0 and assume the result for smaller values of l. We use induction*
* on n
to prove the inequality in Theorem 3.3.
The case n = 0 is trivial. So we now let n > 0 and assume that the inequality
holds with smaller values of n. Observe that
_ ! _ ! l
X n k - r
(-1)k _____ff
k r (mod pff)k p
_ _ ! _ ! ! _ ! l
X n - 1 n - 1 k - r
= + (-1)k _____ff
k r (mod pff) k k - 1 p
_ ! _ ! l
X n - 1 k - r
= (-1)k _____ff
k r (mod pff)k p
_ ! _ ! l
X n - 1 0 k0- (r - 1)
- 0 (-1)k __________ff.
k0 r-1(mod pff)k p
In view of this, if pffdoes not divide n, then, by the induction hypothesis for*
* n - 1,
we have
_ _ ! _ ! l!
X n k - r
ordp (-1)k _____ff
k r (mod pff)k p
_$ %! _$ %!
n - 1 n
ordp _____ = ordp __ .
pff pff
Below we let pff| n and set m = n=pff.
A NUMBER-THEORETIC APPROACH TO HOMOTOPY EXPONENTS OF SU(n) 11
Case 1. r 0 (mod pff). In this case,
_ ! _ ! _ ! l-1
_1_ X n k k r k - r
(-1) __- __ _____
m! k r (mod pff)k pff pff pff
_ ! _ ! l-1
n=pff X n - 1 k k - r
= _____ (-1) _____
m! k r (mod pff)k - 1 pff
_ ! _ ! l-1
r=pff X n k k - r
-____ (-1) _____
m! k r (mod pff)k pff
_ ! _ !l-1
1 X n - 1 k+1 k - (r - 1)
= ____________ (-1) __________
b(n - 1)=pffc!k r-1(mod pff)k pff
_ ! _ ! l-1
r=pff X n k k - r
-_______ (-1) _____ .
bn=pffc!k r (mod pff)k pff
Thus, by the induction hypothesis for l - 1,
_ ! _ ! l
_1_ X n k k - r
(-1) _____
m! k r (mod pff)k pff
is a p-integer (i.e., its denominator is relatively prime to p) and hence the d*
*esired
inequality follows.
P i0j i
Case 2. r 6 0 (mod pff). Note that i r (mod pff)i(-1) = 0. Also,
_ _ ! ! _ !
X n n
ordp (-1)k ordp ____ff-1!= m + ordp(m!)
k r (mod pff)k p
by Lemma 3.1. Thus, in view of Lemma 3.2, it suffices to show that if 0 < j < n*
* then
the p-adic order of
_ ! _ ! _ ! _ !
n X j i X n - j - 1 k k - rj
(-1) (-1) f ______
j pff|i-ri pff|k-rj k pff
is at least ordp(m!), where rj = r - j + pff- 1 and f(x) = xl.
Let 0 < j n - 1 and write j = pffs + t, where s, t 2 N and t < pff. Note th*
*at
$ % $ % $ %
j_ n - j - 1 t + 1
= s and ________ = m - s - _____= m - s - 1.
pff pff pff
12 DONALD M. DAVIS AND ZHI-WEI SUN
Since f(x) = (x+1)l-xl= P l-1ilji
i=0i x , by Lemma 3.1 and the induction hypothesis
we have __ ! _ ! _ ! _ !!
n X j i X n - j - 1 k k - rj
ordp (-1) (-1) f ______
j pff|i-ri pff|k-rj k pff
_ ! _ _ ! !
n X j i
= ordp + ordp (-1)
j i r (mod pff)i
_ _ ! _ !!
X n - j - 1 k - rj
+ ordp (-1)k f ______ff
k rj(mod pff) k p
_ !
n
ordp + (s + ordp(s!)) + ordp((m - s - 1)!)
j
_ !
n
= ordp + s + ordp(s!) - ordp(m(m - 1) . .(.m - s)) + ordp(m!)
j
_ ! _ !
pffm m
= ordp - ordp + s - ordp(m - s) + ordp(m!).
pffs + t s
Observe_that! _ !
pffm Xff pffm pffs pff(m - s)
ordp = ____ - ___- _________
pffs i=1 pi pi pi
1X _ $pffm% $pffs% $pff(m - s)%!
+ ____i- ___i- _________i
i=ff+1 p p p
X1 _ $m % $s % $m - s%! _m !
= __i- __i- _____i = ordp .
i=1 p p p s
Define ordp(a=b) = ordp(a) - ordp(b) if a, b 2 Z and a is not divisible by b. T*
*hen
_ ! _ !
pffm m
ordp - ordp
pffs + t s
ipffmj
ffs+t (pffs)!(pff(m - s))!
