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. 



j-invariant fixed point?
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).


