Examples of Differentials

In the previous post we studied differentials in depth; now we examine examples of this.

Curves and Velocity Vectors on Manifolds

Definition 1 For a manifold \(M\), we call a \(C^\infty\) function \(\gamma:(a,b)\rightarrow M\) defined on \(M\) a \(C^\infty\) curve, and for any \(t\in (a,b)\), we call

\[\dd{\gamma_t}\left(\frac{d}{\dd{r}}\bigg\vert_t\right)\]

the velocity vector of this curve at the point \(\gamma(t)\), and denote it by \(\gamma'(t)\).

The vector \(\gamma'(t)\), as an element of \(T_{\gamma(t)}M\), acts on each element of \(\mathcal{C}^\infty_{M,\gamma(t)}\), \(f\); expanding the definition of the differential, one sees that

\[\gamma'(t)f=\dd{\gamma_t}\left(\frac{d}{\dd{r}}\bigg\vert_t\right)f=\frac{d}{\dd{r}}\bigg\vert_t (f\circ\gamma)=\frac{\dd{(f\circ \gamma)}}{\dd{r}}(t)=(f\circ\gamma)'(t)\]

holds.

In fact, when defining \(T_pM\), there is no real problem in regarding it as the collection of curves passing through the point \(p\) that are \(C^\infty\)¹. We prove the following proposition, which is part of this claim, as it 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, passing through the point \(p\) and having velocity vector at \(p\) equal to \(v\), a \(C^\infty\) curve \(\gamma\).

Proof

It suffices to find, centered at the point \(p\) and satisfying the equation

\[v=\dd{\varphi}^{-1}_{\varphi(p)}\left(\frac{\partial}{\partial r^1}\bigg\vert_0\right)\]

, a coordinate system \((U,\varphi)\). This is because \(v\) then becomes the velocity vector of the \(C^\infty\) curve

\[\gamma: t\mapsto \varphi^{-1}(t, 0,\cdots, 0)\]

at \(t=0\). Note that finding a coordinate system satisfying the above condition is very easy: after choosing an arbitrary coordinate system \((U,\psi)\), one constructs, containing the vector to which \(\dd{\psi_p}(v)\) is mapped, a new basis of \(\mathbb{R}^n\), and then composes the original \(\psi\) with the change of basis thus obtained.

In the special case \(M=\mathbb{R}^m\), a basis of \(T_{\gamma(t)}M\) is given by

\[\frac{\partial}{\partial r^1}\bigg\vert_{\gamma(t)},\cdots,\frac{\partial}{\partial r^m}\bigg\vert_{\gamma(t)}\]

, so

\[\gamma'(t)=\sum_{i=1}^m\frac{\dd{(r^i\circ \gamma)}}{\dd{r}}(t)\frac{\partial}{\partial r^i}\bigg\vert_{\gamma(t)}=\frac{\dd{\gamma}^1}{\dd{r}}\frac{\partial}{\partial r^1}\bigg\vert_{\gamma(t)}+\cdots+\frac{\dd{\gamma}^m}{\dd{r}}\frac{\partial}{\partial r^m}\bigg\vert_{\gamma(t)}\]

, and since in Euclidean space these \(\partial/\partial r^i\) are equal to the \(i\)th standard basis vectors, this can be regarded as

\[\left(\frac{\dd{\gamma}^1}{\dd{r}},\ldots, \frac{\dd{\gamma}^m}{\dd{r}}\right)\]

. Since this equation is identical to the usual derivative

\[\gamma'(t)=\lim_{h\rightarrow 0}\frac{\gamma(t+h)-\gamma(t)}{h}\]

, we can verify that our defined notions of tangent vector and velocity vector coincide in Euclidean space with the (already familiar) velocity vector of a curve.

Between two manifolds \(M,N\), suppose we are given a \(C^\infty\) function \(F:M\rightarrow N\), and consider a \(C^\infty\) function \(\gamma:(a,b)\rightarrow M\). Computing the differential of \(F\circ\gamma\) at \(t\) gives

\[\dd{(F\circ\gamma)_t}=\dd{F_{\gamma(t)}}\circ \dd{\gamma_t}\]

, and therefore

\[\dd{(F\circ\gamma)_t}\left(\frac{d}{\dd{r}}\bigg\vert_t\right)=(\dd{F_{\gamma(t)}}\circ \dd{\gamma_t})\left(\frac{d}{\dd{r}}\bigg\vert_t\right)=\dd{F_{\gamma(t)}}(\gamma'(t))\]

. Since the left-hand side can be regarded, in \(N\), as the velocity vector of the \(C^\infty\) curve \(F\circ\gamma\) at time \(t\), the above equation can be written as

\[(F\circ\gamma)'(t)=\dd{F_{\gamma(t)}}(\gamma'(t))\]

.

