# Tangent Spaces and the Tangent Bundle: Three Ways to Define Velocity on a Curved Space
> We build the tangent space three ways — curves, coordinate components, derivations — prove them naturally isomorphic via Hadamard's lemma, and construct the tangent bundle.
> https://rikai.mugen-giken.com/en/mathematics/manifolds/tangent-spaces
## 0. Key points
- The tangent space $T_pM$ of a manifold $M$ at a point $p$ admits three definitions: as **equivalence classes of curves**, as **coordinate components together with a transformation law**, and as **derivations**. All three are naturally isomorphic, and the isomorphisms do not depend on any choice of chart.
- The analytic heart of the coincidence is **Hadamard's lemma**. Because a smooth function decomposes as $f(x) = f(a) + \sum_i (x^i - a^i) g_i(x)$, the space of derivations is $n$-dimensional and $\partial/\partial x^1|_p, \ldots, \partial/\partial x^n|_p$ form a basis.
- A smooth map $F : M \to N$ induces at each point a linear map $dF_p : T_pM \to T_{F(p)}N$. In coordinates its matrix is precisely the Jacobian matrix, and the chain rule becomes the one-line identity $d(G \circ F)_p = dG_{F(p)} \circ dF_p$.
- Assembling all tangent spaces into $TM = \bigsqcup_{p \in M} T_pM$ produces, in a natural way, a smooth manifold of dimension $2n$. What makes this work is that the chart transitions take the form $(x, v) \mapsto (\tau(x), D\tau(x)v)$: points move by the chart transition, vectors by its Jacobian.
- What physicists describe by saying "a contravariant vector is a quantity transforming as $V'^{\mu} = \frac{\partial x'^{\mu}}{\partial x^{\nu}} V^{\nu}$" is exactly Definition 2, the definition by coordinate components. The four-velocity and the metric tensor of general relativity are objects living on $T_pM$.
## 1. Motivation: putting arrows on a curved space
For a particle moving inside $\mathbb{R}^n$, the velocity vector causes no trouble at all. Differentiate each component of the position $\gamma(t) \in \mathbb{R}^n$ and set $\gamma'(t) = \lim_{h \to 0} \frac{\gamma(t+h) - \gamma(t)}{h}$. This works because $\mathbb{R}^n$ is a vector space, so that the difference $\gamma(t+h) - \gamma(t)$ of two distant points means something.
Attempt the same thing for a point moving on the sphere $S^2$ and you are stuck immediately. The "difference" of two points of the sphere does not lie on the sphere. If we regard $S^2$ as a subset of $\mathbb{R}^3$, then $\gamma(t+h) - \gamma(t) \in \mathbb{R}^3$ does make sense, but only by borrowing the ambient space outside the sphere. A manifold need not come presented inside some large Euclidean space, and even when it is embedded we want an intrinsic definition that does not depend on the embedding.
There is a second problem, naive but troublesome. On a manifold there is no preferred coordinate system. If two charts $(U, \varphi)$ and $(V, \psi)$ cover a neighborhood of the same point $p$, then a velocity has two component descriptions, $v = (\varphi \circ \gamma)'(0)$ and $w = (\psi \circ \gamma)'(0)$. These are different tuples of numbers, yet they must describe the same physical situation. So the question "what is a tangent vector?" has to be answered in one of two ways:
1. define it without using coordinates, or
2. allow coordinates, but identify descriptions according to the transformation law that relates them.
Historically, the tradition of Riemann and of Ricci and Levi-Civita took route 2: a tensor was "a quantity carrying indices and transforming by a prescribed rule". The modern treatment, settled in the middle of the twentieth century, takes route 1, and moreover defines a tangent vector not as "the velocity of a curve" but as "an operator that differentiates functions". This third viewpoint looks strange at first, but its algebraic convenience is unmatched, and it leads directly into the theory of Lie groups and vector fields.
In this article we construct all three definitions and prove that they are naturally isomorphic. The goal is not to memorize one of them but to move freely among the three: curves when geometric intuition is wanted, coordinate components when a concrete computation is wanted, derivations when a proof is to be written.
## 2. Preliminaries: notation and germs of functions
Throughout, $M$ is an $n$-dimensional $C^{\infty}$ manifold (second countable and Hausdorff). We take the definition and the handling of atlases from [Smooth manifolds](/en/mathematics/manifolds/smooth-manifolds) () as known. For a chart $(U, \varphi)$ we write $\varphi = (x^1, \ldots, x^n)$ and call $x^i : U \to \mathbb{R}$ the **coordinate functions**. We follow the physicists' convention of placing indices upstairs, but $x^i$ is not the $i$-th power of $x$. The partial derivative with respect to the $i$-th variable on $\mathbb{R}^n$ is written $\partial_i$.
Let us say at once why we restrict to the $C^{\infty}$ category. The definition by derivations, given below, **fails** in the $C^k$ setting for $1 \le k < \infty$: the space of derivations on the algebra of germs of $C^k$ functions turns out to be infinite-dimensional (see and the Appendix). The definitions by curves and by coordinates work for $C^k$ with $k \ge 1$, but all three definitions agree only in the $C^{\infty}$ world.
A tangent vector is a local object, determined by information near $p$ alone. To state this precisely we first set up a framework for "functions defined only near $p$".
Let $p \in M$. On the collection of all pairs $(U, f)$, where $U$ is an open set containing $p$ and $f \in C^{\infty}(U)$, define a relation $\sim$ by
$$
(U, f) \sim (V, g) \iff \text{there is an open $W \subset U \cap V$ with $p \in W$ and } f|_W = g|_W .
$$
Then $\sim$ is an equivalence relation; an equivalence class is called a **germ** at $p$ of a $C^{\infty}$ function, written $[f]_p$ or simply $f$. The set of all germs is denoted $C^{\infty}_p(M)$.
Let us check that $\sim$ is an equivalence relation. Reflexivity and symmetry are of an obvious shape, but written out: taking $W = U$ gives $(U,f) \sim (U,f)$, and symmetry follows because the definition is symmetric in $f$ and $g$. For transitivity, if $W_1$ witnesses $(U,f) \sim (V,g)$ and $W_2$ witnesses $(V,g) \sim (V', h)$, then $W_1 \cap W_2$ is an open set containing $p$ on which $f = g = h$.
Sums, products and scalar multiples are defined on $C^{\infty}_p(M)$ representative by representative, on the intersection of the domains (independence of the representatives follows by intersecting the sets $W$ above), and this makes $C^{\infty}_p(M)$ a commutative algebra over $\mathbb{R}$. Moreover $[f]_p \mapsto f(p)$ is a well-defined algebra homomorphism $C^{\infty}_p(M) \to \mathbb{R}$.
## 3. The first definition: the velocity of a curve
We begin with the most geometric definition. A curve through $p$ is a $C^{\infty}$ map $\gamma : (-\varepsilon, \varepsilon) \to M$ with $\varepsilon > 0$ and $\gamma(0) = p$. We define what it means for two curves to "have the same velocity at $p$" using a chart. Of course this is a definition only once we have checked that it does not depend on the chart.
Let $p \in M$ and let $\gamma_1, \gamma_2$ be $C^{\infty}$ curves through $p$. If $(U, \varphi)$ and $(V, \psi)$ are charts both containing $p$, then
$$
(\varphi \circ \gamma_1)'(0) = (\varphi \circ \gamma_2)'(0)
\iff
(\psi \circ \gamma_1)'(0) = (\psi \circ \gamma_2)'(0).
$$
Since each $\gamma_j$ is continuous, after shrinking $\varepsilon$ we may assume $\gamma_j((-\varepsilon,\varepsilon)) \subset U \cap V$. The map $\tau = \psi \circ \varphi^{-1}$ is a $C^{\infty}$ diffeomorphism on $\varphi(U \cap V)$ (this is the compatibility condition in the atlas), and
$$
\psi \circ \gamma_j = (\psi \circ \varphi^{-1}) \circ (\varphi \circ \gamma_j) = \tau \circ (\varphi \circ \gamma_j).
$$
Since $\varphi \circ \gamma_j : (-\varepsilon,\varepsilon) \to \mathbb{R}^n$ is $C^{\infty}$, the gives
$$
(\psi \circ \gamma_j)'(0) = D\tau\big(\varphi(p)\big)\,(\varphi \circ \gamma_j)'(0).
$$
Here $D\tau(\varphi(p))$ is the Jacobian matrix of $\tau$, and it is **invertible** because $\tau$ is a diffeomorphism. Multiplying by an invertible matrix is injective, so $(\varphi\circ\gamma_1)'(0) = (\varphi\circ\gamma_2)'(0)$ and $(\psi\circ\gamma_1)'(0) = (\psi\circ\gamma_2)'(0)$ are equivalent.
