Subject: Re: lambda_n = lambda_{2n+1}
Date: Fri, 9 Apr 2004 06:51:07 -0500
From: Bill Richter <richter@math.northwestern.edu>
To: dmd1@lehigh.edu
CC: palmieri@math.washington.edu, markmah@mac.com,
     hikida@bus.hiroshima-pu.ac.jp

Here's a proof of John Palmieri's conjecture posted on Mar 12:

> take the Lambda algebra at the prime 2, and mod out by the ideal J
> generated by (lambda_n + lambda_{2n+1} : n >= 0).  It seems that in
> Lambda/J, every product lambda_i lambda_j is trivial.

Thanks to Mark Mahowald and Mizuho Hikida for encouragement.  The
proof borrows techniques from the easy part of Hikida's Hopf paper.

   My proof was suggested by Scheme programming, which showed John's
   conjecture was true for stem degree <= 76.  I much prefer DrScheme
   (includes nice book <http://www.htdp.org/2002-09-22/Book>) to
   Maple.  I re-used DrScheme code from last year that showed the
   Curtis algorithm is "identical" to the Tangora's preliminary
   algorithm, for t <= 45: using tags is the same as using full tags.
   I also have a proof that Curtis = Prelim, which completes Marty's
   Memoir sketch proof of Curtis algorithm finiteness.  Anyway, Scheme
   led me to the convention of writing (x y) for lambda_x lambda_y.

Proof:
As John posted, it suffices to consider even terms (a b).  The letter
a, b & n will always refer to even integers >=0.  We see immediately
that the "diagonal" vanishes:

(b b) = (b 2b+1) = 0

my the primary Adem relation.  I wasn't able to directly show any
other terms were zero.

It surprised me that the admissible Adem relations are more useful than
the original symmetric Adem relations of the 6 authors
sum_{i+j=k} <k choose i> (p+i 2p+1+j)  = 0.
This stands in sharp contrast to the proof of the Lambda admissible
monomial basis (see my Hopf preprint, where I rediscovered an old
unpublished proof of Pete Bousfield's), where the  symmetric rels are
vastly superior to the  admissible rels.  So below, Adem rel will
always mean an admissible Adem relation.

The proof is by induction on the "stem degree" sigma(a b) = a+b.  That
is, for any (a b), we'll show that (a b) is equivalent in John's ideal
to a sum of monomials (a' b') with a'+b' < a+b.  Let's write that in
shorthand as sigma(a b) < a+b.  Similarly, let's write

                     (a b) = (x y) mod sigma < n
                                  if
          (a b) = (x y) + sum (a' b'),   with all a'+b' < n.

That is, working mod the subvectorspace in Lambda/J spanned by the
terms in Lambda with stem degree < n.  I'll use the phrase "stem
degree" only in the following way: (a b) has stem degree a+b.

So we'll show that sigma(a b) < a+b for all terms (a b) with a+b > 0,
and this implies by induction that (a b) = 0.  Partly the argument is
itself inductive over the stem degree.

Here's a sketch of the argument:

The Adem relations
(even odd)  = ...
show that
sigma(a b) < a+b for a > b.  Lemma 1 below in improve this estimate.
In fact, the Adem relation
(b 2b+1+n) = (b+n 2b+1) + ...
shows sigma(b+n b) < 2b+n.

The Adem relations
(even even) = ...
rewrite inadmissible (a b) recursively as admissible (a b).   Very
important is the Adem relation for (n 2b), for b > a, which shows that

(n 2b) = (b b+n) mod sigma < 2b+n.

Hence we're reduced to showing

sigma(a b) < a+b for a < b <= 2a.

This is a lengthy induction on n = b-a, using the Adem rels
(odd even) = ...

The Adem rels
(odd odd) = ...
are redundant, because Sq^0 preserves Adem relations, as Wang shows.
Following Wang, we'll call theta = Sq^0, and extend it to integers, so
theta(x) = 2x+1.

As to the (odd even) = ... Adem rels, we consider various congruences
of even and odd integers.   The Adem rels
(2n+1 2b) = ...
show by induction on n and the case a > b that

sigma(b b+n) < 2b+n,  if b > 2n,

Sketch: The (odd even) Adem rel shows that
sigma(n 2b) = sigma(2n+1 2b) < 2b+n.
A proof is even included below as Lemma 7.
Then use the above (n 2b) = (b b+n) mod sigma < 2b+n,

b > 2n is the condition that (2n+1 2b) be inadmissible, and (b b+n) is
admissible iff n <= b.  So we're left with only a finite number of
cases for each n: n <= b <= 2n.  This is the difficult part of the
argument, so let's give actual proofs now.  No one would have any
trouble supplying proofs for the above claims, and harder versions of
them arise below.

The case b = n is the hardest, and let's leave it for last.  First we
need to improve the estimate that got rid of (a b) for a > b:

Lemma 1: sigma(b+n b) < b + 2n/3, for even b, n > 0.

Everything follows from Lemma 1 and studying various Adem relations.

Lemma 2: (2x-1 4n+2) = (n 2n+2x) mod sigma < 3n+2x, if n >= x.

This implies

Proposition 3: sigma(b b+n) < 2b+n, for n < b <= 2n.

Now turning to the case b = n, we have

Lemma 4:  (1 4b) = (b 2b) mod sigma < 3b, for b >= 2 even.

Lemma 5:   sigma(3 4b) < 3b, for b >= 2 even.

These two Lemmas then imply the n = b case:

Proposition 6:  sigma(b 2b) < 3b, for b >= 2 even.

Now let's prove these 6 results, restating them for convenience.
First some preliminaries.  Following the easy part of Hikida's Hopf
preprint (accepted by Hiroshima J of Math), we'll write Adem relations

(f theta(f)+p) = sum_{k in F(p-1)} (f+p-k theta(f)+k)

where F(x) := {k : <x-k choose k> = 1, with 0 <= k <= x/2 }, and we
codify Adem relation tricks used by Wang, Mahowald & Singer as

F(theta(x)) = 2 F(x)

F(2x) = 2 F(x) + theta F(x-1)

which imply

F(4x) = 4 F(x) + 2 theta F(x-1) + theta 2 F(x-1)

Here + means disjoint union: In Latex, \amalg should be used instead
of \coprod, as I learned from John Klein, but amalg looks odd in text.

Two facts about theta are useful for the Arithmetic-challenged:

theta(x) + 2k = theta(x+k),    2k - theta(x) = theta(k-1-x),

More interesting fact, rarely used below, are that any x >=0 is
written uniquely as x = theta^t(b), and theta^t(y) = 2^t(y+1) - 1.

In Adem relations below, we'll generally be given (f g), and then
solve for g = theta(f)+p, and then we'll use f + p = g - 2(f+1).

Lemma 1: sigma(b+n b) < b + 2n/3, for n > 0.  Equivalently,

sigma(a b) < 2/3(a + 2b), for a > b >= 0.

Proof:
We'll prove this by induction on the stem degree 2b+n.
So look at the Adem relation

(b theta(b)+n) = sum_{k in F(n-1)} (b+n-k theta(b)+k)

Write n=2m.  Then n-1 = theta(m-1), and theta(b)+n = theta(b+m), so

(b+n b) = (b b+m) + sum_{ k in 2 F(m-1) - {0} } (b+n-k theta(b)+k)

and sigma(b b+m) = 2b+m < 2b + 2n/3.

That handles the induction base case n = 2.

For the other RHS terms, write k = 2t for t in F(m-1), so t != 0 and
t < m/2 = n/4.  Then

(b+n-k theta(b)+k) = (b+n-2t b+t),

since theta(b)+2t = theta(b+t).  This term is of the form (l m) with
l > m, and l+m < 2b+n, since

b+n-2t > b+t, since n > 3t, since 3t < 3n/4 < n,

and the stem degree is

b+n-2t + b+t = 2b+n-t > 2b+n, since t > 0.

So by induction on the stem degree, we know that

sigma(b+n-2t b+t) < 2/3(b+n-2t + 2(b+t)) = 2/3(3b+n) = b + 2n/3

Actually that's cheating in case t is odd, since the Lemma is only
about even monomials.  But this follows trivially, because

(a theta^t(b)) = (a b), and 2/3(a + 2b) <= 2/3(a + 2theta^t(b))

Thus sigma(b+n b) < b + 2n/3.
\qed

Lemma 2: (2x-1 4n+2) = (n 2n+2x) mod sigma < 3n+2x, if n >= x.

Proof:
We'll look at the admissible Adem relation for (2x-1 4n+2).  So let

p = 4n+2 - theta(2x-1) = 4n+2 - theta^2(x-1),

so p-1 = 4n+2 - 4x = 2(2n-2x+1) = 2 theta(n-x),

and let y = n-x.  Then

(2x-1 4n+2) = sum_{k in F(2 theta(y))} (4n+2-2x-k 4x-1+k)

and F(2 theta(y)) = 2 F(theta(y)) + theta F(2y).

