I was reading [that paper][1] about some quantitative results on simplices in subsets of vector spaces over finite fields. I think that I understood most of the technical details of that paper. However, I did not understand how **Theorem 3.1** implies **Theorem 1.3**.

> **Theorem 1.3.** Let $E\subset \mathbb{F}_q^d, d>\tbinom{k+1}{2},$ such that $|E|\geq Cq^{\frac{k}{k+1}d+\frac{k}{2}}$ with a
> sufficiently large constant $C>0$. Then $E$ contains an isometric copy
> of every $k$-simplex.
> 
> **Theorem 3.1.** Let $E\subset \mathbb{F}_q^d, d>\tbinom{k+1}{2},$ such that $|E|\geq Cq^{\frac{k}{k+1}d+\frac{k}{2}}$ with a
> sufficiently large constant $C>0$. Then for every side length $l_k,\
 l_k\in (\mathbb{F}_q^{\times})^{\tbinom{k+1}{2}}$ we have
 $|\mathcal{T}_{l_k}|>0.$ Furthermore, $|\mathcal{T}_{l_k}|\sim
 |E|^{k+1}q^{-\tbinom{k+1}{2}}.$
> 
> Using this theorem we recover the main result of the paper using the
> following linear algebraic observation. 
> 
> **Lemma 3.2.** Let $P$ be a simplex with vertices $v_0,v_1,\dots v_k, \ v_j\in \mathbb{F}_q^d.$ Let $P'$ be another simplex with vertices
> $v'_0,v'_1,\dots,v'_k$. Suppose that $$\lVert v_i-v_j\rVert=\lVert
 v'_i-v'_j\rVert$$ for all $i,j$. Then there exists an orthogonal,
> affine transformation $O$ on $\mathbb{F}_q^d$ such that $O(P)=P'$.

On page 4 authors write that "*This representation does not, in general, always embody a simplex, as $\mathcal{T}_{l_k}$ is not guaranteed to be in general position. However, as we show below, "legitimate" $k$-simplices are equivalent up to an orthogonal transformation.*"

Question 1: I am really confused how these observations imply Theorem 1.3. 

Question 2: What does mean "*isometric copy of $k$-simplex*"?


  [1]: https://arxiv.org/abs/math/0703504