15.4 Tangent map and submanifolds

中文

The tangent map and pullback of a map

Let $\alpha :M\to N$ be a differentiable map between manifolds $M$ and $N_z$ , where $\dim$ $M=m$ $\dim$ $N=n$ . This induces a map ${\alpha }_{\ast }:T_p(M)\to T_{\alpha (p)}(N)$ , called the tangent map of ${\alpha }_{:}$ whereby the tangent vector $Y_{\alpha (p)}={\alpha }_{\ast }{X}_p$ is defined by

$$Y_{\alpha (p)}f=({\alpha }_{\ast }{X}_p)f=X_p(f\circ \alpha )$$

for any function $f\in {\mathcal{F}}_{\alpha (p)}(N)$ . This map is often called the differential of the map ${\alpha }_{:}$ , bu this may cause confusion with our earlier use of this term.

Let $(U,\varphi ;x^i)$ and $(V,\psi ;y^a)$ be charts at $p$ and $\alpha (p)$ ), respectively. The map $\alpha$ has coordinate representation $\hat{\alpha }=\psi \circ \alpha \circ {\varphi }^{-1}:\varphi (U)\psi (V)$ , written

$$y^a={\alpha }^a(x^1,x^2,\dots ,x^m)(a=1,\dots ,n).$$

To compute the components $Y^a$ of $Y_{\alpha (p)}={\alpha }_{\ast }{X}_p=Y^a{\left({\partial }_{y^a}\right)}_{\alpha (p)}:$ , we perform the following steps:

$$\begin{array}{l}{Y}_{\alpha (p)}f=Y^a\frac{\partial f\circ {\psi }^{-1}}{\partial y^a}{|}_{\alpha (p)}=X^i\frac{\partial }{\partial x^i}{|}_{p}f\circ \alpha \\ =X^i\frac{\partial f\circ \alpha \circ {\varphi }^{-1}}{\partial x^i}{|}_{\varphi (p)}\\ =X^i\frac{\partial f\circ {\psi }^{-1}\circ \hat{\alpha }}{\partial x^i}{|}_{\varphi (p)}\\ =X^i\frac{\partial f\circ {\psi }^{-1}({\alpha }^1(\mathbf{x}),{\alpha }^2(\mathbf{x}),\dots ,{\alpha }^n(\mathbf{x}))}{\partial x^i}{|}_{\mathbf{x}=\varphi (p)}\\ =X^i\frac{\partial y^a}{\partial x^i}{|}_{\varphi (p)}\frac{\partial f\circ {\psi }^{-1}}{\partial y^a}{|}_{\hat{\alpha }(\varphi (p))}\\ =X^i\frac{\partial y^a}{\partial x^i}{|}_{\varphi (p)}\frac{\partial }{\partial y^a}{|}_{\alpha (p)}f.\end{array}$$

Hence

$$Y^a=X^i\frac{\partial y^a}{\partial x^i}{|}_{\varphi (p)}.\text{\hspace{1em}(15.16)}$$

Exercise

If $\alpha :M\to N$ and $\beta :K\to M$ are differentiable maps between manifolds, show that

$$(\alpha \circ \beta {)}_{\ast }={\alpha }_{\ast }\circ {\beta }_{\ast }.\text{\hspace{1em}(15.17)}$$

The map $\alpha :M\to N$ also induces a map ${\alpha }^{\ast }$ between cotangent spaces, but in this case it acts in the reverse direction, called the pullback induced by ${\alpha }_{;}$

$${\alpha }^{\ast }:T_{\alpha (p)}^{\ast }(N)\to T_p^{\ast }(M).$$

The pullback of a 1-form ${\omega }_{\alpha (p)}$ is defined by requiring

$$⟨{\alpha }^{\ast }{\omega }_{\alpha (p)},X_p⟩=⟨{\omega }_{\alpha (p)},{\alpha }_{\ast }{X}_p⟩\text{\hspace{1em}(15.18)}$$

for arbitrary tangent vectors $X_p$

Exercise

Show that this definition uniquely defines the pullback ${\alpha }^{\ast }{\omega }_{\alpha (p)}$

Exercise

Show that the pullback of a functional composition of maps is given by

$$(\alpha \circ \beta {)}^{\ast }={\beta }^{\ast }\circ {\alpha }^{\ast }.\text{\hspace{1em}(15.19)}$$

