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December 7, 2010

Pictures of Modular Curves (VI)

Posted by Guest

guest post by Tim Silverman

And here I am again, back with more of “A Child’s Garden of Modular Curves”.

Where We’ve Got to and Where We’re Going Next

We’ve been looking at the modular curves X(N)X(N) by way of their tilings by regular NN-gons, each NN-gon being labelled with one of the “fractions reduced mod NN”. So far, we’ve been trying to understand their structure by looking at the action of the subgroup of PSL(2, N)PSL(2, \mathbb{Z}_N) consisting of matrices of the form (a 0 0 d)\left(\array{a&0\\0&d}\right) (mod multiplication by {1,1}\{1,-1\}) where aa and dd are units with ad=1a d=1. This is obviously isomorphic to the group of units of N\mathbb{Z}_N mod change of sign: N */{1,1}{\mathbb{Z}_N}^*/\{1, -1\}. (I think I’ll call this latter the projective group of units for short.)

In the post before last (which stood for a while as a lone promontory in an almost postless desert), we divided up some tilings using colour to distinguish denominators. In the last post (which, by contrast, hid like a small woodland flower in a luxuriant forest of other posts), we saw how the projective group of units acts on the coloured tilings, but only for prime NN. You might want to glance back at that to refresh yourself with colour before continuing with the present comparatively dry post.

This time, we will continue looking at the action of the projective group of units, but extending our examination to composite NN.

Last time we discovered that when NN is an odd prime, the curve divides up into N12\frac{N-1}{2} blocks, corresponding to the elements of the projective group of units. Each block consists of a central NN-gon with a fraction of the form u0\frac{u}{0}, surrounded by the NN possible NN-gons whose fractions are of the form pv\frac{p}{v}, where uu and vv are units, uv=1u v=1 and pp can be anything.

N = 7: one block
N = 7: one block

However, when NN is not prime, things aren’t quite so simple. I’ll makes some general remarks, and then we’ll look at a few examples to get the idea.

Some General Remarks

The first and nicest thing to say is that the group of projective units acts freely on the tiles, and so we can always split up X(N)X(N) into identical separate blocks, one for each projective unit, with each block consisting of whole tiles. We never need to divide a tile between two blocks, or identify one part of a tile with another part of the same tile, or anything like that.

• For suppose a unit uu stabilises some fraction pq\frac{p}{q}, that is, (v 0 0 u)(p q)=(p q)\left(\array{v&0\\0&u}\right)\left(\array{p\\q}\right)=\left(\array{p\\q}\right). Then we have vp=pv p=p and uq=qu q=q. (We could also have vp=pv p=-p and uq=qu q=-q, but then we could just use u-u instead of uu.) But v=1uv=\frac{1}{u}, so, multiplying the first equation by uu, we have p=upp=u p.

Lifting up to the integers, this implies up=aN+pu p = a N+p and uq=bN+qu q = b N+q for some aa and bb, hence u=aNp+1u=\frac{a N}{p}+1 and u=bNq+1u=\frac{b N}{q}+1. Equating the two formulas for uu, and multiplying by pqp q, we get aNq=bNpa N q=b N p, so aq=bpa q=b p. Now, as we said when introducing mod-NN-reduced fractions, pp and qq cannot share any factors that are factors of NN. But aq=bpa q=b p. So any factors of NN that are in qq must be in bb, since they have to appear on the RHS somewhere, and they cannot be in pp. Hence in bNq\frac{b N}{q}, any factors of NN that are removed by its division by qq will be restored by the multiplication by bb, so bNq\frac{b N}{q} is actually an integer multiple of NN, hence is 00 mod NN. Since u=bNq+1u=\frac{b N}{q}+1, this gives u=1u=1 mod NN. QED •

The second, and also very nice, thing to say is that, to get a block that includes one fraction from each orbit of the projective group of units—which is what we need if our blocks are to reflect that action and also join up to make the whole surface—we don’t need to go very far out from the central point with denominator 00. To be precise: we can construct one block to consist of all the fractions with denominator 11 and some of the fractions in contact with them, and nothing more.

