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RoseHulman Undergraduate Mathematics Journal Relations in the Dyer-Laslof Algebra for Morava E-theory Louis Atsaves a Volume 13, No. 2, Fall 2012 Sponsored by 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 Rose-Hulman Undergraduate Mathematics Journal 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 , · · · , vn−1 ]]. The sequence v0 = p, v1 , · · · , vn−1 is regular by construction and vn has invertible image in R/(v0 , v1 , · · · , 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 , · · · , vn−1 ]][β ±1 ], where β has degree 2. The cohomology theory E(n) is called the Morava E-theory. In [1], 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·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 · 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 < k < 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 · 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 · Cn,0 + x0 p · Cn,0 2 = x0 n + x0 p · Cn,0 from(2.3) and (2.5) ! n X 2 2 = x0 n + x0 p · 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 · 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 · Cn,0 = x0 p · 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 · 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 · 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 · Cn,k = − = − n X j=1 n X x0 (n−j)p 2 +kp+j + x0 · 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 < k < p − 1 from (2.3) and (2.5) RHIT Undergrad. Math. J., Vol. 13, No. 2 Page 184 x0 p Cn,k = x0 p · 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]·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 · 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 ) · Cjp+k,n−j−1 j=0 = x0 (n−1)p+k + n−1 X (−x0 ) · Cjp+k,n−j−1 j=0 = x0 (n−1)p+k − x0 · n−1 X Cjp+k,n−j−1 j=0 = x0 (n−1)p+k − x0 · 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 · n−1 X Cjp+k,n−j−1 j=0 = −x0 p · Ãn,k Verification of D̃n,k D̃n,k = − n X A(n−j)p+k,j j=1 = − n X (−x0 ) · C(n−j)p+k,j j=1 n X = x0 · C(n−j)p+k,j j=1 p = x0 · 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 · 0 = 0 j=0 This is true because lijk = 0 for 0 ≤ k ≤ p and 1 ≤ i ≤ p. Case x2 p+k 0<k<p for 1 ≤ i ≤ p: Subcase i = 1: p X x0 j l1jk (x0 ) = −Bk,1 + x0 · 0 + . . . + x0 k−1 · 0 + x0 k · 1 + x0 k+1 · 0 + . . . + x0 p · 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 · 0 = 0 j=0 Subcase i = p: p X x0 j lpjk (x0 ) = −Ck,1 + x0 · 0 + . . . + x0 p−1 · 0 + x0 p · Ck,0 j=0 = −x0 p Ck,0 + 0 + x0 p · 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 · 0 + . . . + x0 p−1 · 0 + x0 p · 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 · 0 + . . . + x0 p−1 · 0 + x0 p · 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 · 0 + . . . + x0 p−1 · 0 + x0 p · 1 j=0 = −x0 p + 0 + x0 p = 0 Subcase n ≤ i ≤ p − 1: p X x0 j lijk (x0 ) = −x0 p + x0 · 0 + . . . + x0 p−1 · 0 + x0 p · 1 j=0 = −x0 p + 0 + x0 p = 0 Subcase i = p: p X x0 j lpjk (x0 ) = C̃n,0 + x0 · 0 + . . . + x0 p−1 · 0 + x0 p · Ãn,0 j=0 = −x0 p · Ãn,0 + 0 + x0 p · Ãn,0 = 0 from (2.9) Case x2 np+k 2≤n≤p,0<k<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 · 0 + . . . + x0 p · 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 · 0 + . . . + x0 p−1 · 0 + x0 p · 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 · 0 + . . . + x0 k−1 · 0 + x0 k · 1 + x0 k+1 · 0 + . . . + x0 p · 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 · 0 = 0 j=0 Subcase i = p: p X x0 j lpjk (x0 ) = C̃n,k + x0 · 0 + . . . + x0 p−1 · 0 + x0 p · Ãn,k j=0 = −x0 p · Ãn,k + 0 + x0 p · 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 · 0 + . . . + x0 p−1 · 0 + x0 p · 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 · 0 + . . . + x0 p−1 · 0 + x0 p · 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 · 0 + . . . + x0 p−1 · 0 + x0 p · Ãp+1,0 + 1 j=0 = C̃p+1,0 − x0 p + 0 + x0 p · Ãp+1,0 + x0 p = −x0 p · Ãp+1,0 − x0 p + x0 p · 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 [1] Charles Rezk, Modular Isogeny Complexes, ArXiv. [2] Munkres, James. Topology, 2nd edition. Upper Saddle River, N.J.: Prentice Hall, 2000.