The notion of tangent map or pullback of a map can be extended to totally contravarian or totally covariant tensors, such as $r$-vectors or $r$-forms, but is only available for mixed tensors if the map is a diffeomorphism (see Problem 15.13). The tangent map does not in general apply to vector fields, for if it is not one-to-one the tangent vector may not be uniquely defined at the image point $\alpha (p)$ (see Example 15.8 below). However, no such ambiguity in the value of the pullback ${\alpha }^{\ast }\omega$ can ever arise at the inverse image point ${}^{p,}$ even if $\omega$ is a covector field, since its action on every tangent vector $X_p$ is well-defined by Eq. (15.18). The pullback can therefore be applied to arbitrary differentiable 1-forms; the map $\alpha$ need not be either injective or surjective. This is one of the features that makes covector fields more attractive geometrical objects to deal with than vector fields. The following example should make this clear.

Example 15.8

Let $\alpha :{\dot{\mathbb{R}}}^3={\mathbb{R}}^3-\{(0,0,0)\}{\mathbb{R}}^2$ be the differentiable map

$$\alpha :(x,y,z)\mapsto (u,v)\text{ where }u=x+y+z,v=\sqrt{x^2+z^2}.$$

This map is neither surjective, since the whole lower half plane $v<0$ is not mapped onto, nor injective, since, for example, the points $\mathbf{p}=(1,y,0)$ and $\mathbf{q}=(0,y,1)$ ) are both mapped to the point $(y+1,1)$ . Consider a vector field

$$X=X^i\frac{\partial }{\partial x^i}=X^1\frac{\partial }{\partial x}+X^2\frac{\partial }{\partial y}+X^3\frac{\partial }{\partial z}$$

and the action of the tangent map ${\alpha }_{\ast }$ at any point $(x,y,z)$ is

$${\alpha }_{\ast }X=\left(X^1+X^2+X^3\right)\frac{\partial }{\partial u}+\left(X^1\frac{x}{\sqrt{x^2+z^2}}+X^3\frac{z}{\sqrt{x^2+z^2}}\right)\frac{\partial }{\partial v}.$$

While this map is well-defined on the tangent space at any point $\mathbf{x}=(x,y,z)$ , it does not in general map the vector field $X$ to a vector field on ${\mathbb{R}}^2$ . For example, no tangent vecto can be assigned at $\mathbf{u}=\alpha (\mathbf{p})=\alpha (\mathbf{q})$ as we would need

$$\left(X^1+X^2+X^3\right)(\mathbf{p})\frac{\partial }{\partial u}{|}_{\mathbf{u}}+X^1(\mathbf{p})\frac{\partial }{\partial v}{|}_{\mathbf{u}}=\left(X^1+X^2+X^3\right)(\mathbf{q})\frac{\partial }{\partial u}{|}_{\mathbf{u}}+X^3(\mathbf{q})\frac{\partial }{\partial v}{|}_{\mathbf{u}}.$$

There is no reason to expect these two tangent vectors at $u$ to be identical.

However, $\mathrm{f}\omega =w_1\mathrm{d}u+w_2$ $\mathrm{d}v$ is a differentiable 1-form on ${\mathbb{R}}^2$ it induces a differentiable 1-form on ${\dot{\mathbb{R}}}^3$ , on substituting $\mathrm{d}u$ $=(\partial u/\partial x)\mathrm{d}x+(\partial u/\partial y)\mathrm{d}y+(\partial u/\partial z)$ $\mathrm{d}z$, etc.

$${\alpha }^{\ast }\omega =\left(w_1+w_2\frac{x}{\sqrt{x^2+z^2}}\right)dx+w_{1}dy+\left(w_1+w_2\frac{z}{\sqrt{x^2+z^2}}\right)dz,$$

which is uniquely determined at any point $(x,y,z)\ne (0,0,0)$ by the components $w_1(u,v)$ and $w_2(u,v)$ of the differentiable 1-form at $(u,v)=\alpha (x,y,z)$