Write $\mathcal{C}_p$ for the set of all $C^{\infty}$ curves through $p$. Declare $\gamma_1 \sim_{\mathrm{c}} \gamma_2$ when $(\varphi\circ\gamma_1)'(0) = (\varphi\circ\gamma_2)'(0)$ for some chart $(U,\varphi)$ containing $p$. By this does not depend on the chart chosen, and it is evidently an equivalence relation. The quotient $\mathcal{C}_p/\sim_{\mathrm{c}}$ is denoted $T^{\mathrm{c}}_pM$, and its elements $[\gamma]$ are called **tangent vectors** at $p$.
The virtue of this definition is that intuition works. Its defect is that sums and scalar multiples are not available on the spot: there is no operation on a manifold that "adds" two curves. To define the sum one has to choose a chart, add the components there, and pass back to a curve — and then check separately that the result is independent of the chart. This inconvenience is precisely the motivation for the next two definitions.
## 4. The second definition: coordinate components and the transformation law
We now rewrite the classical style of tensor analysis in modern language. The definition reads: a vector is a rule assigning $n$ numbers to each coordinate system, in such a way that a change of coordinates transforms them by the Jacobian matrix.
Let $\mathcal{A}_p$ be the set of all pairs $(\varphi, v)$ consisting of a chart $(U,\varphi)$ containing $p$ and a vector $v \in \mathbb{R}^n$.
For $(\varphi, v), (\psi, w) \in \mathcal{A}_p$ set
$$
(\varphi, v) \sim_{\mathrm{o}} (\psi, w)
\iff
w = D(\psi \circ \varphi^{-1})\big(\varphi(p)\big)\, v .
$$
The quotient set $\mathcal{A}_p/\sim_{\mathrm{o}}$ is denoted $T^{\mathrm{o}}_pM$.
$\sim_{\mathrm{o}}$ is an equivalence relation on $\mathcal{A}_p$.
To shorten the notation write $J_{\psi\varphi} = D(\psi\circ\varphi^{-1})(\varphi(p))$.
**Reflexivity.** $\varphi \circ \varphi^{-1}$ is the identity on $\varphi(U)$, so $J_{\varphi\varphi} = I$ (the identity matrix) and $v = Iv$.
**Symmetry.** Since $(\psi\circ\varphi^{-1})^{-1} = \varphi\circ\psi^{-1}$, the formula for the derivative of an inverse map on $\mathbb{R}^n$ (apply the chain rule to $\varphi\psi^{-1}\circ\psi\varphi^{-1} = \mathrm{id}$) gives $J_{\varphi\psi} = J_{\psi\varphi}^{-1}$. Hence $w = J_{\psi\varphi}v$ implies $v = J_{\varphi\psi} w$.
**Transitivity.** For a third chart $(W,\chi)$ we have $\chi\circ\varphi^{-1} = (\chi\circ\psi^{-1})\circ(\psi\circ\varphi^{-1})$ on $\varphi(U\cap V\cap W)$, so the chain rule yields $J_{\chi\varphi} = J_{\chi\psi}J_{\psi\varphi}$. Therefore $w = J_{\psi\varphi}v$ and $u = J_{\chi\psi}w$ imply $u = J_{\chi\varphi}v$.
Note that all three properties follow from identities valid on intersections of chart domains, and those intersections are open sets containing $p$, hence nonempty.
The set $T^{\mathrm{o}}_pM$ carries a natural vector space structure. Fixing one chart $\varphi$, the map $v \mapsto [(\varphi, v)]$ is a bijection $\mathbb{R}^n \to T^{\mathrm{o}}_pM$ (surjective by definition, injective because $J_{\varphi\varphi} = I$), and we transport the linear structure of $\mathbb{R}^n$ through it. The transported structure is independent of the chart because the transition $v \mapsto J_{\psi\varphi}v$ defining $\sim_{\mathrm{o}}$ is **linear**. This "gluing by linear maps" is also the reason why the tangent bundle will turn out to be a vector bundle.
The sentence found in textbooks of general relativity — "a contravariant vector is a quantity that transforms under a change of coordinates as $V'^{\mu} = \dfrac{\partial x'^{\mu}}{\partial x^{\nu}} V^{\nu}$" — is nothing but $\sim_{\mathrm{o}}$ of written with the Einstein summation convention. Here $\dfrac{\partial x'^{\mu}}{\partial x^{\nu}}$ is the $(\mu,\nu)$ entry of $J_{\psi\varphi}$. A covariant vector (a $1$-form) is a quantity transforming by the inverse matrix, which is to say an element of the dual space of $T_pM$.
## 5. The third definition: derivations
The third definition regards a tangent vector as "an operator taking directional derivatives of functions". Recall the directional derivative $D_v f(a) = \sum_i v^i \partial_i f(a)$ on $\mathbb{R}^n$: it is a real-valued linear functional $f \mapsto D_vf(a)$ satisfying the product rule. Conversely, we shall prove that any functional satisfying those two conditions is a directional derivative.
An $\mathbb{R}$-linear map $X : C^{\infty}_p(M) \to \mathbb{R}$ satisfying
$$
X(fg) = X(f)\,g(p) + f(p)\,X(g)
$$
for all $f, g \in C^{\infty}_p(M)$ (the **Leibniz rule**) is called a **derivation** at $p$. The set of all derivations is denoted $T^{\mathrm{d}}_pM$.
That $T^{\mathrm{d}}_pM$ becomes a real vector space under $(X+Y)(f) = X(f)+Y(f)$ and $(cX)(f) = cX(f)$ is immediate, since both defining conditions behave linearly. Indeed, $(X+Y)(fg) = X(fg)+Y(fg) = (X(f)+Y(f))g(p) + f(p)(X(g)+Y(g))$.
Let $X \in T^{\mathrm{d}}_pM$ and, for $c \in \mathbb{R}$, write $\underline{c}$ for the germ of the constant function $c$. Then $X(\underline{c}) = 0$.
Applying the Leibniz rule to $\underline{1} \cdot \underline{1} = \underline{1}$ gives
$$
X(\underline{1}) = X(\underline{1})\cdot 1 + 1 \cdot X(\underline{1}) = 2X(\underline{1}),
$$
so $X(\underline{1}) = 0$. For general $c$, write $\underline{c} = c\,\underline{1}$ and use linearity: $X(\underline{c}) = cX(\underline{1}) = 0$.
The next result is the analytic lemma at the core of the theory. Think of it as the version of Taylor's theorem in which the remainder is written as a smooth function.
Let $A \subset \mathbb{R}^n$ be an open set that is star-shaped about $a \in A$ (that is, for every $x \in A$ the segment $\{a + t(x-a) : t \in [0,1]\}$ lies in $A$), and let $h \in C^{\infty}(A)$. Then there exist $g_1, \ldots, g_n \in C^{\infty}(A)$ such that for all $x \in A$
$$
h(x) = h(a) + \sum_{i=1}^{n} (x^i - a^i)\, g_i(x),
\qquad
g_i(a) = \partial_i h(a).
$$
Fix $x \in A$ and put $u(t) = h\big(a + t(x-a)\big)$. Star-shapedness gives $a+t(x-a) \in A$ for $t \in [0,1]$, so $u$ is defined on $[0,1]$; it is $C^{\infty}$ because $h$ is and by the chain rule, and
$$
u'(t) = \sum_{i=1}^{n} (x^i - a^i)\,\partial_i h\big(a+t(x-a)\big).
$$
By the we have $u(1) - u(0) = \int_0^1 u'(t)\,dt$, that is,
$$
h(x) - h(a) = \sum_{i=1}^{n} (x^i - a^i) \int_0^1 \partial_i h\big(a+t(x-a)\big)\,dt.
$$
So setting
$$
g_i(x) = \int_0^1 \partial_i h\big(a + t(x-a)\big)\,dt
$$
gives the desired decomposition. Each $g_i$ is $C^{\infty}$ because the integrand is a $C^{\infty}$ function of $(t,x) \in [0,1]\times A$ and the interval $[0,1]$ is compact, so differentiation under the integral sign is legitimate to every order. Finally, substituting $x = a$ makes the integrand the constant $\partial_i h(a)$, whence $g_i(a) = \partial_i h(a)$.
