For embedded sub-manifolds in Euclidean space, a tangent vector is the velocity vector (as a vector in the ambient Euclidean space) of a smooth curve lying on the sub-manifold. In order to make this definition, we need to use the ambient Euclidean structure. For a general manifold such a structure is not available, hence we need a way to generalise tangent vectors. The approach taken here is to observe that many curves may represent the same tangent vector, and our definition is to essentially define a tangent vector to be all curves with the same velocity vector in a chart! Next we can define the differential of a smooth function by mapping curves to curves by the function. In a chart, this notion recovers the usual notion of directional derivative.
Let \(M \subseteq \mathbb{R}^n\) be a \(k\)-dimensional, embedded sub-manifold.
A tangent vector at \(p \in M\) is a vector \(X \in \mathbb{R}^n\) such that \[ X = \partial_t|_{t=0} c(t) = c'(0) \] where \(c : (-\epsilon, \epsilon) \to M\) is a \(C^1\) map such that \(c(0) = p\).
The set of tangent vectors at \(p\) is called the tangent space to \(M\) at \(p\) and is denoted \(T_pM\).
A \(C^1\) map \(f : N \to M\) between manifolds is a map such that \(\varphi \circ f \circ \psi^{-1}\) is \(C^1\) for any \(C^{\infty}\) charts \(\varphi, \psi^{-1}\) for \(M, N\) respectively. For curves \(c : (-\epsilon, \epsilon) \to M\) we take \(N = (-\epsilon, \epsilon) \subseteq \mathbb{R}\) and the identify function as a single chart so that \(\varphi \circ c : (-\epsilon, \epsilon) \to \mathbb{R}^k\) is \(C^{\infty}\).
In the case \(M\) is a sub-manifold of \(\mathbb{R}^n\), if \(\varphi\) is a local parametrisation, then a chart is \(\varphi^{-1}\). Thus \(c\) is \(C^1\) if and only if \(\varphi^{-1} \circ c\) is \(C^1\) for every local parametrisation of \(M\).
Let \(\varphi : U \subseteq \mathbb{R}^k \to V \subseteq M\) be a local parametrisation. Then for \(p \in V\), and \(X \in T_pM\) there exists a unique \(Y \in \mathbb{R}^k\) such that \[ X = d\varphi_x(Y) \] where \(x = \varphi^{-1}(p)\).
Recall that the charts covering \(M\) can be taken as the inverses, \(\varphi^{-1}\) of local parametrisations. Since \(c\) is \(C^1\), the map \(\varphi^{-1} \circ c\) is a \(C^1\) function \((-\epsilon, \epsilon) \to \mathbb{R}^k\). Then by the chain rule, \[ c'(0) = (\varphi \circ \varphi^{-1})' \circ (0) = d\varphi_{\varphi^{-1}(c(0))} (\varphi^{-1} \circ c)' (0). \] But \(\varphi^{-1}(c(0)) = \varphi^{-1}(p) = x\). Thus letting \(Y = (\varphi^{-1} \circ c)'(0)\) we have \[ X = c'(0) = d\varphi_x (Y). \]
That \(Y\) is unique follows since \(d\varphi_x\) is injective.
The tangent space \(T_x M\) is a \(k\)-dimensional vector subspace of \(\mathbb{R}^n\).
The previous lemma implies that \(T_pM = d\varphi_x (\mathbb{R}^k)\) where \(x = \varphi^{-1}(p)\). But \(d\varphi_x : \mathbb{R}^k \to \mathbb{R}^n\) is a linear map, hence it's range, \(T_pM\) is a vector subspace of \(\mathbb{R}^n\). Since \(d\varphi_x\) is injective, by rank-nullity, the dimension of it's range is \(k\).
Let \(M = \mathbb{S}^n \subseteq \mathbb{R}^{n+1}\). Then \[ T_p \mathbb{S}^n = p^{\perp} := \{X \in \mathbb{R}^{n+1} : \langle p, X\rangle = 0\}. \]
Let \(X = c'(0) \in T_p \mathbb{S}^n\) where \(c : (\epsilon, \epsilon) \to \mathbb{S}^n\) is \(C^1\). Then since \(c(t) \in \mathbb{S}^n\) for every \(t\) we have \(\langle c(t), c(t)\rangle = 1\) for every \(t\). Differentiating and using the product rule for the Euclidean inner-product gives
\begin{align*} 0 &= \partial_t|_{t=0} 1 = \partial_t|_{t=0} \langle c(t), c(t) \rangle \\ &= 2 \langle \partial_t|_{t=0} c(t), c(t)\rangle \\ &= 2 \langle X, p\rangle. \end{align*}Thus \(X \in p^{\perp}\) and \(T_p \mathbb{S}^n \subseteq p^{\perp}\). But \(\operatorname{dim} T_p \mathbb{S}^n = n = \operatorname{dim} p^{\perp}\) hence \(T_p\mathbb{S}^n = p^{\perp}\).
