e-1
IMMERSIONS OF RP 2
DONALD M. DAVIS, GIORA DULA, AND MARK MAHOWALD
This paper is dedicated to Michael Barratt on the occasion of his 80th b*
*irthday.
Abstract.We prove that RP2e-1can be immersed in R2e+1-e-8
provided e 7. If e 14, this is 2 better than previously known
immersions. Our method is primarily an induction on geometric
dimension, incorporating also sections obtained from the Radon-
Hurwitz theorem.
1.Statement of result and background
Our main result is the following immersion theorem for real projective spaces.
e-1 2e+1-e-8
Theorem 1.1. If e 7, then RP 2 can be immersed in R .
This improves, in these cases, by 2 dimensions upon the result of Milgram ([9*
*]),
who proved, by constructing bilinear maps, that if n 7 mod 8, then RP ncan be
immersed in R2n-ff(n)-4, where ff(n) denotes the number of 1's in the binary ex*
*pansion
of n. In [2, 1.2], the first and third authors used obstruction theory to prov*
*e that
if n 7 mod 8, then RP ncan be immersed in R2n-D, where D = 14, 16, 17, 18 if
ff(n) = 7, 8, 9, 10. That result, with n = 2e - 1, is 1 or 2 dimensions stro*
*nger
than ours for 7 e 11. If e 13, then our result improves on the result of *
*[2] by
e - 12 dimensions. Thus Theorem 1.1 improves on all known results by 2 dimensio*
*ns
if e 14.
e-1 2e+1-2e-ffi
In [6], James proved that RP 2 cannot be immersed in R where ffi =
e-1
3, 2, 2, 4 for e 0, 1, 2, 3 mod 4. In [5], an immersion result for RP 2 was *
*announced
in dimension 1 greater than that of James' nonimmersion, which would have been
optimal. However, a mistake in the argument of [5] was pointed out by Crabb and
Steer. We hope that a slight improvement in our argument might enable us to pro*
*ve an
__________
Date: January 19, 2007.
2000 Mathematics Subject Classification. 57N35, 55S40.
Key words and phrases. immersion, projective space.
1
2 DONALD M. DAVIS, GIORA DULA, AND MARK MAHOWALD
immersion result in dimension 1 greater than that of James' nonimmersion (possi*
*bly
2 greater if e 3 mod 4). We will point out in Remark 2.14 what would be requi*
*red
for this improvement.
2.Outline of proof
In this section we outline the proof of Theorem 1.1. In subsequent sections, *
*we will
fill in details.
If ` is a vector bundle over a compact connected space X, we define the geome*
*tric
dimension of `, denoted gd(`), to be the fiber dimension of ` minus the maximum
number of linearly independent sections of `. Equivalently, if dim(`) = n, then*
* gd(`)
equals the smallest integer k such that the map X -`!BO(n) which classifies ` f*
*actors
through BO(k). The following lemma is standard (See e.g. [10, 4.2]). Here and
throughout, ,n denotes the Hopf line bundle over RP n.
Lemma 2.1. Let OE(n) denote the number of positive integers i satisfying i n *
*and
i 0, 1, 2, 4 mod 8. Suppose n > 8. Then RP ncan be immersed in Rn+k if and o*
*nly
if gd((2OE(n)- n - 1),n) k.
Thus Theorem 1.1 will follow from the following result, to the proof of which*
* the
remainder of this paper will be devoted.
e-1-1 e e
Theorem 2.2. If e 7, then gd((22 - 2 ),2e-1) 2 - e - 7.
The bulk of the work toward proving Theorem 2.2 will be a determination of up*
*per
bounds for gd(2e,n) for all n 7 mod 8 by induction on e. A similar method cou*
*ld be
employed for all n, but we restrict to n 7 mod 8 to simplify the already form*
*idable
arithmetic. We let Ak = RP 8k+7, and denote gd(m,8k+7) by gd(m, k).
The classifying map for 2e,8k+7will be viewed as the following composite.
[ fxf
Ak d-!(Ak x Ak)(8k+7),! Ajx Ak-j- -! BO2e-1x BO2e-1! BO2e
j (2.3)
Here d is a cellular map homotopic to the diagonal map, X(n)denotes the n-skele*
*ton of
X, and f classifies 2e-1,. We write BOm for BO(m) for later notational convenie*
*nce.
As a first step, we would like to use (2.3) to deduce that
gd(2e, k) max{gd(2e-1, j) + gd(2e-1, k - j) : 0 j k}.
IMMERSIONS OF RP2e-1 3
In order to make this deduction, we need to know that the liftings of the vario*
*us
2e-1,8j+7to various BOm have been made compatibly.
Definition 2.4. If ` is a vector bundle over a filtered space X0 . . .Xk, we *
*say
that
gd(`|Xi) dicompatibly fori k
if there is a commutative diagram
X0 ---! X1 ---! . .-.--! Xk
?? ? ?
?y ??y ??y
BOd0 ---! BOd1 ---! . .-.--! BOdk ---! BOdim(`)
where the map Xk ! BOdim(`)classifies `, and the horizontal maps are the usual
inclusions.
Remark 2.5. In our filtered spaces, we always assume that the inclusions are co*
*fi-
brations.
Remark 2.6. Isomorphism classes of n-dimensional vector bundles over X corre-
spond to homotopy classes of maps of X into BOn. Thus one would initially say
that the diagram in Definition 2.4 commutes up to homotopy. However, by Lemma
2.7, we may interpret this diagram, and other homotopy commutative diagrams that
occur later, as being strictly commutative. To apply the lemma, we will often, *
*at the
outset, replace maps BOn ! BOn+k by homotopy equivalent fibrations.
Lemma 2.7. If
f
A - --! E
?? ??
i?y ?yp
X - --!gB
commutes up to homotopy and p is a fibration, then f is homotopic to a map f0 s*
*uch
that p O f0= g O i.
Proof.Let H : A x I ! B be a homotopy from p O f to g O i. By the definition of
fibration, there exists fH: A x I ! E such that p O fH= H and fH|A x 0 = f. Then
fH|A x 1 is our desired f0. ||
4 DONALD M. DAVIS, GIORA DULA, AND MARK MAHOWALD
If X0 . . .Xk and Y0 . . .Yk are filtered spaces, we define, for 0 i *
*k,
[i
(X x Y )i:= Xjx Yi-j.
j=0
Then (X x Y )0 . . .(X x Y )k is clearly a filtered space. We will prove the
following general result in Section 3.
Proposition 2.8. Suppose gd(`|Xi) di compatibly for i k and gd(j|Yi) d0i
compatibly for i k. For 0 j k, let ej = max (di+ d0j-i: 0 i j). Then
gd(`xj|(X xY )j) ej compatibly for j k. Moreover, if X = Y and ` = j, then *
*the
f
maps (X xX)j -!BOejcan be chosen to satisfy f OT = f, where T : X xX ! X xX
interchanges factors.
We will begin an induction using some known compatible bounds for gd(16, i).
Proposition 2.8 will, after restriction under the diagonal map, allow us to pro*
*ve
P e P e P
gd(( 2 i),n) max { gd(2 i,mi) : mi = n}. These bounds are not yet strong
enough to yield new immersion results. Next, we must improve the bounds by taki*
*ng
advantage of paired obstructions. The following result will be proved in Sectio*
*n 3.
Proposition 2.9. Let BOn[ae] denote the pullback of BOn and the (ae - 1)-connec*
*ted
cover BO[ae] over BO, and let s = min(ae + 2m - 1, 4m - 1).
(1)There are equivalences c01and c02such that the following diagram
commutes.
q1 (s)c01 2m
BO2m[ae](s)---! (BO2m[ae]=BO2m-1[ae]) ---! S
?? ?? ??
p2?y p02?y i?y
q2 (s)c02 2m
BO2m+1[ae](s)---! (BO2m+1[ae]=BO2m-1[ae])---! P2m-1.
Preparatory to the next two parts, we expand this diagram as
follows, with ci= c0iO qiand (X, A) a finite CW pair.
f1 (s)
A ---! BO2m-1[ae]
?? ??
j?y p1?y
X BO2m[ae](s)-c1--! S2m
?? ??
p2?y i?y
BO2m+1[ae](s)c2---! P22mm-1.
