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Revision difference : orthonormal basis
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\section*{orthonormal basis} Let $X$ be a tree, and $\Gamma$ a group acting freely and faithfully by group automorphisms on $X$. Then $\Gamma$ is a free group.
Let $X$ be an inner product space over a field $F$ and \begin{proof}
\item $\{x_{\alpha}\}_{\alpha\in J} \subset X$ be a set of orthonormal vectors in the space. If we can write any vector in our space as the sum of vectors from the set multiplied by elements of the field, or in symbols Since $\Gamma$ acts freely on $X$, the quotient graph $X/\Gamma$ is well-defined, and $X$ is the universal cover of $X/\Gamma$ since $X$ is contractible. Thus $\Gamma\cong\pi_1(X/\Gamma)$. Since any graph is homotopy equivalent to a wedge of spheres, and the fundamental group of such a space is free by Van Kampen's theorem, $\Gamma$ is free.
\item $\forall x\in X:\exists \{a_{\alpha}\}_{\alpha\in J}\subset F:x=\sum_{\alpha\in J} a_{\alpha} x_{\alpha}$ then we say that $\{x_{\alpha}\}$ form an orthonormal basis for $X$. \end{proof}