The class number of integral binary quadratic forms of discriminant $D$ correspond to the class number of $\mathbb Q(\sqrt{D})$. In this case the class number formula is the classical one of Dirichlet.

For the class number of binary hermitian forms of discriminant $D$, on the other hand, there is no such correspondence I am aware of. Still, there is a similar formula proved by G. Humbert around 1920, but it does not give $h(D)$ itself (see here). In a special case: $$ \sum_{j=1}^{h(D)}\frac{D}{4}\sigma(F_j)=\frac{\pi}{4}\prod_{p|D,p\neq2}(1+\Big(\frac{-1}{p}\Big)\frac{1}{p}) $$ where $\sigma(F_j)$ is a certain collection of fundamental domains.

I'd like to know if there is a more precise formula available.


May be you can consult Latimer's paper On ideals in generalized quaternion algebras and Hermitian forms., Trans. Amer. Math. Soc. 38 (1935), no. 3, 436–446.

The main theorem (Theorem 3) establishes a bijection between what he calls "regular classes of ideals" in the maximal order of a certain quaternion algebra, and the classes of binary hermitian forms of discriminant $\alpha$ (in his notation) with coefficients in $\mathbb{Z}$. Actually, he works in greater generality, over a number field $K$ and replacing $\mathbb{Z}$ with $\mathcal{O}_K$ (which he calls $G$). Latimer himself says in the Introduction that what he's doing is a generalization of the classical relation between quadratic forms and class groups of quadratic fields.

  • $\begingroup$ jstor.org/stable/1989805 $\endgroup$ – Chris Wuthrich Jan 18 '17 at 12:37
  • $\begingroup$ Thanks for the reference. I'm not quite sure what to do with this bijection though, esp. with regards to obtaining a closed formula for the class number. $\endgroup$ – Tian An Jan 18 '17 at 19:09
  • $\begingroup$ Well, what do you mean by "closed formula"? In which sense is Dirichlet Class Number Formula a closed one? Or are you looking for an analytic formula, say expressing this quaternionic class group as special value of some $L$-function, for instance? $\endgroup$ – Filippo Alberto Edoardo Jan 18 '17 at 22:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.