SOME NEW IMMERSION RESULTS FOR COMPLEX
PROJECTIVE SPACE
DONALD M. DAVIS
Abstract.We prove the following two new optimal immersion
results for complex projective space. First, if n 3 mod 8 but
n 6 3 mod 64, and ff(n) = 7, then CPn can be immersed in
R4n14. Second, if n is even and ff(n) = 3, then CPn can be
immersed in R4n4. Here ff(n) denotes the number of 1's in the
binary expansion of n. The first contradicts a result of Crabb,
which said that such an immersion does not exist, apparently due
to an arithmetic mistake. We combine Crabb's method with that
developed by the author and Mahowald.
1.Main theorems
We prove the following two new optimal immersion results for 2ndimensional c*
*om
plex projective space CP n.
Theorem 1.1. If n 3 mod 8 and n 6 3 mod 64, and ff(n) = 7, then CP ncan be
immersed in R4n14.
Theorem 1.2. If n is even and ff(n) = 3, then CP ncan be immersed in R4n4.
Here and throughout, ff(n) denotes the number of 1's in the binary expansion *
*of
n. Theorem 1.1 contradicts a result of Crabb ([2]). In Section 2, we prove Theo*
*rem
1.1 by an adaptation of Crabb's argument, and point out what we believe to be h*
*is
mistake, apparently arithmetic. We prove Theorem 1.2 in Section 3.
We now summarize what we feel to be the status of the immersion question for
CP n. In addition to incorporating the two new immersion results above, we list
as unresolved one immersion result claimed by Crabb. We will discuss our reason
for doing so in Section 4. Despite our feeling that two of Crabb's many results*
* are
__________
Date: February 21, 2006.
Key words and phrases. Immersions, complex projective space, obstruction
theory.
2000 Mathematics Subject Classification: 57R42, 55S35.
1
2 DONALD M. DAVIS
flawed, we feel that his overall approach, combining Ktheory with obstruction *
*theory,
is sound; we have checked the details of his immersion results cited in Theorem*
* 1.8.
Now we begin our summary.
There are three families of results that apply to all values of ff(n). All k*
*nown
nonimmersion results follow from the first two.
Theorem 1.3. ([13]) CP ncannot be immersed in R4n2ff(n)+ffl, where
8
>><0 if n is even and ff(n) 1 (mod 4)
ffl = >1 if n is even and ff(n) 2, 3 (mod 4)
>:
1 otherwise.
Theorem 1.4. ([14]) If CP nimmerses in R4n2ff(n), then
(cff(n)1) = (cff(n)) < (ci) for alli < ff(n)  1,
P i 2n+1ff(n)
where cit = ((log(1 + t))=t) .
Here and throughout, () denotes the exponent of 2 dividing an integer. The sp*
*ecific
results obtainable from Theorem 1.4 were determined for ff(n) 5 in [14] and f*
*or
ff(n) = 6 and 7 (with a mistake when ff(n) = 7) in [2]. In Section 5, we derive*
* these
for ff(n) = 8; the results have been incorporated into Tables 1.7 and 1.9.
For large values of ff(n), the best immersion results are obtained in the fol*
*lowing
embedding theorem, which relied on earlier results of Milgram.
Theorem 1.5. ([11]) CP ncan be embedded in R4nff(n), and, if n > 1 is odd, CP n
can be embedded in R4nff(n)1.
For small ff(n), better immersion results are obtained by [6] and [2]. Crabb*
* did
not consider even values of n, and so, when n is even, the immersions are from *
*[6]
and Theorem 1.2, and the nonimmersions from [13] (stated here as 1.3) and from *
*[14]
(stated here as 1.4).
Theorem 1.6. If n is even, then CP nimmerses in R4ndand does not immerse in
R4ne, where d and e are given in Table 1.7.
IMMERSIONS OF COMPLEX PROJECTIVE SPACE 3
Table 1.7. Immersions and nonimmersions when n is even
_ff(n)_________________________d___e_
2 2 3
3 4 5
4 n 6 10 (mod 16) 7 8
4 n 10 (mod 16) 7 9
5 9 10
6 10 11
7 11 13
8 n 6 22 (32) and 6 132 (256)1416
8 n 22 (32) or 132 (256)1417
> 8 14
Thus, when n is even, the only unresolved case for ff(n) 6 occurs when ff(n) *
*= 4
and n 10 mod 16.
