This post was translated from Korean by LLM (Kimi). The translation may contain errors or awkward sentences. The Korean original is the source of truth.
In the previous post we studied differentials in depth; now we examine some examples.
Curves and Velocity Vectors on Manifolds
Definition 1 For a manifold \(M\), we call a \(C^\infty\) function \(\gamma:(a,b)\rightarrow M\) a \(C^\infty\) curve on \(M\), and for any \(t\in (a,b)\) we call
\[d\gamma_t\left(\frac{d}{dr}\bigg|_t\right)\]the velocity vector of this curve at the point \(\gamma(t)\), denoting it by \(\gamma'(t)\).
As an element of \(T_{\gamma(t)}M\), the vector \(\gamma'(t)\) acts on each element \(f\) of \(\mathcal{C}^\infty_{M,\gamma(t)}\); expanding the definition of the differential yields
\[\gamma'(t)f=d\gamma_p\left(\frac{d}{dr}\bigg|_t\right)f=\frac{d}{dr}\bigg|_t (f\circ\gamma)=\frac{d(f\circ \gamma)}{dr}(t)=(f\circ\gamma)'(t).\]In fact, when defining \(T_pM\) we may just as well regard it as the collection of \(C^\infty\) curves passing through the point \(p\)1. We prove the following proposition, which is part of this claim and will be used frequently.
Proposition 2 Fix a manifold \(M\) and a point \(p\in M\). For any nonzero \(v\in T_pM\), there exists a \(C^\infty\) curve \(\gamma\) passing through \(p\) whose velocity vector at \(p\) equals \(v\).
Proof
It suffices to find a coordinate system \((U,\varphi)\) centered at \(p\) satisfying
\[v=d\varphi^{-1}_{\varphi(p)}\left(\frac{\partial}{\partial r^1}\bigg|_0\right).\]Then \(v\) is the velocity vector at \(t=0\) of the \(C^\infty\) curve
\[\gamma: t\mapsto \varphi^{-1}(t, 0,\cdots, 0).\]Finding such a coordinate system is straightforward: choose an arbitrary coordinate system \((U,\psi)\), construct a new basis of \(\mathbb{R}^n\) containing the translated vector \(d\psi_p(v)\), and compose the original \(\psi\) with the resulting change of basis.
In the special case \(M=\mathbb{R}^m\), a basis of \(T_{\gamma(t)}M\) is given by
\[\frac{\partial}{\partial r^1}\bigg|_{\gamma(t)},\cdots,\frac{\partial}{\partial r^m}\bigg|_{\gamma(t)},\]so
\[\gamma'(t)=\sum_{i=1}^m\frac{d(r^i\circ \gamma)}{dr}(t)\frac{\partial}{\partial r^i}\bigg|_{\gamma(t)}=\frac{d\gamma^1}{dr}\frac{\partial}{\partial r^1}\bigg|_{\gamma(t)}+\cdots+\frac{d\gamma^m}{dr}\frac{\partial}{\partial r^m}\bigg|_{\gamma(t)},\]and since in Euclidean space these \(\partial/\partial r^i\) coincide with the \(i\)th standard basis vectors, we may identify this with
\[\left(\frac{d\gamma^1}{dr},\ldots, \frac{d\gamma^m}{dr}\right).\]This coincides with the usual derivative
\[\gamma'(t)=\lim_{h\rightarrow 0}\frac{\gamma(t+h)-\gamma(t)}{h},\]so we can verify that our notions of tangent vector and velocity vector agree in Euclidean space with the familiar velocity vector of a curve.
Now let \(F:M\rightarrow N\) be a \(C^\infty\) function between two manifolds \(M,N\), and let \(\gamma:(a,b)\rightarrow M\) be a \(C^\infty\) curve. Computing the differential of \(F\circ\gamma\) at \(t\) gives
\[d(F\circ\gamma)_t=dF_{\gamma(t)}\circ d\gamma_t,\]and therefore
\[d(F\circ\gamma)_t\left(\frac{d}{dr}\bigg|_t\right)=(dF_{\gamma(t)}\circ d\gamma_t)\left(\frac{d}{dr}\bigg|_t\right)=dF_{\gamma(t)}(\gamma'(t)).\]Since the left-hand side is the velocity vector at time \(t\) of the \(C^\infty\) curve \(F\circ\gamma\) in \(N\), the above equation can be written as
\[(F\circ\gamma)'(t)=dF_{\gamma(t)}(\gamma'(t)).\]Refining this slightly, we see that for a given \(C^\infty\) function \(F:M\rightarrow N\), in order to determine the value \(dF_p(v)\) of the differential at any \(v\in T_pM\), we need only choose any curve having velocity vector \(v\) at the point \(p\)2, and then compute the velocity vector of \(F\circ\gamma\) at time \(t\) for this curve \(\gamma\). That is,
Proposition 3 Let \(M,N\) be two manifolds and \(F:M\rightarrow N\) a \(C^\infty\) function. For any \(v\in T_pM\), any \(C^\infty\) curve \(\gamma:(a,b)\rightarrow M\) satisfying \(\gamma(0)=p\) and \(\gamma'(0)=v\) also satisfies
\[dF_p(v)=(F\circ\gamma)'(0).\]Tangent Spaces of Vector Spaces
In §Examples of Differentiable Manifolds, ⁋Example 2, we saw that any \(m\)-dimensional \(\mathbb{R}\)-vector space \(V\) carries the structure of an \(m\)-dimensional manifold. Hence for any point \(x\in V\), the tangent space \(T_xV\) at \(x\) has the same dimension as the manifold \(V\), so \(\dim T_xV=m\). Therefore \(V\cong T_xV\).