Star-shapedness is essential. When we use this lemma on a manifold we first replace the image of the chart by an open ball centered at $\varphi(p)$ (shrinking the domain loses no information, since we are dealing with germs).
Let $p \in M$, let $(U,\varphi)$ be a chart containing $p$ with $\varphi = (x^1,\ldots,x^n)$, and put $a = \varphi(p)$. For each $i$ define
$$
\left.\frac{\partial}{\partial x^i}\right|_p (f) := \partial_i\big(f \circ \varphi^{-1}\big)(a)
\qquad (f \in C^{\infty}_p(M)).
$$
Then $\partial/\partial x^i|_p \in T^{\mathrm{d}}_pM$, and $\left(\partial/\partial x^1|_p, \ldots, \partial/\partial x^n|_p\right)$ is a basis of $T^{\mathrm{d}}_pM$. In particular $\dim T^{\mathrm{d}}_pM = n$, and every $X \in T^{\mathrm{d}}_pM$ is uniquely expressed as
$$
X = \sum_{i=1}^{n} X(x^i)\left.\frac{\partial}{\partial x^i}\right|_p .
$$
First we check that the right-hand side does not depend on the representative of the germ. If $f$ and $\tilde f$ agree on a neighborhood $W$ of $p$, then $f\circ\varphi^{-1}$ and $\tilde f\circ\varphi^{-1}$ agree on $\varphi(W\cap U)$, an open neighborhood of $a$, so their partial derivatives at $a$ coincide.
**(1) It is a derivation.** Linearity follows from linearity of partial differentiation. For the Leibniz rule, use $(fg)\circ\varphi^{-1} = (f\circ\varphi^{-1})\cdot(g\circ\varphi^{-1})$ and the product rule on $\mathbb{R}^n$:
$$
\partial_i\big((fg)\circ\varphi^{-1}\big)(a)
= \partial_i(f\circ\varphi^{-1})(a)\,(g\circ\varphi^{-1})(a) + (f\circ\varphi^{-1})(a)\,\partial_i(g\circ\varphi^{-1})(a),
$$
and since $(f\circ\varphi^{-1})(a) = f(p)$ and $(g\circ\varphi^{-1})(a) = g(p)$, this is the asserted identity.
**(2) Linear independence.** Suppose $\sum_i c^i \partial/\partial x^i|_p = 0$ and apply it to the germ of the coordinate function $x^j \in C^{\infty}(U)$. Since $x^j \circ \varphi^{-1}$ is the $j$-th coordinate function on $\varphi(U)$, we have $\partial_i(x^j\circ\varphi^{-1}) = \delta^j_i$ (the Kronecker delta). Hence
$$
0 = \sum_i c^i\,\partial_i(x^j\circ\varphi^{-1})(a) = \sum_i c^i \delta^j_i = c^j
$$
for every $j$, so $c^1 = \cdots = c^n = 0$.
**(3) They span.** Take $X \in T^{\mathrm{d}}_pM$ and set $c^i = X(x^i)$. Let $f \in C^{\infty}_p(M)$ be arbitrary; shrinking the domain of a representative, we may assume that $\varphi(U)$ is an open ball $B$ centered at $a$ (a ball is star-shaped). Applying to $h = f\circ\varphi^{-1} \in C^{\infty}(B)$ produces $g_i \in C^{\infty}(B)$ with
$$
h = h(a) + \sum_i (\text{$i$-th coordinate} - a^i)\,g_i \quad\text{on } B,
\qquad g_i(a) = \partial_i h(a).
$$
Composing both sides with $\varphi$ gives the identity
$$
f = \underline{f(p)} + \sum_{i=1}^{n} (x^i - \underline{a^i})\,(g_i\circ\varphi)
$$
on $U$. Read this as an identity of germs and apply $X$. Linearity, and the Leibniz rule give
$$
\begin{aligned}
X(f)
&= X(\underline{f(p)}) + \sum_i \Big[ X(x^i - \underline{a^i})\cdot (g_i\circ\varphi)(p) + \big(x^i(p) - a^i\big)\cdot X(g_i\circ\varphi) \Big] \\
&= 0 + \sum_i \Big[ \big(X(x^i) - 0\big)\, g_i(a) + 0 \Big]
= \sum_i c^i\, \partial_i h(a)
= \sum_i c^i \left.\frac{\partial}{\partial x^i}\right|_p (f).
\end{aligned}
$$
In the second line the second term was killed using $x^i(p) = a^i$ (the $i$-th component of $\varphi(p) = a$). As $f$ was arbitrary, $X = \sum_i c^i \partial/\partial x^i|_p$.
From (2) and (3) we get a basis, and uniqueness of the coefficients together with the computation in (2) pins them down as $c^i = X(x^i)$.
The proof of rests entirely on , and that lemma uses smoothness of $h$ to obtain smoothness of the $g_i$. If one takes the same definition of derivation on the algebra of germs of $C^k$ functions with $k$ finite, the $g_i$ are only guaranteed to be $C^{k-1}$ and the proof collapses. The collapse is genuine: the space of derivations on germs of $C^k$ functions is infinite-dimensional. See the Appendix for details.
Before going on, let us confirm that defining derivations on the global algebra of functions, without germs, gives the same object. This identification is used constantly in practice.
Let $\widetilde{T}_pM$ be the set of all $\mathbb{R}$-linear maps $X : C^{\infty}(M) \to \mathbb{R}$ satisfying the Leibniz rule $X(fg) = X(f)g(p) + f(p)X(g)$. Then the restriction map
$$
\Phi : T^{\mathrm{d}}_pM \to \widetilde{T}_pM, \qquad \Phi(Y)(f) = Y([f]_p)
$$
is a linear isomorphism.
On $M$ there exist **bump functions**: for any open neighborhood $V$ of $p$ one can find $\chi \in C^{\infty}(M)$ with $\chi \equiv 1$ on some neighborhood of $p$ and $\operatorname{supp}\chi \subset V$ (this is the partition-of-unity argument in [Smooth manifolds](/en/mathematics/manifolds/smooth-manifolds)).
Linearity of $\Phi$ is clear from the definition: $\Phi(Y_1+cY_2)(f) = (Y_1+cY_2)([f]_p) = \Phi(Y_1)(f) + c\Phi(Y_2)(f)$. And $\Phi(Y)$ satisfies the Leibniz rule because $[fg]_p = [f]_p[g]_p$.
**Injectivity.** Suppose $\Phi(Y) = 0$. Given a germ $[f]_p$, choose a representative $f \in C^{\infty}(V)$ and, with $\chi$ as above, set $\hat f = \chi f$ (defined to be $0$ outside $V$). Then $\hat f \in C^{\infty}(M)$ and $\hat f = f$ on a neighborhood of $p$, so $[\hat f]_p = [f]_p$. Hence $Y([f]_p) = Y([\hat f]_p) = \Phi(Y)(\hat f) = 0$, and therefore $Y = 0$.
**Surjectivity.** Let $X \in \widetilde{T}_pM$. We first prove **locality**. Suppose $u, v \in C^{\infty}(M)$ agree on a neighborhood $W$ of $p$ and put $w = u - v$, so that $w|_W = 0$. Choose a bump function $\chi$ with $\operatorname{supp}\chi \subset W$ and $\chi\equiv 1$ near $p$; then $\chi w \equiv 0$ (on $W$ we have $w = 0$, outside $W$ we have $\chi = 0$). Since $X(0) = 0$ by linearity, the Leibniz rule gives
$$
0 = X(\chi w) = X(\chi)\,w(p) + \chi(p)\,X(w) = 0 + X(w)
$$
(using $w(p) = 0$ and $\chi(p) = 1$). Hence $X(u) = X(v)$.
Now define $Y([f]_p) := X(\chi f)$, where $\chi$ is a bump function supported in the domain of the representative. By locality this depends on neither the representative nor $\chi$. Linearity is clear. For the Leibniz rule, note that $\chi(fg)$ and $(\chi f)(\chi g)$ agree on the neighborhood where $\chi \equiv 1$, so
$$
Y([f]_p[g]_p) = X\big(\chi fg\big) = X\big((\chi f)(\chi g)\big) = X(\chi f)g(p) + f(p)X(\chi g) = Y([f]_p)g(p) + f(p)Y([g]_p).
$$
Finally $\Phi(Y)(f) = Y([f]_p) = X(\chi f) = X(f)$ by locality, so $\Phi(Y) = X$.