We believe that the following tabulation of results and earliest proofs is ac*
*curate
when n is odd. Note that the case discussed in Proposition 4.1 is the only unre*
*solved
case when n is odd and ff(n) 7.
Theorem 1.8. If n is odd, then CP nimmerses in R4ndand does not immerse in
R4ne, where d and e are given in Table 1.9.
4 DONALD M. DAVIS
Table 1.9. Immersions and nonimmersions when n is odd
_ff(n)_condition_______________d__ref___e_ref_
___2__________________________3___[9]____4[14]_
3 n 1 (mod 4) 5 [12] 6[14]
___3___n___3_(mod__4)__________6__[16]___7[13]_
4 n 6 7 (mod 8) 7  [6] 8[14]
___4___n___7_(mod__8)__________8__[2]____9[13]_
5 n 1 (mod 4) 9 [6] 11[13]
5 n 3 (mod 8) 9 [6] 10[14]
___5___n___7_(mod__8)_________10__[2]____11[13]_
6 n 1 (mod 4) 11 [6] 12[14]
6 n 3 (mod 16) 11 [6] 12[14]
6 n 11 (mod 16) 12 [2] 13[13]
___6___n___7_(mod__8)_________12__[2]____13[13]_
7 n 1 (mod 4) 13 [6] 14[14]
7 n 3 (mod 64) 13 [6] 14[14]
7 n 3 (8), 6 3 (64) 14 Thm.1.115[13]
___7___n___7_(mod__8)_________14__[2]____15[13]_
8 n 6 15 (16) and 6 37 (64)15[6] 16[14]
___8___n___15_(16)_or___37_(64)15_[6]____17[13]_
__>_8_________________________15__[6]_________
2. Proof of Theorem 1.1
In this section we prove Theorem 1.1 and describe what we believe was Crabb's
mistake when he asserted a nonimmersion in this situation.
Let HCndenote the Hopf bundle over CP n. It is standard that the immersion of
1.1 is equivalent to showing that the stable normal bundle (n + 1)HCnis stably
equivalent to a bundle of dimension 2n  14. We let n = 8p + 3 with ff(p) = 5.
In [2, x3, esp. (3.2)], Crabb showed that a necessary condition for the immersi*
*on is
p __ p __
that, if ~(T ) = (sinh1( T)= T)2, and (~8p)idenotes the coefficient of T iin*
* ~(T )8p,
then there exists an integer e such that e(~8p)3 64 mod 128, e(~8p)2 0 mod *
*32,
e(~8p)1 0 mod 8, and e(~8p)0 0 mod 2. (For the reader wishing to compare wi*
*th
Crabb's notation, his l = 14, j = 3, and k = 4p + 1.)
Working mod (T 4), we have
~(T ) = 1  1_3T + 8_45T 2 4_35T 3,
IMMERSIONS OF COMPLEX PROJECTIVE SPACE 5
and
~(T )8 = 1  8_3T + 68_15T 2 1192_189T 3.
We then have
~(T )8p= 1 + u18pT + u24pT 2+ u38pT 3,
where each uiis an odd fraction. The first and last of Crabb's necessary condit*
*ions
stated above require ep 8 mod 16 and e 0 mod 2. These (and the other con
ditions) can be satisfied if and only if p 6 0 mod 8. Crabb's Lemma 3.4 makes*
* it
clear that he believed that his conditions could not be satisfied in the cases *
*which we
address here.
Now we prove that the immersion exists when p 6 0 mod 8. We use modified
Postnikov towers (MPTs), as introduced in [8] and employed in many papers such *
*as
[7], [6], and, more recently, [15]. We consider the lifting question
BgSp(16p  8)
??
q?y (2.1)
f
CP 8p+3h! HP 4p+1 ! BSp
where f classifies the stable bundle (4p + 2)HSp4p+1over the quaternionic proj*
*ective
space. The space BgSp(m) is the classifying space for symplectic vector bundle*
*s of
real geometric dimension m. It is the pullback of BO(m) and BSp over BO. We let
BgSp(16p  8) = E8 ! E7 ! . .!.E1 ! BSp
denote the MPT through dimension 16p + 6. In this range, the fiber of q is the *
*stable
stunted real projective space P16p8= RP 1=RP 16p9, whose homotopy groups in t*
*his
range are displayed in [10, Table 8.9]. We reproduce them in Diagram 2.2, index*
*ed
as ss*( P16p8), which is their dimensions as kinvariants in the MPT.