• The argument goes like this: first, any two adjacent fractions must have mutually coprime denominators. Remember that two fractions ab\frac{a}{b} and cd\frac{c}{d} are adjacent iff |adbc|=1\vert a d-b c\vert=1, and obviously this can’t happen if bb and dd share any factor larger than 11. Second, by the mediancy relations, if ab\frac{a}{b} is adjacent to pq\frac{p}{q}, then the other fractions adjacent to pq\frac{p}{q} will all be of the form a+mpb+mq\frac{a+m p}{b+m q} for some mm. But since bb and qq are coprime relative to NN, the numbers b+mqb + m q must include a unit of N\mathbb{Z}_N among them. The action of the inverse of this unit on X(N)X(N) will then carry the fraction with that unit denominator to a fraction with denominator 11, and hence carry the adjacent fraction, pq\frac{p}{q}, to a fraction adjacent to a fraction with denominator 11. So every orbit contains, among its elements, a fraction which is adjacent to a fraction with denominator 11, and when picking representatives of our orbits to form one of the blocks of X(N)X(N), we need only pick fractions with denominator 11 and (some of the) fractions adjacent to fractions with denominator 11. (In fact, the latter set includes the former set!) All other blocks will then be images of that block under the action of the projective group of units. •

So the blocks can be constructed with a certain simplicity—we don’t need to construct blocks with long, wandering tendrils or with parts inside other blocks or anything like that.

Next, to add a bit more detail and perhaps give a feel for the way this works out in practice, I’ll give a detailed example of orbits for one NN, and then we’ll calculate the numbers of orbits in the general case, and then I’ll give a few more examples.

The Blocks for Non-Prime NN

Start with NN is 1515. (1515? Why? Because this is the smallest NN with at least two prime factors, both larger than 22 and not the same as each other—which ensures that it’s interesting, but not too interesting.)

For N=15N=15, the units are 11, 22, 44, 77, 88, 1111, 1313 and 1414, so for representative of projective units (i.e. mod {1,1}\{1, -1\}) we can just take the first four of these. Corresponding to these are the four matrices (1 0 0 1)\left(\array{1&0\\0&1}\right), (8 0 0 2)\left(\array{8&0\\0&2}\right), (4 0 0 4)\left(\array{4&0\\0&4}\right), (13 0 0 7)\left(\array{13&0\\0&7}\right).

First, to get our bearings, we’ll look at just the orbits of denominators. In other words, we’ll look at the orbits of the integers mod 1515 mod {1,1}\{1,-1\} under multiplication by the projective units. Then we’ll look at the orbits of fractions themselves, corresponding to faces.

As I mentioned last time, these orbits correspond to factors of NN: in this case

11
33
55
1515

(Of course, 15=015=0 mod 1515.)

The orbits of denominators are what we get when multiplying each of these by each of the four units, viz. (after eliminating duplicates):

11,2,4,71\rightarrow 1, 2, 4, 7
33,63\rightarrow 3, 6
555\rightarrow 5
000\rightarrow 0

(e.g. 57=3555\cdot 7=35\equiv 5, 32=63\cdot 2=6, 34=1233\cdot 4=12\equiv 3, etc).

What about the orbits of fractions (i.e. faces of the tiling)? For the sake of showing the full pattern, I’ll give a big table of all of them, grouped by orbit of denominators. In the tables below, each orbit occupies a column. To the left, I’ve put the matrix corresponding to the appropriate element of the projective group of units.

Orbits of denominator 11

There are 1515 of these:

