I'm looking for an explanation of the following result:

If D is a maximal order in a definite (i.e. ramified at infinity) quaternion algebra B over $\mathbb{Q}$, and $\phi : B\otimes\mathbb{R}\rightarrow\mathbb{H}$ is an isomorphism over $\mathbb{R}$ (where $\mathbb{H}$ is the Hamiltonian quaternions over $\mathbb{R}$), then the co-volume of $\phi(D)$ in $\mathbb{H}$ (with respect to the Lebesgue measure for the basis 1,i,j,k) is $(1/4) \cdot Disc(D)$ (where the discriminant is the determinant of the trace form on a basis of $D$, or, equivalently since $D$ is maximal, the product of the finite primes where $B$ ramifies).

Right now the only way I have to see this is through an explicit calculation using the generators for maximal orders described in a paper by Ibukiyama available here: http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.nmj/1118787011 , but I feel like there must be a nice way to see this using just properties of maximal orders without having to calculate with an explicit basis. Any ideas or references?

I'm looking for an explanation of the following result:

If D is a maximal order in a definite (i.e. ramified at infinity) quaternion algebra B over $\mathbb{Q}$, and $\phi : B\otimes\mathbb{R}\rightarrow\mathbb{H}$ is an isomorphism over $\mathbb{R}$ (where $\mathbb{H}$ is the Hamiltonian quaternions over $\mathbb{R}$), then the co-volume of $\phi(D)$ in $\mathbb{H}$ (with respect to the Lebesgue measure for the basis 1,i,j,k) is $(1/4) \cdot Disc(D)$ (where the discriminant is the determinant of the trace form on a basis of $D$, or, equivalently since $D$ is maximal, the product of the finite primes where $B$ ramifies).

Right now the only way I have to see this is through an explicit calculation using the generators for maximal orders described in a paper by Ibukiyama available here: Link , but I feel like there must be a nice way to see this using just properties of maximal orders without having to calculate with an explicit basis. Any ideas or references?