This can be seen from the fact that in Euclidean space, the standard basis vectors of \(\mathbb{R}^m\) and the bases
\[\frac{\partial}{\partial r^1}\bigg|_x,\cdots,\frac{\partial}{\partial r^m}\bigg|_x\]of \(T_x\mathbb{R}^m\) are essentially the same. In fact, this isomorphism does not depend on the choice of basis: for any \(v\in\mathbb{R}^m\), the correspondence with the directional derivative
\[D_v|_x: f\mapsto \lim_{h\rightarrow 0}\frac{f(x+tv)-f(x)}{t}\]gives this isomorphism.
Proposition 4 Let \(V\) be an \(m\)-dimensional \(\mathbb{R}\)-vector space equipped with a manifold structure. For any point \(x\in V\), there exists an isomorphism \(V\cong T_xV\) independent of the choice of basis. Moreover, if \(V,W\) are two \(\mathbb{R}\)-vector spaces and \(L:V\rightarrow W\) is a linear map, then the following diagram commutes.
Proof
For the first part, we use the directional derivative formula shown above,
\[(D_v|_x)f=\lim_{t\rightarrow 0}\frac{f(x+tv)-f(x)}{t}.\]Under the correspondence \(v\mapsto D_v\vert_x\), the vector \(v+w\) is sent to
\[\begin{aligned}(D_{v+w}|_x)f&=\lim_{t\rightarrow 0}\frac{f(x+t(v+w))-f(x)}{t}\\ &=\lim_{t\rightarrow 0}\left(\frac{f((x+tw)+tv)-f(x+tw)}{t}+\frac{f(x+tv)-f(x)}{t}\right)\\ &=(D_v|_x)f+(D_w|_x)f, \end{aligned}\]and similarly \(\alpha v\) yields
\[(D_{\alpha v}|_x)f=\lim_{t\rightarrow 0}\frac{f(x+t\alpha v)-f(x)}{t}=\alpha\lim_{t\rightarrow 0}\frac{f(x+t\alpha v)-f(x)}{\alpha t}=\alpha(D_v|_x)f.\]Thus \(v\mapsto D_v\vert_x\) is linear.
Injectivity follows by substituting \(x^1,\ldots, x^m\) for \(f\), and since the two vector spaces have the same dimension, this correspondence must be an isomorphism. Hence we obtain the isomorphism \(V\cong T_xV\).
For the second part, following \(V\rightarrow W\rightarrow T_{L(x)}W\), any \(v\in V\) is sent to
\[v\mapsto L(v)\mapsto D_{L(v)}|_{L(x)}.\]On the other hand, following \(V\rightarrow T_xV\rightarrow T_{L(x)}W\), we first obtain
\[v\mapsto D_v|_x\]via \(V\rightarrow T_xV\), and then using \(\gamma(t)=x+tv\) and Proposition 3 we get
\[dL_x(D_v|_x)=(L\circ \gamma)'(0).\]But
\[(L\circ\gamma)(t)=L(x+tv)=L(x)+tL(v),\]so for any \(f\), \((L\circ\gamma)'(0)\) satisfies
\[(L\circ\gamma)'(0)f=\lim_{t\rightarrow 0}\frac{f(L(x)+tL(v))-f(L(x))}{t}=(D_{L(v)}|_{L(x)})f.\]Therefore the given diagram commutes.