6 DONALD M. DAVIS
Diagram 2.2. Adams spectral sequence of P16p8
               
               
               
 6           r   
                
_____________________________________________________________________*
*

6           r rr  
_____________________________________________________________________*
*
 r         r  r  r  
             r   
_____________________________________________________________________*
*
 rr 
4 r       r   r   
_____________________________________________________________________*
*
 r   r    rr r  rr r    
                  
_____________________________________________________________________*
*
   r 
2 r  r rr  r rrrrrrr   
_____________________________________________________________________*
*
                  
 r r  r  rrr   r rrr r   r   
                
_____________________________________________________________________*
*
   r              
0 r  r    r       
_____________________________________________________________________*
*
16p+ 7 4 0 4 6
The obstructions for lifting from Eito Ei+1are kinvariants in Hj(Ei) corresp*
*ond
ing to dots in position (j, i) of the diagram. All cohomology groups have coeff*
*icients
in Z=2. The bulk of our work will be in proving the following result, which st*
*ates
that f lifts to the fifth stage of the MPT.
f5
Proposition 2.3. In (2.1), f factors through a map HP 4p+1!E5.
Before giving the proof of Proposition 2.3, we use it to complete the proof of
Theorem 1.1. Let ` = f5 O h : CP 8p+3! E5. To get CP 8p+1to lift to BgSp(16p 
8), we need only show that `*(k2) = 0, where k2 2 H16p+2(E5) corresponds to the
dot in position (16p + 2, 5). The diagonal line emanating from this dot sugges*
*ts,
and the computation of the MPT proves, that there is a relation in H*(E5) of the
form Sq2k2 + ak7 = 0, where k7 2 H16p7(E5) corresponds to the dot at height
5 in the initial tower, and a is a combination of Steenrod operations and Stief*
*el
Whitney classes acting on k7. Therefore, since H16p7(CP 8p+3) = 0, we must ha*
*ve
Sq2(`*(k2)) = 0 in H*(CP 8p+3). Since Sq2 acts injectively on H16p+2(CP 8p+3), *
*this
implies `*(k2) = 0, and hence CP 8p+1lifts to gBSp(16p  8).
IMMERSIONS OF COMPLEX PROJECTIVE SPACE 7
By [2, Prop.3.2], the KOtheoretic obstruction to extending this lifting over*
* CP 8p+2
is given by the conditions on ~8p described above, which are satisfied under our
hypotheses, and hence this KOtheoretic obstruction is 0. The total obstruction*
* to
this extension lies in ss16p+4( P16p8) Z=8 Z=8, depicted in Table 2.2, but*
* in this
case the total obstruction is entirely KOtheoretic, as described in [3, Prop 4*
*.6]. See
also row 3 of [3, Table 4.1], which states explicitly that the kernel of reduct*
*ion from
total obstruction to KOtheoretic obstruction is zero. From our viewpoint, the *
*non
KOtheoretic obstructions are irregular classes such as the one in position (16*
*p, 2) in
Diagram 2.2, which must be dealt with in the proof of Proposition 2.3.
Thus CP 8p+2lifts to BgSp(16p  8). Since, by Diagram 2.2, ss16p+6( P16p8) =*
* 0,
this lifting extends over CP 8p+3, as required for Theorem 1.1.
We complete the proof of Theorem 1.1 by proving Proposition 2.3. We will use *
*the
boprimary classifying spaces Bo(m) constructed in [7]. There is a map of fibra*
*tions
through dimension 2m  2
Pm  ! Pm ^ bo
?? ?
?y ??y
gBSp(m) ! Bo(m)
?? ?
?y ??y
BSp  =! BSp ,
and there are natural maps of MPTs for these fibrations. We will consider the m*
*aps
of MPTs for the following spaces over BSp.
gBSp(16p  11)! gBSp(16p  9)
?? ?
?y ??y (2.4)
Bo(16p  11) ! Bo(16p  9).