(1 0 0 1) 01 11 21 31 41 51 61 71 81 91 101 111 121 131 141 (8 0 0 2) 02 82 12 92 22 102 32 112 42 122 52 132 62 142 72 (4 0 0 4) 04 44 84 124 14 54 94 134 24 64 104 144 34 74 114 (13 0 0 7) 07 137 117 97 77 57 37 17 147 127 107 87 67 47 27\array{\arrayopts{\collines{solid}\frame{solid}}\left(\array{1&0\\0&1}\right)& \frac{0}{1}&\frac{1}{1}&\frac{2}{1}&\frac{3}{1}&\frac{4}{1}&\frac{5}{1}&\frac{6}{1}&\frac{7}{1}&\frac{8}{1}&\frac{9}{1}&\frac{10}{1}&\frac{11}{1}&\frac{12}{1}&\frac{13}{1}&\frac{14}{1}\\ \left(\array{8&0\\0&2}\right)& \frac{0}{2}&\frac{8}{2}&\frac{1}{2}&\frac{9}{2}&\frac{2}{2}&\frac{10}{2}&\frac{3}{2}&\frac{11}{2}&\frac{4}{2}&\frac{12}{2}&\frac{5}{2}&\frac{13}{2}&\frac{6}{2}&\frac{14}{2}&\frac{7}{2}\\ \left(\array{4&0\\0&4}\right)& \frac{0}{4}&\frac{4}{4}&\frac{8}{4}&\frac{12}{4}&\frac{1}{4}&\frac{5}{4}&\frac{9}{4}&\frac{13}{4}&\frac{2}{4}&\frac{6}{4}&\frac{10}{4}&\frac{14}{4}&\frac{3}{4}&\frac{7}{4}&\frac{11}{4}\\ \left(\array{13&0\\0&7}\right)& \frac{0}{7}&\frac{13}{7}&\frac{11}{7}&\frac{9}{7}&\frac{7}{7}&\frac{5}{7}&\frac{3}{7}&\frac{1}{7}&\frac{14}{7}&\frac{12}{7}&\frac{10}{7}&\frac{8}{7}&\frac{6}{7}&\frac{4}{7}&\frac{2}{7}}

Orbits of denominator 33

There are 55 of these:

(1 0 0 1) 13 23 43 53 83 (8 0 0 2) 86 16 26 106 46 (4 0 0 4) 113 73 143 103 133 (13 0 0 7) 136 116 76 56 146\array{\arrayopts{\collines{solid}\frame{solid}}\left(\array{1&0\\0&1}\right)& \frac{1}{3}&\frac{2}{3}&\frac{4}{3}&\frac{5}{3}&\frac{8}{3}\\ \left(\array{8&0\\0&2}\right)& \frac{8}{6}&\frac{1}{6}&\frac{2}{6}&\frac{10}{6}&\frac{4}{6}\\ \left(\array{4&0\\0&4}\right)& \frac{11}{3}&\frac{7}{3}&\frac{14}{3}&\frac{10}{3}&\frac{13}{3}\\ \left(\array{13&0\\0&7}\right)& \frac{13}{6}&\frac{11}{6}&\frac{7}{6}&\frac{5}{6}&\frac{14}{6}}

Orbits of denominator 55

There are 33 of these:

(1 0 0 1) 15 25 35 (8 0 0 2) 75 145 65 (4 0 0 4) 45 85 125 (13 0 0 7) 135 115 95\array{\arrayopts{\collines{solid}\frame{solid}}\left(\array{1&0\\0&1}\right)&\frac{1}{5}&\frac{2}{5}&\frac{3}{5}\\ \left(\array{8&0\\0&2}\right)&\frac{7}{5}&\frac{14}{5}&\frac{6}{5}\\ \left(\array{4&0\\0&4}\right)&\frac{4}{5}&\frac{8}{5}&\frac{12}{5}\\ \left(\array{13&0\\0&7}\right)&\frac{13}{5}&\frac{11}{5}&\frac{9}{5}}

Orbits of denominator 1515 (aka 00)

There is just one orbit here:

(1 0 0 1) 10 (8 0 0 2) 80 (4 0 0 4) 40 (13 0 0 7) 130\array{\arrayopts{\collines{solid}\frame{solid}}\left(\array{1&0\\0&1}\right)&\frac{1}{0}\\ \left(\array{8&0\\0&2}\right)&\frac{8}{0}\\ \left(\array{4&0\\0&4}\right)&\frac{4}{0}\\ \left(\array{13&0\\0&7}\right)&\frac{13}{0}}

Observe that there are 1515 orbits based on denominator 11, 55 orbits based on denominator 33, 33 orbits based on denominator 55, and 11 orbit based on denominator 00, aka 1515. Or, in other words, there are Nd\frac{N}{d} orbits based on denominator dd, for each d|Nd\vert N. It would be be nice if this was generally true. Is it? …

Numbers of Orbits

Well, almost. It’s true for squarefree NN (i.e. with no repeated prime factors). If there are higher powers of a prime in NN, it needs a little adjusting.

