## Tangent Vectors and Differentials of Smooth Maps

Tangent vectors of functions are discussed early on in a standard calculus course. They are described either as directional derivatives or as velocities as curves. In manifold theory, we would like to generalize these ideas of calculus on $\mathbb{R}$ to calculus on manifolds. While there is an algebraic and geometric viewpoint of tangent vectors on manifolds, the algebraic realization is often faster to devlope the theory with. However, the geometric realization can be incredibly useful for computations. In the following, we’ll discuss both the algebraic and geometric viewpoints of tangent vectors, prove the equivalence between then, and see how both are useful by discussing the differential of a map.

#### Tangent Vectors Algebraically

The first of the two descriptions of tangent vectors we will discuss is derivations. To define tangent vectors as derivations, we need the notation of a germ. Fix a point $p \in M$. Consider the space of all smooth fuctions $f:U \to \mathbb{R}$ where $U$ is a neighborhood of $p$. We say that two such functions $f:U \to \mathbb{R}$ and $f:V \to \mathbb{R}$ are equivalent if there is some (possibly smaller) neighborhood $W \subseteq U \cap V$ of $p$ such that $f|_{W} = g|_{W}$. In other words, two functions are considered the same if they have the same local behavior around $p$. This turns out to be an equivalence relation on the space of such functions. We abuse notation and write $f$ for the class of $f$ since all elements of the class have the same local behavior, around $p$, as $f$. Moreover, we call $f$ a germ at $p$. Denote the space of germs at $p$ by $C_{p}^{\infty}(M)$. Addition and multiplication of functions makes $C_{p}^{\infty}(M)$ into a ring; and scalar multiplication by $\mathbb{R}$ makes $C_{p}^{\infty}(M)$ into an associative unital algebra over $\mathbb{R}$.

We are almost ready to define tangent vectors at $p$. We define a derivation at $p$ to be a linear map $D:C_{p}^{\infty}(M) \to \mathbb{R}$ such that

$D(fg) = (Df)g(p)+f(p)Dg$

This above identity is called the Leibniz rule (it generlaizes the product rule for derivatives). We define a tanget vector at $p$ to be a derivation at $p$. The space of tangent vectors at $p$, denoted $T_{p}M$ is a vector space over $\mathbb{R}$ with pointwise addition and scalar multiplication.

Describing tangent vectors as derivations generalizes directional derivatives (tangent vectors on $\mathbb{R}^{n}$). To see this, suppose we have a smooth function $f:\mathbb{R}^{n} \to \mathbb{R}$ and a vector $\mathbf{v} \in \mathbb{R}^{n}$. Then the directional derivative of $f$ at $p \in \mathbb{R}^{n}$ in the direction of $\mathbf{v}$ is

$(D_{\mathbf{v}}f)(p) = \displaystyle{\sum_{i = 1}^{n}v^{i}\frac{\partial f}{\partial r^{i}}\bigg|_{p}}$

where $\mathbf{v} = (v^{1},\ldots,v^{n})$ and $r^{1},\ldots,r^{n}$ are the standard coordiantes on $\mathbb{R}^{n}$. Since the derivative is a local operator, the direction derivative is constant on smooth functions that agree about $p$. This induces an operator

$D_{\mathbf{v}}:C_{p}^{\infty}(\mathbb{R}^{n}) \to \mathbb{R}$.

In fact, it is a derivation at $p$ since partial derivatives are linear and satisfy the Leibniz (product) rule. Since the directional derivative only depends on $\mathbf{v} \in \mathbb{R}^{n}$, the defintion implies that $\{\partial/\partial r^{i}|_{p}\}_{1 \le i \le n}$ is a basis for the space of tanget vectors at $p$. Conversely, all derivations on $\mathbb{R}^{n}$ at $p$ can be shown to be directional derivatives. In the general setting of a smooth manifold, we have a coordinate neighborhood $(U,\phi) = (U,x^{1},\ldots,x^{n})$ about $p \in M$. With respect to this coordinate neighborhood, we can show $\{\partial/\partial x^{i}|_{p}\}_{1 \le i \le n}$ is a basis for $T_{p}M$ so we can view every tangent vector as a directional derivative.