IMMERSIONS OF RP2e-1 5
f (s)
(2)Suppose dim(X) < s, and we are given X -! BO2m[ae] such
g 2m
that f O j = p1 O f1 and c1 O f factors as X ! X=A -! S
with [g] divisible by 2 in [X=A, S2m].1 Then p2O f lifts to a map
X -`!BO2m-1[ae](s)whose restriction to A equals f1.
(3)Suppose, on the other hand, dim(X) s, and we are given
f0 (s) 0 0
X -! BO2m+1[ae] such that f O j = p2 O p1 O f1 and c2 O f
g0 2m 0
factors as X ! X=A -! P2m-1 with [g ] divisible by 2 in
[X=A, P22mm-1]. Then f0 is homotopic rel A to a map which
lifts to BO2m[ae](s).
In Section 4, we will implement Propositions 2.8 and 2.9 to prove that the la*
*st part
of the following important result follows from the first five parts, while in S*
*ection 5,
we will establish the first five parts. Here and throughout, (-) denotes the e*
*xponent
of 2 in an integer.
Theorem 2.10. There is a function g(e, k) defined for e 4 and k 0 satisfyin*
*g:
(1)If k 2e-3, then g(e, k) = 2e.
(2)If e > 4k + 2, then g(e, k) = 0, while if e 4k + 2 and k > 1,
then g(e, k) 4k + 4.
(3)If 0 ` k, then g(e, k) g(e - 1, `) + g(e - 1, k - `) - 1.
(4)If [(e + 1)=4] 2` < 2e-3, then g(e, 2`) 2g(e - 1, `) + 1.
(5)Either g(e, k) = g(e, k - 1) or g(e, k) g(e, k - 1) + 2.
(6)gd(2e, k) g(e, k) compatibly for all k.
By restricting the lifting of P 8k+7to P 8k+ifor 0 i 6, we may use this res*
*ult to
obtain compatible liftings of 2e,n for all n.
The function g will be semiexplicitly defined in (5.2), 5.3, and 5.4. In Tabl*
*e 2.11,
we list its values for small values of the parameters. We prefer not to tabulat*
*e the
values g(e, k) = 2e when k > 2e-3. The numbers in boldface will be given speci*
*al
attention at the beginning of Section 5.
Table 2.11. Values of g(e, k) when e 15 and k 16.
__________
1Note that [X=A, S2m] is in the stable range, from which it gets its group
structure.
6 DONALD M. DAVIS, GIORA DULA, AND MARK MAHOWALD
| k
||
______|1___2___3___4___5___6__7___8___9__10__11__12___13___14___15__16_
4 ||716
5 ||615 22 32
e 6 ||514 21 31 37 46 53 64
7 ||013 20 30 36 45 52 63 68 77 84 94 100 109 116 128
8 ||012 19 29 35 44 51 62 67 76 83 93 99 108 115 127
9 ||012 18 28 34 43 50 61 66 75 82 92 98 107 114 126
10 ||012 17 27 33 42 49 60 65 74 81 91 97 106 113 125
11 ||0 0 16 26 32 41 48 59 64 73 80 90 96 105 112 124
12 ||0 0 16 25 31 40 47 58 63 72 79 89 95 104 111 123
13 ||0 0 16 24 30 39 46 57 62 71 78 88 94 103 110 122
14 ||0 0 16 23 29 38 45 56 61 70 77 87 93 102 109 121
15 ||0 0 0 22 28 37 44 55 60 69 76 86 92 101 108 120
To obtain the best results, we must insert one more bit of sectioning informa*
*tion_
linear vector fields on Sn yield vector fields on P nand hence sections of (n +*
* 1),n =
o(P n) ffl. Let
ae(4a + b) = 8a + 2b if0 b 3.
Eckmann ([4]) used the Radon-Hurwitz theorem to show that Sn has ae( (n + 1)) -*
* 1
linearly independent linear fields of tangent vectors and hence (n+1),n has ae(*
* (n+1))
linearly independent sections. We obtain the following well-known result.
Proposition 2.12. For e 2, gd(2e,2e-1) 2e- ae(e).
If we wish to incorporate these into any subsequent induction argument, it is*
* nec-
essary that the liftings be compatible with the liftings already obtained on th*
*e lower
skeleta. All we can easily assert is the following.
Proposition 2.13. Let
8
>><0 ifn ae(e)
de,n= >g(e, [n_8])ifae(e) < n < 2e- ae(e)
>:e e e
2 - ae(e) if2 - ae(e) n < 2 .
Then gd(2e,n) de,ncompatibly for n < 2e.
IMMERSIONS OF RP2e-1 7
Proof.Since both composites stabilize to 2e,, the obstruction to commutativity *
*of
e-ae(e)-1
P 2 ---! BOde,2e-ae(e)-1
?? ?
?y ??y
e-ae(e)
P 2 ---! BO2e-ae(e)
e-ae(e)-1
is a map P 2 ! V2e-ae(e), which is trivial for dimensional reasons. Here V*
*n is the
fiber of BOn ! BO, and is (n - 1)-connected. The top map in this diagram comes
from 2.10.(6), while the bottom map comes from 2.12. ||
Remark 2.14. If we could assert compatibility of the Eckmann liftings with those
of Theorem 2.10.(6) on a larger skeleton, we might improve our immersion result*
* to
the extent mentioned in Section 1.
Remark 2.15. If one inserts the Eckmann lifting earlier in the inductive determ*
*i-
nation of gd(2e,n), one obtains weaker lifting results than those of 2.10.(6). *
* For
example, one can replace g(6, 7) by 52 = 64 - ae(6), but then, by 2.13, one mus*
*t also
use g(6, 6) = 52. If these values are maintained, then values of g(7, k) will h*
*ave to be
increased for k = 6 and 8 k 14.
Finally, in Section 6, we apply the basic induction argument, Proposition 2.8*
*, and
the results for gd(2e,) in Proposition 2.13 to prove the following result by in*
*duction
on t.
Proposition 2.16. For e 7 and t 1, gd((2e+2e+1+. .+.2e+t),2e-1) 2e-e-7.
This clearly implies Theorem 2.2, and hence the immersion theorem 1.1.
3.Proof of general lifting results
In this section, we prove Propositions 2.8 and 2.9. For the first one, we fin*
*d it more
convenient to work with sections rather than geometric dimension.
Theorem 3.1. Let X0 . . .Xk and Y0 . . .Yk be filtered spaces, and let `
(resp. j) be a vector bundle over Xk (resp. Yk). Suppose given m0 (resp. n0) se*
*ctions
of `|Xk (resp. j|Yk), of which the first mi(resp. ni) are linearly independent *
*(l.i.) on
Xi(resp. Yi) for 0 i k. Let
pj = min(mi+ nj-i: 0 i j).
8 DONALD M. DAVIS, GIORA DULA, AND MARK MAHOWALD
Let
[j
Wj = Xix Yj-i.
i=0
Then there are p0 sections of `xj on Wk of which the first pj are linearly inde*
*pendent
on Wj for 0 j k. Moreover, if ` + i j and m`+ ni pj, then the first pj
sections are l.i. on X`x Yi.
Note that we have m0 . . .mk, n0 . . .nk, and p0 . . .pk.
The following result will be used in the final step of the proof of Theorem 3*
*.1.
Lemma 3.2. Suppose ` is an n-dimensional trivial vector bundle over a space X
with l.i. sections t1, . .,.tn. Suppose s1, . .,.sr are l.i. sections of `, ea*
*ch of which
is a linear combination with constant coefficients of the ti. Then there is a *
*set
s1, . .,.sr, s0r+1, . .,.s0nof linearly independent sections of `.
Proof.Because of the constant-coefficient assumption, this is just a consequenc*
*e of
the result for vector spaces, that a basis for a subspace can be extended to a *
*basis for
the whole space. ||
Note that the assumption about constant coefficients was required. For exampl*
*e,
the section s(x) = (x, x) of S2 x R3 cannot be extended to a set of three l.i. *
*sections.