So, generally, how many orbits are there for a given factor d|Nd\vert N?

I’d like to do the calculation, but it’s a bit involved, so I’ll just give the final answer here, and put the calculations at the end.

Suppose that the prime decomposition of NN is N= ip i n iN=\prod_i p_i^{n_i}. Suppose we want to know the number of orbits based on a factor d|Nd\vert N. (That is, with denominator equal to some projective unit times dd). Let the prime composition of dd be given by d= ip i m id=\prod_i p_i^{m_i}. Then the number of orbits is the product of a factor for each prime, viz.:

For each ii, a factor of p i n im i p i n im ip i n im i1 p i n im i}\thickspace\left.\array{p_i^{n_i-m_i}\\p_i^{n_i-m_i}-p_i^{n_i-m_i-1}\\p_i^{n_i-m_i}}\right\} if {m i=0 0<m i<n i m i=n i\left\{\array{m_i=0\\0\lt m_i\lt n_i\\m_i=n_i}\right..

Examples of Orbits in the Case of Higher Powers of Primes

Let’s take a look at a couple of examples of higher powers of primes.

For example, 8=2 38=2^3, so we have one prime (22), we have n 1=3n_1=3, and m 1m_1 can be 00, 11, 22, or 33, giving the following numbers of orbits for corresponding factors:

factor orbits 1 2 30=8 2 2 312 311=2 4 2 322 321=1 0 2 33=1\array{factor&orbits\\ 1&2^{3-0}=8\\ 2&2^{3-1}-2^{3-1-1}=2\\ 4&2^{3-2}-2^{3-2-1}=1\\ 0&2^{3-3}=1}

Likewise, for 9=3 29=3^2 we expect:

factor orbit 1 3 20=9 3 3 213 211=2 0 3 22=1\array{factor&orbit\\ 1&3^{2-0}=9\\ 3&3^{2-1}-3^{2-1-1}=2\\ 0&3^{2-2}=1}

To check this, let’s do the same exercise as we did for N=15N=15, but this time for N=8N=8 and N=9N=9.

For N=8N=8:

The projective units are {1,3}\{1, 3\}. Convenient matrices are (1 0 0 1)\left(\array{1&0\\0&1}\right) and (3 0 0 3)\left(\array{3&0\\0&3}\right). The factors are {1,2,4,8}\{1, 2, 4, 8\}.

Orbits of denominator 11:

(1 0 0 1) 01 11 21 31 41 51 61 71 (3 0 0 3) 03 33 63 13 43 73 23 53\array{\arrayopts{\collines{solid}\frame{solid}}\left(\array{1&0\\0&1}\right)& \frac{0}{1}&\frac{1}{1}&\frac{2}{1}&\frac{3}{1}&\frac{4}{1}&\frac{5}{1}&\frac{6}{1}&\frac{7}{1}\\ \left(\array{3&0\\0&3}\right)& \frac{0}{3}&\frac{3}{3}&\frac{6}{3}&\frac{1}{3}&\frac{4}{3}&\frac{7}{3}&\frac{2}{3}&\frac{5}{3}}

Orbits of denominator 22:

(1 0 0 1) 12 32 (3 0 0 3) 52 72\array{\arrayopts{\collines{solid}\frame{solid}}\left(\array{1&0\\0&1}\right)& \frac{1}{2}&\frac{3}{2}\\ \left(\array{3&0\\0&3}\right)& \frac{5}{2}&\frac{7}{2}}

Orbits of denominator 44:

(1 0 0 1) 14 (3 0 0 3) 34\array{\arrayopts{\collines{solid}\frame{solid}}\left(\array{1&0\\0&1}\right)& \frac{1}{4}\\ \left(\array{3&0\\0&3}\right)& \frac{3}{4}}

