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If we view the j-invariant of a lattice as a map from the upper-half plane to the complexes by $\tau\mapsto j([1,\tau])$, then it is surjective, holomorphic, and has quite a number of other wonderful properties (see the third part of Cox's Primes of the Form for a great introductory reference).

My question is: does $j$ have any fixed points? If so, do we know what any/all of them are?

I'm in particular curious what goes into the proof. Specifically, whether the answer is immediate from some complex analysis, or whether you need to have a good handle on $j$ itself (or both!). A professor I asked suggested thinking about $j(\tau)-\tau$ on the compactification of the fundamental domain of $SL(2,\mathbb{Z})$, but we weren't able to clean it up.

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Isn't the $j$-invariant really only "natural" up to the constant term? In other words can't one envisage a mathematical world where we defined the $j'$-invariant, by $j'(x)=j(x)-744$ or $j(x)+53$ or whatever, and this function $j'$ was our "canonical" isomorphism of $Y_0(1)$ with the affine line. This makes your question sound very weird. What I'm saying is that perhaps the $j$-function is not sufficiently natural to make the question "interesting"... –  Kevin Buzzard Apr 12 '12 at 17:34
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@Kevin: is it not possible for the $j$-invariant to have a fixed point ALWAYS (that is, independently of the choice of constant term)? –  Igor Rivin Apr 12 '12 at 17:46

3 Answers 3

up vote 7 down vote accepted

As Kevin Buzzard pointed out in the comments, the $j$-invariant has many possible normalizations. Automorphisms of the affine line naturally act on the target space, so any $aj+b$ with invertible $a$ is a decent replacement.

That said, any normalization has infinitely many fixed points. I'll give an argument for the usual normalization $q^{-1} + 744 + \cdots$, first.

Consider the standard fundamental domain $U$ in the upper half-plane (namely, the one bounded by the lines $i\mathbb{R} \pm 1/2$ and the unit circle), and let $\partial U$ be the boundary. The function $j$ takes $\partial U$ to the real ray $(-\infty,1728]$. Because the absolute value of $j$ increases exponentially toward cusps, there is a closed disc $D$ of radius strictly greater than 1728, centered at the origin, such that $j$ takes the tail of the cusp $U \setminus (U \cap D)$ into the complement of $D$. Let $\tilde{D}$ be the compact analytic set formed by making a branch cut of $D$ along $j(\partial U)$, i.e., $\tilde{D} \to D$ is surjective and generically one-to-one, but is 2-to-1 over $D \cap (-\infty, 1728)$. Because $D$ contains 1728, $\tilde{D}$ is homeomorphic to a closed disc. We may then define $j^{-1}$ as a continuous function from $\tilde{D}$ to $U \cap D$ by analytic continuation, and since $U \cap j(\partial U) = \emptyset$, this can be lifted to a continuous function to $\tilde{D}$ that lands in the lift of $U \cap D$. We therefore have a continuous function from a space homeomorphic to a disc into itself, so by Brouwer's fixed-point theorem, it has a fixed point. The image of this point in $U$ is then a fixed point for $j$.

We may do the same for any other $SL_2(\mathbb{Z})$-translate of the fundamental domain $U$, since none of them intersect the branch locus. If we choose an alternative normalization of $j$, we can do a similar trick for any fundamental domain that does not intersect the image of its boundary.

The $q$-expansion of $j$ converges reasonably quickly, so it is not hard to find fixed points numerically. As far as I can tell, the fixed points don't have any particularly interesting arithmetic properties (but I would be interested to hear otherwise).

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@GH : I think Scott's argument provides a fixed point in the upper half-plane. Just rephrasing his argument, we have $|j(\tau)| \sim \exp(2\pi \Im(\tau))$ when $\Im(\tau) \to \infty$. So we can find a compact set $K \subset U$ such that $j(K) \supset K$ and by Brouwer's theorem applied to $j^{-1}$ we find $\tau \in K$ such that $j(\tau)=\tau$. –  François Brunault Apr 13 '12 at 23:57
    
@François: Thank you, I should have read the proof carefully. I will delete my original comment. –  GH from MO Apr 14 '12 at 0:49
    
I'm glad the confusion was cleared up, but I suppose I should have made my exposition easier to understand. –  S. Carnahan Apr 14 '12 at 10:49

It has many periodic points, no matter how you normalize it. About fixed points I am not sure. Dynamics of such maps was studied in Adam Epstein's thesis, http://pcwww.liv.ac.uk/~lrempe/adam/thesis.pdf He extended some basic facts of Fatou-Julia theory to a class of maps that contains j-invariant.

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If there exists a fixed point for $j\(\tau\)$ then the modularity of $j$ will demand that there exist a lattice of fixed points. Since any point in the complex plane is an image of $j$ this is not possible.

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If $j(\tau) = \tau$ for some $\tau$, how does modularity give you another fixed point? For example, $j(\tau+1) = j(\tau) = \tau \neq \tau+1$. –  S. Carnahan Aug 31 '12 at 9:23

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