Proof of Theorem 3.1.Let r1, . .,.rm0 be the given sections of `|Xk, and s1, . *
*.,.sn0
the given sections of j|Yk. These are considered as sections of ` x j by using *
*0 on
the other component. Clearly {r1, . .,.rm0, s1, . .,.sn0} is a set of p0 sectio*
*ns on Wk
which is linearly independent on W0. The proof will proceed by finding p1 line*
*ar
combinations, always with constant coefficients, of these sections which are l.*
*i. on
W1, then p2 linear combinations of these new sections which are l.i. on W2, etc*
*.,
until going into the last stage we have pk-1 sections which are l.i. on Wk-1, a*
*nd we
find pk linear combinations of them which are l.i. on Wk. Now we apply the lemma
repeatedly, starting with the last pk sections. At the first step, we extend t*
*his set
to a set of pk-1 sections l.i. on Wk-1, and continue until going into the last *
*stage we
have p1 sections which are combinations of the original p0 sections and satisfy*
* the
conclusion of the theorem for 1 i k. We apply the lemma one last time to ex*
*tend
the set of p1 sections to the desired set of p0 sections.
IMMERSIONS OF RP2e-1 9
Here is an explicit algorithm for the sections described in the first half of*
* the
preceding paragraph. We may assume without loss of generality that m0 n0.
For j from 0 to k,
oFor i from 1 to pj- n0 (resp. pj- m0), let r(j)i= ri(resp. s(j)i=
si). (Note that if n0 pj, then nothing happens at this step.)
oFor i from max(1, pj- n0+ 1) to min(m0, pj), let both r(j)iand
s(j)pj+1-iequal r(j-1)i+ s(j-1)pj+1-i.
oThen the sections r(j)iand s(j)iconstructed in the two previous
steps give the sections which are l.i. on Wj. (Each section con-
structed in the second step can be counted as an r or an s, but
is only counted once.)
We must show that these have the required linear independence. Before doing
so, we illustrate with an example, computed by Maple. Let k = 4, [m0, . .,.m4] =
[11, 6, 4, 1, 0] and [n0, . .,.n4] = [10, 8, 3, 2, 0]. Then [p0, . .,.p4] = [2*
*1, 16, 14, 9, 7].
The 16 sections l.i. on W1 are
r1, . .,.r6, r7+ s10, r8+ s9, r9+ s8, r10+ s7, r11+ s6, s5, . .,.s1.
The 14 sections l.i. on W2 are
r1, r2, r3, r4, r5+ r7+ s10, r6+ r8+ s9, r7+ r9+ s10+ s8, r8+ r10+ s9+ s7,
r9+ r11+ s8+ s6, r10+ s7+ s5, r11+ s6+ s4, s3, s2, s1.
The 9 sections l.i. on W3 are
r1+ r6+ r8+ s9, r2+ r7+ r9+ s10+ s8, r3+ r8+ r10+ s9+ s7,
r4+ r9+ r11+ s8+ s6, r5+ r7+ r10+ s10+ s7+ s5, r6+ r8+ r11+ s9+ s6+ s4,
r7+ r9+ s10+ s8+ s3, r8+ r10+ s9+ s7+ s2, r9+ r11+ s8+ s6+ s1.
10 DONALD M. DAVIS, GIORA DULA, AND MARK MAHOWALD
The 7 sections l.i. on W4 are
r1+ r3+ r6+ 2r8+ r10+ 2s9+ s7,
r2+ r4+ r7+ 2r9+ r11+ s10+ 2s8+ s6,
r3+ r5+ r7+ r8+ 2r10+ s10+ s9+ 2s7+ s5,
r4+ r6+ r8+ r9+ 2r11+ s9+ s8+ 2s6+ s4,
r5+ 2r7+ r9+ r10+ 2s10+ s8+ s7+ s5+ s3,
r6+ 2r8+ r10+ r11+ 2s9+ s7+ s6+ s4+ s2,
r7+ 2r9+ r11+ s10+ 2s8+ s6+ s3+ s1.
Now we continue with the proof. The property described in the first paragraph*
* of
the proof, that the sections claimed to be l.i. on Wj are linear combinations o*
*f those
on Wj-1, is clear from their inductive definition.
Next we easily show that if i > pj- n0, then
X X
r(j)i= s(j)pj+1-i= ri+ c`r`+ spj+1-i+ d`s`
`>i `>pj+1-i
with c` and d` integers. The point here is that the additional terms have subsc*
*ript
greater than i or pj+ 1 - i. The proof is immediate from the inductive formula
r(j)i= r(j-1)i+ s(j-1)pj+1-i
and the fact that pj pj-1. Indeed, from r(j-1)iwe obtain terms r i and s pj-1*
*+1-i,
and from s(j-1)pj+1-iwe obtain terms s pj+1-iand r pj-1-pj+i.
Finally we show that the asserted sections are l.i. on Wj. Let x 2 X`x Yj-`. *
*Note
that {r1(x), . .,.rm`(x)} is l.i., as is {s1(x), . .,.snj-`(x)}, and that pj *
*m`+ nj-`.
If we form a matrix with columns labeled
r1, . .,.rm0, sn0, . .,.s1,
and rows which express the sections, ordered as
r(j)1, . .,.r(j)min(m0,pj), s(j)pj-m0, .(.,.s(j)1,3.3)
in terms of the column labels, then, by the previous paragraph, the number of c*
*olumns
is (usually strictly greater than) the number of rows, the entry in position *
*(i, i) is
1 for i min(m0, pj), and all entries to the left of these 1's are 0. If i > m*
*in(m0, pj),
then all entries in the r-portion of row i are 0. Moreover an analogous stateme*
*nt is
IMMERSIONS OF RP2e-1 11
true if the order of the rows and of the columns are both reversed. Thus there *
*are
1's on the diagonal running up from the lower right corner of the original matr*
*ix (for
min(n0, pj) positions) and 0's to their right.
If a linear combination of our sections applied to x is 0, then the triangula*
*r form
of the matrix implies that the first m` coefficients are 0, while the triangula*
*r form
looking up from the lower right corner implies that the last nj-` coefficients *
*are
0. Since pj m`+ nj-`, this implies that all coefficients are 0, hence the de*
*sired
independence.
The same argument works for the last statement of the proposition. For k sati*
*sfying
j k ` + i, replace Wk by Wk[ (X`x Yi) Then everything goes through as above.
||
Proof of Proposition 2.8.Let D = dim(`) and D0 = dim(j). Then di, d0i, ei, and
(X x Y )iof Proposition 2.8 correspond to D - mi, D0- ni, D + D0- pi, and Wiof
Theorem 3.1, respectively. The compatible gd bounds may be interpreted as vector
bundles `i over Xi of dimension di and isomorphisms `|Xi `i (D - di) and
`i|Xi-1 `i-1 (di- di-1). The trivial subbundles yield, for all i, D - dil.i. *
*sections
of ` on Xi such that the restrictions of the sections on Xi to Xi-1are a subset*
* of
the sections on Xi-1. Each of the sections on X0 has a largest Xifor which it i*
*s one
of the given l.i. sections. By [1, 1.4.1], this section on Xi can be extended o*
*ver Xk
(although probably not as part of a linearly independent set). Analogous statem*
*ents
are true for sections of j|Yi.
By Theorem 3.1, there are D + D0- e0 l.i. sections of ` x j on W0 of which the
first D + D0- eiare l.i. on Wi. Taking orthogonal complements of the spans of t*
*he
sections yields the desired compatible bundles on Wi of dimension ei, yielding *
*the
first part of Proposition 2.8.
For the second part, first note that in the algorithm in the proof of Theorem
3.1, if the r's and s's are equal, then the set of sections constructed on each*
* Wi is
invariant under the interchange map T . Thus the same will be true of the ortho*
*gonal
complement of their span. ||
12 DONALD M. DAVIS, GIORA DULA, AND MARK MAHOWALD
Proof of Proposition 2.9.(1) Let F1 = S2m-1 denote the fiber of BO2m-1[ae] ! BO*
*2m[ae].
There is a relative Serre spectral sequence for
(CF1, F1) ! (BO2m[ae], BO2m-1[ae]) ! BO2m[ae].(3.4)
The fibration V2m ! BO2m[ae] ! BO[ae] shows that the bottom class of BO2m[ae]
is in dimension min(ae, 2m). The spectral sequence of (3.4) shows that H*(S2m) !