Orbits of denominator 88 (aka 00):

(1 0 0 1) 10 (3 0 0 3) 30\array{\arrayopts{\collines{solid}\frame{solid}}\left(\array{1&0\\0&1}\right)& \frac{1}{0}\\ \left(\array{3&0\\0&3}\right)& \frac{3}{0}}

And for N=9N=9:

The projective units are {1,2,4}\{1, 2, 4\}. We can represent these with matrices (1 0 0 1)\left(\array{1&0\\0&1}\right), (5 0 0 2)\left(\array{5&0\\0&2}\right) and (7 0 0 4)\left(\array{7&0\\0&4}\right). The factors of 99 are of course 11, 33 and 99, and here are the corresponding orbits:

Orbits of denominator 11:

(1 0 0 1) 01 11 21 31 41 51 61 71 81 (5 0 0 2) 02 52 12 62 22 72 32 82 42 (7 0 0 4) 04 74 54 34 14 84 64 44 24\array{\arrayopts{\collines{solid}\frame{solid}}\left(\array{1&0\\0&1}\right)& \frac{0}{1}&\frac{1}{1}&\frac{2}{1}&\frac{3}{1}&\frac{4}{1}&\frac{5}{1}&\frac{6}{1}&\frac{7}{1}&\frac{8}{1}\\ \left(\array{5&0\\0&2}\right)& \frac{0}{2}&\frac{5}{2}&\frac{1}{2}&\frac{6}{2}&\frac{2}{2}&\frac{7}{2}&\frac{3}{2}&\frac{8}{2}&\frac{4}{2}\\ \left(\array{7&0\\0&4}\right)& \frac{0}{4}&\frac{7}{4}&\frac{5}{4}&\frac{3}{4}&\frac{1}{4}&\frac{8}{4}&\frac{6}{4}&\frac{4}{4}&\frac{2}{4}}

Orbits of denominator 33:

(1 0 0 1) 13 23 (5 0 0 2) 43 83 (7 0 0 4) 73 53\array{\arrayopts{\collines{solid}\frame{solid}}\left(\array{1&0\\0&1}\right)&\frac{1}{3}&\frac{2}{3}\\ \left(\array{5&0\\0&2}\right)&\frac{4}{3}&\frac{8}{3}\\ \left(\array{7&0\\0&4}\right)&\frac{7}{3}&\frac{5}{3}}

Orbits of denominator 99 (aka 00):

(1 0 0 1) 10 (5 0 0 2) 50 (7 0 0 4) 70\array{\arrayopts{\collines{solid}\frame{solid}}\left(\array{1&0\\0&1}\right)&\frac{1}{0}\\ \left(\array{5&0\\0&2}\right)&\frac{5}{0}\\ \left(\array{7&0\\0&4}\right)&\frac{7}{0}}

So our formula is confirmed: the extreme cases (11 and 88 for N=8N=8, and 11 and 99 for N=9N=9) had Nd\frac{N}{d} orbits, but the intermediate cases (22 and 44 for N=8N=8, and 33 for N=9N=9 had, respectively, 22, 11 and 11 fewer orbits than expected, corresponding to the term p i n im i1p_i^{n_i-m_i-1}.

How did we get there?

Calculation of the Numbers of Orbits

Finally, we’ll calculate those formulas we gave above. How many orbits are there for a given d|Nd\vert N?

• Well, the number of orbits is the total number of fractions whose denominators are multiples of dd by (projective) units, divided by the total number of projective units. And the number of fractions is the number of possible denominators, multiplied by the number of possible numerators. So let’s go through these in turn.

• First, the projective units. Suppose the prime decomposition of NN is N= ip i n iN=\prod_i{p_i^{n_i}}. There are p i n ip i n i1p_i^{n_i}-p_i^{n_i-1} elements in the unit group of p i n i\mathbb{Z}_{p_i^{n_i}}, and the unit group of N\mathbb{Z}_N is the direct product of these, so there are i(p i n ip i n i1)\prod_i{(p_i^{n_i}-p_i^{n_i-1})} elements in the unit group, and half this projectively.

