I have gotten stuck at an exercise that is concerned with the submanifold graph.
Here is what I have:
Define $q:=f(p)$. Because $M,N$ are smooth manifolds we find open sets $U,V$ in $M,N$ respectively such that $p \in M, q \in N$ and
charts
$$
\varphi: U \to \mathbb{R}^m \\
\psi: V \to \mathbb{R}^n.
$$
Define
$$
\delta: U \times V \to \mathbb{R}^m \times \mathbb{R}^n, \
(x,y) \mapsto (\varphi(x), \psi(y)-(\psi \circ f \circ \varphi^{-1})(y))
$$.
This is injective and maps $U \times V$ to $\delta(U \times V)$ and is hence invertible. Moreover, because $f$ is smooth it follows that $\psi \circ f \circ \varphi^{-1}$ is smooth and hence continous. Because $\varphi, \psi$ are continous, $\delta$ is also continous. So $\delta$ is a chart on $M \times N$. Let $\tilde{U}:=\varphi(U)$. By construction $\delta(\Gamma_f \cap (U \times V))=\tilde{U} \times \{0\}$ which is an open set in $\mathbb{R}^m \times \mathbb{R}^n$. Thus $\Gamma_f$ is a submanifold.
But now I am unsure about the tangent space. The tangent space consists of linear maps
$$
X: C^{\infty}(\Gamma_f) \to \mathbb{R}, \
g \mapsto Xg
$$
such that there is a smooth curve $\gamma: (-\varepsilon, \varepsilon) \to \Gamma_f$ with $\gamma(0)=p$ and $Xg=(g \circ \gamma)'(0)$.
The coordinate charts $\varphi, \psi$ are of the form $\varphi=(x_1,…,x_m), \psi=(y_1,…,y_n)$. Take $g \in C^{\infty}(\Gamma_f)$ and define $X_i,X_j$ via
$$
X_i g:=\partial_i (g \circ \delta^{-1}) (\delta((p,q))), \
i \in \{1,…,m\} \tag{1} \\
$$
$$
X_j g:=\partial_j (g \circ \delta^{-1}) (\delta((p,q))), \
j \in \{1,…,n\} \tag{2}.
$$
where $\delta$ denotes the chart from above. The functions of the form (1) and (2) form a basis of the tangent space. So, if I can find an explicit formula for the vectors I know what the tangent space looks like. However I am a bit at a loss about computing the vectors, because $\delta^{-1}$ is made up of homeomorphisms wich aren't necessarily differentiable by themselves. Is there perhaps another way to find a description of the tangent space?
Best Answer
You're overcomplicating it with charts. The only thing you need to know is that tangent vectors at a point are initial velocities of curves starting at that point. The claim is that $$T_{(p,f(p))}\Gamma_f = \{ (v,{\rm d}f_p(v)) \mid v \in T_pM\}.$$
First, let $(v,w) \in T_{(p,f(p))}\Gamma_f$. Then there is $\gamma\colon (-\epsilon,\epsilon)\to \Gamma_f$ such that $\gamma(0) = (p,f(p))$ and $\gamma'(0) = (v,w)$. Writing $\gamma(t) = (\alpha(t),\beta(t))$, the fact that $\gamma(t)\in \Gamma_f$ says that $\beta(t) = f(\alpha(t))$. So $$(v,w) = \gamma'(0) = (\alpha'(0),(f\circ \alpha)'(0)) = (\alpha'(0), {\rm d}f_{\alpha(0)}(\alpha'(0))),$$and thus $$v = \alpha'(0) \implies w = {\rm d}f_{\alpha(0)}(\alpha'(0)) = {\rm d}f_p(v).$$Conversely, for any $v\in T_pM$, let's show that $(v,{\rm d}f_p(v))\in T_{(p,f(p))}\Gamma_f$. Let $\alpha\colon (-\epsilon,\epsilon) \to M$ be such that $\alpha(0)=p$ and $\alpha'(0)=f(p)$. Now, define $\gamma\colon (-\epsilon,\epsilon)\to \Gamma_f$ by $\gamma(t) = (\alpha(t),f(\alpha(t)))$. Then $\gamma(0) = (p,f(p))$ says that $\gamma'(0)\in T_{(p,f(p))}\Gamma_f$. But what is $\gamma'(0)$? Compute $$\gamma'(0) = (\alpha'(0),(f\circ\alpha)'(0)) = (\alpha'(0),{\rm d}f_{\alpha(0)}(\alpha'(0))) = (v, {\rm d}f_p(v)),$$as required.