We depict in Diagram 2.5 the portion of the Adams spectral sequences in dimensi*
*ons
0 mod 4 (which is all that is relevant for maps from HP 4p+1) for P16p11, *
*P16p9,
P16p11^ bo, and P16p9^ bo, which correspond to the kinvariants for lifting*
*s to
each of the spaces in (2.4).
8 DONALD M. DAVIS
Diagram 2.5. Possible obstructions for liftings
r r
r r r
P16p11 r r P16p9 r
r r ! r
r r r r
r r r r r rr
r r r r r
______________________rr ______________________rr
16p  8 4 0 4 16p  8 4 0 4
# #
r r
r r r
P16p11^ bo r r P16p9^ bo r
r r ! r r
r r r r r r
r r r r r r r
r r r r r r r
______________________rrrr ______________________rrrr
16p  8 4 0 4 16p  8 4 0 4
i j
By [7, 1.8], nHSptlifts to Bo(m) if and only if for all i t, ( ni) (ss4*
*i( Pm ^bo)).
By standard methods, one finds
8
>>>ff(p)  1 ffl = 2
i j >><2 + ff(p) + (p)ffl = 1
( (4p+2)4p+ffl) = >
>>>ff(p) ffl = 0
>:
ff(p) + 1 ffl = 1.
We have ff(p) = 5. Thus (4p+2)HSp4p1lifts to Bo(16p11), and (4p+2)HSp4plift*
*s to
Bo(16p9). By considering the induced map of MPTs for Bo(16p11) ! Bo(16p9),
we deduce that (4p + 2)H4plifts to E5 in the MPT for Bo(16p  11). Then, since*
* all
kinvariants for gBSp(16p  11) which are relevant for HP 4pmap injectively to *
*those
of Bo(16p  11), we infer that HP 4plifts to E5 of the MPT for BgSp(16p  11). *
*We
follow this into E5 of the MPT for gBSp(16p  9). Since this MPT has no kinvar*
*iants
in dimension 16p + 4 in filtration less than 5, the map HP 4p! E5(16p  9) exte*
*nds
over HP 4p+1, establishing Proposition 2.3.
3. Proof of Theorem 1.2
Proof of Theorem 1.2.Let n = 2` with ff(`) = 3. We must show that the map
f C
CP 2`!BU !BSO which classifies the stable normal bundle (2` + 1)H2`factors
IMMERSIONS OF COMPLEX PROJECTIVE SPACE 9
through BSO(4`  4). The fiber in P4`4! BSO(4`  4) ! BSO has an ASS chart
that looks like the first four dimensions of Diagram 2.2, except that if ` is e*
*ven, 2 times
the bottom class in what appears in that chart as dimension 16p  4 equals the *
*sum
of the two dots in the box above it. We will show that the map lifts to level 3*
* in the
MPT for this fibration, and that the level3 kinvariant is in primary indeterm*
*inacy,
which implies that the lifting exists. The reason that we did not notice this r*
*esult
in [6] is apparently that we were hesitant to consider liftings to BSO(m) when *
*m is
divisible by 4 and the bundle is an odd multiple of the complex Hopf bundle. (S*
*ee
Tables 1.8 and 1.9 of [6].)
We let gBU(m) denote the classifying space for stably almost complex vector b*
*undles
of real geometric dimension m; i.e., it is the pullback of BU and BO(m) over BO.
As in [6], we use spaces gBU(m) ! Bom! Bumover BU with fibers through dimension
2m2 given by Pm ! Pm ^bo ! Pm ^bu. We need the following charts of homotopy
groups.
Diagram 3.1. Some homotopy groups
ss4`+x( P4`5^ bo) ss4`+x( P4`3^ bo)
r

r r r
 
r r ! r r
 
______________rr __________rr
4 0 2 0
# #
ss4`+x( P4`5^ bu) ss4`+x( P4`3^ bu)
r

r r ! r
  
______________rrr __________rr
4 0 2 0
i j
We also need the easy fact that, for ffl = 0, 1, 2, ( (2`+1)2`ffl) = 3, 2, 2*
*, respectively. We
use [6, 1.7b],iwhichjstates that, if p is odd, then pHCnlifts toiBomifjand only*
* if for all
i n, ( pi) (ss2i( Pm ^ bu)), and, for all even i n, ( pi) (ss2i( Pm*
* ^ bo)).