Example 15.9

If $\gamma :\mathbb{R}M$ is a curve on $M$ and $p=\gamma (t_0)$ , the tangent vector to the curve at $p$ is the image under the tangent map induced by $\gamma$ of the ordinary derivative on the real line,

$${\dot{\gamma }}_p={\gamma }_{\ast }\frac{\mathrm{d}}{\mathrm{d}t}{|}_{t_0},$$

for if $f:M\to \mathbb{R}$ is any function differentiable at $p$ then

$${\gamma }_{\ast }\frac{\mathrm{d}}{\mathrm{d}t}{|}_{t_0}(f)={\frac{\mathrm{d}f\circ \gamma }{\mathrm{d}t}|}_{t_0}={\dot{\gamma }}_p(f).$$

By a curve with endpoints we shall mean the restriction of a parametrized curve $\gamma :(a,b)\to M$ to a closed subinterval of $\gamma :[t_1,t_2]M$ where $a<t_1<t_2<b$ . The integral of a 1-form $\alpha$ on the curve with end points $\gamma$ is defined as

$${\int }_{\gamma }\alpha ={\int }_{t_1}^{t_2}\alpha (\dot{\gamma })\mathrm{d}t.$$

In a coordinate representation $x^i={\gamma }^i(t)$ and $\alpha ={\alpha }_i\mathrm{d}{x}^i$ ,

$${\int }_{\gamma }\alpha ={\int }_{t_1}^{t_2}{\alpha }_i(x(t))\frac{\mathrm{d}{x}^i}{\mathrm{d}t}\mathrm{d}t.$$

Let ${\gamma }^{\prime }=\gamma \circ f$ be the curve related to $\gamma$ by a change of parametrization $t^{\prime }=f(t)$ where $f:\mathbb{R}\to \mathbb{R}$ is a monotone function on the real line. Then

$${\int }_{\gamma }\alpha ={\int }_{{\gamma }^{\prime }}\alpha$$

for, by the standard change of variable formula for a definite integral,

$$\begin{array}{l}{\int }_{{\gamma }^{\prime }}\alpha ={\int }_{t_1}^{t_2}{\alpha }_i(x(t))\frac{\mathrm{d}{x}^i(t(t^{\prime }))}{\mathrm{d}{t}^{\prime }}\frac{\mathrm{d}{t}^{\prime }}{\mathrm{d}t}\mathrm{d}t\\ ={\int }_{t_1^{\prime }}^{t_2^{\prime }}{\alpha }_i(x(t(t^{\prime })))\frac{\mathrm{d}{x}^{\prime i}(t^{\prime })}{\mathrm{d}{t}^{\prime }}\mathrm{d}{t}^{\prime }\\ ={\int }_{{\gamma }^{\prime }}\alpha .\end{array}$$

Hence the integral of a 1-form is independent of the parametrization on the curve $\gamma$

The integral of $\alpha$ along $\gamma$ is zero if its pullback to the real line vanishes, ${\gamma }^{\ast }(\alpha )=0,$ , for

$${\int }_{\gamma }\alpha ={\int }_{t_1}^{t_2}⟨\alpha ,{\gamma }_{\ast }\frac{\mathrm{d}}{\mathrm{d}t}⟩\mathrm{d}t={\int }_{t_1}^{t_2}⟨{\gamma }^{\ast }(\alpha ),\frac{\mathrm{d}}{\mathrm{d}t}⟩\mathrm{d}t=0.$$

If $\alpha$ is the differential of a scalar field $\alpha =\mathrm{d}f$ it is called an exact 1-form. The integral of an exact 1-form is independent of the curve connecting two points $p_1=\gamma (t_1)$ and $p_2=\gamma (t_2)$ , for

$${\int }_{\gamma }\mathrm{d}f={\int }_{t_1}^{t_2}⟨\mathrm{d}f,\dot{\gamma }⟩\mathrm{d}t={\int }_{t_1}^{t_2}\frac{\mathrm{d}f(\gamma (t))}{\mathrm{d}t}\mathrm{d}t=f(\gamma (t_2))-f(\gamma (t_1)),$$

which only depends on the value of $f$ at the end points. In particular, the integral of an exact 1-form vanishes on any closed circuit, since $\gamma (t_1)=\gamma (t_2)$

For general 1-forms, the integral is usually curve-dependent. For example, let $\alpha =x\mathrm{d}y$ on the manifold ${\mathbb{R}}^2$ with coordinates $(x,y)$ . Consider the following two curves connecting $p_1=(-1,0)$ to $p_2=(1,0)$ 1

$$\begin{array}{ll}{\gamma }_1:x=t,y=0& t_1=-1,t_2=1,\\ {\gamma }_2:x=\cos t,y=\sin t& t_1=-\pi ,t_2=0.\end{array}$$

