1 Fundamental notions

We start by introducing some notions that are fundamental for the study of curved spaces. Doing so will lead to a deeper understanding of some concepts from Linear Algebra and Analysis.

1.1 Points, vectors and the tangent space

Recall that we define \(\mathbb{R}^n\) as ordered \(n\)-tuples \(p=(x_1,\ldots,x_n)\) of scalars \(x_i \in \mathbb{R},\) \(1\leqslant i\leqslant n.\) We also consider column vectors of length \(n\) with real entries \[\vec{v}=\begin{pmatrix} x_1 \\ \vdots \\ x_n \end{pmatrix}.\] We write \(M_{m,n}(\mathbb{R})\) for the set of \((m\times n)\)-matrices with real entries. A column vector of length \(n\) may be thought of as an \((n\times 1)\)-matrix, hence we write \(M_{n,1}(\mathbb{R})\) for the set of such column vectors. Clearly we have a bijective map \[\Psi_n : \mathbb{R}^n \to M_{n,1}(\mathbb{R}), \quad (x_1,\ldots,x_n) \mapsto \begin{pmatrix} x_1 \\ \vdots \\ x_n \end{pmatrix}\] which writes the entries of an \(n\)-tuple into a column vector. Because of this map, we may avoid a distinction between \(\mathbb{R}^n\) and \(M_{n,1}(\mathbb{R})\) and pretend they are the same thing. This was done so in Linear Algebra. In geometry, it turns out to be useful to think of \(\mathbb{R}^n\) and \(M_{n,1}(\mathbb{R})\) as different sets. The elements of \(\mathbb{R}^n\) are interpreted as points and will be denoted by \(p,q,r\ldots.\) The elements of \(M_{n,1}(\mathbb{R})\) are interpreted as vectors in \(\mathbb{R}^n\) that are attached to the origin \(0_{\mathbb{R}^n}=(0,0,\ldots,0)\in \mathbb{R}^n.\) They will be denoted by \(\vec{u},\vec{v},\vec{w},\ldots .\)

Already in elementary geometry the situation occurs where we consider vectors in \(\mathbb{R}^n\) that are not attached to the origin \(0_{\mathbb{R}^n},\) but rather to some other point \(p \in \mathbb{R}^n.\) Think for instance of the normal vector of a plane in \(\mathbb{R}^3\) not containing the origin \(0_{\mathbb{R}^3}.\)

Figure 1.1: A vector \(\vec{v}\) attached at the origin and at the point \(p.\)

In order to deal with vectors that are not attached to the origin, but to a point \(p \in \mathbb{R}^n,\) we introduce the so-called tangent space of \(\mathbb{R}^n\) at \(p,\) \[T_{p}\mathbb{R}^n=\left\{\vec{v}_{p}\,|\, \vec{v} \in M_{n,1}(\mathbb{R})\right\}.\] The element \(\vec{v}_{p} \in T_p\mathbb{R}^n\) is to be interpreted as attaching the vector \(\vec{v} \in M_{n,1}(\mathbb{R})\) at the basepoint \(p \in \mathbb{R}^n.\) The elements of \(T_p\mathbb{R}^n\) are called tangent vectors with basepoint \(p\). Observe that for all \(p \in \mathbb{R}^n\) the tangent space \(T_p\mathbb{R}^n\) is a vector space over \(\mathbb{R}\) when equipped with vector addition \(+_{T_p\mathbb{R}^n} : {T_p\mathbb{R}^n} \times {T_p\mathbb{R}^n} \to {T_p\mathbb{R}^n}\) defined by the rule \[\vec{v}_{p}+_{T_p\mathbb{R}^n}\vec{w}_{p}=(\vec{v}+_{M_{n,1}(\mathbb{R})}\vec{w})_{p}\] for all \(\vec{v}_{p},\)\(\vec{w}_{p} \in T_p\mathbb{R}^n\) and scalar multiplication \(\cdot_{T_p\mathbb{R}^n} : \mathbb{R}\times {T_p\mathbb{R}^n} \to {T_p\mathbb{R}^n}\) defined by the rule \[s\cdot_{T_p\mathbb{R}^n}\vec{v}_{p}=(s\cdot_{M_{n,1}(\mathbb{R})}\vec{v})_{p}\] for all \(s \in \mathbb{R}\) and all \(\vec{v}_{p} \in T_p\mathbb{R}^n.\) Here \(+_{M_{n,1}(\mathbb{R})}\) denotes usual component-wise addition of column vectors and \(\cdot_{M_{n,1}(\mathbb{R})}\) denotes usual component-wise scalar multiplication of a column vector by a scalar. Clearly, for all \(p \in \mathbb{R}^n\) we have a vector space isomorphism \[T_p\mathbb{R}^n \to M_{n,1}(\mathbb{R}), \quad \vec{v}_{p} \mapsto \vec{v}\] which simply “forgets“ the basepoint \(p \in \mathbb{R}^n.\) We can thus think of \(T_p\mathbb{R}^n\) as a copy of \(M_{n,1}(\mathbb{R})\) attached to \(p \in \mathbb{R}^n.\) The union of all these copies of \(\mathbb{R}^n\) is known as the tangent bundle of \(\mathbb{R}^n\) \[T\mathbb{R}^n=\bigcup_{p \in \mathbb{R}^n}T_p\mathbb{R}^n=\bigcup_{p \in \mathbb{R}^n}\left\{\vec{v}_{p}\,|\, \vec{v} \in M_{n,1}(\mathbb{R})\right\}.\] At this point the name tangent space is a bit confusing, since it is unclear to what \(T_p\mathbb{R}^n\) is tangent to. This will be clarified later on. If \(U\subset \mathbb{R}^n\) is an open subset, we define likewise \[TU=\bigcup_{p \in U}T_p\mathbb{R}^n.\] Observe that for each \(p \in \mathbb{R}^n\) the tangent space \(T_p\mathbb{R}^n\) is equipped with an ordered basis \[\mathbf{e}_p^{(n)}=\big([\vec{e}_1]_{p},\ldots,[\vec{e}_n]_{p}\big),\] where \(\{\vec{e}_1,\ldots,\vec{e}_n\}\) denotes the standard basis of \(M_{n,1}(\mathbb{R}).\) For all \(p \in \mathbb{R}^n\) we call \(\mathbf{e}^{(n)}_p\) the ordered standard basis of \(T_p\mathbb{R}^n.\)