More generally there is the case of level sets of regular values.
Let \(F : U \subseteq \mathbb{R}^n \to \mathbb{R}^k\) be a smooth function with \(U\) open. Let \(c \in \mathbb{R}^k\) be a regular value so that \(M = F^{\ast}(\{c\})\) is an embedded sub-manifold. Show that for every \(p \in M\), \[ T_p M = \nabla F(p)^{\perp}. \]
A tangent vector of a Euclidean sub-manifold is the velocity vector of a curve lying on the sub-manifold. Note that in order to differentiate a curve (and thus obtain the velocity vector) we think of a curve as a map into the ambient Euclidean space. For a general manifold, there is no ambient space so we need a different approach, although it is motivated by the Euclidean sub-manifold case. That is, we need a suitable interpretation of velocity vectors of curves. As usual for manifolds, the definition is via the charts and we need to account for the non-uniqueness of curves with given velocity vector.
Let \(M\) be a \(C^{\infty}\) manifold and let \(p \in M\). A tangent vector to \(M\) at \(p\) is an equivalence class of \(C^1\) curves in \(M\) where \(c_1 \sim c_2\) if \(c_1(0) = c_2(0) = p\) and \((\varphi \circ c_1)'(0) = (\varphi \circ c_2)'(0)\) for every chart \(\varphi : U \to \mathbb{R}^n\) on \(M\) with \(p \in U\).
The set of all tangent vectors at \(p\) is called the tangent space at \(p\) and is denoted \(T_p M\).
Since we have no way to differentiate functions into \(M\), we defer to the charts. Thus if \(c : (-\epsilon, \epsilon) \to M\) is a \(C^1\) curve, then by definition \(\varphi \circ c : (-\epsilon, \epsilon) \to \mathbb{R}^n\) is a \(C^1\) function hence \((\varphi \circ c)'(0)\) is defined. For general manifolds, deferring to charts is the basic tool we have to deal with anything regarding differentiation!
The idea behind defining tangent vectors as equivalence classes is that, in Euclidean space for example, many curves through a given point have the same velocity vector. For example we can always take \(c(t) = p + tV\) as a curve through the point \(p\) with velocity vector \(V\) at \(p\). But we could also say \(c(t) = p + e^t V\) and many more different possibilities. In other words, all curves through \(p\) with velocity vector \(V\) represent the same tangent vector and so we define a tangent vector simply to be all curves that pass through the same point with the same velocity vector!
Recall that an equivalence relation is a relation that is
Show that the definition of \(c_1 \sim c_2\) above is an equivalence relation on \(C^1\) curves \(c : (-\epsilon, \epsilon) \to M\).
A familiar example of equivalence relation is that of rational numbers. We can represent a rational number as a pair of integers - the numerator and denominator - where the denominator is non-zero. But this representation is not unique; for example \(1/2 = 2/4 = (-7)/(-14)\) and so on. One way then to define rational numbers is as equivalence classes of pairs of integers. This is the basic idea behind equivalence relations: when we wish to define some object but have different ways of representing that object, we can define said object to be the set of all possible representations!
Our equivalence relation partitions the set of all curves into \(M\) into equivalence classes.
An equivalence class according to our equivalence relation is a maximal subset of curves where every curve in an equivalence class is equivalent to every other curve in the same equivalence class.
For any \(C^1\) curve \(c\), the equivalence class of \(c\), denoted \([c]\) is the subset, \[ [c] = \{d : d \sim c\} = \{d : c \sim d\}. \]
The second equality of sets follows by symmetry. Since \(c \sim c\), we have \(c \in [c]\). By transitivity if \(d_1, d_2 \in [c]\) then \(d_1 \sim c\) and \(c \sim d_2\) implies that \(d_1 \sim d_2\); that is each pair of elements in an equivalence class are equivalent. Moreover, if \(d_1 \in [c]\) and \(d_2 \sim d_1\), then \(d_2 \in [c]\) as well, so that every curve equivalent to \(d_1\) is also in \([c]\).
Show that \([c] = [d]\) if and only if \(c \sim d\).
The upshot of all this is that we define tangent vectors via the charts as velocity vectors. But since there are no canonical ways to represent vectors as velocity vectors, we simply take the set of all possible such velocity vectors as our definition.
Let us now consider the role of charts in the definition. We require \((\varphi \circ c_1)'(0) = (\varphi \circ c_2)'(0)\) for every chart. In fact, it is sufficient to check this requirement in just a single chart.
