# Relations in the Dyer-Laslof Algebra for Morava E-theory Rose- Hulman ```RoseHulman
Mathematics
Journal
Relations in the Dyer-Laslof
Algebra for Morava E-theory
Louis Atsaves
a
Volume 13, No. 2, Fall 2012
Rose-Hulman Institute of Technology
Department of Mathematics
Terre Haute, IN 47803
Email: [email protected]
http://www.rose-hulman.edu/mathjournal
a Massachusetts
Institute of Technology
Volume 13, No. 2, Fall 2012
Relations in the Dyer-Laslof Algebra for
Morava E-theory
Louis Atsaves
Abstract. In this paper, we prove that a map u between two polynomial rings, each with an associated Adem relation, is injective. We prove
injectivity of u, by first finding formulas for elements within each ring
polynomial, and then by computing the map with our associated formulas. After having computed the mapping of u, we then use our computations to show that the kernel of u only contains the zero vector, which
proves that the map u is injective. Then having proved that the map u
is injective, we then use it to find a basis for u∗ , the dual map of u.
Acknowledgements: I would like to thank Mark Behrens for his mentoring and support
while I took part in this research project. I would also like to thank Julie Wayzani.
RHIT Undergrad. Math. J., Vol. 13, No. 2
Page 176
1
Introduction
Morava E-theory is a cohomology theory in algebraic topology which was first introduced
by Jack Morava in 1970. Morava E-theory is defined as follows: For every prime number p, there is a theory E(n), for n ≥ 0, each of which is a ring spectrum. Let k be
a perfect field of characteristic p, and suppose we are given a formal group f of height
n over k. The universal deformation of f is classified by the Labirinth-Tate ring R =
W (k)[[v1 , &middot; &middot; &middot; , vn−1 ]]. The sequence v0 = p, v1 , &middot; &middot; &middot; , vn−1 is regular by construction and vn
has invertible image in R/(v0 , v1 , &middot; &middot; &middot; , vn−1 ) ' k, because of our original assumption that
the formal group law has height n. We can construct an even periodic spectrum E(n) with
π∗ E(n) ' W (k)[[v1 , &middot; &middot; &middot; , vn−1 ]][β &plusmn;1 ], where β has degree 2. The cohomology theory E(n) is
called the Morava E-theory.
In , Charles Rezk computes the relations in the mod p Dyer-Laslof Algebra for height
2 Morava E-theory. This amounts to computing the kernel of the dual of a certain map of
rings:
u : A2 → A1,1
Rezk uses elliptic curves. Here we use an elementary computation of the map u and the
kernel of u∗ . Let p be an odd prime and let k = Z/p. Let A2 and A1,1 be polynomial rings
such that
A2 =
k[x0 ,x1 ]
(x0 p2 −x1 )(x0 p −x1 p )(x0 −x1 p2 )
and A1,1 =
k[x0 ,x1 ,x2 ]
(x2 p+1 +x1 p+1 −x1 x2 −x1 p x2 p =0 , x1 p+1 +x0 p+1 −x1 x0 −x1 p x0 p =0)
In section 2, we will compute the map u in terms of explicit bases for A1,1 and A2
(Proposition 1). In section 3, we will use these explicit formulas to rederive Rezk’s relations
(i.e. compute the kernel of u∗ , Theorem 1).
2
Computation of the map u
Let u be the ring map
A2
u
→ A1,1
such that u (x0 ) = x0 and u (x1 ) = x2 .
By the definition of A2 :
A2 =
(x0
p2
k[x0 , x1 ]
− x1 )(x0 p − x1 p )(x0 − x1 p2 )
RHIT Undergrad. Math. J., Vol. 13, No. 2
Page 177
so then we have that
2
2
(x0 p − x1 )(x0 p − x1 p )(x0 − x1 p ) = 0
which can be expanded as
−x1 p
2 +p+1
+x0 p
2 +p+1
2 +p
+x1 p
2
2 +p
x0 p −x0 p
2
x1 p +x1 p
2 +p
2 +1
x0 p −x0 p
x1 p +x1 p+1 x0 −x0 p+1 x1 = 0
and then rewritten as
x1 p
2 +p+1
2 +p+1
= (−x0 p
)+(x0 p+1 )x1 +(x0 p
2 +1
)x1 p −(x0 )x1 p+1 +(x0 p
2 +p
2
2
2 +p
)x1 p −(x0 p −x0 p )x1 p
.
