,y,z \in \mathbb{R}^n any three points, then

ϕ y,z∘ϕ x,y=ϕ x,z:T xX⟶T yY. \phi_{y,z} \circ \phi_{x,y} = \phi_{x,z} \;\colon\; T_x X \longrightarrow T_y Y \,.

Moreover, by lemma , each tangent space is identified with ℝ n\mathbb{R}^n itself, and this identification in turn is compatible with all the above identifications:

ℝ n ≃↙ ↘ ≃ T xX ⟶ϕ x,y≃ T yY. \array{ && \mathbb{R}^n \\ & {}^{\mathllap{\simeq}}\swarrow && \searrow^{\mathrlap{\simeq}} \\ T_x X && \underoverset{\phi_{x,y}}{\simeq}{\longrightarrow} && T_y Y } \,.

Therefore it makes sense to canonically identify all the tangent spaces of Euclidean space with that Euclidean space itself. As a result, the collection of all the tangent spaces of Euclidean space is naturally identified with the Cartesian product

Tℝ n=ℝ n×ℝ n T \mathbb{R}^n = \mathbb{R}^n \times \mathbb{R}^n

equipped with the projection on the first factor

Tℝ n=ℝ n×ℝ n ↓ π=pr 1 ℝ n, \array{ T \mathbb{R}^n = \mathbb{R}^n \times \mathbb{R}^n \\ \downarrow^{\mathrlap{\pi = pr_1}} \\ \mathbb{R}^n } \,,

because then the pre-image of a singleton {x}⊂ℝ n\{x\} \subset \mathbb{R}^n under this projection are canonically identified with the above tangent spaces:

π −1({x})≃T xℝ n. \pi^{-1}(\{x\}) \simeq T_x \mathbb{R}^n \,.

This way, if we equip Tℝ n=ℝ n×ℝ nT \mathbb{R}^n = \mathbb{R}^n \times \mathbb{R}^n with the product space topology, then Tℝ n⟶πℝ nT \mathbb{R}^n \overset{\pi}{\longrightarrow} \mathbb{R}^n becomes a trivial topological vector bundle.

This is called the tangent bundle of the Euclidean space ℝ n\mathbb{R}^n regarded as a differentiable manifold.

Definition

(tangent bundle of a differentiable manifold)

Let XX be a differentiable manifold with atlas {ℝ n→≃ϕ iU i⊂X} i∈I\left\{ \mathbb{R}^n \underoverset{\simeq}{\phi_i}{\to} U_i \subset X\right\}_{i \in I}.

Equip the set of all tangent vectors (def. ), i.e. the disjoint union of the sets of tangent vectors

TX≔⊔x∈XT xXAAAas underlying sets T X \;\coloneqq\; \underset{x \in X}{\sqcup} T_x X \phantom{AAA} \text{as underlying sets}

with a topology τ TX\tau_{T X} by declaring a subset U⊂TXU \subset T X to be an open subset precisely if for all charts ℝ n→≃ϕ iU i⊂X\mathbb{R}^n \underoverset{\simeq}{\phi_i}{\to} U_i \subset X we have that its preimage under

ℝ 2n≃ℝ n×ℝ n ⟶dϕ TX (x,v→) ↦AAA [t↦ϕ(ϕ −1(x)+tv→)] \array{ \mathbb{R}^{2n} \simeq \mathbb{R}^n \times \mathbb{R}^n & \overset{d \phi}{\longrightarrow} & T X \\ (x, \vec v) &\overset{\phantom{AAA}}{\mapsto}& [ t \mapsto \phi(\phi^{-1}(x) + t \vec v) ] }

is open in the Euclidean space ℝ 2n\mathbb{R}^{2n} with its metric topology.

Equipped with the function

TX ⟶AAp xAA X (x,v) ↦AAAA x \array{ T X &\overset{\phantom{AA}p_x \phantom{AA}}{\longrightarrow}& X \\ (x,v) &\overset{\phantom{AAAA}}{\mapsto}& x }

this is called the tangent bundle of XX.