H*(BO2m[ae]=BO2m-1[ae]) has cokernel beginning in dimension s+1, and so the map*
* is
an s-equivalence. Thus the inclusion of the s-skeleton of BO2m[ae]=BO2m-1[ae] f*
*actors
through S2m to yield the map c01, which is an equivalence.
g
The second map is obtained similarly. A map P22mm-1-!BO2m+1[ae]=BO2m-1[ae] is
obtained as the inclusion of a skeleton of CF2=F2, where F2 = V2m+1,2is the fib*
*er of
BO2m-1[ae] ! BO2m+1[ae]. The relative Serre spectral sequence of
(CF2, F2) ! (BO2m+1[ae], BO2m-1[ae]) ! BO2m+1[ae](3.5)
implies that coker(g*) begins in dimension s + 1, determined by H2m(CF2, F2)
Hmin(ae,2m+1)(BO2m+1[ae]) and the first "product" class in H4m( V2m+1,2). The o*
*btain-
ing of c02now follows exactly as for c01.
(2) Let Q := BO2m+1[ae]=BO2m-1[ae] and E := fiber(BO2m+1[ae] ! Q). The com-
mutative diagram of fibrations
V2m+1,2---! BO2m-1[ae]---! BO2m+1[ae]
?? ? ?
?y ??y ??y
Q ---! E ---! BO2m+1[ae]
implies that the quotient E=BO2m-1[ae] has the same connectivity as Q=V2m+1,2,
which is 1 less than that determined from (3.5); that is, E=BO2m-1[ae] is (s - *
*1)-
connected. Thus, since dim(X) < s, the vertical maps in
BO2m-1[ae]---! BO2m+1[ae]---! P22mm-1
?? ? ?
?y ??y ??y
E ---! BO2m+1[ae]---! Q
are equivalences in the range relevant for maps from X, A, and X=A. Since the
bottom row is a fibration, we may consider the top row to be one, too, as far a*
*s X is
concerned.
IMMERSIONS OF RP2e-1 13
Since g is divisible by 2, and 2ss2m( P22mm-1) = 0, we deduce that the compos*
*ite
g 2m i 2m
X=A -!S -! P2m-1
represents the 0 element of [X=A, P22mm-1]; i.e. the map is null-homotopic rel*
* *. There
is a commutative diagram as below with the left sequence a cofiber sequence and*
* the
right sequence a fiber sequence in the range of dim(X).
f1
A ---! BO2m-1[ae]
?? ??
j1?y j2?y
f
X ---! BO2m+1[ae] (3.6)
?? ??
q?y ?y
iOg 2m
X=A ---! P2m-1
We have just seen that there is a basepoint-preserving homotopy
H : X=A x I ! P22mm-1
from i O g to a constant map. There is a commutative diagram
X x 0 [ A x I -! BO2m+1[ae]
?? ?
?y ??y
qxI H 2m
X x I ---! X=A x I ---! P2m-1
where the top map is f on Xx0 and j2Of1on each Ax{t}. By the Relative Homotopy
Lifting Property of a fibration, there exists a map fH: X x I ! BO2m+1[ae] maki*
*ng
both triangles commute. When t = 1, it maps into BO2m-1[ae], since it projects *
*to
the constant map at the basepoint of P22mm-1.
(3) We use the fact that 2 . 1 P2m2m-1factors as
j 2m 2m
P22mm-1col-!S2m+1 -!S ,! P2m-1
to deduce that the composite
g0 2m col 2m+1
X=A -! P2m-1 -! S
is null-homotopic since g0 is divisible by 2. An argument similar to the one i*
*n the
beginning of the proof of (2) shows that BO2m[ae] ! BO2m+1[ae] ! S2m+1 is a fib*
*ration
through dimension min(ae + 2m, 4m + 1) s + 1. Since dim(X) s + 1, the lifti*
*ng
14 DONALD M. DAVIS, GIORA DULA, AND MARK MAHOWALD
follows as in the proof of (2). However, we need dim(X) s because the map c2 *
*in
(1) only exists on the s-skeleton. ||
4. Inductive determination of a bound for gd(2e, k)
In this section, we prove that part (6) of Theorem 2.10 follows from its firs*
*t five
parts, together with initial values of g(e, k) given in Table 2.11 when k = 1 o*
*r e = 4.
We begin by noting that 2.10.(6) is true for e = 4, since, by [7, 6.1] or Pro*
*position
2.12, gd(16,15) 7. The compatibility requirement is trivially satisfied becau*
*se there
are only three values for the number of sections involved_no sections, full sec*
*tions
(i.e. trivial bundle), and one intermediate value. Indeed 16,n has no section*
*s for
n 16, 16 sections for n 8, and at least (and in fact exactly) 9 sections for
8 n 15. These values, gd(16, 0) = 0, gd(16, 1) = 7, and gd(16, 2) = 16 agre*
*e with
the values of g(e, k) tabulated in Table 2.11.
Let ae = ae[e - 1]. Assume that we have obtained compatible liftings of 2e-1,*
*8k+7to
BOg(e-1,k)[ae] for all k. For 0 k 2e-3, define
g1(e, k) := max{g(e-1, i)+g(e-1, k-i) : max(0, k-2e-4) i [k=2]}.
Note that by 2.10.(3),
g(e, k) g1(e, k) - 1. (4.1)
Recall Ak = P 8k+7, and let
[k
(A x A)k = Aix Ak-i.
i=0
Then by Proposition 2.8 there are compatible symmetric liftings `k of 2e-1, x2e*
*-1, on
(AxA)k to BOg1(e,k)[ae] for all k. We precede by compatible maps dk : Ak ! (AxA*
*)k,
`kOdk
cellular maps homotopic to the diagonal. The composites Ak ---! BOg1(e,k)[ae] a*
*re
compatible liftings of 2e,8k+7for all k.
By decreasing induction on k starting with k = 2e-3, we will construct compat*
*ible
factorizations through BOg(e,k)[ae] of the maps `kOdk. Assume inductively that,*
* for all
j > k, compatible factorizations, up to homotopy rel Ak, of `jO dj through BOg(*
*e,j)[ae]
have been attained. If g(e, k) g1(e, k), then no factorization of `k O dk is *
*required,
and so our induction on k is extended. So we may assume g(e, k) = g1(e, k) - 1.
IMMERSIONS OF RP2e-1 15
Let h = [k=2]. Let k0be the largest integer less than k such g(e, k0) < g(e, *
*k). By
2.10.(5), g(e, k0) < g(e, k) - 1, and hence by (4.1)
g1(e, k0) g(e, k) - 1. (4.2)
By (4.2), 2.10.(4), and the last part of Proposition 3.1 (which is required for*
* compat-
ibility of the lifts of (A x A)k0and Ah x Ah to BOg(e,k)-1), we have the commut*
*ative
diagram below, similar to (3.6).
0
Ak0 --d-! (A x A)k0[ Ah x Ah ---! BOg(e,k)-1[ae](8k+7)
??
?y
| | BOg(e,k)[ae](8k+7)
?? ? ?
?y ??y ??y
dk `k (8k+7)
Ak ---! (A x A)k ---! BOg(e,k)+1[ae]
?? ? ?
?y ??y c??y
_ _
Ak=Ak0 --d-! (A x A)k=((A x A)k0[ Ah x Ah)-`--! C,
where C = Sg(e,k)+1if g(e, k) is odd, and C = Pgg(e,k)(e,k)-1if g(e, k) is eve*
*n. The
maps labeled d are cellular maps homotopic to the diagonal. The map c is obtain*
*ed
similarly to the first paragraph of the proof of 2.9, using 2.10.(2)2 to conclu*
*de that
g(e, k) 4k + 4, provided k > 1. We will deal with the case k = 1 at the end o*
*f this
proof.
The quotient (A x A)k=(Ah x Ah) equals B _ T (B), where T reverses the order
of the factors, and B is the union of all cells eix ej with i < j. By the symme*
*try
_ _ _ _ _ _
property of `k, `|T (B) = (`|B) O T . Since T O d' d, we conclude that `O dis d*
*ivisible
by 2. Indeed, with rB denoting the retraction onto B,
_ _ _ _ _ _
[`O d] = [(`|B) O rB O d] + [(`|T (B)) O rT(B)O d]
and we have
_ _ _ _ _ _
[(`|T (B)) O rT(B)O d] = [(`|T (B)) O T O rB O d] = [(`|B) O rB O d].
__________
2By 2.10.(2), if g(e, k) < 4k + 4, then e > 4k + 2, and hence gd(2e,8k+7) = 0*
*, in
which case we are done.