To summarise: for each ii, a factor of p i n ip i n i1p_i^{n_i}-p_i^{n_i-1}.

• Secondly, how many denominators? Although the denominators will all be multiples of dd by a projective unit, we can’t simply take the total number of projective units, because some multiples will be the same as each other. E.g. see above for N=15N=15, d=5d=5, where multiplying 55 by the various projective units always gives the same result, viz. 55. So instead, we’ll consider all multiples kdk d of dd, 0kNd0\le k\le \frac{N}{d}, and eliminate the ones we don’t want, namely the ones that are larger factors of NN than dd itself is.

What we want is for kk to not contain any factors of NN (hence, a fortiori, no prime factors of NN). Since some of these factors may already have been “used up” by dd—mod NN, kdk d obviously can’t contain more prime factors than are already present in NN—what we really want is for kk to be coprime to Nd\frac{N}{d}, i.e. a unit in Nd\mathbb{Z}_{\frac{N}{d}}.

So, let the prime decomposition of dd be d= ip i m id=\prod_i{p_i^{m_i}}, where m in im_i\le n_i. Then Nd= ip i n im i\frac{N}{d}=\prod_i{p_i^{n_i-m_i}}, and in this case the number of denominators that we want to keep is i:m i<n i(p i n im ip i n im i1)\prod_{i: m_i\lt n_i}{(p_i^{n_i-m_i}-p_i^{n_i-m_i-1})}. (Notice the restriction in the product to the values of ii with m i<n im_i\lt n_i. Obviously if m i=n im_i=n_i then the corresponding factor in Nd\frac{N}{d}, viz. p n im i=p 0=1p^{n_i-m_i}=p^0=1, is trivial, and gives 11 unit, not p 0p 1p^0-p^{-1} units!) Actually, since we are working projectively, we should halve the number of denominators. This will cancel the factor of 12\frac{1}{2} in the number of projective units.

To summarise: for each ii, a factor of p i n im ip i n im i1 1}\thickspace\left.\array{p_i^{n_i-m_i}-p_i^{n_i-m_i-1}\\1}\right\} if {m i<n i m i=n i\left\{\array{m_i\lt n_i\\m_i=n_i}\right..

• Finally, for the numerators. These need to be coprime to dd, but otherwise can be anything N\le N. So, e.g., if dd is even, we remove all multiples of 22 (of which there are N2\frac{N}{2}); if dd is a multiple of 33, we remove the multiples of 33 (of which there are N3\frac{N}{3}); and if dd is a multiple of both 22 and 33 then we remove both but, by an inclusion-exclusion principle, need to add back on the multiples of both (of which there are N6\frac{N}{6}), which will have got removed twice. In short, for each prime appearing in the decomposition of dd, i.e. each p ip_i with m i>0m_i\gt 0, we have a factor of (p i n ip i n i1)(p_i^{n_i}-p_i^{n_i-1}); and for each prime p ip_i with m i=0m_i=0, we have a factor of p i n ip_i^{n_i} (we don’t need to specially remove multiples of these, since they don’t appear in the decomposition of dd).

To summarise the numerators: for each ii, a factor of p i n i p i n ip i n i1}\thickspace\left.\array{p_i^{n_i}\\p_i^{n_i}-p_i^{n_i-1}}\right\} if {m i=0 m i>0\left\{\array{m_i=0\\m_i\gt 0}\right..

Let’s put all this together. We’ll look at the case for a single prime factor p ip_i, since for multiple prime factors we can simply multiply the single-factor results together.

There are three cases: m i=0m_i=0, m i=n im_i=n_i, and 0<m i<n i0\lt m_i \lt n_i. We’ll look at these in turn.

If m i=0m_i=0, we have m i<n im_i\lt n_i, giving a factor of (p i n im ip i n im i1)(p_i^{n_i-m_i}-p_i^{n_i-m_i-1}) for the denominator. We have a factor of p i n ip_i^{n_i} from the numerator, and divide by a factor of (p i n ip i n i1)(p_i^{n_i}-p_i^{n_i-1}) for the units. (There are factors of 12\frac{1}{2} because we are doing things projectively, but they cancel.) So we get (p i n im ip i n im i1)p i n i(p i n ip i n i1)=p i n im i1p i n i1p n i=p n im i\frac{(p_i^{n_i-m_i}-p_i^{n_i-m_i-1})p_i^{n_i}}{(p_i^{n_i}-p_i^{n_i-1})}=\frac{p_i^{n_i-m_i-1}}{p_i^{n_i-1}}p^{n_i}=p^{n_i-m_i}.

