How can we write explicitly the length, in the relativistic formalism, as orthogonality relations? In the last chapter of W. Greiner's book, Relativistic Quantum Mechanics, orthogonal transformations are defined as follows :
$$
x^{\prime\mu}=a^{\mu}{}_{\nu}x^{\nu}
$$
When proceeding to find $a^{\mu}{}_{\nu}$'s components, he says that the equation defining the length :
$$
s=\sqrt {g_{\mu}{}_{\nu}x^{\mu}x^{\nu}},
$$
is actually ten orthogonality relations that contrain the components of $a^{\mu}{}_{\nu}$. I fail to see how.
 A: Since the invariant interval $s$ (I think it's a little misleading to call it length) is invariant under the coordinate transformation, you require that $$s^2 = g_{\mu\nu}x^\mu x^\nu = g_{\rho\sigma} x'^\rho x'^\sigma,$$ where I have changed the dummy indices in the second equality to avoid confusion. Now, you can write $x'^\rho = {a^\rho_{}}_{\mu} x^\mu$ and $x'^\sigma = {a^\rho_{}}_{\nu} x^\nu$, so that the above relation reduces to $$g_{\rho\sigma} {a^\rho_{}}_{\mu} {a^\rho_{}}_{\nu} = g_{\mu\nu}.$$
This relation is a tensor relation, and thus has to hold true for every value of $\mu$ and $\nu$. Thus, we have as many independent relations as there are independent components of $g_{\mu\nu}$. Being a symmetric $4\times4$ tensor (since $g_{\mu\nu} = g_{\nu\mu}$), this number  is not 16, but rather 10.
Note: You cannot choose all the elements of a symmetric matrix arbitrarily. For an $n\times n$ matrix, only the elements above (or below) the diagonal and on it may be chosen independently, since the symmetry will decide the values of the other components. It should be easy to see that this number is $n (n+1)/2$, which in our case is 10.
A: This is the fact that the length $s$ must be invariant under an orthogonality transformation that provides constraints on the components of ${a^\mu}_\nu$, equation (16.4) of your book : $${a^\mu}_\sigma{a^\tau}_\mu = {\delta^\tau}_\sigma$$
Since $(\tau, \sigma = 0, 1, 2, 3)$, and avoiding double counting, this equation provides the $10$ constraints