This is the faster method of defining tangent vectors as it requires less work to introduce and is easier to use in the general theory. However, it is not as intutive as the next description.

#### Tangent Vectors Geometrically

We can also view tangent vectors at $p \in M$ as velocities of curves through $p$. Let $(U,\phi) = (U,x^{1},\ldots,x^{n})$ be a coordinate neighborhood about $p$. Also, let $\gamma:[-\epsilon,\epsilon] \to M$ for some $\epsilon > 0$ be a smooth map such that $\gamma(0) = p$. We say $\gamma$ is a curve initalized at $p$. Consider the space of all curves initalized at $p$ such that $\gamma$ maps into $U$ (this condition is satisfied for sufficiently small $\latex \epsilon > 0$). We say two such curves $\gamma_{1}$ and $\gamma_{2}$ are equivalent if $(\phi \circ \gamma_{1})'(0) = (\phi \circ \gamma_{2})'(0)$ with respect to the local coordinate neighborhood. In other words, in terms of local coodinates, $\gamma_{1}$ and $\gamma_{2}$ have the same velocity vector at $p$. This turns out to be an equivalence relation on the space of such curves. We abuse notation and write $\gamma'(0)$ for the class of $\gamma$ since, in terms of local coordinates, all elements of the class have $\gamma'(0)$ for a velocity vector at $p$. We call $\gamma'(0)$ a tangent vector at $p$. It can be shown that $\gamma'(0)$ is independent of the coordinate neighborhood.

This definition has a geometric description. With respect to the coordinate neighborhood, we can view $U$ as $\mathbb{R}^{n}$. Then we can view $\gamma'(0)$ as the velocity vector at $p$ sitting inside $\mathbb{R}^{n}$. If we consider a general surface, and identify $U$ and $\mathbb{R}^{2}$ by placing $\mathbb{R}^{2}$ ontop of $p$, then the situation looks as follows:

While intuitive, with this description it is not easy to show that tangent vectors at $p$ form a vector space over $\mathbb{R}$. Instead, we will show that the two defintions of tangent vectors are equivalent. This will let us transfer the vector space structure the algebraic description to the geometric one.

#### Equivalence of Descriptions

Having two definitions of tangent vectors, we would like to show that they are equivlaent:

Theorem: The space of tangent vectors $D$ and the space of tangent vectors $\gamma'(0)$ are in bijective correspondence.

Proof sketch. If we have a tangent vector $\gamma'(0)$ at $p$, define an operator

$D_{\gamma}:C_{p}^{\infty} \to \mathbb{R} \qquad f \mapsto (f \circ \gamma)'(0)$.

This operator is well-defined since it is independent of the representatives of $f$ and $\gamma'(0)$ by using the chain rule. It is linear and satisfies Leibniz’s rule because the derivative does and function composition respects these properties. Now let $D$ be a tangent vector at $p$, and $(U,\phi) = (U,x^{1},\ldots,x^{n})$ be a coordinate neighborhood initalized at $p$ (i.e, $\phi(p) = 0$). Recall $\{\partial/\partial x^{i}|_{p}\}_{1 \le i \le n}$ is a basis for $T_{p}M$ so that

$D = \displaystyle{\sum_{i = 1}^{n}a^{i}\frac{\partial}{\partial x^{i}}\bigg|_{p}}$

for some constants $a^{1},\ldots,a^{n}$. In particular, $D(x^{i}) = a^{i}$. In these coordinates consider the curve

$c:[-\epsilon,\epsilon] \to \mathbb{R}^{n} \qquad t \mapsto (ta^{1},\ldots,ta^{n})$

for any $\epsilon > 0$. Then $\gamma_{D} = \phi^{-1} \circ c$ will be a curve initalized at $p$ with image in $U$. This induces a class $\gamma_{D}'(0)$. We are left to show that these are inverse operations. Indeed,

$D_{\gamma_{D}'(0)}(x^{i}) = (x^{i} \circ \gamma_{D})'(0) = (ta^{i})'(0) = a^{i}$,

because $x^{i} = r^{i} \circ \phi$. The coefficients $a^{i}$ uniquely determine $D$ since $\{\partial/\partial x^{i}|_{p}\}_{1 \le i \le n}$ is a basis. Thus $D_{\gamma_{D}'(0)} = D$. Conversely, given $\gamma'(0)$ we have in local coordinates