= ordp_p____ipffmj= ordp______________________ffff
pffs (p s + t)!(p (m - s) - t)!
pff(m - s) Y pff(m - s) - i
= ordp_________+ ordp ____________.
pffs + t 0* 2, hence n=2 + 1 is even and*
* not
larger than n - 1. As first noted in [1, 1.1] and restated in [9, 6.5], for k =*
* 2L + n - 1,
d3 : E1,2k+13! E4,2k+33is nonzero if and only if
e2(n, 2L + n - 1) = e2(n - 1, 2L + n - 1) + n - 1.
We show at the end of the section that
e2(n - 1, 2L + n - 1) = ord2((n - 1)!). (4.1)
Thus, if the above d3 is nonzero, then e2(n, 2L + n - 1) = n - 1 + ord2((n - 1)*
*!) and
hence
exp2(SU (n)) e2(n, 2L+n-1)-1 = n-1+ord2((n-1)!)-1 n-1+ord2(bn=2c!),
as claimed in Theorem 1.1.
Proof of (4.1).Putting p = 2, ff = L, l = h = 1 and m = n - 1 in the first part*
* of
Theorem 1.3, we get that
i j
ord2 (n - 1)!S(n - 1, n - 1) - (n - 1)!S(2L + n - 1, n - 1)
`~n - 1 '
n - 1 + ord2 _____ ! n - 1 > ord2((n - 1)!).
2
Therefore ord2((n - 1)!S(2L + n - 1, n - 1)) = ord2((n - 1)!). On the other han*
*d, by
the second part of Theorem 1.3, ord2(m!S(2L + n - 1, m)) n - 1 + ord2(bn=2c!)*
* for
all m n. So we have (4.1). ||
5.Strengthening and sharpness of Theorem 3.3
In this section, we give an example illustrating the extent to which Theorem *
*3.3 is
sharp when r = 0, which is the situation that is used in our application to top*
*ology.
Then we show in Theorem 5.1 that the lower bound in Theorem 3.3 can sometimes
be increased slightly.
We begin with a typical example of Theorem 3.3. Let p = ff = 2, r = 0 and
n = 100. Then bn=pffc = 25 and ordp(bn=pffc!) = 22. For l 25, set
_ _ ! _ !l!
X n k
ffi(l) = ord2 __ - 22.
k 0(mod 4)k 4
A NUMBER-THEORETIC APPROACH TO HOMOTOPY EXPONENTS OF SU(n) 15
The range l bn=pffc = 25 is that in which we feel Theorem 3.3 to be very stro*
*ng.
(See Remark 5.3(2).) Clearly ffi(l) measures the amount by which the actual p-a*
*dic
order of the sum in Theorem 3.3 exceeds our bound for it. The values of ffi(l)*
* for
25 l 45 are given in order as
0, 0, 0, 0, 2, 3, 2, 4, 1, 1, 1, 1, 2, 2, 4, 1, 0, 0, 0, 0, 3.
When r = 0 and in many other situations, Theorem 3.3 appears to be sharp for
infinitely many values of l.
Before presenting our strengthening of Theorem 3.3 we need some notations. For
a 2 Z and m 2 Z+, we let {a}m denote the least nonnegative residue of a modulo
m. Given a prime p, for any a, b 2 N we let op(a, b) represent the number of ca*
*rries
occurring in the addition of a and b in base p; actually
1X_ $a + b% $a % $b %! _a + b!
op(a, b) = _____i- __i- __i = ordp
i=1 p p p a
as observed by E. Kummer.
Here is our strengthening of Theorem 3.3. The right hand side is the amount by
which the bound in Theorem 3.3 can be improved. This amount does not exceed ff,
by the definition of op. In Table 2, we illustrate this amount when p = 3 and f*
*f = 2.
Theorem 5.1. Let p be a prime, and let ff, l, n 2 N. Then, for all r 2 Z, we ha*
*ve
_ _ ! _ ! l! _$ % !
X n k - r n
ordp (-1)k _____ff - ordp __ff!
k r (mod pff)k p p
_ !
{r}pff+ {n - r}pff
op({r}pff, {n - r}pff) = ordp .
{r}pff
Proof.We use induction on n.
In the case n = 0, whether r 0 (mod pff) or not, the desired result holds *
*trivially.
Now let n > 0 and assume the corresponding result for n - 1. Suppose that
op({r}pff, {n - r}pff) > 0. Then neither r nor n - r is divisible by pff.
Set _ ! _ ! l
1 X n k k - r
R = _______ (-1) _____
bn=pffc!k r (mod pff)k pff
16 DONALD M. DAVIS AND ZHI-WEI SUN
and
_ ! _ ! l
n=pff X n - 1 k k - (r - 1)
R0= _______ (-1) __________ .
bn=pffc!k r-1(mod pff)k pff
Clearly
_ ! _ ! l
n=pff X n - 1 k k - r
R0 = -_______ (-1) _____
bn=pffc!k r (mod pff)k - 1 pff
_ ! _ !l
1 X n k k k - r
= -_______ (-1) __ _____ ,
bn=pffc!k r (mod pff)k pff pff
and thus
_ ! _ ! l+1
r_ 0 1 X n k k - r
R + R = -_______ (-1) _____ .
pff bn=pffc!k r (mod pff)k pff
This is a p-integer by Theorem 3.3; therefore ordp(rR + pffR0) ff.