Whenever \(n\) is clear from the context we simply write \(\mathbf{e}_p\) instead of \(\mathbf{e}^{(n)}_p.\)

Remark 1.1

Since \(M_{1,1}(\mathbb{R})\) is one-dimensional, so is \(T_t\mathbb{R}\) for all \(t \in\mathbb{R}\) and the ordered standard basis of \(T_t\mathbb{R}\) consists of a single vector which we denote by \(1_{t}.\)

Recall that a pair \((V,\langle\cdot{,}\cdot\rangle)\) consisting of a vector space \(V\) over \(\mathbb{R}\) and an inner product \(\langle\cdot{,}\cdot\rangle: V\times V \to \mathbb{R}\) is called a Euclidean space.1 We can turn each tangent space into a Euclidean space:
Definition 1.2 • Standard inner product

For all \(p \in \mathbb{R}^n,\) the standard inner product on \(T_p\mathbb{R}^n\) is the unique inner product \(\langle\cdot{,}\cdot\rangle_p\) for which \(\mathbf{e}_p\) is an orthonormal basis, that is, we have \[\langle [\vec{e}_i]_{p},[\vec{e}_j]_{p}\rangle_p=\delta_{ij}=\left\{\begin{array}{ll} 1, & i=j, \\ 0, & i\neq j.\end{array}\right.\]

We will henceforth always assume that \(T_p\mathbb{R}^n\) is equipped with \(\langle\cdot{,}\cdot\rangle_p.\) Whenever no confusion can arise about the point \(p\) at which \(\langle\cdot{,}\cdot\rangle_p\) is computed, we will usually simply write \(\langle\cdot{,}\cdot\rangle.\)

1.2 Smooth maps, diffeomorphisms and the differential

We recall some facts from Analysis II, but now with a slightly more geometric perspective.

For \(n \in \mathbb{N}\) we let \(\{e_1,\ldots,e_n\}\) – here interpreted as points – denote the standard basis of \(\mathbb{R}^n,\) that is \(e_1=(1,0,0,\ldots,0),\) \(e_2=(0,1,0,\ldots,0)\) and so on. Let \(U\subset\mathbb{R}^n\) be an open set and consider a map \(f : U \to \mathbb{R}^m.\) Recall that for all \(p \in U\) and \(1\leqslant i\leqslant n\) we define the partial derivative of \(f\) in the coordinate direction \(i\) as \[\partial_if(p)=\lim_{h \to 0} \frac{1}{h}\left(f(p+he_i)-f(p)\right),\] provided the limit exists. Recall also from Analysis II that the map \(f : U \to \mathbb{R}^m\) is continuously differentiable if and only if2 for all \(1\leqslant i\leqslant n\)
  1. the partial derivative \(\partial_if(p)\) exists for all \(p \in U\);

  2. the map \(\partial_i f : U \to \mathbb{R}^m,\) \(p \mapsto \partial_if(p)\) is continuous.

