Skip to main content

Section 1.8 Endomorphisms and the Tate module (Berke)

Motivation.
\begin{align*} f \colon \PP^n\subseteq V_1 \amp\to V_2 \subseteq \PP^m,\,V_i = V(I_i)\\ P\amp \mapsto \cdots \end{align*}
\begin{equation*} f = \lb f_1 : \cdots : f_m\rb ,\,f_i\in \overline K (V_1) \end{equation*}

this feels quite restrictive, an isogeny is even more so, rational, regular, homomorphism, surjective, finite kernel. It feels like there won't be too many but we have multiplication by \(n\) etc. so we should ask how many are there that will surprise us? I.e. what is

\begin{equation*} \rank_\ZZ \Hom(A,B) = ? \end{equation*}

Notation: \(A,B,C , A_i, B_i\) are all abelian varieties. \(l \ne \characteristic k\text{,}\) \(\sim\) is isogeny.

Subsection 1.8.1 Poincaré's complete reducibility theorem

Choose \(\sheaf L\) ample on \(A\)

\begin{equation*} \xymatrix{ B\ar[r]^i \ar[d]_{\phi_{i*\sheaf L}} & A \ar[d]_\sim^{\phi_{\sheaf L}} \\ \hat B & \hat A \ar[l]_{\hat i} } \end{equation*}

\(C\) is defined to be the connected component of \(\phi^{-1}_{\sheaf L}(\ker \hat i)\) in \(A\)

\begin{equation*} \dim C = \dim \ker \hat i \ge \dim \hat A - \dim \hat B = \dim A - \dim B\text{.} \end{equation*}

\(B \cap C\) finite, \(z\in B\text{,}\) \(z\in B\cap \phi_{\sheaf L^{-1}} (\ker \hat i) = T_z^* \sheaf L \otimes \sheaf L^{-1} |_B\) is trivial if and only if \(z\in K(\sheaf L|_B)\text{.}\) So \(\sheaf L|_B\) ample implies \(K(\sheaf L|_B)\) finite and so \(B\cap C\) is finite. So \(B\times C \to A\) has finite kernel and

\begin{equation*} \dim (B\times C) = \dim B + \dim C \ge \dim A \end{equation*}

and surjective implies its an isogeny.

Definition 1.8.2. Simple abelian varieties.

\(A\) is called simple if there does not exists \(B\subseteq A\) other than \(B = 0,A\text{.}\)

\(\alpha(A) \subseteq B\) which implies \(\alpha(A) = B\) or \(0\text{.}\) The connected component of 0 of \(\ker \alpha\) will be an abelian subvariety of \(A\text{,}\) denote it \(C\) If \(C = 0\) then \(\ker \alpha\) is finite, if \(C = A\) then \(\alpha = 0\text{.}\) So \(\alpha\) is an isogeny or \(0\text{.}\)

Definition 1.8.6.
\begin{equation*} \End^0(A) = \End(A) \otimes \QQ\text{.} \end{equation*}

\(\alpha\) an isogeny implies \(\ker \alpha\) is finite. So there exists \(n\) with \(n \ker \alpha = 0\text{.}\) \(\ker\alpha \subseteq \ker n_A\)

\begin{equation*} \xymatrix{ & A \ar[dl]_\alpha \ar[d]\ar[r]^{n_A} & A \\ B\ar[r]_\sim \ar@/_3.0pc/[urr]_{\exists | \beta} & A/\ker\alpha \ar[d]\ar[ur]_{\circ} &\\ & A/n_A & } \end{equation*}

so \(\beta\circ \alpha = n_A\text{,}\) also \(\alpha \circ \beta = n_B\text{.}\)

\begin{align*} \End(A) \otimes \QQ \amp \simeq \prod_{i,j} \Hom(A_i^{n_i}, A_j^{n_j}) \otimes \QQ\\ \amp \simeq \prod_{i} \End(A_i)^{n_i^2} \otimes \QQ\\ \amp \simeq \prod_{i} \End^0(A_i)^{n_i^2} \end{align*}

If \(m | n\) then \(|\ker (m_A)| = m^{2g}\text{,}\) then use structure theorem.