This implies that our map f : CP 2`! BU lifts to Bo4`3, and fCP 2`1lifts to *
*Bo4`5.
Thus f lifts to level 3 in the MPT for Bo4`5. By [10, Tables 8.4,8.12], for **
* 4`,
10 DONALD M. DAVIS
ss*( P4`5) ! ss*( P4`5^ bo) is surjective with kernel consisting of a single *
*class in
* = 4`  1. Since H4`1(CP 2`) = 0, we conclude that f lifts to level 3 in the *
*MPT
for gBU(4`  5), and hence also in those for gBU(4`  4) and BSO(4`  4), using*
* the
maps of MPTs induced by gBU(4`  5) ! gBU(4`  4) ! BSO(4`  4).
We now employ a standard indeterminacy argument, as explained clearly in [15],
to show that the final obstruction in H4`(E3) can be varied, if necessary. Let*
* Ei
denote the spaces in the MPT of BSO(4`  4) ! BSO. The fiber F of E3 ! E2 is
K4`4x K4`2x K4`1x K4`1, corresponding to elements at height 2 in Diagram 2.*
*2,
desuspended once. Here Ki = K(Z=2, i). If f3 : CP 2`! E3 sends the kinvariant
k3 2 H4`(E3) trivially, then the lifting to BSO(4`  4) exists, since there are*
* no more
evendimensional kinvariants. If f*3(k3) 6= 0, then we will show that the comp*
*osite
'4`4xf3 ~
CP 2`! F x E3 !E3, (3.2)
which is also a lifting of f, sends k3 to 0, and hence the lifting to BSO(4`  *
*4) exists.
Here ~ denotes the action of the fiber on the total space in the principal fibr*
*ation, and
'4`4: CP 2`! F is the map which is nontrivial into the first factor of F . Thi*
*s will
follow because a computation of the relations in the MPT, performed below, shows
that
~*(k3) = 1 x k3+ Sq1'04`1x 1 + Sq2'4`2x 1 + '4`2x w2
+ Sq4'4`4x 1 + '4`4x (w4+ w22). (3.3)
Thus the composite (3.2) sends k3 to
f*3(k3) + Sq4(x2`2) + x2`2. (w4+ w22)((2` + 1)H).
Here x denotes the generator of H2(CP 2`). Since w2((2`+1))H = x, and Sq4(x2`*
*2)
and w4((2` + 1)) are either both nonzero (` even) or both zero (` odd), we ded*
*uce
that f*3(k3) can be varied, if necessary, establishing the lifting.
We conclude the proof by listing the relations in the MPT of BSO(4`4) ! BSO,
the last of which yields the crucial fact (3.3). These are computed by the met*
*hod
initiated in [8] and utilized in such papers as [6], [12], and [15]. It is a m*
*atter of
building a minimal resolution using MasseyPeterson algebras. In this table, ff*
*l = 1 if
` is even, and 0 if ` is odd.
IMMERSIONS OF COMPLEX PROJECTIVE SPACE 11
___________________________________________________________
 w4`3 
 w 
 4`2 
 w 
__4`______________________________________________________
 k1 :Sq1 w 
 4`3 4`3 
 k1 :(Sq2+w )w 
 4`2 2 4`3 
 k1 :(Sq2+w )w 
 4`1 2 4`2 
 k1 : Sq1 w + (Sq2+w ) Sq1w 
 4` 4` 2 4`2 
 ek1: ffl Sq1w + (Sq4+w )w + (w Sq1 +w )w 
___4`_________4`________4__4`3____2______3__4`2_________ 
 k2 :Sq1 k1 
 4`3 4`3 
 k2 :(Sq2+w )k1 + (Sq3+w )k1 
 4`1 2 4`2 3 4`3 
 k2 : Sq1 k1 + (Sq2+w )k1 
 4` 4` 2 4`1 
 ek2: Sq1 ek1+ (Sq2Sq 1+w )k1 + (Sq4+w + w2 + w Sq2)k1 
___4`_______4`___________3__4`2________4___2____2_____4`3_
 k3 :Sq1 k2 
 4`3 4`3 
 k3 : Sq1 ek2+ (Sq2+w )k2 + (Sq4+w + w2)k2 
__4`________4`_______2__4`1________4___2__4`3___________
4. Discussion of one of Crabb's proofs
In [2, 0.2], Crabb presented many new immersions of complex projective spaces.