2
This shows that a basis as a k[x0 ] module of A2 consists of 1, x1 , x1 2 , . . . , x1 p +p since every x1 q q≥p2 +p+1 may be expressed (according to the above relation) into powers of x1 smaller
than p2 + p + 1.
Similarly, by the definition of A1,1 :
A1,1 =
x2
p+1
+ x1
p+1
− x1 x2 −
k[x0 , x1 , x2 ]
= 0, x1 p+1 + x0 p+1 − x1 x0 − x1 p x0 p = 0
x1 p x 2 p
so then we have that
x2 p+1 + x1 p+1 − x1 x2 − x1 p x2 p = 0 , x1 p+1 + x0 p+1 − x1 x0 − x1 p x0 p = 0
which can be rewritten as
x2 p+1 = −x1 p+1 + x1 x2 + x1 p x2 p , x1 p+1 = −x0 p+1 + x1 x0 + x1 p x0 p .
(2.1)
If an element in A1,1 contains an x2 q or x1 q where q ≥ p + 1, then it must be reduced into
powers of x1 and x2 which are smaller than p + 1, according to (2.1). This then shows that
a basis for A1,1 consists of all x1 i x2 j where 0 ≤ i, j ≤ p.
u(x1 k ) = u(x1 )k = x2 k



x2 k
if 0 ≤ k ≤ p
p
p
X
X
=
lijk (x0 )x1 i x2 j if k ≥ p + 1.


i=0 j=0
The lijk ’s are explicitly found using (2.1), the modding out relations of A1,1 .
RHIT Undergrad. Math. J., Vol. 13, No. 2
Page 178
Proposition 1.
x1 np+k = An,k (x0 ) + Bn,k (x0 )x1 k + Cn,k (x0 )x1 p
where

n
X
2



−
x0 (n−j)p +(kp+j) if k ≥ 1, n ≥ 2


An,k =
j=1
n−1
X





−
Bn,k =
n
X
x0
(n−j)p2 +j
(2.3)
if k = 0, n ≥ 2
j=1
Cn,k =
if 1 ≤ n ≤ p and 0 ≤ k ≤ p (2.2)
x0 n if k ≥ 1, n ≥ 2
0
if k = 0, n ≥ 2.
x0 (n−j)p
2 +kp+(j−1)
if k ≥ 0, n ≥ 2
(2.4)
(2.5)
j=1
n−1
X
x2 np+k = Ãn,k (x0 )x1 p x2 p +
x0 k+(j−1)p x1 n−j x2 p + B̃n,k (x0 )x2 p + x1 n x2 k + C̃n,k (x0 )x1 p
j=1
−
n
X
x0 k+(j−1)p x1 n−j+1 + D̃n,k (x0 )
if 1 ≤ n ≤ p and 0 ≤ k ≤ p
(2.6)
j=1
where Ãn,k =
k−1
X
l=0
n&middot;x0
(n−1)p+(k−1)+l(p2 −1)
p−1
n−2 X
X
2
+
(n−j−1)x0 (n−1)p+(k−1)+(k+l+jp)(p −1) if k ≥ 0, n ≥ 2
j=0 l=0
(2.7)
B̃n,k = −x0 &middot; Ãn,k + x0 (n−1)p+k
C̃n,k = −xp0 Ãn,k
D̃n,k = x0 p+1 Ãn,k
if k ≥ 0, n ≥ 2
if k ≥ 0, n ≥ 2
if k ≥ 0, n ≥ 2
(2.8)
(2.9)
(2.10)
Proof:
The above formula for xq1q≥p+1 can be expressed as three separate equations:
x1 p+k 1≤k≤p−1 = (−x0 kp+1 ) + (x0 )x1 k + (x0 kp )x1 p
(2.11)
p
x1 np
2≤n≤p+1 = An,0 (x0 ) + Cn,0 (x0 )x1
(2.12)
x1 np+k 2≤n≤p,1≤k≤p−1 = An,k (x0 ) + Bn,k (x0 )x1 k + Cn,k (x0 )x1 p
(2.13)
RHIT Undergrad. Math. J., Vol. 13, No. 2
Page 179
We will prove each of these formulas individually and, by doing so, we will have proven
the general formula for x1 q q≥p+1 .
From (2.1), we can use the relation
x1 p+1 = −x0 p+1 + x1 x0 + x1 p x0 p
in order to derive a formula for x1 q q≥p+1 . Multiply both sides of the above relation by
x1 k−1 1≤k≤p+1 which yields
x1 p+k = −x0 p+1 x1 k−1 + x0 x1 k + x0 p x1 p+(k−1) .
Now let gk := x1 p+k , for k ≥ 0, which yields
gk = −x0 p+1 x1 k−1 + x1 k x0 + x0 p gk−1
, k≥1
and letting k → k − 1 (assuming k ≥ 2), we obtain a formula for gk−1
gk−1 = −x0 p+1 x1 k−2 + x0 x1 k−1 + x0 p gk−2 .
Now plugging the formula for gk−1 into gk and simplifying we get
gk = −x0 2p+1 x1 k−2 + x0 x1 k + x0 2p gk−2 .
Now repeating this procedure j − 1 times of plugging our recursion into itself, we obtain
gk = −x0 jp+1 x1 k−j + x0 x1 k + x0 jp gk−j , j ≤ k .
(2.14)
Now setting j = k in (2.14), we obtain that
gk = −x0 kp+1 x1 0 + x0 x1 k + x0 kp g0 .
Since gk = x1 p+k then this implies that
x1 p+k = −x0 kp+1 + x0 x1 k + x0 kp x1 p for 0 &lt; k &lt; p.
This proves (2.11). In order to prove (2.12), we will do a proof by induction.
Base Case: n = 2
First, we must prove the base case true in which
x1 np |n=2 = x1 2p = A2,0 + C2,0 x1 p .
where
A2,0 = −x0 p
2 +1
2
, C2,0 = x0 + x0 p .
(2.15)
RHIT Undergrad. Math. J., Vol. 13, No. 2
Page 180
Using (2.15) and evaluating this formula at k = p − 1, then we obtain
x1 p+k |k=p−1 = x1 p+p−1 = −x0 (p−1)p+1 + x0 x1 p−1 + x0 (p−1)p x1 p .
Now mulitplying both sides by x1 of the above equation, then we obtain
x1 2p = −x0 (p−1)p+1 x1 + x0 x1 p + x0 (p−1)p x1 p+1 .
(2.16)
Now substituting (2.1), our relation for x1 p+1 , into the right-hand side of (2.16) yields
x1 2p = −x0 (p−1)p+1 x1 + x0 x1 p + x0 (p−1)p (−x0 p+1 + x1 x0 + x1 p x0 p )
p2 +1
p2
(2.17)
p
(2.18)
(2.19)
x1 np = An,0 + Cn,0 x1 p
(2.20)
= −x0
+ (x0 + x0 )x1
= A2,0 + C2,0 x1 p
This proves the base case.
Inductive hypothesis:
Lets assume
holds for some n, in which n ≥ 2.
Inductive step:
Lets see if this equation holds for n + 1. Multiply both sides of the previous equation
by x1 p , then we obtain
x1 (n+1)p = An,0 x1 p + Cn,0 x1 2p .
(2.21)
Now plugging in (2.18), the formula for x1 2p , into the right-hand side of (2.21), we obtain
x1 (n+1)p = An,0 x1 p + Cn,0 (−x0 p
2 +1
2
+ (x0 + x0 p )x1 p ).
(2.22) can be rewritten as
2 +1
x1 (n+1)p = −x0 p
2
Cn,0 + (An,0 + (x0 + x0 p )Cn,0 )x1 p
and it suffices to prove that
An+1,0 = −x0 p
2 +1
2
Cn,0 and Cn+1,0 = An,0 + (x0 + x0 p )Cn,0 .
if
x1 (n+1)p = An+1,0 + Cn+1,0 x1 p .
(2.22)
RHIT Undergrad. Math. J., Vol. 13, No. 2
Page 181
To prove the first relation:
2 +1
−x0 p
Cn,0 = −x0 p
2 +1
&middot;
n
X
!
x0 (n−j)p
2 +(j−1)
from (2.5)
j=1
= −
= −
n
X
j=1
n
X
2 +j
x0 (n−j+1)p
x0 ((n+1)−j)p
2 +j
j=1
= An+1,0
To prove the second relation:
2
2
An,0 + (x0 + x0 p )Cn,0 = An,0 + x0 &middot; Cn,0 + x0 p &middot; Cn,0
2
= x0 n + x0 p &middot; Cn,0
from(2.3) and (2.5)
!
n
X
2
2
= x0 n + x0 p &middot;
x0 (n−j)p +(j−1)
from (2.5)
j=1
= x0 n +
n
X
x0 (n−j+1)p
2 +(j−1)
j=1
= x0 n +
n
X
x0 ((n+1)−j)p
2 +(j−1)
j=1
=
n+1
X
x0 ((n+1)−j)p
2 +(j−1)
j=1
= Cn+1,0
This proves the formula for (2.12). In order to prove (2.13), we will do a proof by induction over k.