First take k = 2t even, so the k-th term becomes

(4n+2-2x-2t 4x-1+2t) = (2(2n+1-x-t) 2x-1+t),

since 4x-1+2t = theta(2x-1) + 2t = theta(2x-1+t), and by  Lemma 1,

sigma(2(2n+1-x-t) 2x-1+t) < 4/3(2n+1-x-t + 2x-1+t) = 4/3[2n+x] < 3n+2x

Turning to k odd, we have k = theta(t), for t in F(2y), so t <= y,
and y in F(2y).  Then the k-th term becomes

(4n+2-2x-k 4x-1+k) = (2n-x-t 4x+2t)

since 2(2n+1-x)-theta(t) = theta(2n-x-t), and its stem degree is

2n-x-t + 4x+2t = 2n+3x+t <= 2n+3x+n-x = 3n+2x,

and we achieve this maximum iff t = n-x, and then the term is

(2n-x-t 4x+2t) = (n 2n+2x).
\qed

We deduce Proposition 3 from two applications of Lemma 2:

Proposition 3: sigma(b b+n) < 2b+n, for n < b <= 2n.

Proof:
As we claimed above, (n 2b) = (b b+n) mod sigma < 2b+n, for b > n.
[The proof uses an same k even/odd argument as in Lemma 2, but a much
 simpler one: k even gives a lower stem degree, since the term is odd,
 and for k odd, we merely note that increasing k gives higher "n".]

So let b = n+2x, for x > 0.  n >= 2x, since b <= 2n.  So by Lemma 2,

(4x-1 4n+2) = (n 2n+4x) mod sigma < 3n+4x

Since n > x, Lemma 2 also says that

sigma(2x-1 4n+2) <= 3n+2x < 3n+4x

Since (2x-1 4n+2) = (4x-1 4n+2),

(n 2n+4x) = 0 mod sigma < 3n+4x.

Since 2n+4x = 2b,  sigma(n 2b) < 2b+n.  By our above result,

sigma(b b+n)  < n+2b.
\qed

Now only the case n = b remains.  For this we need Lemma 4 and 5,
whose proofs use the same technique as the proof of Lemma 2, but the
congruences are a little different, so we give the entire arguments.

Lemma 4:  (1 4b) = (b 2b) mod sigma < 3b, for b >= 2 even.

Proof:
We look at the admissible Adem relation for (1 4b).  Letting p = 4b-3,
we have p-1 = 4(b-1), so let x = b-1.  Then

(1 4b) = sum_{k in F(4x)} (4b-2-k 3+k)

and as we showed above,

F(4x) = 4 F(x) + 2 theta F(x-1) + theta 2 F(x-1)

So there are 3 types of terms on the RHS, and two of them involve k
even.  So if k = 2t, then

(4b-2-k 3+k) = (4b-2-2t t+1)

since 2t+3 = theta(t+1).  By Lemma 1,

sigma(4b-2-2t t+1) < 2/3(4b-2-2t + 2(t+1)) = 2/3(4b) < 3b.

We're left with terms with k = theta(2t), for t in F(x-1) = F(b-2), so
2t <= b-2, i.e. 2(t+1) <= b, and we can achieve equality, since b is
even.  Since 4b-2-k = 2(2b-1)-theta(t) = theta(2b-2-2t),

(4b-2-k 3+k) = (2b-2(t+1) 4(t+1)),

which has stem degree (without using Lemma 1)

2b + 2(t+1) <= 2b + b = 3b,

with equality achieved only when 2(t+1) = b, so the other term have
lower stem degree.  And  with the term 2(t+1) = b, we have

(2b-2(t+1) 4(t+1)) = (b 2b).
\qed

Our Proposition now follows from (1 4b) = (3 4b) and the following
result, whose proof is similar to the above:

Lemma 5:   sigma(3 4b) < 3b, for b >= 2 even.

Proof.
The case b = 2 is trivial, since

(3 8) = (4 7) = (4 0)

So assume b >= 4.  We need the Adem relation for (3 4b).

Let p = 4b-7, so p-1 = 4b-8 = 4(b-2), so let x=b-2.  Then

(3 4b) = sum_{k in F(4x)} (4b-4-k 7+k)

and as in the proof above,

F(4x) = 4 F(x) + 2 theta F(x-1) + theta 2 F(x-1)

First let k =2t be even.  Then

 (4b-4-k 7+k) = (2(2b-2-t) 3+t),

since 7+2t = theta(3+t).  By Lemma 1,

sigma(2(2b-2-t) 3+t) < 4/3(2b-2-t + 3+t) = 4/3(2b+1) <= 3b,

since 8b+4 <= 9b, since 4 <= b.

