8
$\begingroup$

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.

Improve this question
$\endgroup$
3
  • 2
    $\begingroup$ 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"... $\endgroup$
    – Kevin Buzzard
    Commented Apr 12, 2012 at 17:34
  • 1
    $\begingroup$ @Kevin: is it not possible for the $j$-invariant to have a fixed point ALWAYS (that is, independently of the choice of constant term)? $\endgroup$
    – Igor Rivin
    Commented Apr 12, 2012 at 17:46
  • $\begingroup$ @Igor Yes, see my answer below. $\endgroup$
    – Adam Epstein
    Commented Feb 24, 2016 at 15:31

4 Answers 4

Reset to default
9
$\begingroup$

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).

Improve this answer
$\endgroup$
3
  • $\begingroup$ @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 $j(\tau)=\tau$. $\endgroup$
    – François Brunault
    Commented Apr 13, 2012 at 23:57
  • $\begingroup$ @François: Thank you, I should have read the proof carefully. I will delete my original comment. $\endgroup$
    – GH from MO
    Commented Apr 14, 2012 at 0:49
  • $\begingroup$ I'm glad the confusion was cleared up, but I suppose I should have made my exposition easier to understand. $\endgroup$
    – S. Carnahan ♦
    Commented Apr 14, 2012 at 10:49
3
$\begingroup$

As already pointed out by Alexandre Eremenko, the $j$ function is a finite type map in the sense of my thesis. It follows that any open set intersecting the domain boundary (here the extended real line) contains infinitely many fixed points which are repelling in the sense that the derivative there has modulus greater than 1.

The normalization of $j$ is irrelevant: for any Mobius transformation $M$, the composition $M\circ j$ is also a finite type map.

Improve this answer
$\endgroup$
3
  • $\begingroup$ That's very cool! (is the repelling aspect a consequence of finite-typeness [I haven't read the thesis, otherwise I would know...])? $\endgroup$
    – Igor Rivin
    Commented Feb 24, 2016 at 15:34
  • $\begingroup$ It's a consequence of a nondynamical statement - that near the domain boundary the map has an Islands property (like Picard's Theorem, but with discs rather than points in the target). The proof in general requires some harmonic measure considerations (this a clarification a bit after the thesis) but actually for the $j$ function you can do this by hand, given the tiling of the domain. $\endgroup$
    – Adam Epstein
    Commented Feb 24, 2016 at 16:56
  • $\begingroup$ Cool, I will need to meditate on this... $\endgroup$
    – Igor Rivin
    Commented Feb 24, 2016 at 17:36
2
$\begingroup$

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.

Improve this answer
$\endgroup$
0
$\begingroup$

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.

Improve this answer
$\endgroup$
1
  • 2
    $\begingroup$ If $j(\tau) = \tau$ for some $\tau$, how does modularity give you another fixed point? For example, $j(\tau+1) = j(\tau) = \tau \neq \tau+1$. $\endgroup$
    – S. Carnahan ♦
    Commented Aug 31, 2012 at 9:23

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .