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Tag: Lenstra

A suit with shorts

I’m retiring in two weeks so I’m cleaning out my office.

So far, I got rid of almost all paper-work and have split my book-collection in two: the books I want to take with me, and those anyone can grab away.

Here’s the second batch (math/computer books in the middle, popular science to the right, thrillers to the left).

If you’re interested in some of these books (click for a larger image, if you want to zoom in) and are willing to pay the postage, leave a comment and I’ll try to send them if they survive the current ‘take-away’ phase.

Here are two books I definitely want to keep. On the left, an original mimeographed version of Mumford’s ‘Red Book’.

On the right, ‘Een pak met een korte broek’ (‘A suit with shorts’), a collection of papers by family and friends, presented to Hendrik Lenstra on the occasion of the defence of his Ph.D. thesis on Euclidean number-fields, May 18th 1977.

If the title intrigues you, a photo of young Hendrik in suit and shorts is included.

This collection includes hilarious ‘papers’ by famous people including

  • ‘A headache-causing problem’ by Conway (J.H.), Paterson (M.S.), and Moscow (U.S.S.R.)
  • ‘A projective plain of order ten’ by A.M. Odlyzko and N.J.A. Sloane
  • ‘La chasse aux anneaux principaux non-Euclidiens dans l’enseignement’ by Pierre Samuel
  • ‘On time-like theorems’ by Michiel Hazewinkel
  • ‘She loves me, she loves me not’ by Richard K. Guy
  • ‘Theta invariants for affine root systems’ by E.J.N. Looijenga
  • ‘The prime of primes’ by F. Lenstra and A.J. Oort
  • (and many more, most of them in Dutch)

Perhaps I can do a couple of posts on some of these papers. It might break this clean-up routine.

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How to play Nimbers?

Nimbers is a 2-person game, winnable only if you understand the arithmetic of the finite fields $\mathbb{F}_{2^{2^n}} $ associated to Fermat 2-powers.

It is played on a rectangular array (say a portion of a Go-board, for practical purposes) having a finite number of stones at distinct intersections. Here’s a typical position

The players alternate making a move, which is either

  • removing one stone, or
  • moving a stone to a spot on the same row (resp. the same column) strictly to the left (resp. strictly lower), and if there’s already a stone on this spot, both stones are removed, or
  • adding stones to the empty corners of a rectangle having as its top-right hand corner a chosen stone and removing stones at the occupied corners

Here we illustrate two possible moves from the above position, in the first we add two new stones and remove 2 existing stones, in the second we add three new stones and remove only the top right-hand stone.

As always, the last player able to move wins the game!

Note that Nimbers is equivalent to Lenstra’s ‘turning corners’-game (as introduced in his paper Nim-multiplication or mentioned in Winning Ways Chapter 14, page 473).

If all stones are placed on the left-most column (or on the bottom row) one quickly realizes that this game reduces to classical Nim with Nim-heap sizes corresponding to the stones (for example, the left-most stone corresponds to a heap of size 3).

Nim-addition $n \oplus m $ is defined inductively by

$n \oplus m = mex(n’ \oplus m,n \oplus m’) $

where $n’ $ is any element of ${ 0,1,\ldots,n-1 } $ and $m’ $ any element of ${ 0,1,\ldots,m-1 } $ and where ‘mex’ stands for Minimal EXcluded number, that is the smallest natural number which isn’t included in the set. Alternatively, one can compute $n \oplus m $ buy writing $n $ and $m $ in binary and add these binary numbers without carrying-over. It is well known that a winning strategy for Nim tries to shorten one Nim-heap such that the Nim-addition of the heap-sizes equals zero.

This allows us to play Nimber-endgames, that is, when all the stones have been moved to the left-column or the bottom row.

To evaluate general Nimber-positions it is best to add another row and column, the coordinate axes of the array

and so our stones lie at positions (1,3), (4,7), (6,4), (10,3) and (14,8). In this way all legal moves follow the rectangle-rule when we allow rectangles to contain corners on the added coordinate axes. For example, removing a stone is achieved by taking a rectangle with two sides on the added axes, and, moving a stone to the left (or the bottom) is done by taking a rectangle with one side at the x-axes (resp. the y-axes)

However, the added stones on the coordinate axes are considered dead and may be removed from the game. This observation allows us to compute the Grundy number of a stone at position (m,n) to be

$G(m,n)=mex(G(m’,n’) \oplus G(m’,n) \oplus G(m,n’)~:~0 \leq m’ < m, 0 \leq n’ < n) $

and so by induction these Grundy numbers are equal to the Nim-multiplication $G(m,n) = m \otimes n $ where

$m \otimes n = mex(m’ \otimes n’ \oplus m’ \otimes n \oplus m \otimes n’~:~0 \leq m’ < m, 0 \leq n’ < n) $

Thus, we can evaluate any Nimbers-position with stone-coordinates smaller than $2^{2^n} $ by calculating in a finite field using the identification (as for example in the odd Knights of the round table-post) $\mathbb{F}_{2^{2^n}} = \{ 0,1,2,\ldots,2^{2^n}-1 \} $

For example, when all stones lie in a 15×15 grid (as in the example above), all calculations can be performed using

Here, we’ve identified the non-zero elements of $\mathbb{F}_{16} $ with 15-th roots of unity, allowing us to multiply, and we’ve paired up couples $(n,n \oplus 1) $ allowing u to reduce nim-addition to nim-multiplication via

$n \oplus m = (n \otimes \frac{1}{m}) \otimes (m \oplus 1) $

In particular, the stone at position (14,8) is equivalent to a Nim-heap of size $14 \otimes 8=10 $. The nim-value of the original position is equal to 8

Suppose your opponent lets you add one extra stone along the diagonal if you allow her to start the game, where would you have to place it and be certain you will win the game?