Equivalently this means that the tangent bundle TXT X is the topological vector bundle which is glued (via this example) from the transition functions g ij≔d(ϕ j −1∘ϕ i) ϕ −1(−)g_{i j} \coloneqq d(\phi_j^{-1} \circ \phi_i)_{\phi^{-1}(-)} from lemma :

TX≔(⊔i∈IU i×ℝ n)/({d(ϕ j −1∘ϕ i) ϕ i −1(−)} i,j∈I). T X \;\coloneqq\; \left( \underset{i \in I}{\sqcup} U_i \times \mathbb{R}^n \right)/\left( \left\{ d( \phi_j^{-1} \circ \phi_i )_{\phi^{-1}_i(-)} \right\}_{i, j \in I} \right) \,.

(Notice that, by examples , each U i×ℝ n≃TU iU_i \times \mathbb{R}^n \simeq T U_i is the tangent bundle of the chart U i≃ℝ nU_i \simeq \mathbb{R}^n.)

The co-projection maps of this quotient topological space construction constitute an atlas

{ℝ 2n→≃TU i⊂TX} i∈I. \left\{ \mathbb{R}^{2n} \underoverset{\simeq}{}{\to} T U_i \subset T X \right\}_{i \in I} \,.

Proof

First to see that TXT X is Hausdorff:

Let (x,v→),(x′,v→′)∈TX(x,\vec v), (x', \vec v') \in T X be two distinct points. We need to produce disjoint openneighbourhoods of these points in TXT X. Since in particular x,x′∈Xx,x' \in X are distinct, and since XX is Hausdorff, there exist disjoint open neighbourhoods U x⊃{x}U_x \supset \{x\} and U x′⊃{x′}U_{x'} \supset \{x'\}. Their pre-images π −1(U x)\pi^{-1}(U_x) and π −1(U x′)\pi^{-1}(U_{x'}) are disjoint open neighbourhoods of (x,v→)(x,\vec v) and (x′,vectv′)(x',\vect v'), respectively.

Now to see that TXT X is paracompact.

Let {U i⊂TX} i∈I\{U_i \subset T X\}_{i \in I} be an open cover. We need to find a locally finite refinement. Notice that π:TX→X\pi \colon T X \to X is an open map (by this example) so that {π(U i)⊂X} i∈I\{\pi(U_i) \subset X\}_{i \in I} is an open cover of XX.

Let now {ℝ n→≃ϕ jV j⊂X} j∈J\{\mathbb{R}^n \underoverset{\simeq}{\phi_j}{\to} V_j \subset X\}_{j \in J} be an atlas for XX and consider the open common refinement

{π(U i)∩V j⊂X} i∈I,j∈J. \left\{ \pi(U_i) \cap V_j \subset X \right\}_{i \in I, j \in J} \,.

Since this is still an open cover of XX and since XX is paracompact, this has a locally finite refinement

{V′ j′⊂X} j′∈J′ \left\{ V'_{j'} \subset X\right\}_{j' \in J'}

Notice that for each j′∈J′j' \in J' the product topological space V′ j′×ℝ n⊂ℝ 2nV'_{j'} \times \mathbb{R}^n \subset \mathbb{R}^{2n} is paracompact (as a topological subspace of Euclidean space it is itself locally compact and second countable and since locally compact and second-countable spaces are paracompact). Therefore the cover

{π −1(V′ j′)∩U i⊂V′ j′×ℝ n} (i,j′)∈I×J′ \{ \pi^{-1}(V'_{j'}) \cap U_i \subset V'_{j'} \times \mathbb{R}^n \}_{(i,j') \in I \times J'}

has a locally finite refinement

{W k j′⊂V′ j′×ℝ n} k j′∈K j′. \{W_{k_{j'}} \subset V'_{j'} \times \mathbb{R}^n \}_{k_{j'} \in K_{j'}} \,.

We claim now that

{W k j′⊂TX} j′∈J′,k j′∈K j′ \{ W_{k_{j'}} \subset T X \}_{j' \in J', k_{j'} \in K_{j'}}

is a locally finite refinement of the original cover. That this is an open cover refining the original one is clear. We need to see that it is locally finite.

