The tangent space.

Most of the theory of calculus on manifolds needs the idea of tangent vectors and tangent spaces. The name tangent vector' comes of course from examples like $S^2 \subset \mathbb{R}^3$ where a tangent vector at $x \in S^2$ is a vector in $\mathbb{R}^3$ tangent to the sphere which in that particular case means orthogonal to $x$. However in the case of a general manifold $M$ it does not come to us sitting inside some $\mathbb{R}^N$ and we have to work a little harder to develop a notion of tangent vector.

Although we do not have a notion of tangent vector yet we do have the notion of a smooth path in a manifold. Let us we what this does for is in $\mathbb{R}^n$. In that case if $f \colon \mathbb{R}^n \to \mathbb{R}$ and a path $\gamma \colon \mathbb{R}\to \mathbb{R}^n$ with $\gamma(0) = x$ then we can consider

$$(f \circ \gamma)'(0)$$

the rate of change of $f$ along $\gamma$ as we go through 0. By the chain rule we can write this as

$$(f \circ \gamma)'(0) = df(x)(\gamma'(0))$$

where $\gamma'(0)$ is the tangent vector to $\gamma$ at $t=0$. Notice that this equation tells us the $(f \circ \gamma)'(0)$ depends on $\gamma$only through $\gamma'(0)$, that is if we replace $\gamma$ by another path $\rho$ with $\rho(0) =x$ and $\rho'(0)=\gamma'(0)$ then

$$(f \circ \gamma)'(0) = (f \circ \rho)'(0).$$

On a manifold we do not have the vector space structure of $\mathbb{R}^n$ so we cannot, immediately, differentiate a path. However we can compose a smooth path $\gamma$ and a smooth function $f$ to obtain a function

$$f\circ \gamma \colon (-\epsilon, \epsilon) \to \mathbb{R}.$$

Moreover if we choose co-ordinates $(U, \psi)$ with $\gamma((-\epsilon, \epsilon)) \subset U$ then we have that

$$f\circ \gamma=(f\circ\psi^{-1}) \circ (\psi\circ \gamma)$$

so that $f \circ \gamma$, being the composition of two smooth functions, is smooth and it makes sense to consider

$$(f \circ \gamma)'(0).$$

If we insert the co-ordinates again and apply the chain rule this is

$$(f \circ \gamma)'(0) = d(f \circ \psi^{-1})(\psi(x)) (\psi\circ\gamma)'(0).$$

Now we would like $(f \circ \gamma)'(0)$ to be the rate of change of $f$ in the direction $\gamma'(0)$ but because we are on a manifold we do not know what $\gamma'(0)$ is. To avoid this problem we just define $\gamma'(0)$ to be the set of all paths which should have the same tangent vector $\gamma'(0)$. We do this as follows.

Definition 4.1 (Tangency)   Let $\gamma \colon (-\epsilon, \epsilon) \to M$ and $\rho \colon (-\Delta, \Delta) \to M$ be paths though a point $x$. We say that $\gamma$ and $\rho$ are tangent at $t=0$ if there is a co-ordinate chart $(U, \psi)$ with $\gamma((-\epsilon, \epsilon)) \subset U$ and $\rho ((-\Delta, \Delta))\subset U$ and

$$(\psi\circ\gamma)'(0) = (\psi\circ\rho)'(0).$$

Again we have the usual lemma.

Lemma 4.1   Let $\gamma \colon (-\epsilon, \epsilon) \to M$ and $\rho \colon (-\Delta, \Delta) \to M$ be paths though a point $x$ which are tangent at $t=0$. Then if $(V, \chi)$ is a chart with $\gamma ( (-\epsilon, \epsilon) ) \subset V$ and $\rho ((-\Delta, \Delta))\subset V$ then

$$\chi\circ\gamma'(0) = \chi\circ\rho'(0).$$

Proof. Chain rule and compatibility. $\qedsymbol$

It is easy to see that tangency is an equivalence relation on the set of all paths through the point $x$. The equivalence classes are called tangent vectors (although we have not yet shown that they are vectors). The equivalence class containing a path $\gamma$ is denoted by $\gamma'(0$ or $t_0(\gamma)$. If $X$ is a tangent vector and $\gamma \in X$ then we usually say that $X$ is tangent to $\gamma$ rather than that $\gamma$ is an element of $X$. The set of all tangent vectors at $x$ we denote by $T_xM$. We want to show now that $T_xM$ has the structure of a vector space.