Let \(c_i : (-\epsilon_i, \epsilon_i) \to M\), \(i = 1,2\) be \(C^1\) functions. Then for any charts \(\varphi : U \to \mathbb{R}^n\), \(\psi : V \to \mathbb{R}^n\) with \(p \in U\) and \(p \in V\) we have \((\varphi \circ c_1)'(0) = (\varphi \circ c_2)'(0)\) if and only if \((\psi \circ c_1)'(0) = (\psi \circ c_2)'(0)\).
Let \(\tau = \varphi \circ \psi^{-1}\) denote the transition function and note that \(\tau\) is a diffeomorphism. Then
\begin{align*} (\varphi \circ c_i)'(0) &= (\varphi \circ \psi^{-1} \circ \psi \circ c_1)'(0) \\ &= (\tau \circ \psi \circ c_i)'(0) \\ &= d\tau [(\psi \circ c_i)'(0)]. \end{align*}Thus \((\varphi \circ c_1)'(0) = (\varphi \circ c_2)'(0)\) if and only if \(d\tau[(\psi \circ c_1)'(0)] = d\tau [(\psi \circ c_2)'(0)]\). But since \(\tau\) is a diffeomorphism, \(d\tau\) is an isomorphism. Thus \(d\tau[(\psi \circ c_1)'(0)] = d\tau [(\psi \circ c_2)'(0)]\) if and only if \((\psi \circ c_1)'(0) = (\psi \circ c_2)'(0))\). The result follows by transitivity of "if and only if".
The lemma tells us that either \((\varphi \circ c_1)'(0) = (\varphi \circ c_2)'(0)\) for every chart or \((\varphi \circ c_1)'(0) \neq (\varphi \circ c_2)'(0)\) for every chart. In particular it implies that we need only check equality in a single chart and the outcome applies in every chart.
Let \(M = \mathbb{R}^n\) and take \(\{\operatorname{Id} : \mathbb{R}^n \to \mathbb{R}^n\}\) as a single chart covering \(M\). Then a curve \(c\) into \(M\) is \(C^1\) if and only if \(c = \operatorname{Id} \circ c\) is \(C^1\)!
One would expect that for any \(p \in M\), \(T_p M\) is just \(\mathbb{R}^n\). We can exhibit a bijection \(\mathbb{R}^n \to T_p M\) as follows: given \(V \in \mathbb{R}^n\), define the curve \[ c_V (t) = p + tV. \] and corresponding tangent vector \[ X = [c_V]. \] Note that \(c_V(0) = p\) so \([C]\) is indeed a tangent vector at \(p\). Thus we have the function \[ V \in \mathbb{R}^n \mapsto [c_V] \in T_p M. \] To see that this is a bijection, first we show it is surjective: if \(X \in T_p M\), then \(X\) is an equivalence class of curves. Let \(d \in X\) be an arbitrary curve. Then \(d\) is a \(C^1\) function \((-\epsilon, \epsilon) \to M\) with \(d(0) = p\). Now let \(V = d'(0)\). Then \(c_V(t) = p + tV\) satisfies \(c_V(0) = p = d(0)\) and \(c_V'(0) = V = d'(0)\) hence \(c \sim d\). Thus \([c_V] = [d] = X\) and the function \(V \mapsto [c_V]\) is surjective.
For injectivity, suppose that \([c_V] = [c_W]\). Then \(c_V \sim d_V\). That is, \(c_V(0) = c_W(0)\) and \(c_V'(0) = c_W'(0)\). But \(c_V'(0) = V\) and \(c_W'(0) = W\). Thus \(V = W\) and hence the function \(V \mapsto [c_V]\) is injective.
Thus we can think of \(T_p \mathbb{R}^n\) as \(\mathbb{R}^n\) via the bijection \(V \mapsto [c_V]\).
We have successfully taken inspiration from Euclidean sub-manifolds to come up with a suitable definition of tangent vector on general manifolds. But Euclidean sub-manifolds are in particular manifolds and we should verify that our new definition is in fact a generalisation of the old definition. That is, we should check that in the case of Euclidean sub-manifolds, this new definition agrees with the old definition.
Let \(M^k \subseteq \mathbb{R}^n\) be a \(k\)-dimensional sub-manifold. There is a bijection between \(T_p M\) as the tangent space to \(M\) as a sub-manifold and \(T_p M\) as a manifold.
Recall that our previous definition of \(T_p M\) is \[ T_p M = \{c'(0) \> | \> c : (-\epsilon, \epsilon) \to M, \> c(0) = p\}. \] Let us temporarily write \[ E_p M = \{[c] \> | \> c : (-\epsilon, \epsilon) \to M, \> c(0) = p\} \] for the set of equivalence classes of curves; that is for the new definition from this section.