Base case: k = 1
x1 np+1 =
=
=
=
=
=
x1 x1 np
x1 (An,0 + Cn,0 x1 p )
from (2.20)
x1 An,0 + Cn,0 x1 p+1
x1 An,0 + Cn,0 (−x0 p+1 + x0 x1 + x1 p x0 p )
from (2.1)
p+1
p p
−x0 Cn,0 + (An,0 + x0 Cn,0 )x1 + Cn,0 x0 x1
−x0 p+1 Cn,0 + x0 n x1 + Cn,0 x0 p x1 p
from (2.5) and (2.3)
RHIT Undergrad. Math. J., Vol. 13, No. 2
Page 182
and it suffices to prove that
−x0 p+1 Cn,0 = An,1 , x0 n = Bn,1 , x0 p Cn,0 = Cn,1
to show that
x1 np+1 = An,1 + Bn,1 x1 + Cn,1 x1 p .
To prove the first relation:
−x0
p+1
Cn,0 = −x0
p+1
&middot;
n
X
x0 (n−j)p
2 +(j−1)
from (2.5)
j=1
= −
n
X
2 +p+j
x0 (n−j)p
j=1
= An+1,0 .
The second relation holds true because by definition
Bn,1 = x0 n .
from (2.4)
To prove the third relation:
x0 p &middot; Cn,0 = x0 p &middot;
n
X
!
x0
(n−j)p2 +(j−1)
from (2.5)
j=1
=
n
X
x0 (n−j)p
2 +p+(j−1)
j=1
= Cn,1 .
Having proved these three relations, then this proves our base case.
Inductive Hypothesis:
Lets assume that
x1 np+k = An,k + Bn,k x1 k + Cn,k x1 p
holds true for some k, where 1 ≤ k ≤ p − 2.
Inductive Step:
Lets see if the formula holds true for k + 1. Multiply both sides of our equation by x1 then
x1 np+(k+1) = An,k x1 + Bn,k x1 k+1 + Cn,k x1 p+1
RHIT Undergrad. Math. J., Vol. 13, No. 2
Page 183
Now plugging (2.1), the relation for x1 p+1 , into the right-hand side of x1 np+(k+1) and grouping
terms we obtain
x1 np+(k+1) = −x0 p+1 Cn,k + (An,k + x0 Cn,k ) x1 + Bn,k x1 k+1 + x0 p Cn,k x1 p .
It suffices to show that
−x0 p+1 Cn,k = An,k+1 , An,k + x0 &middot; Cn,k = 0 , Bn,k = Bn,k+1 , x0 p Cn,k = Cn,k+1
to prove that
x1 np+(k+1) = An,k+1 + Bn,k+1 xk+1
+ Cn,k+1 xp1 .
1
Proving the first relation:
−x0 p+1 Cn,k = −x0 p+1 &middot;
n
X
!
x0
(n−j)p2 +(kp+j−1)
from (2.5)
j=1
= −
n
X
x0 (n−j)p
2 +(k+1)p+j
j=1
= An,k+1
Proving the second relation:
An,k + x0 &middot; Cn,k = −
= −
n
X
j=1
n
X
x0 (n−j)p
2 +kp+j
+ x0 &middot;
n
X
!
x0 (n−j)p
2 +kp+(j−1)
j=1
x0
(n−j)p2 +kp+j
j=1
+
n
X
x0 (n−j)p
2 +kp+j
j=1
= 0.
By definition of (2.4):
Bn,k+1 = Bn,k = x0 n
which proves the third relation.
Proving the fourth relation:
for 0 &lt; k &lt; p − 1
from (2.3) and (2.5)
RHIT Undergrad. Math. J., Vol. 13, No. 2
Page 184
x0 p Cn,k = x0 p &middot;
n
X
!
x0
(n−j)p2 +kp+(j−1)
from (2.5)
j=1
=
n
X
x0 (n−j)p
2 +(k+1)p+(j−1)
j=1
= Cn,k+1 .
Having proved all four relations, then this proves (2.13).
Now to prove the formulas for xq2q≥p+1 .The formula for xq2q≥p+1 can be expressed as three
separate equations:
x2 p+k = Ck,0 (x0 )x1 p x2 p + Ak,0 (x0 )x2 p + x1 x2 k − Ck,1 (x0 )x1 p − Bk,1 (x0 )x1 − Ak,1 (x0 ) (2.23)
x2 np = Ãn,0 (x0 )x1 p x2 p + x1 n−1 x2 p +
n−2
X
x0 jp x1 n−j−1 x2 p + B̃n,0 (x0 )x2 p + C̃n,0 (x0 )x1 p
j=1
−
n−1
X
x0 jp x1 n−j + D̃n,0 (x0 ) (2.24)
j=1
x2
np+k
p
p
= Ãn,k (x0 )x1 x2 +
n−1
X
x0 k+(j−1)p x1 n−j x2 p + B̃n,k (x0 )x2 p + x1 n x2 k + C̃n,k (x0 )x1 p
j=1
−
n
X
x0 k+(j−1)p x1 n−j+1 + D̃n,k (x0 ) (2.25)
j=1
We will prove each of these formulas individually and, by doing so, we will have proven
the general formula for x2 q q≥p+1 .
We can use (2.1), our modding out relations, in order to proof the formulas for (2.23),
(2.24), and (2.25). Since the modding out relation
x1 p+1 = −x0 p+1 + x0 x1 + x0 p x1 p is symmetric to x2 p+1 = −x1 p+1 + x1 x2 + x1 p x2 p
RHIT Undergrad. Math. J., Vol. 13, No. 2
Page 185
we can let
(x0 , x1 ) → (x1 , x2 )
and use the formulas for x1 q q≥p+1 to find the formulas for x2 q q≥p+1 . By plugging in x1 for
x0 and x2 for x1 into (2.11), (2.12), and (2.13), we obtain
x2 p+k = (−x1 kp+1 ) + (x1 )x2 k + (x1 kp )x2 p
x2 np = An,0 (x1 ) + Cn,0 (x1 )x2 p
(2.26)
(2.27)
x2 np+k = An,k (x1 ) + Bn,k (x1 )x2 k + Cn,k (x1 )x2 p
(2.28)
Now reducing (2.26) in x1 according to (2.11), (2.12), and (2.13) yields
x2 p+k = (−x1 kp+1 ) + (x1 )x2 k + (x1 kp )x2 p
= −(Ak,1 (x0 ) + Bk,1 (x0 )x1 + Ck,1 (x0 )x1 p ) + (x1 )x2 k + (Ak,0 (x0 ) + Ck,0 (x0 )x1 p )x2 p .
The previous relation is exactly (2.6) evaluated at n = 1, so this proves (2.23). We will
now prove (2.24) and (2.25) together, namely the formula for xnp+k
2
2≤n≤p,0≤k≤p .
x2 np+k = An,k (x1 ) + Bn,k (x1 )x2 k + Cn,k (x1 )x2 p
(2.29)
Lets now reduce the coefficients An,k (x1 ), Bn,k (x1 ), Cn,k (x1 ) according to the relations derived for x1 q q≥p+1 :
An,k (x1 ) = −
= −
n
X
j=1
n
X
x1 [(n−j)p+k]&middot;p+j
A(n−j)p+k,j + B(n−j)p+k,j x1 j + C(n−j)p+k,j x1 p
j=1
= −
n
X
!
A(n−j)p+k,j
j=1
−
n
X
B(n−j)p+k,j x1 j −
j=1
The relation for Bn,k is simple since there is no reduction
Bn,k (x1 ) = x1 n .
Now for the last coefficient Cn,k :
n
X
j=1
!
C(n−j)p+k,j
x1 p
RHIT Undergrad. Math. J., Vol. 13, No. 2
Page 186
Cn,k (x1 ) =
=
=
n
X
j=1
n
X
j=1
n
X
x1 (n−j)p
2 +kp+(j−1)
x1 [(n−j)p+k]p+(j−1)
A(n−j)p+k,j−1 + B(n−j)p+k,j−1 x1 j−1 + C(n−j)p+k,j−1 x1 p
A(n−j)p+k,j−1 + B(n−j)p+k,j−1 x1 j−1 + C(n−j)p+k,j−1 x1 p
j=1
=
n
X
j=1
=
!
n
X
A(n−j)p+k,j−1
+
j=1
n
X
B(n−j)p+k,j−1 x1 j−1 +
j=1
n
X
!