## 6. Equivalence of the three definitions
All the players are now on stage. We show that the three constructions are naturally isomorphic, "naturally" meaning that the maps giving the isomorphisms do not depend on a choice of chart.
curves carrying a velocity"] -->|"Λ: send a curve to its components (φ∘γ)'(0)"| B["Coordinate components T_p^o M n numbers transforming by the Jacobian"]
B -->|"Θ: send components v to Σ vⁱ ∂/∂xⁱ"| C["Derivations T_p^d M functionals satisfying the Leibniz rule"]
A -->|"Ξ: send a curve to f ↦ (f∘γ)'(0)"| C`} />
Let $p \in M$. Each of the following maps is a bijection, and $\Xi = \Theta \circ \Lambda$.
1. $\Lambda : T^{\mathrm{c}}_pM \to T^{\mathrm{o}}_pM$, $\Lambda([\gamma]) = \big[(\varphi, (\varphi\circ\gamma)'(0))\big]$, where $(U,\varphi)$ is any chart containing $p$.
2. $\Theta : T^{\mathrm{o}}_pM \to T^{\mathrm{d}}_pM$, $\Theta\big([(\varphi,v)]\big) = \sum_{i=1}^n v^i \left.\dfrac{\partial}{\partial x^i}\right|_p$.
3. $\Xi : T^{\mathrm{c}}_pM \to T^{\mathrm{d}}_pM$, $\Xi([\gamma])(f) = (f\circ\gamma)'(0)$.
Moreover $\Theta$ is a linear isomorphism, and if $T^{\mathrm{c}}_pM$ is given the linear structure transported by $\Xi$, then all three are isomorphic as vector spaces. From now on we identify them and write $T_pM$.
**$\Lambda$ is well defined and bijective.** If $\gamma_1 \sim_{\mathrm{c}} \gamma_2$, then by definition their velocities agree in the same chart, so the value of $\Lambda$ is independent of the representative. Independence of the chart follows from the identity $(\psi\circ\gamma)'(0) = J_{\psi\varphi}(\varphi\circ\gamma)'(0)$ established in the proof of : it says exactly that $(\varphi, (\varphi\circ\gamma)'(0)) \sim_{\mathrm{o}} (\psi, (\psi\circ\gamma)'(0))$. Injectivity: reading $\Lambda([\gamma_1]) = \Lambda([\gamma_2])$ through the representative in the chart $\varphi$ gives $(\varphi\circ\gamma_1)'(0) = (\varphi\circ\gamma_2)'(0)$, i.e. $\gamma_1\sim_{\mathrm{c}}\gamma_2$. Surjectivity: given $[(\varphi,v)]$, put $a = \varphi(p)$ and define
$$
\gamma(t) = \varphi^{-1}(a + tv) \qquad (|t| < \varepsilon).
$$
Since $\varphi(U)$ is open, $\varepsilon > 0$ can be taken small enough that $a+tv \in \varphi(U)$; then $\gamma$ is $C^{\infty}$ as a composition of $C^{\infty}$ maps, $\gamma(0) = p$, and $(\varphi\circ\gamma)(t) = a+tv$ gives $(\varphi\circ\gamma)'(0) = v$.
**$\Theta$ is well defined.** Suppose $(\varphi,v)\sim_{\mathrm{o}}(\psi,w)$, i.e. $w = J_{\psi\varphi}v$. Write $\psi = (y^1,\ldots,y^n)$, $\tau = \psi\circ\varphi^{-1}$ and $a = \varphi(p)$, so that the $(j,i)$ entry of $J_{\psi\varphi}$ is $\partial_i\tau^j(a)$. For any germ $f$, applying the chain rule on $\mathbb{R}^n$ to $f\circ\varphi^{-1} = (f\circ\psi^{-1})\circ\tau$ gives
$$
\left.\frac{\partial}{\partial x^i}\right|_p (f)
= \partial_i\big((f\circ\psi^{-1})\circ\tau\big)(a)
= \sum_{j=1}^n \partial_j(f\circ\psi^{-1})\big(\tau(a)\big)\,\partial_i\tau^j(a)
= \sum_{j=1}^n \partial_i\tau^j(a) \left.\frac{\partial}{\partial y^j}\right|_p(f)
$$
(using $\tau(a) = \psi(p)$). Therefore
$$
\sum_i v^i \left.\frac{\partial}{\partial x^i}\right|_p
= \sum_j \Big(\sum_i \partial_i\tau^j(a)\,v^i\Big) \left.\frac{\partial}{\partial y^j}\right|_p
= \sum_j w^j \left.\frac{\partial}{\partial y^j}\right|_p ,
$$
so the value of $\Theta$ is independent of the representative.
**$\Theta$ is a linear isomorphism.** Fixing a chart $\varphi$, the correspondence $T^{\mathrm{o}}_pM \ni [(\varphi,v)] \leftrightarrow v \in \mathbb{R}^n$ is a linear isomorphism (this is exactly how the linear structure was defined in §4), and under it $\Theta$ becomes $v \mapsto \sum_i v^i \partial/\partial x^i|_p$. By the family $\{\partial/\partial x^i|_p\}$ is a basis of $T^{\mathrm{d}}_pM$, so this map is a linear isomorphism.
**$\Xi = \Theta\circ\Lambda$.** Let $\gamma$ be a curve through $p$ and $v = (\varphi\circ\gamma)'(0)$. For any germ $f$ we have $f\circ\gamma = (f\circ\varphi^{-1})\circ(\varphi\circ\gamma)$, so the chain rule gives
$$
(f\circ\gamma)'(0) = \sum_i \partial_i(f\circ\varphi^{-1})\big(\varphi(p)\big)\cdot \big((\varphi\circ\gamma)^i\big)'(0)
= \sum_i v^i \left.\frac{\partial}{\partial x^i}\right|_p(f).
$$
The right-hand side is $\Theta(\Lambda([\gamma]))(f)$. Incidentally this also shows that $\Xi([\gamma])$ is a derivation, being in the image of $\Theta\circ\Lambda$. That $\Xi$ is a bijection follows, as a composition, from the bijectivity of $\Lambda$ and $\Theta$.
Equip $M = \mathbb{R}^n$ with the atlas whose only chart is the identity map $\mathrm{id}$. The coordinate functions are the usual $x^i$, and $\partial/\partial x^i|_a$ is the ordinary partial derivative $f \mapsto \partial_i f(a)$. By , $T_a\mathbb{R}^n$ is $n$-dimensional, and the map sending $v \in \mathbb{R}^n$ to $\sum_i v^i\partial/\partial x^i|_a$ is a linear isomorphism $\mathbb{R}^n \xrightarrow{\ \sim\ } T_a\mathbb{R}^n$. This isomorphism is canonical — no chart has to be chosen — so from now on we identify $T_a\mathbb{R}^n$ with $\mathbb{R}^n$. On the side of curves it is the familiar correspondence $[\gamma] \mapsto \gamma'(0)$.
Take $M = \mathbb{R}^2$ and $U = \{(x,y) : x > 0\}$, and put the polar chart $\psi = (r,\theta)$ on $U$ by prescribing its inverse
$$
\psi^{-1}(r,\theta) = (r\cos\theta,\ r\sin\theta), \qquad r > 0,\ -\tfrac{\pi}{2} < \theta < \tfrac{\pi}{2}.
$$
We use the transformation formula from the proof of , with polar coordinates as the source and Cartesian coordinates as the target. The Jacobian matrix of $\tau = (\text{Cartesian})\circ(\text{polar})^{-1}$ is
$$
D\tau(r,\theta) = \begin{pmatrix} \cos\theta & -r\sin\theta \\ \sin\theta & r\cos\theta \end{pmatrix},
$$
so that
$$
\left.\frac{\partial}{\partial r}\right|_p = \cos\theta \left.\frac{\partial}{\partial x}\right|_p + \sin\theta \left.\frac{\partial}{\partial y}\right|_p,
\qquad
\left.\frac{\partial}{\partial \theta}\right|_p = -r\sin\theta \left.\frac{\partial}{\partial x}\right|_p + r\cos\theta \left.\frac{\partial}{\partial y}\right|_p.
$$
Let us compute concretely at $p = (1,1)$, that is, $r = \sqrt{2}$ and $\theta = \pi/4$. Since $\cos\theta = \sin\theta = 1/\sqrt{2}$,
$$
\left.\frac{\partial}{\partial r}\right|_p = \frac{1}{\sqrt2}\left(\left.\frac{\partial}{\partial x}\right|_p + \left.\frac{\partial}{\partial y}\right|_p\right),
\qquad
\left.\frac{\partial}{\partial\theta}\right|_p = -\left.\frac{\partial}{\partial x}\right|_p + \left.\frac{\partial}{\partial y}\right|_p .
$$
As a check, take $f(x,y) = x^2+y^2$. In polar coordinates $f\circ\psi^{-1}(r,\theta) = r^2$, so $\partial/\partial r|_p (f) = 2r = 2\sqrt2$ and $\partial/\partial\theta|_p(f) = 0$. Computing instead with the right-hand sides,
$$
\frac{1}{\sqrt2}\big(2x + 2y\big)\Big|_{(1,1)} = \frac{4}{\sqrt2} = 2\sqrt2,
\qquad
\big(-2x + 2y\big)\Big|_{(1,1)} = 0,
$$
in agreement. Note that $\partial/\partial\theta|_p$ is not a "unit vector". We have not yet introduced any notion of length, but its Cartesian components are $(-1,1)$, of Euclidean length $\sqrt2 = r$. That a coordinate basis need not be orthonormal is the first thing to get used to when working with curvilinear coordinates.