About his proof, he wrote "Details will be omitted," although sketched arguments
for each case were presented. We have checked the details of his arguments, and
found what appears to be a flaw in one case. This case was of particular intere*
*st to
us, because, if true, it would have implied a new immersion result for real pro*
*jective
space which would be an addition to [5]. We present here our analysis of this c*
*ase.
Proposition 4.1. The argument for the portion of [2, 0.2] which states that if k
is even and ff(k) = 4 then CP 2k+1immerses in R8k6is invalid. The question of
whether this immersion exists is unresolved.
Proof.Let k = 2K. To obtain the immersion, one must prove that (4K + 2)HC4K+1
lifts to gBSp(8K 8). The KOtheoretic obstruction for lifting this bundle, cal*
*culated
similarly to the one in Section 1, is 0. However, there are several elements in*
* the kernel
of the reduction from the total obstruction to the KOtheoretic obstruction whi*
*ch
cannot be ruled out. In [3, Table 4.1], these are the three Z=2's in row 1, col*
*umn 9.
In our Diagram 2.2, which, after a slight reindexing, serves as the obstruction*
*s for
12 DONALD M. DAVIS
this lifting question as well as the one considered in Section 1, these corresp*
*ond to
three of the many dots in column 16p + 2.
Crabb realized that these could cause a problem, and so he hoped to utilize t*
*he
factorization through HP 2K, similarly to what we did in Section 1. In fact, he*
* wrote
in his proof on page 166 that in several cases, including this one, it can be s*
*hown that
the bundle over the quaternionic projective space is stably equivalent to a bun*
*dle of
the desired dimension. In this case, he would be saying that (2K + 1)HSp2Klift*
*s to
BgSp(8K  8). However, this lifting does not exist; its KOtheoretic obstructi*
*on is
nonzero.
This can be seen similarly to our calculation in Section 1. We use [2, Prop 2*
*.5].
With ~(T ) as in Section 1, the necessary condition is that there exists an int*
*eger e
such that e(~4K1)2 16 mod 32, e(~4K1)1 8 mod 16, and e(~4K1)0 0 mod 2.
Since
(1  1_3T + 8_45T 2)4K1 = 1 + od. T + od. T 2
mod (T 3), where od denotes an odd fraction, we require e to satisfy both e 1*
*6 mod
32, and e 8 mod 16, which is clearly impossible.
We close by commenting on the relationship between Crabb's necessary conditio*
*ns
p __ p __
for immersion involving powers of ~(T ) = (sinh1( T)= T)2, which we have used
above, and the SigristSuter necessary condition involving powers of log(1 + t)*
*=t in
Theorem 1.4. These conditions involving power series can be directly related to*
* one
another by a slight extension of [1, 1.5], which was proved using a proof sugge*
*sted to
the authors by Crabb, and based on his earlier topological work described in [4*
*, x4].
5. Evaluation of a SigristSuter condition
In Theorem 1.4, a general statement of a necessary condition for CP nto immer*
*se
in R4n2ff(n)is presented. We evaluate this explicitly when ff(n) = 8 in the fo*
*llowing
result, which we have incorporated into Tables 1.7 and 1.9.
Proposition 5.1. If ff(n) = 8 and CP nimmerses in R4n2ff(n), then (n  7) = 3
or (n  6) = 4 or (n  5) = 5 or (n  4) = 7.
This follows readily from Theorem 1.4 and the following lemma.
IMMERSIONS OF COMPLEX PROJECTIVE SPACE 13
Lemma 5.2. Let (log(1 + t)=t)m = P cm,iti. Then
(cm,7) = (cm,8) < (cm,i) for alli < 7
if and only if (m  7) = 4 or (m  5) = 5 or (m  3) = 6 or (m  1) = 8.
Proof.Let
v(m) = (v0(m), . .,.v1(m)) = ( (cm,0), . .,. (cm,8)).
Also define a function, for which some of its values are only specified to sati*
*sfy an
inequality, by 8
< (k) if (k) < e
(k, e) = :
e if (k) e.