C(n−j)p+k,j−1 x1 p
j=1
Now plugging in the relations for An,k (x1 ), Bn,k (x1 ), Cn,k (x1 ) into the right-hand side of
(2.29), the formula for x2 np+k , yields
x2 np+k =
n
X
!
C(n−j)p+k,j−1
n
n
X
X
j−1 p
p p
A(n−j)p+k,j−1
B(n−j)p+k,j−1 x1 x2 +
x1 x2 +
−
n
X
!
C(n−j)p+k,j
x1 p −
n
X
B(n−j)p+k,j x1 j −
j=1
j=1
x2 p +x1 n x2 k
j=1
j=1
j=1
!
n
X
!
A(n−j)p+k,j
. (2.30)
j=1
(2.30) must be equivalent to our formula for xnp+k
2
x2
np+k
p
p
= Ãn,k (x0 )x1 x2 +
n−1
X
x0 k+(j−1)p x1 n−j x2 p + B̃n,k (x0 )x2 p + x1 n x2 k + C̃n,k (x0 )x1 p
j=1
−
n
X
j=1
so then these relations must hold if (2.25) is true:
Ãn,k =
n
X
C(n−j)p+k,j−1
j=1
B̃n,k =
n
X
j=1
A(n−j)p+k,j−1
x0 k+(j−1)p x1 n−j+1 + D̃n,k (x0 )
RHIT Undergrad. Math. J., Vol. 13, No. 2
C̃n,k = −
n
X
Page 187
C(n−j)p+k,j
j=1
D̃n,k = −
n
X
A(n−j)p+k,j
j=1
Now we will prove that the above relations hold:
Verification of Ãn,k
Ãn,k =
n
X
C(n−j)p+k,j−1
j=1
=
n−1
X
Cjp+k,n−j−1
j=0
=
n−1 jp+k
X
X
x0 (jp+k−q)p
2 +(n−j−1)p+(q−1)
j=0 q=1
=
k−1
X
n &middot; x0
(n−1)p+(k−1)+l(p2 −1)
+
p−1
n−2 X
X
(n − j − 1)x0 (n−1)p+(k−1)+(k+l+jp)(p
j=0 l=0
l=0
Verification of B̃n,k
B̃n,k =
n
X
A(n−j)p+k,j−1
j=1
=
n−1
X
Ajp+k,n−j−1
j=0
= A(n−1)p+k,0 +
n−2
X
Ajp+k,n−j−1
j=0
= −x0 C(n−1)p+k,0 + x0
(n−1)p+k
+
n−2
X
(−x0 ) &middot; Cjp+k,n−j−1
j=0
= x0 (n−1)p+k +
n−1
X
(−x0 ) &middot; Cjp+k,n−j−1
j=0
= x0
(n−1)p+k
− x0 &middot;
n−1
X
Cjp+k,n−j−1
j=0
= x0 (n−1)p+k − x0 &middot; Ãn,k
2 −1)
RHIT Undergrad. Math. J., Vol. 13, No. 2
Page 188
Verification of C̃n,k
C̃n,k = −
n
X
C(n−j)p+k,j
j=1
= −
n−1
X
Cjp+k,n−j
j=0
= −
n−1 jp+k
X
X
2 +(n−j)p+(q−1)
x0 (jp+k−q)p
j=0 q=1
p
= −x0 &middot;
n−1
X
Cjp+k,n−j−1
j=0
= −x0 p &middot; Ãn,k
Verification of D̃n,k
D̃n,k = −
n
X
A(n−j)p+k,j
j=1
= −
n
X
(−x0 ) &middot; C(n−j)p+k,j
j=1
n
X
= x0 &middot;
C(n−j)p+k,j
j=1
p
= x0 &middot; x0 Ãn,k
= x0 p+1 Ãn,k
Since we have verified the coefficient relations for Ãn,k , B̃n,k , C̃n,k , D̃n,k , this then proves
(2.25). Having verified the formulas for x1 q q≥p+1 and x2 q q≥p+1 , then we have completely
proven Proposition 1.
3
Computation of the kernel of u∗
Proposition 2. The map u is injective.
Proof:
Now we will give a proof by contradiction that u is injective by using the general formula
for x2 qq≥p+1 . Let z A2 then
RHIT Undergrad. Math. J., Vol. 13, No. 2
Page 189
z = a0 (x0 ) + a1 (x0 )x1 + . . . + ap2 +p (x0 )x1 p
2 +p
Suppose u(z) = 0 and z 6= 0.
Take m maximal s.t.
z = x0 m z
0
0
where z =
There exists a j where j = np + k 1≤k≤p,
X
0
ai (x0 )x1 i .
s.t.