16 DONALD M. DAVIS, GIORA DULA, AND MARK MAHOWALD
Thus, by Proposition 2.9, `k O dk is homotopic rel Ak0to a map which lifts to
BOg(e,k)[ae]. Note that the lifting into BOg(e,k)-1[ae] was not needed if g(e, *
*k) is odd.
Hence, once we handle the case k = 1 postponed above, we will have extended our
inductive lifting hypothesis, and so will have proved that there are compatible*
* liftings
of Ak to BOg(e,k)[ae] for all k. This extends the induction on e and proves The*
*orem
2.10.(6), assuming the first five parts of 2.10.
The case k = 1 was postponed above. We consider it here. The subtlety is that*
* we
are asserting a lifting outside the stable range. We consider primarily the cas*
*e e = 5.
The case e = 4, discussed at the beginning of the section, has yielded a commut*
*ative
diagram
P 7---! BO0[9]
?? ?
?y ??y
P 15---! BO7[9]
?? ?
?y ??y
P 23---! BO16[9].
f
The maps factor through maps on P923, P915, and P97= *. The map P915-!BO7[9] is
obtained (see [8, 3.2]) as a compression of
2g
P815r-!S8 -! BO[9],
which lifts to BO7[9] by [3, 2.1]. Note that we could have chosen any lifting o*
*f the
stable map 16, to BO7, and we chose this one. By [3, 2.1], [2f] lifts to BO6, a*
*nd [4f]
to BO5.
Proposition 2.8 yields a commutative diagram
P915-d1--!(P915x P915)(15)---!BO7[9]
?? ? ?
?y ??y ??y
P923-d2--!(P923x P923)(23)---!BO16[9]
?? ? ?
?y ??y ??y
P931-d3--!(P931x P931)(31)---!BO23[9]
?? ? ?
?y ??y ??y
P939-d4--!(P939x P939)(39)---!BO32[9],
IMMERSIONS OF RP2e-1 17
where the horizontal maps are liftings of 16, x 16,, and di are cellular maps h*
*omo-
topic to the diagonal. Proposition 2.9 then allows an improvement to a commutat*
*ive
diagram
P915---! P915_ P915---! BO7[9]
?? ?
?y ??y
P923 -! BO15[9]
?? ?
?y ??y
P931 -! BO22[9]
?? ?
?y ??y
P939 -! BO32[9].
We could not use 2.9 to lift the top map to BO6 because the dimensional conditi*
*ons
were not satisfied. However, the class of this map is 2 times the map f descri*
*bed
in the preceding paragraph, and hence, by the argument there, it lifts to BO6, *
*as
desired. A very similar argument works when e = 6 to lift to BO5.
5.The function g(e, k)
In this section, we define the function g(e, k) which has been used in the pr*
*evious
sections, and prove the first five parts of Theorem 2.10, its numerical propert*
*ies which
were already used to prove 2.10.(6), its important geometrical property.
Let lg(k) = [log2(k)]. Except for an irregularity when k = 1, g is determined*
* by
g(k, lg(k) + 4), which we will denote by f(k). For k 16, this function f(k) o*
*f one
variable has the values indicated in boldface in Table 2.11. The companion equa*
*tions
relating f and g are
f(k) = g(lg(k) + 4, k), (5.1)
and, if k > 1,
8
>><2e e lg(k) + 3
g(e, k) = >max (4k + 4, f(k) - e + lg(k) +l4)g(k) + 4 e 4k + 2
>:
0 e > 4k + 2. (5.2)
For k = 1, the values of g are as in Table 2.11. The reason for the irregularit*
*y when
k = 1 is that in the previous section we used special considerations to get lif*
*tings of
18 DONALD M. DAVIS, GIORA DULA, AND MARK MAHOWALD
bundles over P 15beyond the stable range. It was important to do this to get a *
*good
start on the induction.
The formula for f is too complicated to write explicitly, largely due to requ*
*irement
2.10.(5). It utilizes, among other things, the following auxiliary function.
Definition 5.3.
ffi(n) = max( (n) - 1, min(2, ffi0(n))),
where
ffi0(n) = max{ (n - d) - 4d + 3 : 2 d < n}.
Recall that (-) denotes the exponent of 2. For example, ffi(n) = (n) - 1 if n*
* 0
mod 8, while 8
>><1if (n - 2) = 6
ffi(n) = >2 if (n - 2) 7
>:
0 if (n - 3) = 9
are the only cases with n < 310+ 3 for which ffi(n) 6= (n) - 1.
A first approximation to f is given by
f00(n) = 8n - lg(n) + ffi(n).
We explain now the rationale behind the definition of ffi. The ( (n)-1)-part is*
* present
just to make the basic induction work and to agree with some initial values. If*
* it were
not for two delicate matters, we could just define f(n) = 8n - lg(n) + (n) - 1*
*. The
first of these delicate matters is the stability requirement in 2.10.(2), illus*
*trated by
the last two 12's in column 2 and the last three 16's in column 3 of Table 2.11*
*. The
smallest n for which ffi(n) 6= (n) - 1 is n = 66; ffi(66) = 1, using d = 2 in *
*5.3. The
need for this is seen in
522+ffi(66) = 8.66-6+ffi(66) = f(66) = g(10, 66) g(9, 64)+g(9, 2)-1 = 512+12-*
*1.
Note from Table 2.11 that the 12 = g(9, 2) is 1 greater than it would have been*
* were
it not for the stability considerations. This "1" is intimately related to ffi(*
*66) = 1.
The other delicate matter is that 2.10.(5) requires f(n) 6= f(n - 1) + 1. Th*
*is
is the cause of most of the complications. Recall that the reason that this is*
* so
important goes back to Proposition 2.9.(3), which requires that if you want to *
*utilize
IMMERSIONS OF RP2e-1 19
paired obstructions to lift from 2m + 1 to 2m, compatibly with a given lifting *
*on a
subcomplex, then that lifting must be to 2m - 1.
Now we give the definition of f.
Definition 5.4. For n 1, let f00(n) = 8n - lg(n) + ffi(n) and
f0(n) = max{f00(m) : m n}.
Note that f0 is an increasing function of n. Define s(n) and f(n) := f0(n) + s*
*(n)
inductively by f(0) = 0 and for n 1
8
<1 iff0(n) = f(n - 1) 1
s(n) = :
0 otherwise.
Note that the - and + parts of are present for different reasons. The - is
present to make f increasing, while the + occurs to prevent f(n) - f(n - 1) = 1*
*. For
example, if f(n - 2) = f0(n - 2) = A, f0(n - 1) = A + 1, and f0(n) = A + 1, then
s(n - 1) = 1 for the latter reason, yielding f(n - 1) = A + 2, and so s(n) = 1 *
*for the
former reason.
Then f is an increasing function such that f(n) - f(n - 1) 6= 1. Before provi*
*ng
that f (actually the associated g) satisfies the required properties of 2.10, w*
*e give a
few examples.
Example 5.5. Let u always denote an odd positive integer, and A an arbitrary po*
*s-
itive integer.
os(n) = 0 if n < 27+ 1, because f00(n) - f00(n - 1) 8 - (n - 1).
os(u . 27+ 1) = 1 for most odd integers u. When u = 1, this is
due to 8
>>>-15 ffl = -1
>><
-1 ffl = 0
f00(27+ ffl) = 210+ >
>>>0 ffl = 1
>:
9 ffl = 2.
However, if e 29 + 1, then s(2e+ 27 + 1) = 0 due to ffi(2e+
27+ 1) > -1.
20 DONALD M. DAVIS, GIORA DULA, AND MARK MAHOWALD
oIf k 2 and mk = max( (k) - 1, 2), then for most odd integers
u, s(u . 28k+mk+ k) = 1. When k = 2, this is due to
8
>>>-26 ffl = -1
>>>
>><-1 ffl = 0
f00(218+ ffl) = 221+ >-11 ffl = 1
>>>
>>>0 ffl = 2
>:
5 ffl = 3.
10+1 8 8
os(u . 22 + 2 + 1) = 1 for most u, using d = 2 + 1 in ffi(n) in
11+2 9 212+3 10
5.3. Similarly, s(22 +2 +1) = 1 and s(A.2 +2 +1) =
1.
20+16 17
os(22 + 2 + ffl) = 1 for 0 ffl 2, since
8
>>-1 0 t < 217
20+16 220+19 < 17 17
f0(22 + t) = 2 + >0 2 t 2 + 1
>: 17
2 t = 2 + 2.