Our proposed bijection \(E_p M \to T_p M\) is then \[ X \mapsto c'(0) \] where \(c\) is any curve into \(M\) such that \([c] = X\); that is \(c \in X\). Note here that for such a curve \(c : (-\epsilon, \epsilon) \to M\), when we take \(c'(0)\) we are thinking of \(c : (-\epsilon,\epsilon) \to \mathbb{R}^n\) which we can do since \(M \subseteq \mathbb{R}^n\).
The proof that this function is a bijection is relatively straight forward, provided the function is well defined. So let's first tackle the latter point. In order to define our candidate bijection we made an arbitrary choice of \(c \in X\). But what if we made a different choice? Do we get the same result, and hence that the function \(X \mapsto c'(0)\) is well defined independently of our choice? The answer is indeed yes, and to see this let \(c, d \in X\) so that \(c \sim d\). Then \(c(0) = d(0) = p\), and for any chart \(\varphi\) we have \((\varphi \circ c)'(0) = (\varphi \circ d)'(0)\). Since \(M\) is a sub-manifold, \(\varphi^{-1}\) is a local parametrisation hence
\begin{align*} c'(0) &= (\varphi^{-1} \circ \varphi \circ c)'(0) \\ &= d\varphi^{-1} [(\varphi \circ c)'(0)] \\ &= d\varphi^{-1} [(\varphi \circ d)'(0)] \\ &= (\varphi^{-1} \circ \varphi \circ d)'(0) \\ &= d'(0). \end{align*}That is, \(c'(0) = d'(0)\) and hence \(X \mapsto c'(0) \in \mathbb{R}^n\) gives the same result whether we choose \(c\) or \(d\), and thus \(X \mapsto c'(0)\) is a well defined function \(E_p M \to \mathbb{R}^n\). The range of this function does in fact lie in \(T_p M\) since \(c\) is a curve lying in \(M\), passing through \(p\) at \(0\). The velocity vectors of such curves are precisely the elements of \(T_pM\). Thus \(X \mapsto c'(0)\) is a well defined function \(E_p M \to T_p M\).
Now we prove it's a bijection.
To prove surjectivity, let \(V \in T_p M\) so that \(V = c'(0)\) for some \(C^1\) curve \(c : (-\epsilon, \epsilon) \to M\). Then letting \(X = [c]\) we have \(X \mapsto c'(0) = V\) and the function is surjective.
To prove injectivity, suppose \(X \mapsto V \in T_p M\) and \(Y \mapsto V \in T_p M\). Thus \(V = c'(0) = d'(0)\) where \(c \in X\) and \(d \in Y\). Since \(X, Y\) both map to \(T_p M\) we have \(c(0) = d(0) = p\). We also have \(c'(0) = d'(0) = V\) hence \(c \sim d\), hence \(X = [c] = [d] = Y\).
Define the inverse function and prove directly (i.e. without using the lemma) that it's a bijection. Also show that it's a bijection by showing that it is indeed the inverse function - note you must show it's both a left and right inverse!
We already saw that \[ T_p \mathbb{S}^n = p^{\perp}. \]
Let us verify directly that the new definition of \(T_p \mathbb{S}^n\) in terms of equivalence classes agrees with this. That is, the function \(X \mapsto c'(0)\) from the previous lemma identifies \(T_p \mathbb{S}^n\) (as newly defined) with \(p^{\perp}\). Thus we need to show that for every \(X \in T_p \mathbb{S}^n\) and one (hence every) \(c \in X\), we have \(c'(0) \perp p\), and conversely that for every \(V \in p^{\perp}\), there is an \(X \in T_p \mathbb{S}^n\) with \(c'(0) \perp p = c(0)\) for one (hence all) \(c \in X\).
For the former, let \(X \in T_p \mathbb{S}^n\) and let \(c \in X\). Then \(c : (-\epsilon, \epsilon) \to \mathbb{S}^n\) with \(c(0) = p\). Since \(c(t) \in \mathbb{S}^n\) for \(t \in (-\epsilon, \epsilon)\) we have \(\|c(t)\| \equiv 1\) and differentiating shows that \(c'(0) \perp c(0) = p\). Thus \(X \mapsto c'(0)\) has range contained in \(p^{\perp}\).
Conversely, if \(V \in p^{\perp}\), \(c(t) = (\cos t) p + (\sin t) V\) is a curve lying in \(\mathbb{S}^n\) with \(c(0) = p\) and \(c'(0) = V\). Taking \(X = [c]\) we then have \(X \mapsto c'(0) = V\) and hence \(p^{\perp}\) is contained in the range of this function.
Thus the function \(X \mapsto c'(0)\) has range precisely equal to \(p^{\perp}\) and we've already seen that this function is injective, thus a bijection with it's range \(p^{\perp}\).