## 7. The differential of a map
The greatest dividend of introducing tangent spaces is that a smooth map can be linearized at each point. With the definition by derivations, the differential can be written down with astonishing brevity.
Let $F : M \to N$ be a $C^{\infty}$ map, $p \in M$ and $q = F(p)$. For $X \in T_pM$ define $dF_p(X)$ by
$$
\big(dF_p(X)\big)(g) := X(g\circ F) \qquad \big(g \in C^{\infty}_q(N)\big).
$$
The map $dF_p : T_pM \to T_qN$ is called the **differential** of $F$ at $p$ (also the pushforward, written $F_{*p}$).
Let us first check that the definition makes sense. If $g$ is a $C^{\infty}$ function on a neighborhood $V$ of $q$, then $F^{-1}(V)$ is an open neighborhood of $p$ by continuity of $F$, and $g\circ F \in C^{\infty}(F^{-1}(V))$, so $[g\circ F]_p \in C^{\infty}_p(M)$ is well defined. Replacing the representative of $g$ on a neighborhood of $q$ does not change $g\circ F$ on a neighborhood of $p$, so we obtain a map of germs $F^{*} : C^{\infty}_q(N) \to C^{\infty}_p(M)$, $[g] \mapsto [g\circ F]$, and it is an algebra homomorphism.
In , $dF_p(X)$ is a derivation at $q$, and $dF_p : T_pM \to T_qN$ is linear.
Linearity of $dF_p(X)$ holds because $F^{*}$ is linear and $X$ is linear. For the Leibniz rule, use that $F^{*}$ preserves products, $(gh)\circ F = (g\circ F)(h\circ F)$, together with the Leibniz rule for $X$:
$$
dF_p(X)(gh) = X\big((g\circ F)(h\circ F)\big) = X(g\circ F)\,h(F(p)) + g(F(p))\,X(h\circ F)
= dF_p(X)(g)\,h(q) + g(q)\,dF_p(X)(h)
$$
(where we used $(g\circ F)(p) = g(q)$). Linearity of $X \mapsto dF_p(X)$ is immediate from $dF_p(X+cY)(g) = (X+cY)(g\circ F) = X(g\circ F) + cY(g\circ F)$.
In the situation above, if $X = \Xi([\gamma]) \in T_pM$ then $dF_p(X) = \Xi([F\circ\gamma])$. In the language of curves, then, the differential is nothing but the operation of pushing a curve forward by $F$.
Being a composition of $C^{\infty}$ maps, $F\circ\gamma$ is a $C^{\infty}$ curve through $q$. For any $g \in C^{\infty}_q(N)$,
$$
\Xi([F\circ\gamma])(g) = \big(g\circ (F\circ\gamma)\big)'(0) = \big((g\circ F)\circ\gamma\big)'(0) = \Xi([\gamma])(g\circ F) = X(g\circ F) = dF_p(X)(g).
$$
Nothing beyond associativity of composition has been used.
Let $F : M \to N$ and $G : N \to P$ be $C^{\infty}$ maps and $p \in M$. Then
$$
d(G\circ F)_p = dG_{F(p)} \circ dF_p : T_pM \to T_{G(F(p))}P .
$$
Moreover $d(\mathrm{id}_M)_p = \mathrm{id}_{T_pM}$.
Let $X \in T_pM$ and $h \in C^{\infty}_{G(F(p))}(P)$. Using three times,
$$
d(G\circ F)_p(X)(h) = X\big(h\circ(G\circ F)\big) = X\big((h\circ G)\circ F\big) = dF_p(X)(h\circ G) = dG_{F(p)}\big(dF_p(X)\big)(h).
$$
As $h$ is arbitrary, the identity follows. For the identity map, $d(\mathrm{id})_p(X)(f) = X(f\circ\mathrm{id}) = X(f)$.
If $F : M \to N$ is a diffeomorphism, then $dF_p : T_pM \to T_{F(p)}N$ is a linear isomorphism at every $p$, with $(dF_p)^{-1} = d(F^{-1})_{F(p)}$. In particular, corresponding points of diffeomorphic manifolds have tangent spaces of the same dimension.
Applying to $F^{-1}\circ F = \mathrm{id}_M$ and $F\circ F^{-1} = \mathrm{id}_N$ gives
$$
d(F^{-1})_{F(p)}\circ dF_p = \mathrm{id}_{T_pM}, \qquad dF_p \circ d(F^{-1})_{F(p)} = \mathrm{id}_{T_{F(p)}N},
$$
so $dF_p$ is bijective with inverse $d(F^{-1})_{F(p)}$. A linear isomorphism forces equality of dimensions (see [Vector spaces and linear maps](/en/mathematics/linear-algebra/vector-spaces), ).
Let $F : M \to N$ be $C^{\infty}$, let $(U,\varphi)$ be a chart of $M$ containing $p$ (coordinates $x^i$, $\dim M = n$) and $(V,\psi)$ a chart of $N$ containing $q = F(p)$ (coordinates $y^j$, $\dim N = m$), and put $\widehat{F} = \psi\circ F\circ\varphi^{-1}$ and $a = \varphi(p)$. Then
$$
dF_p\!\left(\left.\frac{\partial}{\partial x^i}\right|_p\right) = \sum_{j=1}^{m} \frac{\partial \widehat{F}^j}{\partial x^i}(a) \left.\frac{\partial}{\partial y^j}\right|_q .
$$
That is, the matrix of $dF_p$ with respect to the bases $\{\partial/\partial x^i|_p\}$ and $\{\partial/\partial y^j|_q\}$ is the Jacobian matrix $D\widehat{F}(a)$ of the local representative $\widehat{F}$.
Take $g \in C^{\infty}_q(N)$. By definition,
$$
dF_p\!\left(\left.\frac{\partial}{\partial x^i}\right|_p\right)(g)
= \left.\frac{\partial}{\partial x^i}\right|_p (g\circ F)
= \partial_i\big((g\circ F)\circ\varphi^{-1}\big)(a).
$$
Rewrite $(g\circ F)\circ\varphi^{-1} = (g\circ\psi^{-1})\circ(\psi\circ F\circ\varphi^{-1}) = (g\circ\psi^{-1})\circ\widehat{F}$ and apply the chain rule to the composition $\mathbb{R}^n \to \mathbb{R}^m \to \mathbb{R}$:
$$
\partial_i\big((g\circ\psi^{-1})\circ\widehat F\big)(a)
= \sum_{j=1}^m \partial_j (g\circ\psi^{-1})\big(\widehat F(a)\big)\cdot \partial_i \widehat F^j(a)
= \sum_{j=1}^m \partial_i\widehat F^j(a)\left.\frac{\partial}{\partial y^j}\right|_q(g)
$$
(using $\widehat F(a) = \psi(q)$). Since $g$ is arbitrary, the claim follows.
This proposition guarantees that the abstract is the same thing as the classical "linear approximation by the Jacobian matrix". Writing out through returns the chain rule as a product of matrices, $D(\widehat{G}\circ\widehat{F})(a) = D\widehat{G}(\widehat F(a))\,D\widehat F(a)$.