At the opposite extreme, for m i=n im_i=n_i, we have a factor of 11 for the denominators; we have m i>0m_i\gt 0, giving (p i n ip i n i1)(p_i^{n_i}-p_i^{n_i-1}) for the numerators, and again dividing by (p i n ip i n i1)(p_i^{n_i}-p_i^{n_i-1}) for the units. So we get 1(p i n ip i n i1)(p i n ip i n i1)=1=p n im i\frac{1\cdot(p_i^{n_i}-p_i^{n_i-1})}{(p_i^{n_i}-p_i^{n_i-1})}=1=p^{n_i-m_i} (since n im i=0n_i-m_i=0).

For squarefree NN, where n i=1in_i=1\thickspace\forall i, these are the only cases we need consider, so we always have a factor of p n im ip^{n_i-m_i}, or in other words, the number of orbits is Nd\frac{N}{d}.

However, in other cases, we will have examples of 0<m i<n i0\lt m_i\lt n_i. For these cases, we have (p i n im ip i n im i1)(p_i^{n_i-m_i}-p_i^{n_i-m_i-1}) for the denominators, (p i n ip i n i1)(p_i^{n_i}-p_i^{n_i-1}) for the numerators, and divide by (p i n ip i n i1)(p_i^{n_i}-p_i^{n_i-1}) for the units, giving a factor of (p i n im ip i n im i1)(p_i^{n_i-m_i}-p_i^{n_i-m_i-1}) in the number of orbits. •

To summarise our final result:

For each ii, a factor of p i n im i p i n im ip i n im i1 p i n im i}\thickspace\left.\array{p_i^{n_i-m_i}\\p_i^{n_i-m_i}-p_i^{n_i-m_i-1}\\p_i^{n_i-m_i}}\right\} if {m i=0 0<m i<n i m i=n i\left\{\array{m_i=0\\0\lt m_i\lt n_i\\m_i=n_i}\right..

Well, goshdarn it, I’ve run out of time again, and no pretty pictures yet. I’ll finish off this section next time. For now, I’ll leave you with the following view of the N=15N=15 case to ponder. Look at how the denominators are distributed around the central face.

Pentadecagons around 1/5: N = 15
Pentadecagons around 1/5: N = 15
Posted at December 7, 2010 12:06 PM UTC

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Re: Pictures of Modular Curves (VI)

Beautiful! Not only pictorially, but verbally!

Posted by: jim stasheff on December 7, 2010 1:18 PM | Permalink | Reply to this

Re: Pictures of Modular Curves (VI)

Thanks, Jim! Much appreciated.

Posted by: Tim Silverman on December 7, 2010 8:50 PM | Permalink | Reply to this

Re: Pictures of Modular Curves (VI)

One tiny comment: in many cases it’s the group PGL(n)PGL(n) that really matters, not PSL(n)PSL(n). For example, the obvious symmetry group of an nn-dimensional vector space is GL(n)GL(n), and if you let that group act on the projectized vector space, the action factors through PGL(n)PGL(n). To some extent PSL(n)PSL(n) is an overly esoteric substitute.

One can have lots of fun, of an annoying sort, trying to keep track of the difference in size between GL(n),SL(n),PGL(n)GL(n), SL(n), PGL(n) and PSL(n)PSL(n) for your favorite fields (or in your case, commutative rings).

Posted by: John Baez on December 8, 2010 5:55 AM | Permalink | Reply to this

Re: Pictures of Modular Curves (VI)

One can have lots of fun …

Yes, I did briefly consider looking that those issues, but every time I did so my heart filled with dread and I felt faint … so I didn’t.