Modifying this slightly, we see that for a given \(C^\infty\) function \(F:M\rightarrow N\), in order to determine, for any \(v\in T_pM\), the value of the differential \(\dd{F_p}(v)\), one only needs to choose any curve having at the point \(p\) a velocity vector in the direction of \(v\)², and for this curve \(\gamma\), find the velocity vector of \(F\circ\gamma\) at time \(t\). That is,

Proposition 3 Consider two manifolds \(M,N\) and a \(C^\infty\) function \(F:M\rightarrow N\). For any \(v\in T_pM\), satisfying \(\gamma(0)=p\) and \(\gamma'(0)=v\), a \(C^\infty\) curve \(\gamma:(a,b)\rightarrow M\) also satisfies the equation

\[\dd{F_p}(v)=(F\circ\gamma)'(0)\]

Tangent Spaces of Vector Spaces

Earlier, in §Examples of Differentiable Manifolds, ⁋Example 2, we saw that any \(m\)-dimensional \(\mathbb{R}\)-vector space \(V\) carries an \(m\)-dimensional manifold structure. Then for any point \(x\in V\), the tangent space at the point \(x\), \(T_xV\), also has the same dimension as the manifold \(V\), so \(\dim T_xV=m\) holds. Therefore, \(V\cong T_xV\) must hold.

This can essentially be shown by using the fact that in Euclidean space, the standard basis vectors of \(\mathbb{R}^m\) and the basis vectors of \(T_x\mathbb{R}^m\),

\[\frac{\partial}{\partial r^1}\bigg\vert_x,\cdots,\frac{\partial}{\partial r^m}\bigg\vert_x\]

are the same. In fact, this isomorphism does not depend on the choice of basis, because it is given by associating to each \(v\in V\) the directional derivative defined by

\[D_v\vert_x: f\mapsto \lim_{t\rightarrow 0}\frac{f(x+tv)-f(x)}{t}\]

Proposition 4 Consider an \(m\)-dimensional \(\mathbb{R}\)-vector space \(V\) equipped with a manifold structure. For any point of \(V\), \(x\in V\), 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 can use the formula for the directional derivative shown above:

\[(D_v\vert_x)f=\lim_{t\rightarrow 0}\frac{f(x+tv)-f(x)}{t}\]

Under the correspondence \(v\mapsto D_v\vert_x\), the vector \(v+w\) is sent to

\[\begin{aligned}(D_{v+w}\vert_x)f&=\lim_{t\rightarrow 0}\frac{f(x+t(v+w))-f(x)}{t}\\ &=\lim_{t\rightarrow 0}\left(\frac{f((x+tw)+tv)-f(x+tw)}{t}+\frac{f(x+tw)-f(x)}{t}\right)\\ &=(D_v\vert_x)f+(D_w\vert_x)f \end{aligned}\]

and similarly \(\alpha v\) is obtained from

\[(D_{\alpha v}\vert_x)f=\lim_{t\rightarrow 0}\frac{f(x+t\alpha v)-f(x)}{t}=\alpha\lim_{t\rightarrow 0}\frac{f(x+t\alpha v)-f(x)}{\alpha t}=\alpha(D_v\vert_x)f\]

Therefore, \(v\mapsto D_v\vert_x\) is linear.

That this correspondence is injective can be seen by substituting for the function \(f\) each of \(x^1,\ldots, x^m\), and since the two vector spaces have the same dimension, this correspondence must be an isomorphism. From the above, we obtain the isomorphism \(V\cong T_xV\).