The lemma follows immediately from the following, which we will prove for v(m) *
*by
induction on m. Note that some components are only asserted to satisfy an inequ*
*ality.
(1)If e 3, then v(2e(2a + 1)) = (0, e  1, e  3, e  3, e  6, e 
5, e  7, e  7, e  11).
(2)v(7 + 8k) = (0, 1, 2, 3, 4, 5, 6, 7, 8 + (k, 2)).
(3)v(5 + 8k) = (0, 1, 1, 1, 4, 5, 4, 6, 8 + (k, 3)).
(4)v(3 + 32k) = (0, 1, 2, 3, 1, 4, 4, 5, 6 + (k, 2)).
(5)v(19 + 32k) = (0, 1, 2, 3, 2, 4, 3, 4, 7).
(6)v(11 + 16k) = (0, 1, 2, 3, 3, 3, 5, 6, 8).
(7)v(9 + 16k) = (0, 1, 1, 2, 3, 4, 4, 3, 8).
(8)v(17 + 32k) = (0, 1, 0, 2, 2, 3, 3, 2, 7).
(9)v(1+32k) = (0, 1, 0, 2, 1, 2, 2, 3, 6+ (4, k)).
We begin by using Maple to verify v(m) for m = 8, 7, 5, 3, 19, 11, 9, 17, and*
* 1.
The induction proof for v(2e) is obtained using
X X
c2e+1,iti= ( c2e,iti)2.
We have
vi(2e+1) = 1 + vi(2e), (5.3)
since, as is easily verified, the RHS of (5.3) is strictly less than 1 + vj(2e)*
* + vij(2e)
for 1 j i=2.
14 DONALD M. DAVIS
Next we obtain the claim for v(2e(2a + 1)) from (P c2e,iti)(P c2e+1a,iti). Fr*
*om this,
we obtain vi(2e(2a + 1)) = vi(2e) since the strict minimum of vj(2e) + vij(2e+*
*1a) is
obtained when j = i, essentially the same verification as the previous one.
The cases in which the asserted vi(u + 2ek) is not of the form t or (k, t)*
* are,
except for v7(1+32k) with k odd, obtained from vi(u)+v0(2ek) which, in these ca*
*ses,
is strictly less than other values of vj(u) + vij(2ek). Quite a few verificat*
*ions are
required for this. For example, since v4(2ek) = e + (k)  6, it is relevant th*
*at
8
>><3u = 7, 5
vi(u) < vi4(u)  >2 u = 11, 9
>:
1 u = 3, 19, 17, 1
in the cases in which the asserted value of vi(u) is a single integer. These ca*
*ses are
differentiated according to whether the argument of v is u + 8k, u + 16k, or u *
*+ 32k.
If k is odd, then v7(1 + 32k) = 3 comes from having each of v7(1) + v0(32k),
v3(1) + v4(32k), and v1(1) + v6(32k) equal 3.
In the cases in which the asserted vi(u + 2ek) is of the form t, one verifi*
*es, unless
i = 7 and u = 9 or 17, that vj(u) + vij(2ek) t for all j, with the possibili*
*ty of
equality for several values of j. For example, v6(5 + 8k) 4 comes from v6(5)*
* +
v0(8k) 4, v4(5) + v2(8k) 4, and v2(5) + v4(8k) 4, while vj(5) + v6j(8*
*k) >
4 for j = 0, 1, 3, and 5. If k is odd, the exceptional case v7(9 + 16k) 3 c*
*omes
from v1(9) + v6(16k) = 4 and v3(9) + v4(16k) = 4, with other values 3. The
case v7(17 + 32k) is similar.
Finally, for (k) = 0 or 1, v8(7 + 8k) = 8 + (k) comes from v0(7) + v8(8k),
which is strictly less than all other vj(7) + v8j(8k), while if (k) 2, we h*
*ave
v0(7) + v8(8k) = (k)  8 and v8(7) + v0(8k) 6 with other terms larger. The *
*same
argument works for v8(5 + 8k), v8(3 + 32k), and v8(1 + 32k).
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Department of Mathematics, Lehigh University, Bethlehem, PA 18015, USA
Email address: dmd1@lehigh.edu