Ytukyg

Definition: A subbundle $W subseteq V$ of vector bundle $V rightarrow X$ is a union vector subspaces $bigsqcup_x W_x$ which is locally trivial with the subspace topology.

Lemma: A subbdunel $W subseteq V$ has adapted charts, i.e. for each $x in X$, there is a neighborhood $U$ and a bundle chart $V|_{U} cong U times Bbb K^n$ that sends $W|_U$ to $U times Bbb K^m$

Claim: Let $V subseteq X times Bbb K^n$ be a subbundle, and let $P_x$ be the orthogonal projection onto $V_x$ then $X mapsto Mat_{n,n}(Bbb K)$, $ xmapsto P_x$ is continuous.

The proof of this claim goes as follows:

Proof: The problem is local, so assume $V cong X times Bbb K^m$ using lemma. This isomoprhism defines sections $s_1, ldots, s_m$ of $V$ which are everywhere linearly independent. By Gram-Schimdt, $s_1, ldots, s_m$ defines orthonormal basis $t_1, ldots, t_m$ of $V$.

It is from now on that I do not understand at all:

The inclusion $V rightarrow X times Bbb K^m$ is given by continuous function $A:X rightarrow Mat_{n,m}(Bbb K)$ that takes values in the matrices $A$ such that $A^*A=1_m$. The orthogonal projection is $P=AA^*$.

I believe I understand why $A^*A=1_m$ . As a matrix $A$ has rows the $m$ orthonormal vectors. Hence $A^*A$ is precisely the dot product giving $1_m$. But why is $AA^*$ the orthogonal projection?

EDIT: I believe I have obtain the answer.

Let us suppose that the orthonormal basis is given $$alpha_i = sum_{k=1}^n alpha_{ik} e_k$$ for $i=1,ldots, m$. So to project $e_1$ on to this space we take the inner product,
$$ sum langle e_i, alpha_{i}rangle alpha_i$$

So the matrix $A$ is given by $(alpha_{ki})$. What $AA^* e_1$ does is it gives $$A(alpha_{11} , alpha_{21} , cdots , alpha_{m1} )^T = sum_{j=1}^m alpha_{1j} alpha_j $$
which is what we wanted.

Since the question you're asking isn't really about the bundle stuff, I'll just ignore that part. Also for simplicity of notation, I'll use the field $Bbb{R}$, but everything works identically over the complex numbers.

Simplifying, the part you're asking about says, if we have $Vcong newcommandRR{Bbb{R}}RR^m$ and a metric preserving inclusion $iota : RR^mtoRR^n$,
which has matrix $A$ with respect to the standard bases of $RR^m$ and $RR^n$, why is $AA^*$ the projection?

Well, what is $A^*$? It's the map $RR^nto RR^m$ such that $langle A^*v,wrangle = langle v, Awrangle$, and if $w=e_i$, then $Aw=t_i$ by definition of the map $A$, so
$langle A^*v,e_irangle = langle v, t_irangle$. Hence
$$A^*v = sum_i langle v,t_irangle e_i,$$
and
$$AA^*v = sum_i langle v,t_irangle t_i,$$
which is precisely orthogonal projection onto the subspace spanned by the $t_i$, i.e. $V$.

Edit And I see you've figured it out.