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Lambda-rings for formula-phobics

In 1956, Alexander Grothendieck (middle) introduced $\lambda $-rings in an algebraic-geometric context to be commutative rings A equipped with a bunch of operations $\lambda^i $ (for all numbers $i \in \mathbb{N}_+ $) satisfying a list of rather obscure identities. From the easier ones, such as

$\lambda^0(x)=1, \lambda^1(x)=x, \lambda^n(x+y) = \sum_i \lambda^i(x) \lambda^{n-i}(y) $

to those expressing $\lambda^n(x.y) $ and $\lambda^m(\lambda^n(x)) $ via specific universal polynomials. An attempt to capture the essence of $\lambda $-rings without formulas?

Lenstra’s elegant construction of the 1-power series rings $~(\Lambda(A),\oplus,\otimes) $ requires only one identity to remember

$~(1-at)^{-1} \otimes (1-bt)^{-1} = (1-abt)^{-1} $.

Still, one can use it to show the existence of ringmorphisms $\gamma_n~:~\Lambda(A) \rightarrow A $, for all numbers $n \in \mathbb{N}_+ $. Consider the formal ‘logarithmic derivative’

$\gamma = \frac{t u(t)’}{u(t)} = \sum_{i=1}^\infty \gamma_i(u(t))t^i~:~\Lambda(A) \rightarrow A[[t]] $

where $u(t)’ $ is the usual formal derivative of a power series. As this derivative satisfies the chain rule, we have

$\gamma(u(t) \oplus v(t)) = \frac{t (u(t)v(t))’}{u(t)v(t)} = \frac{t(u(t)’v(t)+u(t)v(t)’}{u(t)v(t))} = \frac{tu(t)’}{u(t)} + \frac{tv(t)’}{v(t)} = \gamma(u(t)) + \gamma(v(t)) $

and so all the maps $\gamma_n~:~\Lambda(A) \rightarrow A $ are additive. To show that they are also multiplicative, it suffices by functoriality to verify this on the special 1-series $~(1-at)^{-1} $ for all $a \in A $. But,

$\gamma((1-at)^{-1}) = \frac{t \frac{a}{(1-at)^2}}{(1-at)} = \frac{at}{(1-at)} = at + a^2t^2 + a^3t^3+\ldots $

That is, $\gamma_n((1-at)^{-1}) = a^n $ and Lenstra’s identity implies that $\gamma_n $ is indeed multiplicative! A first attempt :

hassle-free definition 1 : a commutative ring $A $ is a $\lambda $-ring if and only if there is a ringmorphism $s_A~:~A \rightarrow \Lambda(A) $ splitting $\gamma_1 $, that is, such that $\gamma_1 \circ s_A = id_A $.

In particular, a $\lambda $-ring comes equipped with a multiplicative set of ring-endomorphisms $s_n = \gamma_n \circ s_A~:~A \rightarrow A $ satisfying $s_m \circ s_m = s_{mn} $. One can then define a $\lambda $-ringmorphism to be a ringmorphism commuting with these endo-morphisms.

The motivation being that $\lambda $-rings are known to form a subcategory of commutative rings for which the 1-power series functor is the right adjoint to the functor forgetting the $\lambda $-structure. In particular, if $A $ is a $\lambda $-ring, we have a ringmorphism $A \rightarrow \Lambda(A) $ corresponding to the identity morphism.

But then, what is the connection to the usual one involving all the operations $\lambda^i $? Well, one ought to recover those from $s_A(a) = (1-\lambda^1(a)t+\lambda^2(a)t^2-\lambda^3(a)t^3+…)^{-1} $.

For $s_A $ to be a ringmorphism will require identities among the $\lambda^i $. I hope an expert will correct me on this one, but I’d guess we won’t yet obtain all identities required. By the very definition of an adjoint we must have that $s_A $ is a morphism of $\lambda $-rings, and, this would require defining a $\lambda $-ring structure on $\Lambda(A) $, that is a ringmorphism $s_{AH}~:~\Lambda(A) \rightarrow \Lambda(\Lambda(A)) $, the so called Artin-Hasse exponential, to which I’d like to return later.

For now, we can define a multiplicative set of ring-endomorphisms $f_n~:~\Lambda(A) \rightarrow \Lambda(A) $ from requiring that $f_n((1-at)^{-1}) = (1-a^nt)^{-1} $ for all $a \in A $. Another try?

hassle-free definition 2 : $A $ is a $\lambda $-ring if and only if there is splitting $s_A $ to $\gamma_1 $ satisfying the compatibility relations $f_n \circ s_A = s_A \circ s_n $.

But even then, checking that a map $s_A~:~A \rightarrow \Lambda(A) $ is a ringmorphism is as hard as verifying the lists of identities among the $\lambda^i $. Fortunately, we get such a ringmorphism for free in the important case when A is of ‘characteristic zero’, that is, has no additive torsion. Then, a ringmorphism $A \rightarrow \Lambda(A) $ exists whenever we have a multiplicative set of ring endomorphisms $F_n~:~A \rightarrow A $ for all $n \in \mathbb{N}_+ $ such that for every prime number $p $ the morphism $F_p $ is a lift of the Frobenius, that is, $F_p(a) \in a^p + pA $.

Perhaps this captures the essence of $\lambda $-rings best (without the risk of getting an headache) : in characteristic zero, they are the (commutative) rings having a multiplicative set of endomorphisms, generated by lifts of the Frobenius maps.