1≤n≤p
0
aj = C + (terms involving x0 )
where C 6= 0.
,
Since
0
0
0
0 = u(z) = u(x0 m z ) = x0 m u(z ) ⇒ u(z ) = 0.
0
0
0
If u(z ) = 0, then equivalently aj = 0. If aj = 0, then
0
aj = 0
C + (terms involving x0 ) = 0
C = − (terms involving x0 )
This is a contradiction since the left-hand side is a non-zero constant and the right-hand
side is terms involving x0 . So then u(z) = 0 if and only if z = 0, which is an equivalent
statement as u being injective.
Theorem 1. Consider the dual map
u∗
A1,1 ∗ → A2 ∗
The kernel of u∗ has basis:
Pi P0 + x 0 Pi P1 + . . . + x 0 p Pi Pp
for 1 ≤ i ≤ p
where
Pi P j A∗1,1
m
n
and Pi Pj (x1 x2 ) =
1 if i = m and j = n
0
otherwise
RHIT Undergrad. Math. J., Vol. 13, No. 2
Page 190
Proof:
As was shown previously:
 k

 xp2 p
XX
u(x1 k ) = u(x1 )k = xk2 =
lijk x1 i x2 j


for 0 ≤ k ≤ p
for k ≥ p + 1
i=0 j=0
m
n
Since Pi Pj (x1 x2 ) = 1 if and only if i = m and j = n and 0 otherwise, then
p
(Pi P0 + x0 Pi P1 + . . . + x0 Pi Pp )
p
p
X
X
i
j
lijk x1 x2 =
i=0 j=0
p
X
x0 j lijk (x0 ) = 0
j=0
This must hold true for any k where 0 ≤ k ≤ p2 + p and also for all i s.t. 1 ≤ i ≤ p. We
will prove this true by using case-work analysis.
Case x2 k 0≤k≤p for 1 ≤ i ≤ p:
p
X
x0 j lijk (x0 ) =
j=0
p
X
x0 j &middot; 0 = 0
j=0
This is true because lijk = 0 for 0 ≤ k ≤ p and 1 ≤ i ≤ p.
Case x2 p+k 0&lt;k&lt;p for 1 ≤ i ≤ p:
Subcase i = 1:
p
X
x0 j l1jk (x0 ) = −Bk,1 + x0 &middot; 0 + . . . + x0 k−1 &middot; 0 + x0 k &middot; 1 + x0 k+1 &middot; 0 + . . . + x0 p &middot; 0
j=0
= −x0 k + 0 + x0 k + 0 = 0
Subcase 2 ≤ i ≤ p − 1:
p
X
j=0
j
x0 lijk (x0 ) =
p
X
x0 j &middot; 0 = 0
j=0
Subcase i = p:
p
X
x0 j lpjk (x0 ) = −Ck,1 + x0 &middot; 0 + . . . + x0 p−1 &middot; 0 + x0 p &middot; Ck,0
j=0
= −x0 p Ck,0 + 0 + x0 p &middot; Ck,0 = 0 from (2.5)
RHIT Undergrad. Math. J., Vol. 13, No. 2
Case x2 np 2≤n≤p for 1 ≤ i ≤ p:
Subcase i = 1:
p
X
x0 j l1jk (x0 ) = −x0 (n−1)p + x0 &middot; 0 + . . . + x0 p−1 &middot; 0 + x0 p &middot; x0 (n−2)p
j=0
= −x0 (n−1)p + 0 + x0 (n−1)p = 0
Subcase 2 ≤ i ≤ n − 2:
p
X
x0 j lijk (x0 ) = −x0 (n−k)p + x0 &middot; 0 + . . . + x0 p−1 &middot; 0 + x0 p &middot; x0 (n−k−1)p
j=0
= −x0 (n−k)p + 0 + x0 (n−k)p = 0
Subcase i = n − 1:
p
X
x0 j l(n−1)jk (x0 ) = −x0 p + x0 &middot; 0 + . . . + x0 p−1 &middot; 0 + x0 p &middot; 1
j=0
= −x0 p + 0 + x0 p = 0
Subcase n ≤ i ≤ p − 1:
p
X
x0 j lijk (x0 ) = −x0 p + x0 &middot; 0 + . . . + x0 p−1 &middot; 0 + x0 p &middot; 1
j=0
= −x0 p + 0 + x0 p = 0
Subcase i = p:
p
X
x0 j lpjk (x0 ) = C̃n,0 + x0 &middot; 0 + . . . + x0 p−1 &middot; 0 + x0 p &middot; Ãn,0