We now show the second part. Any \(v\in V\), following \(V\rightarrow W\rightarrow T_{L(x)}W\), is sent to

\[v\mapsto L(v)\mapsto D_{L(v)}\vert_{L(x)}\]

On the other hand, following \(V\rightarrow T_xV\rightarrow T_{L(x)}W\), we first obtain from \(V\rightarrow T_xV\)

\[v\mapsto D_v\vert_x\]

and after that, using \(\gamma(t)=x+tv\) and applying Proposition 3, we obtain

\[\dd{L_x}(D_v\vert_x)=(L\circ \gamma)'(0)\]

However, since

\[(L\circ\gamma)(t)=L(x+tv)=L(x)+tL(v)\]

\((L\circ\gamma)'(0)\) satisfies, for any \(f\),

\[(L\circ\gamma)'(0)f=\lim_{t\rightarrow 0}\frac{f(L(x)+tL(v))-f(L(x))}{t}=(D_{L(v)}\vert_{L(x)})f\]

Therefore, the given diagram commutes.

The isomorphism \(V\cong T_xV\) constructed in the proposition above does not depend on the choice of basis, but if, for \(V\), some basis \(e_1,\ldots, e_m\) and its dual basis \(r^1,\ldots, r^m\) are given, one can verify that this isomorphism is given by

\[\sum a_ie_i\leftrightarrow\sum a_i\frac{\partial}{\partial r^i}\bigg\vert_x\]

Example 5 The collection of \(n\times n\) matrices \(\Mat_n(\mathbb{R})\) is an \(n^2\)-dimensional \(\mathbb{R}\)-vector space. Therefore, the tangent space at any point of \(\Mat_n(\mathbb{R})\) is the same as \(\Mat_n(\mathbb{R})\).

In particular, considering an open submanifold of \(\Mat_n(\mathbb{R})\), namely \(\GL(n;\mathbb{R})\), 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 covector

Let an arbitrary manifold \(M\) and a \(C^\infty\) function \(f:M\rightarrow\mathbb{R}\) be given. Then for every point \(p\in M\), the differential \(\dd{f_p}:T_pM\rightarrow T_{f(p)}\mathbb{R}\) is well-defined. By the preceding Proposition 4, as 1-dimensional \(\mathbb{R}\)-vector spaces, there exists an isomorphism between \(\mathbb{R}\) and its tangent space \(T_{f(p)}\mathbb{R}\). Then now, via

\[T_pM\overset{\dd{f_p}}{\rightarrow}T_{f(p)}\mathbb{R}\overset{\sim}{\rightarrow}\mathbb{R}\]

we can view \(\dd{f_p}\) as an element of \((T_pM)^\ast\).

Definition 6 For a manifold \(M\) and a point \(p\in M\), the dual space of the \(\mathbb{R}\)-vector space \(T_pM\), \((T_pM)^\ast\), 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.

Therefore, 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, bases of \(T_pM\) define a dual basis of \(T_p^\ast M\).

Let \((U,\varphi)\) be a coordinate system containing the point \(p\), and let \(\varphi=(x^i)_{i=1}^m\). Then the following \(m\) tangent vectors

\[\frac{\partial}{\partial x^1}\bigg\vert_p,\cdots\frac{\partial}{\partial x^m}\bigg\vert_p\]

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

\[(\xi^i \vert_p)\left(\frac{\partial}{\partial x^j}\bigg\vert_p\right)=\delta_{ij}\tag{1}\]

where \(\delta_{ij}\) denotes the Kronecker delta.

Proposition 7 In the above situation, \(\xi^i\vert_p=\dd{x}^i\vert_p\). That is, arising from \((U,\varphi)\), the dual basis elements of \(T_pM\), \((\xi^i \vert_p)\), are in fact equal to the differentials of the coordinate functions \(x^i\) at the point \(p\).

Proof

It suffices to show that the \(\dd{x}^i\) satisfy equation (1) above. By definition,

\[\dd{x}^i\vert_p\left(\frac{\partial}{\partial x^j}\bigg\vert_p\right)=\frac{\partial}{\partial x^j}\bigg\vert_p x^i=\delta_{ij}\]

holds.

The above proof becomes more plausible if we recall §Cotangent Space, ⁋Lemma 1, which we proved when we first introduced the tangent space. That is, passing from the first equality to the second expression is the definition of the differential \(\dd{x}^i\vert_p\), but at the same time, via

\[T_p^\ast M\cong (\mathfrak{m}_p/\mathfrak{m}_p^2)^{\ast\ast}\cong\mathfrak{m}_p/\mathfrak{m}^2_p\]

it is also 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.

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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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1. More precisely, curves having the same velocity vector at the point \(p\) should be treated as identical to give an equivalence relation. ↩

2. By Proposition 2, at least one such curve exists. ↩