Recursively, we can define higher derivatives. For \(k \in \mathbb{N}, k \geqslant 2\) we call \(f : U \to \mathbb{R}^m\) \(k\)-times continuously differentiable if \(\partial_i f : U \to \mathbb{R}^m\) is \((k-1)\)-times continuously differentiable for all \(1\leqslant i\leqslant n.\) We write \[C^k(U,\mathbb{R}^m)=\left\{f : U \to \mathbb{R}^m\, |\, f \text{ is }k\text{-times continuously differentiable}\right\}\] and

Definition 1.3 • Smooth maps

We set \[C^{\infty}(U,\mathbb{R}^m)=\bigcap_{k \in \mathbb{N}}C^k(U,\mathbb{R}^m)\] and call the elements of \(C^{\infty}(U,\mathbb{R}^m)\) smooth maps from \(U\) to \(\mathbb{R}^m\).

Throughout this module we will almost exclusively consider smooth maps.

Remark 1.4 • Smooth maps on non-open domains

It is useful to have a notion of smoothness for maps that are defined on some arbitrary subset \(\mathcal{X}\subset \mathbb{R}^n.\) A map \(f : \mathcal{X} \to \mathbb{R}^m\) is called smooth if there exists an open subset \(U\subset \mathbb{R}^n\) containing \(\mathcal{X}\) and a smooth function \(\hat{f} : U \to \mathbb{R}^m\) so that \(\hat{f}(p)=f(p)\) for all \(p \in \mathcal{X}.\)

Definition 1.5 • Differential of a map

Given \(U\subset \mathbb{R}^n,\) let \(f : U \to \mathbb{R}^m\) be smooth and write \(f=(f_1,\ldots,f_m)\) for real-valued functions \(f_i : U \to \mathbb{R}.\)

  1. The differential of \(f\) at \(p \in U\) is the unique linear map \[f_*|_{p} : T_p\mathbb{R}^n \to T_{f(p)}\mathbb{R}^m\] so that for all \[\vec{v}_p=\begin{pmatrix} v_1 \\ \vdots \\ v_n\end{pmatrix}_p \in T_p\mathbb{R}^n,\] we have \[f_*|_p(\vec{v}_p)=\vec{w}_{f(p)}=\begin{pmatrix} w_1 \\ \vdots \\ w_m\end{pmatrix}_{f(p)}\] with \[\tag{1.1} \begin{pmatrix} w_1 \\ \vdots \\ w_m\end{pmatrix}=\begin{pmatrix} \partial_1f_1(p) && \cdots && \partial_n f_1(p) \\ \vdots && \ddots && \vdots \\ \partial_1f_m(p) && \cdots && \partial_n f_m(p)\end{pmatrix}\begin{pmatrix} v_1 \\ \vdots \\ v_n\end{pmatrix}.\]

  2. Recall that the \((m\times n)\)-matrix on the right in (1.1) is called the Jacobian matrix of \(f\) at \(p\). We denote it by \(\mathbf{J}f(p).\)

  3. The map \(f_*|_{p}\) is thus the unique linear map \(T_p\mathbb{R}^n \to T_{f(p)}\mathbb{R}^m\) whose matrix representation with respect to the ordered basis \(\mathbf{e}_p^{(n)}\) of \(T_p\mathbb{R}^n\) and the ordered basis \(\mathbf{e}_{f(p)}^{(m)}\) of \(T_{f(p)}\mathbb{R}^m\) is given by \(\mathbf{J}f(p),\) that is, \[\boxed{\mathbf{M}\left(f_*|_{p},\mathbf{e}_p^{(n)},\mathbf{e}_{f(p)}^{(m)}\right)=\mathbf{J}f(p).}\] In particular, we have \[\boxed{\operatorname{rank}(f_*|_{p})=\operatorname{rank}(\mathbf{J}f(p))}\] for all \(p \in U.\)

  4. For each \(p \in U\) we obtain a linear map \(f_*|_p : T_p \mathbb{R}^n \to T_{f(p)}\mathbb{R}^m.\) It is useful to think of the family \(\{ f_*|p\}_{p \in U}\) of all such linear maps as a single map \[f_* : TU \to T\mathbb{R}^m\] defined by the rule \[f_*(\vec{v}_p)=\vec{w}_{f(p)}, \qquad \text{where} \qquad \vec{w}=\mathbf{J}f(p)\vec{v}.\] That is, for all \(p \in U,\) the restriction of \(f_*\) to \(T_p\mathbb{R}^n \subset TU\) is given by \(f_*|_p.\) The map \(f_* : TU \to T\mathbb{R}^m\) is called the differential of \(f\).