For k odd, we have k in theta 2 F(x-1).  Write b = 2m, and then

x-1 = b-3 = theta(m-2), so

F(x-1) = F(theta(m-2)) = 2 F(m-2),

so k = theta(4s), for s in F(m-2).  Then 2(s+1) <= m, and

(4b-4-k 7+k) = (b-2s-2 8s+8),

since 4b-4-theta(4s) = 4(b-1)-8s-1 = 4(b-2s-1)-1 = theta^2(b-2s-2).

But this term has stem degree

b-2s-2 + 8s+8 = b+6(s+1) <= b+3m = b + 3b/2 = 5b/2 < 3b.

so all the terms the Adem relation for (3 4b) have sigma < 3b.
\qed

   Remark: Before debugging my proof, I thought I needed to do the
   cases b = 4 & 6 by hand.  Just to cover the bases, let me give
   explicit proofs for b = 4 & 6.  b = 4 is easy, since

   (3 16) = (7 16) = (8 15) = (8 0),  and 8 < 12 = 3b.

   For the case b = 6, we have two Adem relations

   (7 24) = (16 15) (15 16) (14 17) (12 19)

   (8 23) = (14 17) (12 19),

   [+ signs not given in the Scheme output].  By reducing mod J, we have

   (3 24) = (16 0) (0 16) (8 2) (12 4) (12 4),

   so sigma(3 24) <= 16 < 18 = 3b.

Now we give the immediate proof of

Proposition 6:  sigma(b 2b) < 3b, for b >= 2 even.

Proof:

We work mod the subvector space of Lambda/J generated by terms of stem
degree < 3b.  Then by Lemma 4,

(1 4b) cong  (b 2b).

But (1 4b) = (3 4b), and by Lemma 5

(3 4b) cong 0.

Hence (b 2b) cong 0, so

sigma(b 2b) < 3b.
\qed

Yah, I'm done!  I thought that was really hard.  That's all the Math I
did in the last almost-month.  I think it's an acceptable proof, but
it's purely motivated by computer output.  Prior to programming a
solution for stem degree <= 76, I had no clue about how to prove
John's conjecture.  My proof just follows leads of my program output.
So maybe there's a nicer argument.  In particular, it would certainly
be more pleasing to prove John's using the symmetric Adem relations.

[p, k] := sum_{i+j=k} <k choose i> (p+i 2p+1+j)  = 0.

As a start, here's a proof using the symmetric Adem relations for one
of the easy results not proved above.

Lemma 7: (theta(n) 2b) = 0 mod sigma < 2b+n, if b > 2n.

Proof:
(theta(n) 2b) is an outer term of the symmetric Adem relation

[theta(n), theta(y)],

for 2b - theta^2(n) = theta(y), which is of course odd.  But

(theta(n) 2b) = (n 2b) has stem degree 2b + n, and we'll show that
every other term in [theta(n), theta(y)] has sigma < 2b + n.  Since
the Adem relation is zero, the result follows.

A reduction from theta(p) to p takes the stem degree by p+1, and we
want the stem degree of each term, other than (n 2b), to lower by at
least n+2, so that 2b+theta(n) is lowered below 2b+n.  Now

  [theta(n), theta(y)]

= sum_{i+j=theta(y)} <theta(y) choose i> (theta(n)+i   theta^2(n)+j)

= sum_{i+j=y} <y choose i> { (theta(n)+2i   theta^2(n)+theta(j))
                                          +
                             (theta(n)+theta(i)   theta^2(n)+2j) }

since either i is odd, and j is even, or vice versa, and we reduce

= sum_{i+j=y} <y choose i> { (theta(n+i)          theta^2(n)+theta(j))
                                                +
                             (theta(n)+theta(i)   theta(theta(n)+j)) }

= sum_{i+j=y} <y choose i> { (n+i                 theta^2(n)+theta(j))
                                                +
                             (theta(n)+theta(i)   theta(n)+j) }

so all the bottom terms have sigma < 2b+n, since they were lowered by
theta(n)+1+j = 2n+2+j >= n+2.
But the top terms are lowered by
n+1+i >= n+2
except for the first term, with i=0, which is (n 2b).
\qed

Lemma 1 also has a proof using the symmetric Adem relations.  But for
Lem/Props 2--6 I have only admissible Adem rel proofs.

Happy Easter!
Bill