12+8 9
oSimilarly, s(22 + 2 + ffl) = 1 for 0 ffl 1.
Now we proceed to the proof of the relevant portions of 2.10. The reader can
easily check that the formulas (5.2) and 5.4 agree with Table 2.11 in its limit*
*ed range.
Properties (1) and (2) of 2.10 are immediate from (5.2). To establish property *
*(5),
we first note that f(n) 6= f(n-1)+1 by Definition 5.4. This and (5.2) imply 2.1*
*0.(5)
with the only minor worry being about the truncation of g(e, k) at 4k + 4, but *
*this
only occurs when (k) < 4, and from this it follows easily that g(e, k +1) > g(*
*e, k)+1
in these cases. Observation of columns 2, 3, and 4 of Table 2.11 should convinc*
*e the
reader that this is true.
We now proceed to prove 2.10.(3). We first prove that it is true using f0. Fi*
*rst note
that we need not worry that f0 is a max, for the sum of the parameters which yi*
*eld
the max for the terms of the RHS of 2.10.(3) will be admissible for the evaluat*
*ion of
the LHS. It suffices to prove 2.10.(3) when e = lg(k) + 4, in which case it red*
*uces to
f00(k) f00(`) - (lg(k) + 3) + lg(`) + 4 + f00(k - `) - (lg(k) + 3) + lg(k - `*
*) + 4 - 1,
(5.6)
IMMERSIONS OF RP2e-1 21
unless g(e - 1, `) = 4` + 4 due to truncation. We will deal with this possibili*
*ty later.
Using 5.4, (5.6) reduces to
ffi(k) + lg(k) ffi(`) + ffi(k - `) + 1,(5.7)
provided 2 ` k - 2.
The case ` = 1 is easily handled. Then g(lg(k) + 3, `), which occurs in the s*
*impli-
fication to (5.6), is 0 unless lg(k) 3, and 2.10.(3) is easily verified in th*
*ese cases.
Returning to the proof of (5.7), write k = 2a + m with 0 m < 2a. Cases with
a 6 are easily checked directly, and so we assume a 7. Assume without loss *
*of
generality that (`) (k - `). If (`) < 3, then (5.7) is clearly satisfied s*
*ince its
LHS is 6, while its RHS is 5.
Now we may assume (`) 3, and hence ffi(`) = (`) - 1. If k 0 mod 8, then
(k) (k - `), and so (5.7) follows from a (`). Hence we may assume k 6 0
mod 8. Then (k - `) = (k) and ffi(k - `) - ffi(k) 2 - (-1) = 3, so that th*
*e only
way (5.7) might fail is if (`) a - 2; i.e., if ` = 2a, 2a-1, 2a-2, 3 . 2a-2,*
* or possibly
(depending on how large m is) 3 . 2a-1, 5 . 2a-2, and 7 . 2a-2. One easily veri*
*fies that
(5.7) holds in these cases. For example, if ` = 2a, then (5.7) reduces to show*
*ing
ffi(2a + m) ffi(m). This is true because any value of d in 5.3 that causes ff*
*i(m) to be
greater than its minimal value will also cause the same for ffi(2a + m). If ` =*
* 2a-1,
then (5.7) reduces to ffi(2a + m) ffi(2a-1+ m) - 1, which is true with 1 to s*
*pare.
We complete our proof of the f0-version of 2.10.(3) by considering what happe*
*ns
in the postponed case in which g(e - 1, `) = 4` + 4 due to truncation. The defi*
*nition
of the function ffi has been formulated to handle this case. We begin by illust*
*rating
with the case ` = 3, e = 15. Note that g(14, 3) is the lowest 16 in the k = 3 c*
*olumn
of Table 2.11, and is 3 larger than it would have been if the values of g(e, 3)*
* were
allowed to decrease below 16. In the context of (5.7), that would add 3 to the *
*RHS.
Since e = lg(k) + 4 in (5.6), we have lg(k) = 11. So k = 211+ t, 0 t < 211, a*
*nd we
need to verify
ffi(211+ t) + 11 ffi(3) + ffi(211+ t - 3)(+54..8)
Since ffi(3) = -1, (5.8) reduces to
ffi(211+ t) + 8 ffi(211+ t - 3). (5.9)
22 DONALD M. DAVIS, GIORA DULA, AND MARK MAHOWALD
The only way this could fail is if (211+ t - 3) 9. In Table 5.10, we tabulat*
*e the
values of both sides of (5.9) for these values of t, and see that (5.9) holds i*
*n each.
The definition of ffi has been formulated so that it will always work this way.
Table 5.10. Verification of (5.9).
____t____||ffi(211+_t)_+f8fi(211+_t_- 3)
3 || 2 + 8 10
29+ 3 || 0 + 8 8
210+ 3 || 1 + 8 9
3 . 29+ 3 || 0 + 8 8
The general case of 2.10.(3) when truncation occurs is extremely similar. Let*
* ` > 1
be arbitrary.3 The worst case occurs when e = 4` + 3, because then g(e - 1, `) *
*is the
last nonzero entry in its column. The amount of truncation is max(2 - ffi(`), 0*
*). This
is achieved from (5.2) and 5.4 as
4`+4-(f00(`)-(4`+2)+lg(`)+4) = 8`+2-(8`-lg(`)+ffi(`))-lg(`).
Since e = lg(k)+4 in (5.6), we have k = 24`-1+t with 0 t < 24`-1. The analo*
*gue
of (5.7), which we must establish, is
ffi(24`-1+ t) + 4` - 1 ffi(`) + ffi(24`-1+ t - `) + 1 + (2 - ffi(`)),
which reduces to
ffi(24`-1+ t) + 4` ffi(24`-1+ t - `) + 4.(5.11)
This inequality is easily verified, using the definition of ffi, as follows:
LHS 7 RHS if (t - `) 3
LHS 4` - 1 RHS if 3 (t - `) < 4` - 3
LHS (t - `) + 3 RHSif 4` - 3 (t - `) 4` - 1.
Note that (t - `) can be no larger than 4` - 1. The first inequality in the th*
*ird line
follows from 5.3.
Having now verified 2.10.(3) when f0 is used, we next show that it follows th*
*at
this is also valid when f is used. Because 0 f - f0 1, the principal worry *
*is to
show that if equality was attained in (5.7) or (5.11) using f0, then it cannot *
*happen
__________
3There is no truncation when ` = 1.
IMMERSIONS OF RP2e-1 23
that s = 1 on the RHS but not on the LHS, causing the inequality to fail.4 Equa*
*lity
occurs in (5.7) using f0 only when k = 2a + m, 0 m < 2a, and ` = 2a. Thus we
need to show here that
(ffi + s)(2a + m) (ffi + s)(m). (5.12)
Some typical occurrences of s(n) = 1 were given in Example 5.5 and a complete
description of these is given in Lemma 6.7. It follows from this that the only*
* way
that we can have s(m) = 1 while s(2a + m) = 0 is if ffi(2a + m) > ffi(m), as oc*
*curred
for m = 27 + 1 and a 29 + 1 in the second bullet in 5.5. In such cases, (5.12*
*) is
necessarily satisfied because of the increase in ffi. In the notation of Lemma*
* 6.7, if
m = A0 + . .+.At has s(m) = 1, then 2a + m with 2a > m can be written with
A0 replaced by 2a + A0, with (-) unchanged. Then the condition which caused
s(m) = 1 will also cause s(2a + m) = 1 unless ffi(2a + m) 6= ffi(m). However, a*
*dding a
large 2-power such as 2a cannot decrease ffi. Thus ffi(2a+ m) > ffi(m) and henc*
*e (5.12)
is satisfied.
The only cases of equality in (5.11) occur when (t - `) 4` - 4. Thus the (*
*ffi + s)-
version of (5.11) could fail only if s(24`-1+u2e) = 1 with e 4`-4 (and u2e < *
*24`-1).
But Lemma 6.7 shows that s(n) = 1 only when n has at least one long string of 0*
*'s
in its binary expansion, which is not the case for n = 24`-1+ u2e with e 4` -*
* 4 and
u2e < 24`-1. This completes the proof of 2.10.(3).