The pullback of $\alpha$ to the first curve vanishes, ${\gamma }_1^{\ast }x\mathrm{d}y=t\mathrm{d}0=0$ , while the pullback to ${\gamma }_2$ is given by

$${\gamma }_2^{\ast }x\mathrm{d}y=\cos t\mathrm{d}(y\circ {\gamma }_2(t))=\cos t\mathrm{d}(\sin t)={\cos }^{2}t\mathrm{d}t.$$

Hence

$${\int }_{{\gamma }_2}x\mathrm{d}y={\int }_{-\pi }^0{\cos }^{2}t\mathrm{d}t=\frac{\pi }{2}\ne {\int }_{{\gamma }_1}x\mathrm{d}y=0.$$

Submanifolds

Let $\alpha :MN$ be a differentiable mapping where $m$ $\dim$ $M\le n$ $\dim$ $N$. The map is said to be an immersion if the tangent map ${\alpha }_{\ast }:T_p(M)\to T_{\alpha (p)}(N)$ is injective at every point $p\in M;i.e.,{\alpha }_{\ast }$ is everywhere a non-degenerate linear map. From the inverse function theorem, it is straightforward to show that there exist charts at any point $p$ and its image $\alpha (p)$ such that the map $\hat{\alpha }$ is represented as

$$\begin{array}{ll}{y}^i={\alpha }^i(x^1,x^2,\dots ,x^m)=x^i& \text{ for }i=1,\dots ,m,\\ y^a={\alpha }^a(x^1,x^2,\dots ,x^m)=0& \text{ for }a>m.\end{array}$$

A detailed proof may be found in [11].

Example 15.10

In general the image $\alpha (M)\subset N$ of an immersion is not a genuine ‘sub manifold’, since there is nothing to prevent self-intersections. For example the mapping $\alpha :\mathbb{R}{\mathbb{R}}^2$ defined by

$$x={\alpha }^1(t)=t(t^2-1),y={\alpha }^2(t)=t^2-1$$

is an immersion since its Jacobian matrix is everywhere non-degenerate,

$$\left(\begin{array}{cc}\frac{\partial x}{\partial t}& \frac{\partial y}{\partial t}\end{array}\right)=\left(\begin{array}{cc}2t^2-1& 2t\end{array}\right)\ne \left(\begin{array}{cc}0& 0\end{array}\right)\text{ for all }t\in \mathbb{R}.$$

The subset $\alpha (\mathbb{R})\subset {\mathbb{R}}^2$ does not, however, inherit the manifold structure of $R$ since there is a self-intersection at $t=\pm 1$ , as shown in Fig. 15.4.

Figure 15.4 Immersion that is not a submanifold

In order to have a natural manifold structure on the subset $\alpha (M)$ we require that the map $\alpha$ is itself injective as well as its tangent map ${\alpha }_{\ast }$ . The map is then called an embedding, and the pair $(M,\alpha )$ an embedded submanifold of $N$.

Example 15.11

Let $A$ be any open subset ofa manifold $M$. As in Example 15.2, it inherits a manifold structure from $M$, whereby a chart is said to be admissible if it has the form $(U\cap A,\varphi {|}_{U\cap A})$ for some chart $(U,\varphi )$ on $M$. With this differentiable structure, $A$ is said to be an open submanifold of $M$. It evidently has the same dimension as $M$. The pair $(A,\mathrm{id}{|}_A)$ is an embedded submanifold of $M$.