So let (x,v→)∈TX(x,\vec v) \in T X. By local finiteness of {V′ j′⊂X} j′∈J′\{ V'_{j'} \subset X\}_{j' \in J'} there is an open neighbourhood V x⊃{x}V_x \supset \{x\} which intersects only finitely many of the V′ j′⊂XV'_{j'} \subset X. Then by local finiteness of {W k j′⊂V′ j ′}\{ W_{k_{j'}} \subset V'_{j_'}\}, for each such j′j' the point (x,v→)(x,\vec v) regarded in V′ j′×ℝ nV'_{j'} \times \mathbb{R}^n has an open neighbourhood U j′U_{j'} that intersects only finitely many of the W k j′W_{k_{j'}}. Hence the intersection π −1(V x)∩(∩j′U j′)\pi^{-1}(V_x) \cap \left( \underset{j'}{\cap} U_{j'} \right) is a finite intersection of open subsets, hence still open, and by construction it intersects still only a finite number of the W k j′W_{k_{j'}}.

This shows that TXT X is paracompact.

Finally the statement about the differentiability of the gluing functions and of the projections is immediate from the definitions

Proposition

(differentials of differentiable functions between differentiable manifolds)

Let XX and YY be differentiable manifolds and let f:X⟶Yf \;\colon\; X \longrightarrow Y be a differentiable function. Then the operation of postcomposition, which takes differentiable curves in XX to differentiable curves in YY,

Hom Diff(ℝ 1,X) ⟶f∘(−) Hom Diff(ℝ 1,Y) (ℝ 1→γX) ↦AAA (ℝ 1→f∘γY) \array{ Hom_{Diff}(\mathbb{R}^1, X) &\overset{f \circ (-)}{\longrightarrow}& Hom_{Diff}(\mathbb{R}^1, Y) \\ \left( \mathbb{R}^1 \overset{\gamma}{\to} X \right) &\overset{\phantom{AAA}}{\mapsto}& \left( \mathbb{R}^1 \overset{f \circ \gamma}{\to} Y \right) }

descends at each point x∈Xx \in X to the tangency equivalence relation (def. , lemma ) to yield a function on sets of tangent vectors (def. ), called the differential df xd f_x of ff at xx

df| x:T xX⟶T f(x)Y. d f|_{x} \;\colon\; T_x X \longrightarrow T_{f(x)} Y \,.

Moreover:

1. (linear dependence on the tangent vector) these differentials are linear functions with respect to the vector space structure on the tangent spaces from lemma , def. ;

2. (differentiable dependence on the base point) globally they yield a homomorphism of real differentiable vector bundles between the tangent bundles (def. , lemma ), called the global differential dfd f of ff

   df:TX⟶TY. d f \;\colon\; T X \longrightarrow T Y \,.

3. (chain rule) The assignment f↦dff \mapsto d f respects composition in that for XX, YY, ZZ three differentiable manifolds and for

   X⟶AfAY⟶AgAZ X \overset{\phantom{A}f\phantom{A}}{\longrightarrow} Y \overset{\phantom{A}g\phantom{A}}{\longrightarrow} Z

   two composable differentiable functions then their differentials satisfy

   d(g∘f)=(dg)∘(df). d(g \circ f) = (d g) \circ (d f) \,.

Proof

All statements are to be tested on charts of an atlas for XX and for YY. On these charts the statement reduces to that of example .

Definition

(tangent bundle in smooth toposes)

Let (𝒯,(R,+,⋅))(\mathcal{T},(R,+,\cdot)) be a smooth topos and write D={ϵ∈R|ϵ 2=0}D = \{\epsilon \in R| \epsilon^2 = 0\} for the standard infinitesimal interval. For X∈𝒯X \in \mathcal{T} any object (any space in 𝒯\mathcal{T}), the tangent bundle of XX is the morphism

p:TX→X p : T X \to X

with

• TX∶−X DT X \coloneq X^D the internal hom of DD into XX;

• p=ev 0p = ev_0 the evaluation map at the origin of DD

  ev 0:(U→vX D)↦(U×*→Id×0U×D→v¯X)ev_0 : (U \stackrel{v}{\to} X^D) \mapsto (U \times {*} \stackrel{Id \times 0}{\to} U \times D \stackrel{\bar v}{\to} X),

  where v¯\bar v is the hom-adjunct of vv.