Next we prove 2.10.(4). Note that it is similar to 2.10.(3), except it is st*
*ronger
by 2. Note also from Table 2.11 that the claim is false when 2` = 2e-3, for we *
*have
g(e, 2e-3) = 2e = 2g(e - 1, 2e-4). The exclusion on the other side of 2.10.(3),*
* when
[(e+1)=4] > 2`, is because both g(e, 2`) = 0 and g(e-1, `) = 0 in this case. Si*
*milarly
to (5.7), the claim reduces to
(ffi + s)(2`) + lg(2`) 2(ffi + s)(`) +(3.5.13)
Note that for ` < 28,
8
< (`) if ` = 26+ 2, 27+ 1, or 27+ 2
(ffi + s)(`) = :
(`) - 1 otherwise.
__________
4The possibility of two cases of s = 1 on the RHS of (5.7) when the inequality
was satisfied with 1 to spare can be eliminated similarly.
24 DONALD M. DAVIS, GIORA DULA, AND MARK MAHOWALD
For these three special values of `, (5.13) is easily verified, while if (ffi+s*
*)(`) = (`)-1,
then (5.13) reduces to lg(`) (`), which is clearly true. Thus (5.13) is tru*
*e for
` < 28. The reason that this analysis didn't catch the failure of 2.10.(4) to h*
*old when
2` = 2e-3is that the analysis deals with f(-), and the values g(e, 2e-3) = 2esi*
*t above
the f-values in Table 2.11; for example, f(8) = g(7, 8) = 63, not the 64 which *
*sits
above it in the table.
If ` 28, so that lg(2`) 9, and ffi(`) < 3, then (5.13) is certainly true.*
* Here we
use 5.5 or 6.7 to see that s cannot play a significant role here; the second va*
*lue of n
with s(n) > 0 is s(218+ 2) = 1, for which lg(n) will certainly make (5.13) hold*
*. Thus,
we may assume ` 0 mod 8 and then ffi(`) = (`) - 1, and hence (5.13) reduces*
* to
lg(`) + s(2`) (`) + 2s(`). (5.14)
Note that 8
>><0 ` = 2a
lg(`) - (`) = >1 ` = 2a-1
>:
2 otherwise.
Since s(2e) = 0, we deduce that (5.14) holds.
6. Inductive determination of a bound for gd of normal bundle
In this section, we prove the following result, of which Proposition 2.16 is *
*an im-
mediate consequence.
Theorem 6.1. Let e 7 and
8
>>>0 ifn ae(e)
>><
g(e, [n_]) ifae(e) < n < 2e- ae(e)
d0e,n= > 8 n e e e
>>>max(g(e, [_8]) - 1, 2 - ae(e))if2 - ae(e) n 2 - 9
>:e e e
2 - e - 7 if2 - 8 n 2 - 1.
For t 1, gd((2e+ 2e+1+ . .+.2e+t),n) d0e,ncompatibly for n < 2e.
Remark 6.2. The all-important 2e- e - 7 arises from the bound g(e, 2e-4- 1) +
g(e + 1, 2e-4) = 2e-1- e - 5 + 2e-1- 2 for gd((2e+ 2e+1),2e-1).
Example 6.3. We illustrate the argument when e = 7. Here we have
8
>><100if104 n 111
d0e,n= >112 if112 n 119
>:
114 if120 n 127.
IMMERSIONS OF RP2e-1 25
By Proposition 2.13, we can replace the 109 and 116 in the (e = 7)-row of Table*
* 2.11
by 112 and 112. Call the values in this modified row g0(7, k). These are compat*
*ible
bounds for gd(27, k). Apply Propositions 2.8 and 2.9 to this to get a modified *
*(e = 8)-
row, for k 15, with the 108 and 115 replaced by 111 and 114. Call the values *
*in
this new row g0(8, k). The 111 for g0(8, 14) is determined by g0(7, 14) + g0(7,*
* 0) - 1 =
112 + 0 - 1, while the 114 for g0(8, 15) is determined by g0(7, 7) + g0(7, 8) -*
* 1 =
52 + 63 - 1. Now apply Proposition 2.8 to g0(7, k) and g0(8, k) to obtain compa*
*tible
bounds for gd((27 + 28),n), n 127. The value d07,119= 112 is determined by
g0(7, 14) + g0(8, 0) = 112 + 0, while d07,127= 114 is determined by g0(7, 0) + *
*g0(8, 15) =
0 + 114 or g0(7, 7) + g0(8, 8) = 52 + 62. Applying Proposition 2.8 to the d07,n*
*bounds
for gd((27 + 28),n) and the g(9, k) bounds for gd(29,8k+7) maintains the d07,nb*
*ound
for gd((27+ 28+ 29),n), and the addition of larger 2e, is handled in the same w*
*ay.
Proof of Theorem 6.1.The above example when e = 7 was slightly simpler than the
general situation because ae(7) 0 mod 8. Each value g(e, k) gives a bound for
gd(2e,i) for 8k i 8k + 7. If ae(e) 6 0 mod 8, then the skip of d0e,nat n =*
* 2e- ae(e)
occurs in the middle of one of these ranges, forcing a refinement of the filter*
*ing of
e-1 i
P 2 . It becomes convenient to filter it using all skeleta P .
The proof will proceed in five steps.
(1)Use Proposition 2.13 for gd(2e,n) for n < 2e.
(2)Use (1) and Propositions 2.8 and 2.9 to prove
8
>>>0 n ae(e + 1)
>><
g(e + 1, [n_]) ae(e + 1) < n < 2e- ae(e)
gd(2e+1,n) > 8 n e e e
>>>max(g(e + 1, [_8]), 2 - ae(e)2--1)ae(e) n 2 - 9
>:e e e
2 - e - 7 2 - 8 n 2 - 1
compatibly for n < 2e.
(3)Use (1) and (2) and Proposition 2.8 to prove
gd((2e+ 2e+1),n) d0e,n
compatibly for n < 2e.
(4)By induction on t, using (2) to get started and Propositions
2.8 and 2.9, show gd(2e+t,n) has the same bound as in (2),
26 DONALD M. DAVIS, GIORA DULA, AND MARK MAHOWALD
compatibly for n < 2e. We can actually do better than this,
but this is all we need.
(5)By induction on t, using (3) to get started and then also (4) and
2.8, show that gd((2e+ 2e+1+ . .+.2e+t),n) d0e,ncompatibly
for n < 2e, completing the proof of the theorem.
Step (1) is immediate, and steps (4) and (5) are similar to and easier than s*
*teps
(2) and (3), respectively. We now prove step (2).
For n < 2e- ae(e), this is Theorem 2.10.(6), which has already been proven. F*
*or
2e- ae(e) n 2e- 9, we have gd(2e+1,n) max{de,i+ de,n-i- 1 : 0 i n}. We
must show that each de,i+de,n-i-1 is either g(e+1, [n_8]) or 2e-ae(e)-1. For *
*those
i such that de,i= 2e- ae(e), we have de,n-i= 0, and so the desired result is tr*
*ue in
these cases. For other i, the numbers de,i- de,n-i- 1 are among those which yie*
*lded
gd(2e+1,n) g(e + 1, [n_8]) in 2.10.(6), yielding the claim in these cases. Fi*
*nally, using
(1), 2.8, and 2.9,
gd(2e+1,2e-1) max{2e-ae(e)-1, g(e, `)+g(e, 2e-3-1-`)-1 : [ae(e)_8] ` 2e-3*
*-1-[ae(e)_8]}.
We have
g(e, `) + g(e, 2e-3- 1 - `) = 2e- 2e + (ffi + s)(`) + (ffi + s)(2e-3- 1 - `),
which for 2 ` 2e-3- 3 and e 7 has maximum value of 2e- e - 6 when ` = 2e-*
*4.
We will first prove (3) using f0 instead of f, and then explain why it still *
*holds
when f is used. We wish to prove
de,i+ fgd(2e+1,n-i) d0e,n, (6.4)
where fgd(2e+1,n-i) refers to the bound given in (2).
If n < 2e- ae(e), there are two cases depending on whether or not g(e + 1, [n*
*_8]) =
g(e, [n_8]). This equality occurs only for the bottom few nonzero elements in c*
*olumns
in (the extension of) Table 2.11 for which the column number is not divisible b*
*y 8.