Let $S^n = \{u \in \mathbb{R}^{n+1} : \|u\| = 1\}$ be an embedded submanifold of $\mathbb{R}^{n+1}$ and $\iota : S^n \hookrightarrow \mathbb{R}^{n+1}$ the inclusion. Under the identification $T_p\mathbb{R}^{n+1}\cong\mathbb{R}^{n+1}$ of ,
$$
d\iota_p(T_pS^n) = p^{\perp} = \{v \in \mathbb{R}^{n+1} : \langle p, v\rangle = 0\}.
$$
**(One inclusion.)** Given $[\gamma] \in T_pS^n$, the curve $\iota\circ\gamma$ lies in $\mathbb{R}^{n+1}$ and satisfies $\|\gamma(t)\|^2 = 1$. Differentiating in $t$ (using bilinearity of the inner product and the product rule),
$$
0 = \frac{d}{dt}\langle \gamma(t),\gamma(t)\rangle\Big|_{t=0} = 2\langle \gamma(0), (\iota\circ\gamma)'(0)\rangle = 2\langle p, (\iota\circ\gamma)'(0)\rangle,
$$
so by we get $d\iota_p([\gamma]) = (\iota\circ\gamma)'(0) \in p^{\perp}$.
**(The other inclusion.)** Let $v \in p^{\perp}$ with $v \ne 0$, put $e = v/\|v\|$ and $\omega = \|v\|$, and define
$$
\gamma(t) = (\cos\omega t)\,p + (\sin\omega t)\,e .
$$
Since $p$ and $e$ are orthonormal, $\|\gamma(t)\|^2 = \cos^2\omega t + \sin^2\omega t = 1$, so $\gamma$ is a curve in $S^n$; as $S^n$ is an embedded submanifold, it is $C^{\infty}$ also as a map into $S^n$. From $\gamma(0) = p$ and $\gamma'(0) = \omega e = v$ we conclude that $v$ lies in the image. The vector $v = 0$ is obtained as $0 = d\iota_p(0)$.
**(Conclusion.)** The map $d\iota_p$ is linear (), and its image both contains and is contained in $p^{\perp}$, hence equals $p^{\perp}$. Since $\dim T_pS^n = n = \dim p^{\perp}$ (), a surjective linear map between spaces of equal dimension is also injective, so $d\iota_p$ is injective and $T_pS^n \cong p^{\perp}$. The familiar picture of "a plane touching the sphere" is this isomorphism drawn inside $\mathbb{R}^{n+1}$ after translating it so that its base point is $p$.
The set $O(n) = \{A \in \mathrm{Mat}_n(\mathbb{R}) : A^{\mathsf T}A = I\}$ is a manifold of dimension $n(n-1)/2$ (see [Lie groups and Lie algebras](/mathematics/manifolds/lie-groups), ). Inside $\mathrm{Mat}_n(\mathbb{R})\cong\mathbb{R}^{n^2}$ we have
$$
T_I O(n) = \{A \in \mathrm{Mat}_n(\mathbb{R}) : A + A^{\mathsf T} = 0\},
$$
the space of skew-symmetric matrices. Indeed, a curve $\gamma$ in $O(n)$ with $\gamma(0) = I$ satisfies $\gamma(t)^{\mathsf T}\gamma(t) = I$, and differentiating at $t = 0$ gives
$$
\gamma'(0)^{\mathsf T}\gamma(0) + \gamma(0)^{\mathsf T}\gamma'(0) = \gamma'(0)^{\mathsf T} + \gamma'(0) = 0.
$$
Conversely, if $A^{\mathsf T} = -A$ then $\gamma(t) = \exp(tA)$ takes values in $O(n)$, because $\gamma(t)^{\mathsf T} = \exp(tA^{\mathsf T}) = \exp(-tA) = \gamma(t)^{-1}$, and $\gamma(0) = I$, $\gamma'(0) = A$. The space of skew-symmetric matrices has dimension $n(n-1)/2$, matching $\dim O(n)$, so the same dimension count as in settles the equality.
## 8. The tangent bundle
Now that we have a tangent space at each point, we bundle them together over all points. A vector field on a manifold is "a rule assigning to each point a tangent vector at that point", but to discuss its smoothness the set of all tangent vectors must itself be a manifold.
Let $M$ be an $n$-dimensional $C^{\infty}$ manifold. As a set, put
$$
TM = \bigsqcup_{p \in M} T_pM = \{(p, X) : p \in M,\ X \in T_pM\}
$$
and call $\pi : TM \to M$, $\pi(p,X) = p$, the **projection**. We call $TM$ the **tangent bundle** of $M$, and $\pi^{-1}(p) = T_pM$ the **fiber** over $p$.
Let $M$ be an $n$-dimensional $C^{\infty}$ manifold. For a $C^{\infty}$ atlas $\{(U_{\alpha},\varphi_{\alpha})\}$ of $M$, writing $\varphi_{\alpha} = (x^1_{\alpha},\ldots,x^n_{\alpha})$, define
$$
\widetilde{\varphi}_{\alpha} : \pi^{-1}(U_{\alpha}) \to \varphi_{\alpha}(U_{\alpha})\times\mathbb{R}^n \subset \mathbb{R}^{2n},
\qquad
\widetilde{\varphi}_{\alpha}\Big(p, \sum_i v^i \left.\frac{\partial}{\partial x^i_{\alpha}}\right|_p\Big) = \big(\varphi_{\alpha}(p), v\big).
$$
Then there is exactly one topology and $C^{\infty}$ structure on $TM$ having $\{(\pi^{-1}(U_{\alpha}), \widetilde\varphi_{\alpha})\}$ as an atlas, and $TM$ becomes a $C^{\infty}$ manifold of dimension $2n$. Moreover $\pi : TM \to M$ is a surjective $C^{\infty}$ submersion.
**(1) Each $\widetilde\varphi_{\alpha}$ is a bijection.** By , every tangent vector at $p \in U_{\alpha}$ has a unique component expression in the basis $\{\partial/\partial x^i_{\alpha}|_p\}$, so $\pi^{-1}(U_{\alpha}) \to \varphi_{\alpha}(U_{\alpha})\times\mathbb{R}^n$ is bijective.
**(2) Transition maps.** Let $U_{\alpha\beta} = U_{\alpha}\cap U_{\beta} \ne \emptyset$ and put $\tau = \varphi_{\beta}\circ\varphi_{\alpha}^{-1}$. The set $\widetilde\varphi_{\alpha}(\pi^{-1}(U_{\alpha\beta})) = \varphi_{\alpha}(U_{\alpha\beta})\times\mathbb{R}^n$ is open in $\mathbb{R}^{2n}$. By the change-of-basis formula established in the proof of ,
$$
\widetilde\varphi_{\beta}\circ\widetilde\varphi_{\alpha}^{-1}(x, v) = \big(\tau(x),\ D\tau(x)\,v\big).
$$
Since $\tau$ is $C^{\infty}$, every entry of $x \mapsto D\tau(x)$ is $C^{\infty}$, and $D\tau(x)v$ is $C^{\infty}$ in $x$ and linear in $v$, hence $C^{\infty}$ altogether. The inverse map has the same form with $\alpha$ and $\beta$ interchanged, so the transition maps are diffeomorphisms.
**(3) Topology.** We check that $\mathcal{B} = \{\widetilde\varphi_{\alpha}^{-1}(W) : \alpha,\ W \subset \varphi_{\alpha}(U_{\alpha})\times\mathbb{R}^n \text{ open}\}$ is a basis for a topology on $TM$. It covers $TM$. Take a point $\xi$ in the intersection of two members $\widetilde\varphi_{\alpha}^{-1}(W_1)$ and $\widetilde\varphi_{\beta}^{-1}(W_2)$; then $\xi \in \pi^{-1}(U_{\alpha\beta})$, and by (2) the set $\widetilde\varphi_{\alpha}\big(\widetilde\varphi_{\alpha}^{-1}(W_1)\cap\widetilde\varphi_{\beta}^{-1}(W_2)\big) = W_1 \cap (\widetilde\varphi_{\beta}\circ\widetilde\varphi_{\alpha}^{-1})^{-1}(W_2)$ is open; calling it $W_3$ we get $\xi \in \widetilde\varphi_{\alpha}^{-1}(W_3) \subset$ the intersection. So $\mathcal{B}$ is a basis, and in the topology it generates each $\widetilde\varphi_{\alpha}$ is a homeomorphism.