Example 15.12

Let $T^2=S^1\times S^1$ be the 2-torus (see Example 15.6). The space $T^2$ can also be viewed as the factor space $R^2/\mathrm{mod}1$ , where $(x,y)=(x^{\prime },y^{\prime })$ mod 1 if there exis integers $k$ and $l$ such that $x-x^{\prime }=k$ and $y-y^{\prime }=l$ . Denote equivalence classes mod 1 by the symbol $[(x,y)]$ . Consider the curve $\alpha :\mathbb{R}{T}^2$ defined by $\alpha (t)=[(at,bt)]$ . This map is an immersion unless $a=b=0$ . If $a/b$ is a rational number it is not an embedding since the curve eventually passes through $(1,1)=(0,0)$ for some $t$ and $\alpha$ is not injective. For $a/b$ irrational the curve never passes through any point twice and is therefore an embedding. Figure 15.5 illustrates these properties. When $a/b$ is rational the image $C=\alpha (\mathbb{R})$ has the relative topology in $T^2$ ofa circle. Hence there is an embedding $\beta :S^1\to C$ , making $(S^1,\beta )$ an embedded submanifold of $T^2$ . It is left to the reader to explicitly construct the map $\beta$ In this case the subset $C=\beta (S^1)\subset T^2$ is closed.

The set $\alpha$($R$) is dense in $T^2$ when $a/b$ is irrational, since the curve eventually passes arbitrarily close to any point of $T^2$ , and cannot be a closed subset. Hence the relative topology on $\alpha$($R$) induced on it as a subset of $T^2$ is much coarser than the topology it would obtain from $R$ through the bijective map $\alpha .$ . The embedding $\alpha$ is therefore not a homeomorphism from $R$ to $\alpha (\mathbb{R})$ when the latter is given the relative topology.

In general, an embedding $\alpha :M\to N$ that is also a homeomorphism from $M$ to $\alpha$($M$) when the latter is given the relative topology in $N$ is called a regular embedding. A necessary and sufficient condition for this to hold is that there be a coordinate chart $(U,\varphi ;x^i)$

Figure 15.5 Submanifolds of the torus

at every point $p\in \alpha (M)$ such that $\alpha (M)\cap U$ is defined by the ‘coordinate slice

$$x^{m+1}=x^{m+2}=\cdots =x^n=0.$$

It also follows that the set $\alpha (M)$ must be a closed subset of $N$ for this to occur. The proofs of these statements can be found in [11]. The above embedded submanifold $(S^1,\beta )$ is a regular embedding when $a/b$ is rational.

Problems

Problem 15.13

Show that if ${\rho }_p=r_i(\mathrm{d}{x}^i{)}_p={\alpha }^{\ast }{\omega }_{\alpha (p)}$ then the components are given by

$$r_i=\frac{\partial y^a}{\partial x^i}{|}_{\varphi (p)}{w}_a$$

where ${\omega }_{\alpha (p)}=w_a(\mathrm{d}{y}^a{)}_{\alpha (p)}$

If $\alpha$ is a diffeomorphism, define a map ${\alpha }_{\ast }:T_p^{(1,1)}(M)\to T_{\alpha (p)}^{(1,1)}(N)$ by setting

$${\alpha }_{\ast }T({\omega }_{\alpha (p)},X_{\alpha (p)})=T({\alpha }^{\ast }{\omega }_{\alpha (p)},{\alpha }_{\ast }^{-1}{X}_{\alpha (p)})$$

and show that the components transform as

$$({\alpha }_{\ast }T{)}_b^a=T_j^i\frac{\partial y^a}{\partial x^i}\frac{\partial x^j}{\partial y^b}.$$

Problem 15.14

$\mathrm{If}\gamma :\mathbb{R}\to M$ is a curve on $M$ and $p=\gamma (t_0)$ and $\alpha :M\to N$ is a differentiable map show that

$${\alpha }_{\ast }{\dot{\gamma }}_p={\dot{\sigma }}_{\alpha (p)}\text{ where }\sigma =\alpha \circ \gamma :\mathbb{R}\to N.$$

Problem 15.15

Is the map $\alpha :\mathbb{R}\to {\mathbb{R}}^2$ given by $x=\sin t,y=\sin 2t(\mathrm{i})$ an immersion, (ii) an embedded submanifold?

Problem 15.16

Show that the map $\alpha :{\dot{\mathbb{R}}}^2\to {\mathbb{R}}^3$ defined by

$$u=x^2+y^2,v=2xy,w=x^2-y^2$$

is an immersion. Is it an embedded submanifold

Evaluate ${\alpha }^{\ast }(u\mathrm{d}u+v\mathrm{d}v+w\mathrm{d}w)$ and ${\alpha }_{\ast }({\partial }_x{)}_{(a,b)}$ . Find a vector field $X$ on ${\dot{\mathbb{R}}}^2$ for which ${\alpha }_{\ast }X$ is not a well-defined vector field.