IMMERSIONS OF RP2e-1 27
If g(e + 1, [n_]) < g(e, [n_]), then
8 8
de,i+ fgd(2e+1,n-i)=g(e, [_i8]) + g(e + 1, [n-i_8])
g(e, [_i8]) + g(e, [n-i_8])
g(e + 1, [n_8]) + 1 (6.5)
= g(e, [n_8]) = d0e,n.
Here we used 2.10.(3) at the middle step. If, on the other hand, g(e+1, [n_8]) *
*= g(e, [n_8]),
then g(e, j) = g(e + 1, j) = 0 for all j < [n_8]. (See Table 2.11.) Thus in t*
*his case
de,i+ fgd(2e+1,n-i) g(e, [n_8]), since at least one term is 0.
Now assume 2e - ae(e) n 2e - 9. (a) If 2e - ae(e) g(e, [n_8]), then de*
*,i+
fgd(2e+1,n-i) g(e, [n_]) as in the previous paragraph, and this is 2e - ae(*
*e), as
8
claimed. (b) The case in which 2e- ae(e) < g(e, [n_8]) requires a little more a*
*rgument.
If g(e+1, [n-i_8]) < g(e, [n-i_8]), then the desired inequality follows similar*
*ly to (6.5). The
first there becomes <, and so we deduce de,i+ fgd(2e+1,n-i) g(e, [n_8]) - 1*
* = d0e,n.
If, on the other hand, g(e + 1, [n-i_8]) = g(e, [n-i_8]), then g(e, k) with k =*
* [n-i_8] must be
one of the equal bottom nonzero entries in a column k 2, with e = 4k - 1 + r *
*with
(k) r 2. Then (6.4) becomes
4k + 4 g(e, 2e-3- t) - 1 - g(e, 2e-3- t - k)(6.6)
with [n_8] = 2e-3-t. The hypothesis 2e-ae(e) < g(e, 2e-3-t) implies 8t- (t) < e*
*+5
4k + 6. This implies (t + k) lg(2k) and so the RHS of (6.6) is 8k - 1 - lg*
*(2k).
Since 4k + 4 8k - 1 - lg(2k) for k 2, (6.6) is valid, hence so is (6.4) in *
*this case.
Note that the inequalities in this paragraph are quite crude, but are all that *
*we need
here.
Finally, suppose 2e- 8 n 2e- 1. We have
de,0+ fgd(2e+1,2e-1) = 2e- e - 7.
Cases in which de,i= 2e-ae(e) have fgd(2e+1,n-i) = 0, and, since 2e-ae(e) < 2e-*
*e-7,
(6.4) is valid in these cases. In other cases, the LHS of (6.4) equals
g(e, `) + g(e + 1, 2e-3- 1 - `) = 2e- 2e - 1 + ffi(`) + ffi(2e-3- 1 - `).
The largest value of this occurs when ` = 2e-4and is 2e- e - 7.
28 DONALD M. DAVIS, GIORA DULA, AND MARK MAHOWALD
What remains is to show that incorporating positive values of s cannot affect*
* valid-
ity of the above argument. We saw in the paragraph containing (5.12) that 2.10.*
*(3),
which is the primary tool throughout this proof, is valid with s incorporated. *
*The
above argument also required that f(2e-3-1) = f0(2e-3-1); i.e., that s(2e-3-1) *
*= 0.
This is clear from Lemma 6.7, which implies that if s(n) > 0 then n has at leas*
*t one
huge gap (i.e. string of 0's) in its binary expansion, where "huge" is one wit*
*h a
number of 0's nearly eight times as large the value of the number which follows*
* it.
||
We close with a complete account of how s(n) can be nonzero in 5.4.
Lemma 6.7. Suppose s(n) = 1.
(1)If this is due to f0(n) = f(n - 1) + 1 in 5.4, then either n =
A0 + A1 with (A0) = 8A1 + ffi(n), or for some t > 1, n =
A0 + A1 + . .+.At with (A0) = 8A1 + (A1) - 1, (Ai) =
8Ai+1+ (Ai+1)-2 for 1 i < t-1 and (At-1) = 8At+ffi(n)-1.
(2)If this is due to f0(n) = f(n - 1) - 1 in 5.4, then n = n*+ B
with n* as in (1) and (At) 8B + 3.
Example 6.8. We illustrate this with the next-to-last example from 5.5. We have
20+16 17
s(22 + 2 + ffl) = 1 as follows:
20+216 17
oIf ffl = 0, it is type 6.7.(1) with A0 = 22 and A1 = 2 .
oIf ffl = 1, it is type 6.7.(2) with n* = A0 + A1 as in the case
ffl = 0 just considered, and B = 1.
20+16 17
oIf ffl = 2, it is type 6.7.(1) with A0 = 22 , A1 = 2 , and
A2 = 2.
Proof.(1) The inductive definition of s and f in 5.4, without regard for the sp*
*ecific
definition of f0, just the fact that f0 is an increasing function, implies5that*
* if s(n) = 1
due to f0(n) = f(n - 1) + 1, then there is a positive integer t and integers n0*
* < . .<.
__________
5One can formulate and prove a closely-related result about how to form from a
strictly increasing sequence of integers an increasing sequence
with si= 0 or 1 such that mi+1- minever equals 1.
IMMERSIONS OF RP2e-1 29
nt= n such that
8
<1 i = 1
f00(ni) - f00(ni-1) = :
2 2 i t.
(It must also be true that if ni-1< m < ni, then f00(m) < f00(ni).)
Consider first the case t = 1. The difference f00(m) - f00(m - 1) is at leas*
*t 5
(= 8 + (-1) - 2) unless (m - 1) 4. Thus the only way that f00(n1) - f00(n0) *
*might
equal 1 is if n0 = u . 2e with e 4 and n = n1 = u . 2e+ A1 with e = 8A1 + ffi*
*(n),
using 5.4 and ffi(n0) = e - 1. Note that A0 in the lemma equals u . 2e. The cla*
*im of
the lemma when t = 1 is thus established.
Now let t = 2. We must have n0 = A0 and n1 = A0 + A1 as in the previous
paragraph. If n = n2 = A0+ A1+ A2 with f00(n2) = f00(n1) + 2, then
8A2+ ffi(n) = ffi(n1) + 2.
This implies that ffi(n1) > 3 and hence ffi(n1) = (A1) - 1 by 5.3. This yield*
*s the
claim (A1) = 8A2 + ffi(n) - 1 of the lemma when t = 2. Note that the condition
(A0) = 8A1+ffi(A0+A1) will now be given in the more explicit form with ffi(A0+*
*A1)
replaced by (A1) - 1, since we now have the additional information that (A1) *
*> 3.
We will conclude with the case t = 3, after which the pattern for larger valu*
*es of t
will have become clear. We must have n0 = A0, n1 = A0+ A1, and n2 = A0+ A1+ A2
as in the previous paragraph. If n = n2+ A3 satisfies f00(n3) = f00(n2) + 2, th*
*en
8A3+ ffi(n) = ffi(n2) + 2.
This implies that ffi(n2) > 3 and hence ffi(n2) = (A2) - 1, and yields the cla*
*im
(A2) = 8A3 + ffi(n) - 1 of the lemma when t = 3. The condition (A1) = 8A2 +
ffi(A0+ A1+ A2) - 1 has ffi(A0+ A1+ A2) replaced by (A2) - 1.
(2) Cases of s(n) = 1 due to f0(n) = f(n - 1) - 1 must be caused by an n*
as in (1) with n = n* + B and f00(n* + i) f00(n*) for 1 i B. This can
only happen if 8B + ffi(n* + B) ffi(n*). As this implies that ffi(n*) > 3, w*
*e have
ffi(n*) = (n*) - 1 = (At) - 1. Because n* contains large gaps, we must also h*
*ave
ffi(n*+ B) 2, and hence (At) 8B + 3. ||
30 DONALD M. DAVIS, GIORA DULA, AND MARK MAHOWALD
References
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die Komposition quadratischer Formen, Comm Math Helv 15 (1942) 358-366.
[5]S. Gitler and M. Mahowald, Some immersions of real projective spaces, Bol
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[6]I. M. James, On the immersion problem for real projective spaces, Bull Amer
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Lehigh University, Bethlehem, PA 18015, USA
E-mail address: dmd1@lehigh.edu
Netanya College, Netanya 42365, Israel
E-mail address: giora@mail.netanya.ac.il
Northwestern University, Evanston, IL 60208, USA
E-mail address: mark@math.northwestern.edu