Let fi = ordp(n). We consider three cases.
Case 1. fi ff. In this case, bn=pffc!=(n=pff) = b(n - 1)=pffc! and hence R0*
*is a
p-integer by Theorem 3.3. In view of the inequality ordp(rR + pffR0) ff, we h*
*ave
ordp(R) ff - ordp(r) = op({r}pff, {n - r}pff),
where the last equality follows from the definition of op and the condition n *
* 0 6
r (mod pff).
Case 2. ordp(r) fi < ff. Since bn=pffc = b(n - 1)=pffc, the definition of R*
*0implies
that
_ ! _ ! l
pffR0_ 1 X n - 1 k k - (r - 1)
= ____________ (-1) __________ .
n b(n - 1)=pffc!k r-1(mod pff)k pff
Applying the induction hypothesis, we find that
ordp(pffR0)-fi op({r-1}pff, {n-1-(r-1)}pff) = op({r-1}pff, {n-r}pff).
Since {r}pff+ {n - r}pff n 6 0 (mod pff) and
_ ! _ !
{r}pff+ {n - r}pff {r}pff+ {n - r}pff{r}pff+ {n - r}pff- 1
= ________________
{r}pff {r}pff {r}pff- 1
_ !
{r}pff+ {n - r}pff{r - 1}pff+ {n - r}pff
= ________________ ,
{r}pff {r - 1}pff
A NUMBER-THEORETIC APPROACH TO HOMOTOPY EXPONENTS OF SU(n) 17
we have
op({r}pff, {n - r}pff) = op({r - 1}pff, {n - r}pff) + fi - ordp(r).
Thus
ordp(pffR0) ordp(r) + op({r}pff, {n - r}pff).
Clearly op({r}pff, {n - r}pff) ff - ordp(r) by the definition of op, so we al*
*so have
ordp(rR + pffR0) ordp(r) + op({r}pff, {n - r}pff).
Therefore
ordp(R) = ordp(rR) - ordp(r) op({r}pff, {n - r}pff).
Case 3. fi < min{ff, ordp(r)}. In this case, ordp(~r) = fi < ff where ~r= n -*
* r. Also,
_ ! _ ! l _ ! _ !l
X n k - ~r X n r - (n - k)
(-1)k _____ff = (-1)k __________ff
k ~r(mod pff)k p n-k r (mod pff)k p
_ ! _ ! l
X n k - r
= (-1)l+n (-1)k _____ff.
k r (mod pff)k p
Thus, as in the second case, we have
_ _ ! _ ! l!
1 X n k k - ~r
ordp(R) = ordp _______ (-1) _____
bn=pffc!k ~r(mod pff)k pff
op({~r}pff, {n - ~r}pff) = op({r}pff, {n - r}pff).
The induction proof of Theorem 5.1 is now complete. ||
The following conjecture is based on extensive Maple calculations.
Conjecture 5.2. Let p be any prime. And let ff, l 2 N, n 2 Z+ and r 2 Z. Then
equality in Theorem 5.1 is attained if l bn=pffc and
$ % $ %
r n - r i blog(n=pff)cj
l __ + _____ mod (p - 1)p p .
pff pff
Remark 5.3. (1) The conjecture, if proved, would show that Theorem 5.1 would be
optimal in the sense that it is sharp for infinitely many values of l.
(2) Note that the conjecture only deals with equality when l bn=pffc. For s*
*maller
values of l, our inequality is still true, but not so strong. In [14], we obtai*
*n a stronger
inequality when l < bn=pffc.
18 DONALD M. DAVIS AND ZHI-WEI SUN
We close with a table showing the amount by which the bound in Theorem 5.1
improves on that of Theorem 3.3. That is, we tabulate op({r}pff, {n - r}pff) w*
*hen
p = 3 and ff = 2.
Table 2. Values of o3({r}9, {n - r}9)
{r}9
|0 1 2 3 4 5 6 7 8 |
__|_______________________|
0 ||02 2 1 2 2 1 2 2 ||
1 ||00 2 1 1 2 1 1 2 ||
2 ||00 0 1 1 1 1 1 1 ||
{n}9 3 ||01 1 0 2 2 1 2 2 ||
4 ||00 1 0 0 2 1 1 2 ||
5 ||00 0 0 0 0 1 1 1 ||
6 ||01 1 0 1 1 0 2 2 ||
7 ||00 1 0 0 1 0 0 2 ||
_8_||00_0__0__0_0__0_0__0_||
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Department of Mathematics, Lehigh University, Bethlehem, PA 18015, USA
E-mail address: dmd1@lehigh.edu
Department of Mathematics, Nanjing University, Nanjing 210093, People's Repub-
lic of China
E-mail address: zwsun@nju.edu.cn
*