Tangent space to a submanifold

If $Z$ is a submanifold then there is a natural notion of the plane tangent to $Z$ at any point $x$ independent of abstract notions such as equivalence classes of paths and co-ordinates. It is just the subspace of $\mathbb{R}^n$ tangential to $Z$ at $x$. More precisely if $Z = f^{-1}(0)$ it is the kernel of $df(x)$ which we denoted by $K_x$. To relate this to the abstract notion of tangent vector consider a smooth path

$$\gamma\colon (-\epsilon , \epsilon ) \to Z.$$

Because $Z \subset \mathbb{R}^n$ this is naturally a path in $\mathbb{R}^n$. We check first that this is smooth. To do this choose co-ordinates $(U, \psi)$ for $\mathbb{R}^n$ about $x$ satisfying

$$U \cap Z = \{ x \in Z \colon \mid \psi^{d+1}(x)= \dots = \psi^n(x) = 0\}.$$

and denote by $\bar \psi$ the corresponding co-ordinates on $U \cap Z$. Smoothness of $\gamma$ means that the functions $\hat\psi^i \circ \gamma = \psi^i \circ \gamma$ are smooth for each $i = 1, \dots, d$. Because $\gamma$ has image inside $Z$ we also have that $\psi^i \circ \gamma = 0$ for each $i = d+1, \dots, n$ and hence these are also smooth. So $\gamma$ is a smooth path in $\mathbb{R}^n$. Consider the vector $\gamma'(0)$ in $\mathbb{R}^n$. We have that $f\circ \gamma(t) = 0$ for all $t$ so by the chain rule $df(x) (\circ'(0)) = 0$ so that $\gamma'(0) \in K_x$.

We can now define a map $T_xZ \to K_x$. If $X \in T_xZ$ then we choose a path $\gamma$ whose tangent vector at 0 is $X$ and map $X$ to $\gamma'(0) \in K_x$. We have to check first that this is well-defined. Let $\rho$ be another such path and consider the co-ordinates $\bar \psi$. By definition we have

$$(\bar\psi^i\circ\gamma)'(0) = (\bar\psi^i\circ\rho)'(0)$$

for every $i = 1, \dots, d$. Hence we also have

$$(\psi^i\circ\gamma)'(0) = (\psi^i\circ\rho)'(0)$$

for every $i = 1, \dots, d$. But

$$(\psi^i\circ\gamma)'(0) = (\psi^i\circ\rho)'(0) = 0$$

for $i = d+1, \dots, n$ so we have

$$(\psi^i\circ\gamma)'(0) = (\psi^i\circ\rho)'(0)$$

for $i = 1, \dots, n$. Hence $X$ maps to the same element of $K_x$ whether we use $\gamma$ or $\rho$. To show that this map is injective we use a similar argument. It is easy to see that this map is linear. Hence, counting, dimensions we see that this is a linear isomorphism.

We conclude that if $Z \subset \mathbb{R}^n$ is a submanifold and we consider the tangents to all the paths through $x \in Z$, thought of as maps into $\mathbb{R}^n$ then they span the space $K_x$.