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The Dedekind eta function $\eta(\tau)$ can be regarded as a formula which assigns a number to a lattice $L \subset \mathbb{C}$. The algorithm is: rotate the lattice so that one of its basis vectors lies along the real axis, then pick another basis vector $\tau$ in the upper half plane, then compute the usual formula $$ \eta(\tau) = q^\frac{1}{24} \prod_{n=1}^\infty (1-q^n) $$ where $q=e^{2 \pi i \tau}$. It would be nice if there were a more "canonical" way to compute it directly from the lattice $L$ (i.e. without this rotate-and-pick-a-basis-vector story which breaks symmetry). I'm looking for a formula similar to that of the Eisenstein series where one sums over all points of the lattice: $$ G_n (L) = \sum_{\omega \in L, \omega \neq 0} \frac{1}{\omega^n} $$ We have the theorem of Jacobi that the 24th power of $\eta$ computes as the discriminant of the lattice, $$ (2\pi)^{12} \eta(\tau)^{24} = 20G_4(\mathbb{Z} + \tau \mathbb{Z})^3 -49 G_6 (\mathbb{Z} + \tau \mathbb{Z})^2, $$ which is great, since it shows that the 24th power of $\eta$ can be defined canonically via a sum over the lattice points... but how about $\eta$ itself?

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Thanks to those who answered my question about the eta function. I've grabbed the part of the question about trying to "see" the Hopf fibration in the space of lattices and put it into a separate question. –  Bruce Bartlett Oct 17 '10 at 10:05
    
All of the answers are very interesting, but David's response (in short, "not possible") seems to be the one that actually answers your title question. –  S. Carnahan Oct 18 '10 at 2:49
    
Right - fair enough. –  Bruce Bartlett Oct 18 '10 at 8:25
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3 Answers

up vote 4 down vote accepted

The functions $G_k$ and $\Delta = \eta^{24}$ can be regarded as functions on the set of lattices because they're modular of level 1. The $\eta$ function isn't modular of level 1 (its level is 24) so there's no natural way to regard it as a function on lattices -- it's a function on lattices with additional "level structure".

Similarly, in order to regularise $G_2$ to get something with good convergence, you end up having to make it depend on some level structure as well.

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Bruce, the Dedekind function itself is a single sum, $ \prod_{n=1}^{\infty}\left(1-q^n\right)=\sum_{n=-\infty}^{\infty}(-1)^n q^{n (3n+1)/2}$ (Euler). But there is a result about which powers of $\eta$ can be expressed as lattice sums; see for example [Heng Huat Chan, Shaun Cooper, and Pee Choon Toh, The 26th power of Dedekind's $\eta$-function, Adv. Math. 207 (2006), no. 2, 532-–543]. What is your question about?

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Thanks, that's useful, I hadn't known about the Euler identity, and the reference looks relevant. I had indeed wanted an expression which calculated the eta function $\eta(L)$ as a lattice sum, the reason being that then it is easy to see how $\eta$ will transform when you rotate the lattice $\Lambda \mapsto e^{i\theta} \Lambda$. The single sum isn't useful for this since one first has to rotate into standard position,so you can't "see" rotations! I'd like to know if one can see the Hopf fibration in the space of lattices! I've explained this point now further at the end of my question above. –  Bruce Bartlett Oct 17 '10 at 9:17
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The logarithmic derivative of the η function is the Eisenstein series G2, up to elementary factors. The series for G2 does not quite converge when summed over a lattice, but can be regularized in various ways so that it does converge.

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Thanks, I hadn't known this, this is a very useful point. I've clarified the nature of my question above by adding to the last paragraphs at the end. –  Bruce Bartlett Oct 17 '10 at 9:19
    
I tried to compute the log of $\eta$ formally, but I am getting the $E_0$ series instead of $E_2$. Is there some way to get rid of the denominators in the Taylor expansion? –  S. Carnahan Oct 18 '10 at 2:44
    
There was a typo in my post: logarithm should have been logarithmic derivative. –  Richard Borcherds Oct 18 '10 at 5:02
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