The isomorphism \(V\cong T_xV\) constructed above does not depend on the choice of basis, but if a basis \(e_1,\ldots, e_n\) of \(V\) and its dual basis \(r^1,\ldots, r^n\) are given, one can check that this isomorphism is
\[\sum a_ie_i\leftrightarrow\sum a_i\frac{\partial}{\partial r^i}.\]Example 5 The set \(\Mat_n(\mathbb{R})\) of \(n\times n\) matrices is an \(n^2\)-dimensional \(\mathbb{R}\)-vector space. Hence the tangent space at any point of \(\Mat_n(\mathbb{R})\) is the same as \(\Mat_n(\mathbb{R})\).
In particular, for the open submanifold \(\GL(n,\mathbb{R})\) of \(\Mat_n(\mathbb{R})\), the tangent space at any element of \(\GL(n,\mathbb{R})\) coincides with the tangent space of that element viewed as an element of \(\Mat_n(\mathbb{R})\), and therefore equals \(\Mat_n(\mathbb{R})\).
Tangent Covectors
Let \(M\) be an arbitrary manifold and \(f:M\rightarrow\mathbb{R}\) a \(C^\infty\) function. Then for every point \(p\in M\), the differential \(df_p:T_pM\rightarrow T_{f(p)}\mathbb{R}\) is well defined. By Proposition 4, there exists an isomorphism between \(\mathbb{R}\) and its tangent space \(T_{f(p)}\mathbb{R}\) as 1-dimensional \(\mathbb{R}\)-vector spaces. Thus, via
\[T_pM\overset{df_p}{\longrightarrow}T_{f(p)}\mathbb{R}\overset{\sim}{\longrightarrow}\mathbb{R}\]we may regard \(df_p\) as an element of \((T_pM)^\ast\).
Definition 6 For a manifold \(M\) and a point \(p\in M\), the dual space \((T_pM)^\ast\) of the \(\mathbb{R}\)-vector space \(T_pM\) is called the cotangent space, written simply as \(T_p^\ast M\). The elements of \(T_p^\ast M\) are called tangent covectors, or simply covectors.
Thus the preceding discussion can be summarized as saying that any \(C^\infty\) function \(f:M\rightarrow\mathbb{R}\) determines a tangent covector.
Meanwhile, \(T_p^\ast M\) is the dual space of the vector space \(T_pM\), and since \(T_pM\) is a finite-dimensional \(\mathbb{R}\)-vector space, any basis of \(T_pM\) defines a dual basis of \(T_p^\ast M\).
Let \((U,\varphi)\) be a coordinate system containing the point \(p\), and write \(\varphi=(x^i)_{i=1}^m\). Then the following \(m\) tangent vectors
\[\frac{\partial}{\partial x^1}\bigg|_p,\cdots\frac{\partial}{\partial x^m}\bigg|_p\]form a basis of \(T_pM\). Let us temporarily denote their dual basis by \(\xi^i \vert_p\). That is, \(\xi^i \vert_p\) is a linear map from \(T_pM\) to \(\mathbb{R}\), uniquely defined by the formula
\[(\xi^i |_p)\left(\frac{\partial}{\partial x^j}\bigg|_p\right)=\delta_{ij}\tag{1}\]where \(\delta_{ij}\) denotes the Kronecker delta.
Proposition 7 In the above situation, \(\xi^i\vert_p=dx^i\vert_p\). In other words, the dual bases \((\xi^i \vert_p)\) of \(T_pM\) arising from \((U,\varphi)\) are precisely the differentials at \(p\) of the coordinate functions \(x^i\).
Proof
It suffices to show that the \(dx^i\) satisfy equation (1). By definition,
\[dx^i|_p\left(\frac{\partial}{\partial x^j}\bigg|_p\right)=\frac{\partial}{\partial x^j}\bigg|_p x^i=\delta_{ij}.\]This proof becomes more transparent if we recall §Cotangent Space, ⁋Lemma 1, proved when we first introduced the tangent space. That is, passing from the first equality to the second is by definition of the differential \(dx^i\vert_p\), but simultaneously it is the process of naturally identifying the double dual of the finite-dimensional \(\mathbb{R}\)-vector space \(\mathfrak{m}_p/\mathfrak{m}^2_p\) with itself via
\[T_p^\ast M\cong (\mathfrak{m}_p/\mathfrak{m}_p^2)^{\ast\ast}\cong\mathfrak{m}_p/\mathfrak{m}^2_p.\]References
[War] Frank W. Warner. Foundations of Differentiable Manifolds and Lie Groups, Graduate texts in mathematics, Springer, 2013
[Lee] John M. Lee. Introduction to Smooth Manifolds, Graduate texts in mathematics, Springer, 2012
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More precisely, we must impose an equivalence relation by treating curves having the same velocity vector at the point \(p\) as identical. ↩
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By Proposition 2, at least one such curve exists. ↩
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