In Interchapter VII of his biography of Gauss, W.K. Buhler describes Gauss's discovery of one of the important properties that characterize the $j$ invariant (Klein's absolute invariant; Gauss called it "Summatorische Function"): that it assumes every complex value exactly once when calculated on a fundamental domain for the modular group action. He mentions that this result was rediscovered much later by Dedekind.

In other words, this result says that $j(\tau)$ is a surjective function ("onto function") from the fundamental domain into the complex plane. And it also says that $j(\tau)$ is injective, since it maps the fundamental domain to the entire complex plane in a one-to-one correspondence.

I was trying to search for this result in Gauss's werke, and I did find a result that comes close to this statement, but not exactly; in p. 478 of volume 3 of his works Gauss says:

The equation $$\left(\frac{Q(t)}{P(t)}\right)^2 = A$$ always has one and exactly one solution in a certain domain.

Gauss doesn't specify any restriction on $A$ and therefore i assume it means "an arbitrary complex number". This remark is accompanied by a drawing of a certain fundamental domain. $Q(t),P(t)$ are Jacobi's theta functions, and therefore the left side of the equation coincides with the definition of the square root of the "modular lambda function" $\lambda(\tau)$.

Therefore Gauss says that $\sqrt{\lambda(\tau)}$ maps the fundamental domain onto the entire complex plane. Now, I read that $\lambda(\tau)$ and $j(\tau)$ are intimately related, but does the result written by Gauss implies the result discovered later by Dedekind? and if not, can someone refer to the exact place in Gauss's werke where he states this result?

  • $\begingroup$ It is not true that $\lambda(\tau)$ is surjective. It omits $0,1,\infty$. $\endgroup$ Commented Sep 22, 2021 at 1:15
  • $\begingroup$ I do not read German, but Klein (which I read in translation) gives these references: p. 103, p. 105 of Werke, volume 8; p. 386 of vol. 3, and Diary, record of June 3, 1800 (Werke v. 10.1, p. 350). $\endgroup$ Commented Sep 22, 2021 at 1:23

1 Answer 1


I have found a reference for the injectivity property of the $j$ function when it acts on a fundamental domain, and this constitutes the first part of the "special property". However, regarding the surjectivity of $j(\tau)$, I'm still searching.

In a short (unpublished) paragraph in p.386 of vol.3, Gauss introduces a mapping from the set of binary quadratic forms of discriminant $d$ and the upper half plane $H$. To each form $(a,b,c)$ Gauss associates a (complex) point $t = \frac{\sqrt{d}+bi}{a}$. This mapping connects the arithmetical theory of binary quadratic forms (which Gauss has exhaustively dealt with in chapter V of the D.A) with its function-theoretic aspects. Gauss than describes a "summatory function" $f(t)$ (which Klein later called $j(\tau)$), as a function that remains constant under the action of the full modular group. He says:

In particular, if the forms $(a,b,c)$ and $(A,B,C)$ are equivalent, so then set $f(t)=f(u)$ if $\frac{t-u}{i}$ is a whole number or if $t=\frac{1}{u}$.

Gauss says here that $f(t)$ remains constant when applying the two generators of the modular group (and therefore doesnot alter under any combination of these generators). Note that he works with the right half plane (not the upper half), but the idea is similar.

Although this is not explicity stated in Gauss's passage, his arithmetical definition of the unique reduced quadratic form (for a given equivalency class) as one with $2b\le a < c$, corresponds under Gauss's mapping to the fundamental domain for the modular group (and therefore for $j(\tau)$). From the uniqueness of the reduced form one concludes that $f(t)$ is indeed injective on a fundamental domain; this is not surprising, as this was the initial motivation for the construction of a fundamental domain. Gauss drew fundamental domains for several subgroups of $\Gamma$ (the modular group), but I didn't find the one for the full $\Gamma$; however, from his familiarity with this matters, one can conclude that he definitely knew it.

  • $\begingroup$ @Alexandre Eremenko - is what i wrote mathematically correct? $\endgroup$
    – user2554
    Commented Sep 22, 2021 at 17:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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