Example 1.6

Consider the smooth map \[f :\mathbb{R}^2 \to \mathbb{R}^2,\qquad p=(x,y) \mapsto f(p)=(x^2-y^2,xy).\] For the Jacobian we obtain \[\mathbf{J}f(p)=\begin{pmatrix} 2x && -2y \\ y && x \end{pmatrix}\] and hence for \[\vec{v}_p=\begin{pmatrix} u \\ w \end{pmatrix}_{(x,y)}\] we have \[f_*(\vec{v}_p)=\begin{pmatrix} 2xu-2yw \\ yu+xw\end{pmatrix}_{(x^2-y^2,xy)}.\]

Remark 1.7 • Matrices acting on points

Recall that \(\Psi_n : \mathbb{R}^n \to M_{n,1}(\mathbb{R})\) is the map that turns a point into a column vector. We use \(\Psi_n\) to let an \((m\times n)\)-matrix \(\mathbf{A}\) act on points of \(\mathbb{R}^n\) by the rule \[\mathbf{A}p:=\Psi_m^{-1}(\mathbf{A}\Psi_n(p))\] for all \(p \in \mathbb{R}^n,\) where on the right hand side \(\mathbf{A}\) acts on the column vector \(\Psi_n(p)\) by matrix multiplication.

Example 1.8

Let \(\mathbf{A}\in M_{m,n}(\mathbb{R}),\) \(b \in \mathbb{R}^m\) and consider the map \[f_{\mathbf{A},b} : \mathbb{R}^n \to \mathbb{R}^m, \qquad p \mapsto \mathbf{A}p+b.\] Then we have \[(f_{\mathbf{A},b})_*(\vec{v}_p)=(\mathbf{A}\vec{v})_{\mathbf{A}p+b}\] for all \(p \in \mathbb{R}^n\) and \(\vec{v}_p \in T_p\mathbb{R}^n.\)

Animation: We consider the map \(p=(x,y)\mapsto f(p)=(x^2-y^2,xy).\) The initial frame of the animation shows two tangent vectors (in red and blue). The last frame of the animation shows the image of the red and blue vector under the differential \(f_*.\)
Remark 1.9

Let \(U \subset \mathbb{R}\) be open and \(f : U \to \mathbb{R}\) a smooth function. We have the usual derivative from Analysis I \[f^{\prime} : U \to \mathbb{R}, \qquad t \mapsto f^{\prime}(t)=\frac{\mathrm{d}f}{\mathrm{d}t}(t)=\lim_{h \to 0}\frac{1}{h}(f(t+h)-f(t)).\] We also have the differential in the sense of Definition 1.5 which is a map \(f_* : TU \to T\mathbb{R}.\) Now notice that for all \(t \in U\) we have \[\tag{1.2} f_*\left(1_{t}\right)=f^{\prime}(t)1_{f(t)}.\] Recommendation: Pause here and think about (1.2) until you understand it.

Diffeomorphisms are smooths maps that are bijective and admit a smooth inverse:

Definition 1.10 • Diffeomorphism

Let \(U\subset \mathbb{R}^n\) and \(V\subset \mathbb{R}^m\) be open sets and \(f : U \to V\) a smooth map. If \(f\) is bijective and \(f^{-1} : V \to U\) is smooth as well, then \(f : U \to V\) is called a diffeomorphism.

Recall from Analysis that if \(f : U \to V\) is a diffeomorphism, then \(n=m\) and moreover, for all \(p \in U\) the linear map \(f_*|_{p} : T_p\mathbb{R}^n \to T_{f(p)}\mathbb{R}^n\) is invertible.