**(4) Hausdorff.** Let $\xi \ne \eta$ in $TM$. If $\pi(\xi) = \pi(\eta) = p$, choose $\alpha$ with $p \in U_{\alpha}$; then $\widetilde\varphi_{\alpha}(\xi)\ne\widetilde\varphi_{\alpha}(\eta)$, and these can be separated inside $\varphi_\alpha(U_\alpha)\times\mathbb{R}^n$ because $\mathbb{R}^{2n}$ is Hausdorff; the preimages are the required open sets. If $\pi(\xi)\ne\pi(\eta)$, then since $M$ is Hausdorff there are disjoint open sets $U \ni \pi(\xi)$ and $V \ni \pi(\eta)$, and $\pi^{-1}(U)$ and $\pi^{-1}(V)$ are disjoint open sets ($\pi$ is continuous in the topology of (3) because its local representative is the continuous map $(x,v)\mapsto x$).
**(5) Second countability.** As $M$ is second countable, there is a countable subatlas $\{(U_{\alpha_k},\varphi_{\alpha_k})\}_{k\in\mathbb{N}}$. Each $\pi^{-1}(U_{\alpha_k})$ is homeomorphic to an open subset of $\mathbb{R}^{2n}$ and hence second countable, and a space covered by countably many second countable open subspaces is second countable.
**(6) The $C^{\infty}$ structure and $\pi$.** By (2), $\{(\pi^{-1}(U_{\alpha}),\widetilde\varphi_{\alpha})\}$ is a $C^{\infty}$ atlas, and the maximal atlas containing it determines the $C^{\infty}$ structure. Both the topology and the atlas are forced by the requirement that each $\widetilde\varphi_\alpha$ be a diffeomorphism, so they are unique. The local representative of $\pi$ is the projection $\varphi_{\alpha}\circ\pi\circ\widetilde\varphi_{\alpha}^{-1}(x,v) = x$, which is $C^{\infty}$ with surjective differential (its Jacobian matrix is $\begin{pmatrix} I_n & 0\end{pmatrix}$), so $\pi$ is a submersion. Surjectivity follows because every $T_pM$ contains the zero vector.
A $C^{\infty}$ map $\mathcal{X} : M \to TM$ with $\pi\circ\mathcal{X} = \mathrm{id}_M$ is called a **section** of $TM$, and this is the official definition of a vector field. On a chart we may write $\mathcal{X} = \sum_i \mathcal{X}^i \partial/\partial x^i$, and, read in the charts of , smoothness of $\mathcal{X}$ is equivalent to smoothness of the components $\mathcal{X}^i$. Vector fields, and the differential forms dual to them, are treated in [Vector fields and differential forms](/mathematics/manifolds/differential-forms) (the official definition as sections is , the dual side is ).
Let $S^1 \subset \mathbb{R}^2$ and identify $T_pS^1 \cong p^{\perp}$ using , where $p = (p_1,p_2)$. The line $p^{\perp}$ is one-dimensional, with basis $Jp := (-p_2, p_1)$ (indeed $\langle p, Jp\rangle = -p_1p_2 + p_2p_1 = 0$ and $\|Jp\| = 1 \ne 0$). So define
$$
\Psi : S^1\times\mathbb{R} \to TS^1, \qquad \Psi(p,t) = \big(p,\ t\,Jp\big).
$$
Then $\Psi$ is a bijection, since $t \mapsto tJp$ is an isomorphism on each fiber. For smoothness, take the chart $\theta \mapsto (\cos\theta,\sin\theta)$ of $S^1$: the local representative of $\Psi$ becomes $(\theta,t)\mapsto(\theta,t)$. Indeed, in these coordinates $\partial/\partial\theta|_p = (-\sin\theta,\cos\theta) = Jp$, so the component of $tJp$ is exactly $t$. The inverse is $C^{\infty}$ for the same reason, so $\Psi$ is a diffeomorphism and $TS^1 \cong S^1\times\mathbb{R}$. Equivalently, $S^1$ carries a nowhere vanishing vector field, namely $p \mapsto Jp$.
What happens in is not typical of all manifolds. A manifold with $TM \cong M\times\mathbb{R}^n$ is called **parallelizable**, and $S^2$ is not parallelizable. This is a consequence of the hairy ball theorem (Poincaré–Brouwer): every continuous vector field on $S^2$ vanishes somewhere. Among spheres, only $S^1$, $S^3$ and $S^7$ are known to be parallelizable (Bott–Milnor and Kervaire, 1958). That the question of triviality of the tangent bundle contains deep topology is one of the reasons for introducing $TM$ at all.
## 9. Exercises
On $M = \mathbb{R}^2$, besides the usual coordinates $(x,y)$, introduce the global chart $\psi = (u,v)$ given by $u = x+y$ and $v = x-y$. At a point $p$, express $\partial/\partial u|_p$ and $\partial/\partial v|_p$ in terms of $\partial/\partial x|_p$ and $\partial/\partial y|_p$. Then check the expression on the function $f(x,y) = xy$.
Since $\psi^{-1}(u,v) = \left(\frac{u+v}{2}, \frac{u-v}{2}\right)$, the Jacobian matrix of $\tau = \varphi\circ\psi^{-1}$ (with $\varphi$ the Cartesian chart) is
$$
D\tau = \begin{pmatrix} 1/2 & 1/2 \\ 1/2 & -1/2 \end{pmatrix}.
$$
By the transformation formula in the proof of (namely $\partial/\partial u^i = \sum_j \partial_i\tau^j \cdot \partial/\partial x^j$),
$$
\left.\frac{\partial}{\partial u}\right|_p = \frac12\left(\left.\frac{\partial}{\partial x}\right|_p + \left.\frac{\partial}{\partial y}\right|_p\right),
\qquad
\left.\frac{\partial}{\partial v}\right|_p = \frac12\left(\left.\frac{\partial}{\partial x}\right|_p - \left.\frac{\partial}{\partial y}\right|_p\right).
$$
Now the check. In the new coordinates $f(x,y) = xy$ becomes $f\circ\psi^{-1}(u,v) = \frac{u+v}{2}\cdot\frac{u-v}{2} = \frac{u^2-v^2}{4}$, so $\partial/\partial u|_p(f) = u/2$ and $\partial/\partial v|_p(f) = -v/2$. Computing with the right-hand sides gives $\frac12(y + x) = \frac{x+y}{2} = \frac{u}{2}$ and $\frac12(y - x) = -\frac{x-y}{2} = -\frac v2$, in agreement.
Let $X \in T_pM$ be a derivation and let $f, g \in C^{\infty}_p(M)$ satisfy $f(p) = g(p) = 0$. Show that $X(fg) = 0$. Then, setting $\mathfrak{m}_p = \{f \in C^{\infty}_p(M) : f(p) = 0\}$ and $\mathfrak{m}_p^2 = \{\sum_{k=1}^N f_kg_k : f_k,g_k\in\mathfrak{m}_p\}$, show that $X \mapsto X|_{\mathfrak{m}_p}$ gives a linear isomorphism $T_pM \to (\mathfrak{m}_p/\mathfrak{m}_p^2)^{*}$.
The first claim is an immediate consequence of the Leibniz rule: $X(fg) = X(f)g(p) + f(p)X(g) = X(f)\cdot 0 + 0\cdot X(g) = 0$. By linearity, $X$ vanishes on every element of $\mathfrak{m}_p^2$, these being finite sums.
Now the isomorphism. Restricting $X \in T_pM$ to $\mathfrak{m}_p$ gives, by the above, a map vanishing on $\mathfrak{m}_p^2$, hence an induced linear functional $\bar X$ on the quotient $\mathfrak{m}_p/\mathfrak{m}_p^2$. The correspondence $X \mapsto \bar X$ is clearly linear, restriction and passage to a quotient both being linear operations.
Injectivity. Suppose $\bar X = 0$, i.e. $X|_{\mathfrak{m}_p} = 0$. Any $f \in C^{\infty}_p(M)$ decomposes as $f = \underline{f(p)} + (f - \underline{f(p)})$, with the second term in $\mathfrak{m}_p$. By we have $X(\underline{f(p)}) = 0$, so $X(f) = X(f - \underline{f(p)}) = 0$. Hence $X = 0$.
Surjectivity. Given $\lambda \in (\mathfrak{m}_p/\mathfrak{m}_p^2)^{*}$, define $X(f) := \lambda\big([f - \underline{f(p)}]\big)$. Linearity is clear. For the Leibniz rule, decompose
$fg - \underline{f(p)g(p)} = f(p)\,(g - \underline{g(p)}) + g(p)\,(f - \underline{f(p)}) + (f-\underline{f(p)})(g-\underline{g(p)})$
and note that the last term lies in $\mathfrak{m}_p^2$, hence is killed by $\lambda$. That this $X$ induces $\lambda$ is clear from the definition.