$D_{\gamma} = \displaystyle{\sum_{i = 1}^{n}a^{i}\frac{\partial}{\partial x^{i}}\bigg|_{p}}$

where $D_{\gamma}(x^{i}) = (x^{i} \circ \gamma)'(0) = a^{i}$ is the $i$-th coordinate of the velocity vector $\gamma'(0)$. Now $\gamma_{D_{\gamma}}'(0)$ has representative $\phi^{-1} \circ c$ and

$(\phi \circ \phi^{-1} \circ c)'(0) = c'(0) = (a^{1},\ldots,a^{n})$

so that $\gamma_{D_{\gamma}}'(0) = \gamma'(0)$. This finishes the proof.

The intuition behind this argument is as follows: given a smooth curve $\gamma$ initalized at $p$, we can get a derivation at $p$ by pushing foward germs through $\gamma$. Conversely, if we have a derivation $D$ at $p$, then it can be written in terms of a basis via local coordinates. Since any two vector spaces of the same dimension are isomorphic, the coefficients of $D$ corresponds to a vector in $\mathbb{R}^{n}$ and this is the velocity vector we want. Then all we do is cook up a line in $\mathbb{R}^{n}$ with this velocity at $\mathbf{0}$ and use coordinate charts to map this line into $M$.

This lets us give the space of classes $\gamma'(0)$ a vector space structure over $\mathbb{R}$ since $T_{p}M$ has a natural one. We will also now label tangent vectors at $p$ by $X_{p}$ (this is the standard notation).

#### The Differential of a Smooth Map

We have two equivalent description of tangent vectors; that’s great but why do we care? One of the best examples comes from the differential of a smooth map. We will use the algebraic description to define the differential, but the geometric description will be useful for computations.

If we have a smooth map $F:N \to M$ of manifolds, then the differential of $F$ at $p$ is defined by

$F_{\ast,p}T_{p}N \to T_{F(p)}M \qquad (F_{\ast}(X_{p}))f = X_{p}(f \circ F)$

for $f \in C_{F(p)}^{\infty}(M)$. Here we are viewing tangent vectors as derivations. It can be shown that this defintion is indepndent of germ representatives. Let’s analize this defintion a little further. Derivations $X_{p}$ at $p$ are sent to derivations $F_{\ast,p}(X_{p})$ at $F(p)$. The derivation $F_{\ast,p}(X_{p})$ at $F(p)$ acts by taking germs $f$ at $F(p)$, pulling them back by $F$ to germs $f \circ F \in C_{p}^{\infty}(N)$, and then applying the derivation $X_{p}$.

This map turns out to be linear, satisfies a chain rule (generalizing the chain rule from calculus), and is an isomorphism of vector spaces if $F$ is a diffeomorphism of manifolds (in other words, it is functorial).

The drawback with this approach is that its not very clear how we can compute the differential of a smooth map explicitely. However, it so happens that we can use curves to compute the differential. A preliminary proposition is:

Proposition: For any $p \in M$ and velocity vector $X_{p} \in M$, there is a smooth curve $c:[-\epsilon,\epsilon] \to M$ for some $\epsilon > 0$ with $c(0) = p$ and $c'(0) = X_{p}$.

In other words, we can always find small smooth curves initalized at $p$ with with any velocity vector we want. Using this proposition we can prove the theorem:

Theorem: Let $F:N \to M$ be a smooth map of manifolds, $p \in N$, and $X_{p} \in T_{p}N$. If $c$ is a smooth curve starting at $p$ in $N$ with velocity $X_{p}$, then $F_{\ast,p}(X_{p}) = \displaystyle{\frac{d(F \circ c)}{dt}\bigg|_{0}}$

In other words, the derivation $F_{\ast,p}(X_{p})$ is just the velocity vector of $F \circ c$ at $0$, and computing this is easy since $F \circ c$ is a one variable function! All in all, we have shown that tangent vectors (as derivations) on manifolds can be computed using calculus on $\mathbb{R}$ and this required us to view tangent vectors on manifolds as velocity vectors of curves.

#### References

An Introduction to Manifolds – Loring Tu