Let $\gamma$ be a path and $(U, \psi)$ a choice of co-ordinates with $U$ containing the image of $\gamma$. Then

$$\psi\circ \gamma \colon (-\epsilon, \epsilon) \to \mathbb{R}^n$$

is a smooth path in $\mathbb{R}^n$. This has a tangent vector at zero which is the vector

$$(\psi\circ\gamma)'(0)$$

in $\mathbb{R}^n$ at $\psi(x)$. Notice that from the lemma this depends only on $\gamma'(0)$. We define a map

$$d\psi(x) \colon T_xM \to \mathbb{R}^n$$

by

$$d\psi(x)(\gamma'(0)) = (\psi\circ\gamma)'(0).$$

By definition of tangency this map is injective we want to prove

Proposition 4.1   The map $d\psi(x)$ is a bijection.

Proof. As we have already noted it suffices to show that this map is onto. Let $(U, \psi)$ be a chart about $x$. If $v$ is a vector in $\mathbb{R}^n$ then $t \mapsto \psi(x) + tv$ is a path in $\mathbb{R}^n$ with tangent vector $v$. Because $\psi(U)$ is open we can find an $\epsilon > 0$ such that if $\vert t\vert <\epsilon$ then $\psi(x) + tv \in \psi(U)$. Then we can define $\gamma \colon (-\epsilon, \epsilon) \to M$ by

$$\gamma(t) =\psi^{-1}(x + tv).$$

Then we have $\psi\circ\gamma(t) = \psi(x) + tv$ so that $(\psi\circ\gamma)'(0) = v$. $\qedsymbol$

Lemma 4.2   If $\chi$ and $\psi$ are co-ordinates on $M$ and $\gamma$ is a path through $x$ then

$$(\chi\circ\gamma)'(0) = d(\chi\circ \psi^{-1})(\psi(x)) (\psi\circ\gamma)'(0).$$

or

$$d\chi(x) = d(\chi\circ \psi^{-1})(\psi(x)) \circ d\psi(x)$$

Proof. The lemma follows immediately from the chain rule applied to the composition of maps

$$\chi\circ\gamma = (\chi\circ \psi^{-1})\circ(\psi\circ\gamma).$$

Notice that all the maps here are defined on open subsets of $\mathbb{R}^n$ so that we can apply the standard chain rule. $\qedsymbol$

From the discussion in the previous section the maps $d\chi(x)$ define linear co-ordinates on $T_xM$ and hence by Proposition 2.2 $T_xM$ has a unique vector space structure which makes all the maps $d\psi(x)$ linear isomorphisms.

Example 4.1   As always the first example is $M = \mathbb{R}^n$. In that case we have a preferred set of co-ordinates. These are just the identity. So two paths $\gamma$ and $\rho$ are tangent if and only if $\gamma'(0) = \rho'(0)$. In other words two paths are tangent if they have the same tangent vector at $x$. Notice also that if $v$ is any vector there is a preferred path whose tangent vector is $v$. That is the straight line $t \mapsto x + tv$. So in the case of $\mathbb{R}^n$ there is no reason to introduce all the extra machinery of equivalence classes of paths.

Example 4.2   The second example is $M = V$ a finite dimensional vector space. Notice that if $\gamma$ is a path taking values in $V$ then we can make sense of of the derivative of $\gamma$ at 0 directly by

$$\gamma'(0) = \lim_{t \to 0} \frac{\gamma(t) - \gamma(0)}{t}.$$

Of course $\gamma'(0)$ defined in this way is a vector in $V$ whereas above we have define $\gamma'(0)$ as an equivalence class of paths. The relationship is that the equivalence class of paths is the unique one containing the path $t \mapsto x = x + t \gamma'(0)$. Again in this case the extra machinery of equivalence classes of paths adds nothing to what we already know.