Combined with , this shows $\dim \mathfrak{m}_p/\mathfrak{m}_p^2 = n$, and $\mathfrak{m}_p/\mathfrak{m}_p^2$ is identified with the cotangent space $T_p^{*}M$. From this angle, says that every element of $\mathfrak{m}_p$ is a linear combination of the coordinate functions plus an element of $\mathfrak{m}_p^2$.
The set $\mathrm{GL}(n,\mathbb{R}) = \{A \in \mathrm{Mat}_n(\mathbb{R}) : \det A \ne 0\}$ is open in $\mathrm{Mat}_n(\mathbb{R})\cong\mathbb{R}^{n^2}$ and hence a manifold of dimension $n^2$. Under the identification $T_I\mathrm{GL}(n,\mathbb{R})\cong\mathrm{Mat}_n(\mathbb{R})$, show that the differential of $\det : \mathrm{GL}(n,\mathbb{R})\to\mathbb{R}$ at the identity is $d(\det)_I(A) = \operatorname{tr}A$.
Since $\mathrm{GL}(n,\mathbb{R})$ is open in $\mathbb{R}^{n^2}$, taking the inclusion as a chart gives $T_I\mathrm{GL}(n,\mathbb{R}) \cong \mathrm{Mat}_n(\mathbb{R})$ (the same argument as in ). For $A \in \mathrm{Mat}_n(\mathbb{R})$ take the curve $\gamma(t) = I + tA$. As $\det$ is continuous, $\det\gamma(t)\ne 0$ for $|t|$ small enough, so $\gamma$ is a curve in $\mathrm{GL}(n,\mathbb{R})$ with $\gamma(0) = I$ and $\gamma'(0) = A$. By , $d(\det)_I(A) = (\det\circ\gamma)'(0)$.
Expand $\det(I+tA)$. By the definition of the determinant (see [Determinants](/en/mathematics/linear-algebra/determinants), ),
$$
\det(I+tA) = \sum_{\sigma\in S_n}\operatorname{sgn}(\sigma)\prod_{i=1}^{n}\big(\delta_{i\sigma(i)} + tA_{i\sigma(i)}\big).
$$
Collect the terms of degree at most $1$ in $t$. If $\sigma$ is not the identity permutation, then $\sigma(i)\ne i$ for at least two values of $i$ (a permutation moves at least two points), so the corresponding product is divisible by $t^2$. The identity permutation contributes $\prod_i (1 + tA_{ii}) = 1 + t\sum_i A_{ii} + O(t^2)$. Hence
$$
\det(I+tA) = 1 + t\operatorname{tr}A + O(t^2),
$$
and differentiating in $t$ at $t=0$ gives $d(\det)_I(A) = \operatorname{tr}A$.
Let $M$ be a connected $C^{\infty}$ manifold, $N$ a $C^{\infty}$ manifold and $F : M \to N$ a $C^{\infty}$ map. Show that if $dF_p = 0$ for every $p \in M$, then $F$ is constant.
Fix $p_0\in M$ arbitrarily, put $q_0 = F(p_0)$, and show that $S = F^{-1}(q_0)$ is nonempty, open and closed. Connectedness of $M$ (: a nonempty subset that is both open and closed must be the whole space) then gives $S = M$, i.e. $F$ is constant.
$S \ne \emptyset$ because $p_0 \in S$. And $S$ is closed because $\{q_0\}$ is closed ($N$ being Hausdorff) and $F$ is continuous.
Now we show $S$ is open. Take $p \in S$ and choose a chart $(U,\varphi)$ containing $p$ and a chart $(V,\psi)$ containing $F(p) = q_0$ such that $F(U)\subset V$ and $\varphi(U)$ is an open ball, hence connected (by continuity of $F$ it suffices to shrink $U$ to $U\cap F^{-1}(V)$). Put $\widehat F = \psi\circ F\circ\varphi^{-1} : \varphi(U)\to\psi(V)$. By , the condition $dF_x = 0$ for $x\in U$ is equivalent to the vanishing of the Jacobian matrix $D\widehat F(\varphi(x))$. By hypothesis this holds on all of $\varphi(U)$, so all partial derivatives of every component of $\widehat F$ vanish. Since $\varphi(U)$ is a connected open set, $\widehat F$ is constant (componentwise, either join points by segments and apply the mean value theorem, or quote it as a corollary in [Differentiation in several variables](/mathematics/calculus/multivariable-differentiation)). As $\widehat F(\varphi(p)) = \psi(q_0)$, we get $\widehat F \equiv \psi(q_0)$, that is $F|_U \equiv q_0$, so $U \subset S$. Hence $S$ is open.
Dropping the hypothesis that $M$ is connected destroys the conclusion. For instance take $M = \mathbb{R}\sqcup\mathbb{R}$ (the disjoint union of two lines) and $N=\mathbb{R}$, and let $F$ be $0$ on one copy and $1$ on the other: then $dF_p = 0$ at every point, yet $F$ is not constant.
## References
- Matsumoto Yukio, *Tayōtai no Kiso*, University of Tokyo Press, 1988 (in Japanese) — an elementary introduction to tangent vectors and tangent spaces. A representative example of the exposition that starts from the definition by curves.
- J. M. Lee, *Introduction to Smooth Manifolds*, 2nd ed., Springer GTM 218, 2013 — Chapter 3 (Tangent Vectors). The definition by derivations, the differential, and the construction of the tangent bundle are developed in almost the same order as in this article.
- F. W. Warner, *Foundations of Differentiable Manifolds and Lie Groups*, Springer GTM 94, 1983 — Chapter 1. A concise account of the definition by derivations and of why smoothness is needed.
- M. Spivak, *A Comprehensive Introduction to Differential Geometry, Vol. 1*, 3rd ed., Publish or Perish, 1999 — Chapter 3. A detailed side-by-side comparison of the three definitions, with the role of Hadamard's lemma made explicit.
- R. Bott and J. Milnor, "On the parallelizability of the spheres", *Bulletin of the American Mathematical Society* 64 (1958), 87–89 — the result on parallelizability of spheres mentioned in .
## Appendix: Why the space of derivations grows in the $C^k$ setting
**Where the problem lies.** What was essential in the proof of was , that is, the fact that an element of $\mathfrak{m}_p$ falls into $\mathfrak{m}_p^2$ once linear combinations of the coordinate functions are removed. As we saw in , the space of derivations at $p$ is identified with $(\mathfrak{m}_p/\mathfrak{m}_p^2)^{*}$. So the statement "the space of derivations is $n$-dimensional" and the statement "$\mathfrak{m}_p/\mathfrak{m}_p^2$ is $n$-dimensional" are one and the same.
**What goes wrong for $C^1$.** Take $n=1$ and $p=0$, and consider the algebra $C^1_0(\mathbb{R})$ of germs of $C^1$ functions. Let $\mathfrak{m}$ be the germs vanishing at $0$ and $\mathfrak{m}^2$ the finite sums of products of such. For $0 < \alpha < 1$ consider $f_{\alpha}(x) = |x|^{1+\alpha}$. It is $C^1$ (its derivative $f_{\alpha}'(x) = (1+\alpha)\operatorname{sgn}(x)|x|^{\alpha}$ is continuous at $0$ with value $0$) and lies in $\mathfrak{m}$. On the other hand, if $g,h\in\mathfrak{m}$ are $C^1$, then the mean value theorem gives $|g(x)| \le C|x|$ and $|h(x)|\le C|x|$ near $0$, so any finite sum $\sum_k g_kh_k$ is $O(x^2)$. But $|x|^{1+\alpha}/x^2 \to \infty$ as $x\to 0$ when $\alpha<1$, so $f_{\alpha}\notin\mathfrak{m}^2$. Moreover, the same comparison of growth rates shows that the $f_{\alpha}$ for distinct $\alpha$ are linearly independent in $\mathfrak{m}/\mathfrak{m}^2$. Hence $\mathfrak{m}/\mathfrak{m}^2$ is infinite-dimensional, and so is its dual, the space of derivations.
**Conclusion.** In the $C^{\infty}$ world a small miracle occurs: $\mathfrak{m}_p/\mathfrak{m}_p^2$ collapses to exactly $n$ dimensions, and what guarantees this is . When working with $C^k$ manifolds for finite $k$, the standard prescription is to adopt or as the definition of the tangent space and to give up the characterization by derivations. A detailed discussion of this state of affairs can be found in Chapter 1 of Warner.