If \(f : U \to \mathbb{R}^m\) is a smooth and injective map, we say \(f\) is a diffeomorphism onto its image, provided the inverse map \(f^{-1} : \operatorname{Im}(f) \to U\) is smooth as well. Here as usual we define \[\operatorname{Im}(f)=f(U)=\{q \in \mathbb{R}^m | q=f(p),\, p \in U\}.\]

From the chain rule in Analysis II we conclude:

Proposition 1.11 • Chain rule

Let \(U\subset \mathbb{R}^n\) and \(V\subset \mathbb{R}^m\) be open sets and \(f : U \to \mathbb{R}^m\) and \(g : V \to \mathbb{R}^k\) be smooth maps with \(f(U)\subset V.\) Then \(g\circ f : U \to \mathbb{R}^k\) is smooth and for all \(p \in U\) we have \[\tag{1.3} (g\circ f)_*|_{p}=g_*|_{{}f(p)}\circ f_*|_{p}.\] That is, the differential of the composition \(g\circ f\) at \(p\) is given by the composition of the linear map \(f_*|_{p} : T_p\mathbb{R}^n \to T_{f(p)}\mathbb{R}^m\) and the linear map \(g_*|_{{}f(p)} : T_{f(p)}\mathbb{R}^m \to T_{g(f(p))}\mathbb{R}^k.\)

Remark 1.12 • Sums and products of smooth maps

The chain rule tells us that compositions of smooth maps are smooth, so are sums and products. More precisely:

  1. If \(f,g : U \to \mathbb{R}^m\) are smooth, then so is \(f+_{C^{\infty}(U,\mathbb{R}^m)}g : U \to \mathbb{R}^m,\) where \[(f+_{C^{\infty}(U,\mathbb{R}^m)}g)(p)=f(p)+_{\mathbb{R}^m}g(p)\] for all \(p \in U.\)

  2. If \(f,g : U \to \mathbb{R}\) are smooth, then so is \(f\cdot_{C^{\infty}(U,\mathbb{R})}g : U \to \mathbb{R},\) where \[(f\cdot_{C^{\infty}(U,\mathbb{R})}g)(p)=f(p)\cdot_{\mathbb{R}}g(p)\] for all \(p \in U.\)

1.3 Vector fields and the gradient

A vector field attaches a tangent vector \(\vec{v}_p\) to every point \(p\) of its domain of definition. More precisely:

Definition 1.13 • Vector field

A vector field on some open subset \(U\subset \mathbb{R}^n\) is a map \(X : U \to T\mathbb{R}^n\) so that \(X(p) \in T_p\mathbb{R}^n\) for all \(p \in U.\) For a vector field \(X : U \to T\mathbb{R}^n\) there exists unique functions \(X_i : U \to \mathbb{R},\) \(1\leqslant i\leqslant n,\) so that \[X(p)=\begin{pmatrix} X_1(p) \\ \vdots \\ X_n(p)\end{pmatrix}_p\] for all \(p \in U.\) The vector field is called smooth if the functions \(X_i\) are smooth for all \(1\leqslant i\leqslant n.\)

Vector fields appear naturally in physics. For instance, an electromagnetic field is an example of a vector field. Likewise, in the classical Newtonian theory of gravity, the gravitational field is an example of a vector field.

Example 1.14

Write \(p=(x_1,x_2)\) for an element of \(\mathbb{R}^2,\) then \[X : \mathbb{R}^2 \to T\mathbb{R}^2, \qquad p=(x_1,x_2) \mapsto \begin{pmatrix} -x_2 \\ x_1 \end{pmatrix}_{p}\] is a smooth vector field on \(\mathbb{R}^2.\)

Figure 1.2: A visualisation of the vector field \(p=(x_1,x_2) \mapsto \begin{pmatrix} -x_2 \\ x_1\end{pmatrix}_p.\)

Every smooth function gives rise to a vector field:

Definition 1.15 • Gradient

Let \(U\subset \mathbb{R}^n\) and \(f : U \to \mathbb{R}\) be a smooth function. Then the so-called gradient of \(f\) defined by \[\operatorname{grad}f : U \to T\mathbb{R}^n, \qquad p \mapsto \begin{pmatrix} \partial_1 f(p) \\ \vdots \\ \partial_n f(p) \end{pmatrix}_p\] is a smooth vector field on \(U.\)

Example 1.16

Consider the smooth function \(f : \mathbb{R}^2 \to \mathbb{R}\) defined by the rule \[f(p)=(x_1)^2+(x_2)^2,\] where we write \(p=(x_1,x_2).\) Then we have \[\operatorname{grad} f(p)=\begin{pmatrix} 2x_1 \\ 2x_2 \end{pmatrix}_p.\]

Figure 1.3: A visualisation of the gradient of the function \(p=(x_1,x_2)\mapsto (x_1)^2+(x_2)^2.\)
Exercise 1.17

For an open set \(U \subset \mathbb{R}^n\) and a smooth real-valued function \(f : U \to \mathbb{R},\) show that \[f_*(\vec{v}_p)=\langle \operatorname{grad}f(p),\vec{v}_p\rangle 1_{f(p)}\] for all \(\vec{v}_p \in TU.\)

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