Posted by: Tim Silverman on December 8, 2010 11:36 AM | Permalink | Reply to this

Re: Pictures of Modular Curves (VI)

Well, let me think a bit about it for n=2n = 2. I guess GL(2,k)/SL(2,k)GL(2,k)/SL(2,k) is the group of units of the commutative ring kk. I guess the kernel of GL(2,k)PGL(2,k)GL(2,k) \to PGL(2,k) is also the group of units of kk. So SL(2,k)SL(2,k) and PGL(2,k)PGL(2,k) have the same ‘size’, but the first is a subgroup of GL(2,k)GL(2,k) and the second is a quotient group, and I don’t think they’re always isomorphic (that would make life too easy).

The issue comes to a head when we consider PSL(2,k)PSL(2,k), which is a subgroup of PGL(2,k)PGL(2,k) and a quotient group of SL(2,k)SL(2,k). (I’m alluding to a little commutative square of groups which I am too lazy to draw here.) Sometimes PSL(2,k)PSL(2,k) is isomorphic to PGL(2,k)PGL(2,k) and sometimes it’s not: I seem to recall the issue comes to whether you can get a diagonal matrix with any invertible determinant you want, or not. In other words, whether ‘squaring’ maps the group of units of kk onto itself, or not.

Is that right? I’m starting to regret thinking about this — I’m sick with a fever, and this isn’t helping — so I’ll stop here and let someone else take over. It’s basically pretty simple stuff but there’s something annoying about it.

Posted by: John Baez on December 9, 2010 1:56 AM | Permalink | Reply to this

Re: Pictures of Modular Curves (VI)

OK, I will think about this a bit, but I’m ill too, so it may take some time …

Posted by: Tim Silverman on December 9, 2010 10:55 AM | Permalink | Reply to this

Re: Pictures of Modular Curves (VI)

I was thinking,

These denominators might be important in understanding the relation with modular forms.

A tile in the modular tiling may be labeled by p/q, the rational number at which it touches the real axis. The condition for modular form is:

At the same time, this matrix carries p/q to (ap + bq)/(cp + dq).

So if a matrix conserves the denominator of a fraction, then images of any tile by that matrix will map to tiles with the same values of the modular form.

Gerard

Posted by: Gerard Westendorp on December 10, 2010 9:14 AM | Permalink | Reply to this

Re: Pictures of Modular Curves (VI)

Indeed so.

Moreover, the fact that (1 1 0 1)\left(\array{1&1\\0&1}\right) conserves modular forms is basically what ensures they have Fourier expansions, whose coefficients are sometimes of such interest.

Posted by: Tim Silverman on December 10, 2010 12:00 PM | Permalink | Reply to this

Re: Pictures of Modular Curves (VI)

If N = 5*7*23

Then the order of Gamma(N) is
210 * 33 * 5 * 7 * 11 * 23.

This is the correct order for M24.

M24 is a famous sporadic group. Makes you wonder.
There are a couple of other similar cases for other N.
I think Gamma(N) cannot itself be isomorphic to a sporadic group like M24, but maybe by doing some tricks like quotienting out by something, there might be some cool stuff out there. We know at least that the monster group is related to congruence subgroups through Monstrous Moonshine.

Gerard

Posted by: Gerard Westendorp on December 13, 2010 10:52 PM | Permalink | Reply to this

Re: Pictures of Modular Curves (VI)

There are a lot of interrelations between these groups and other small-to-mediums-sized groups, that I don’t pretend to understand in any depth—though I’d like to!

By the way, I think you mean PSL(2, N)PSL(2, \mathbb{Z}_N), rather than Γ(N)\Gamma(N). The Γ(N)\Gamma(N) groups are infinite subgroups of PSL(2,)PSL(2, \mathbb{Z}). We have PSL(2, N)PSL(2,)/Γ(N)PSL(2, \mathbb{Z}_N)\simeq PSL(2, \mathbb{Z})/\Gamma(N). (Or you mean “index” rather than “order”. For a normal subgroup, the index of the subgroup is the order of the quotient.)

Posted by: Tim Silverman on December 14, 2010 12:16 PM | Permalink | Reply to this

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