j=0
= −x0 p &middot; Ãn,0 + 0 + x0 p &middot; Ãn,0 = 0 from (2.9)
Case x2 np+k 2≤n≤p,0&lt;k&lt;p for 1 ≤ i ≤ p:
Page 191
RHIT Undergrad. Math. J., Vol. 13, No. 2
Page 192
Subcase i = 1:
p
X
x0 j l1jk (x0 ) = −x0 k+(n−1)p + x0 &middot; 0 + . . . + x0 p &middot; x0 k+(n−2)p
j=0
= −x0 k+(n−1)p + 0 + x0 k+(n−1)p = 0
Subcase 2 ≤ i ≤ n − 1:
p
X
x0 j lijk (x0 ) = −x0 k+(n−k)p + x0 &middot; 0 + . . . + x0 p−1 &middot; 0 + x0 p &middot; x0 k+(n−k−1)p
j=0
= −x0 k+(n−k)p + 0 + x0 k+(n−k)p = 0
Subcase i = n:
p
X
x0 j lnjk (x0 ) = −x0 k + x0 &middot; 0 + . . . + x0 k−1 &middot; 0 + x0 k &middot; 1 + x0 k+1 &middot; 0 + . . . + x0 p &middot; 0
j=0
= −x0 k + 0 + x0 k + 0 = 0
Subcase n + 1 ≤ i ≤ p − 1:
p
X
p
X
j
x0 lijk (x0 ) =
j=0
x0 j &middot; 0 = 0
j=0
Subcase i = p:
p
X
x0 j lpjk (x0 ) = C̃n,k + x0 &middot; 0 + . . . + x0 p−1 &middot; 0 + x0 p &middot; Ãn,k
j=0
= −x0 p &middot; Ãn,k + 0 + x0 p &middot; Ãn,k = 0 from (2.9)
Case: x2 (p+1)p for 1 ≤ i ≤ p:
Subcase i = 1:
p
X
2 2
x0 j l1jk (x0 ) = −x0 p + x0 &middot; 0 + . . . + x0 p−1 &middot; 0 + x0 p &middot; x0 p −p
j=0
2
2
= −x0 p + 0 + x0 p = 0
RHIT Undergrad. Math. J., Vol. 13, No. 2
Page 193
Subcase 2 ≤ i ≤ p − 1:
p
X
x0 j lijk (x0 ) = −x0 (p+1−k)p + x0 &middot; 0 + . . . + x0 p−1 &middot; 0 + x0 p &middot; x0 (p−k)p
j=0
= −x0 (p+1−k)p + 0 + x0 (p+1−k)p = 0
Subcase i = p:
p
X
x0 j lpjk (x0 ) =
C̃p+1,0 − x0 p + x0 &middot; 0 + . . . + x0 p−1 &middot; 0 + x0 p &middot; Ãp+1,0 + 1
j=0
= C̃p+1,0 − x0 p + 0 + x0 p &middot; Ãp+1,0 + x0 p
= −x0 p &middot; Ãp+1,0 − x0 p + x0 p &middot; Ãp+1,0 + x0 p = 0 from (2.9)
This then shows that
Pi P 0 + x 0 Pi P1 + . . . + x 0 p Pi Pp
for 1 ≤ i ≤ p
∗
are contained in the basis of the kernel of u , but we do not know if there are any other
elements which could be in the basis. Since u is injective, this implies that u∗ is surjective.
The basis for A1,1 ∗ consists of all Pi Pj such that 0 ≤ i, j ≤ p. A1,1 ∗ then contains (p + 1)2
elements in its basis.The basis for A2 ∗ consists of all Pi such that 0 ≤ i ≤ p2 + p. A2 ∗
then contains p2 + p + 1 elements in its basis. So then the basis for the kernel of u∗ has
(p + 1)2 − (p2 + p + 1) = p elements in it. Since we have found p elements which compose a
basis for the kernel of u∗ namely:
Pi P 0 + x 0 Pi P1 + . . . + x 0 p Pi Pp
for 1 ≤ i ≤ p
then these basis elements form a complete basis for the kernel of u∗ .
References
 Charles Rezk, Modular Isogeny Complexes, ArXiv.
 Munkres, James. Topology, 2nd edition. Upper Saddle River, N.J.: Prentice Hall, 2000.
```