In particular if we let \(A\lb l^n\rb = A(k^\sep) \lb l^n\rb\text{,}\) then \(A\lb l^n\rb \simeq (\ZZ/l^n)^{2g}\) Define

\begin{equation*} T_l(A) = \varprojlim_n A[l^n],\, A[l^{n+1}] \xrightarrow{l} A[l] \end{equation*}

\(\alpha \colon A \to B\) induces

\begin{equation*} T_l\alpha\colon T_l(A) \to T_l(B) \end{equation*}
\begin{equation*} (a_1,a_2, \ldots) \mapsto(\alpha(a_1),\alpha(a_2), \ldots) \end{equation*}

Let \(\alpha \in \Hom(A,B)\) and assume \(T_l \alpha = 0\) then

\begin{equation*} \ker (\alpha|_{A_i}) \supseteq A_i[l^n] \forall n \end{equation*}

for any simple component \(A_i\) of \(A\) so \(\alpha =0\) on each \(A_i\) and hence \(\alpha = 0\) on \(A\text{.}\)

Recall we are interested in knowing about \(\rank_\ZZ\Hom(A,B) = ?\text{,}\) can we bound this? If we could show that

\begin{equation*} \Hom(A,B) \otimes \ZZ_l \hookrightarrow \Hom(T_l(A),T_l(B)) \end{equation*}

we could conclude, so:

\begin{equation*} \xymatrix{ \Hom(A,B)\otimes \ZZ_l \ar[d]_{\sim} \ar@{^(->}[r] & \Hom(T_lA,T_lB)\ar[d]_{\sim} \\ \prod_{i,j} (\Hom(A_i, B_j) \otimes \ZZ_l) \ar@{^(->}[r] & \prod_{i,j}\Hom(T_lA_i, T_l B_j) % TODO right arrows hooks, fix } \end{equation*}

\(A_i + B_j = 0\text{,}\) \(A_i\sim B_j\) \(\Hom(A_i, B_j) \hookrightarrow \End(A_i)\text{.}\) Assume \(A= B\) and \(A\) simple, then \(\End(A) \otimes \ZZ_l \hookrightarrow \End(T_l(A))\text{.}\)

Definition 1.8.15.

\(V/k\) then \(f\colon V \to k\) is called a (homogenous) polynomial function of degree \(d\) if \(\forall \{v_1,\cdots, v_m\}\subseteq V\) linearly independent.

\begin{equation*} f(\lambda_1 v_1 + \lambda_2 v_2 +\cdots + \lambda_m v_m) \end{equation*}

is given by a homogenous polynomial of degree \(d\) in \(\lambda_i\) i.e.

\begin{equation*} f(\lambda_1 v_1 + \lambda_2 v_2 +\cdots + \lambda_m v_m) = P(\lambda_1,\ldots, \lambda_m) \end{equation*}

for some \(P\in k\lb X_m\rb\) homogenous of degree \(d\text{.}\)

\begin{equation*} \deg\colon \End(A) \to \ZZ \end{equation*}

\(\alpha\) an isogeny iff \(\deg \alpha\text{,}\) \(\alpha\) not an isogeny iff \(0\text{.}\)

(of above continued)

\begin{equation*} \End(A) \otimes \ZZ_l \hookrightarrow\End(T_l(A)) \end{equation*}

for \(A\) simple iff for any finitely generated \(M \subseteq \End(A)\)

\begin{equation*} M \otimes \ZZ_l \hookrightarrow \End(T_l(A)) \end{equation*}

Claim:

\begin{equation*} M^{\text{div}} = \{f\in \End(A) : nf\in M \text{ for some }n\ge 1\} \end{equation*}

is finitely generated.

Proof: \(M^\text{div} = (M\otimes \QQ) \cap \End(A)\) \(\deg \colon M \otimes \QQ \to \QQ\) is a polynomial so it is continuous.

\begin{equation*} U = \{\phi\in M \otimes \QQ : \deg \phi \lt 1\} \end{equation*}

is open in \(M \otimes \QQ\) but \(U \cap M^\text{div} = 0\) so \(M^\text{div} \) is a discrete subgroup of the finite dimensional \(\QQ\)-vector space \(M \otimes \QQ\) so \(M^\text{div}\) is finitely generated. \(M \hookrightarrow M^\text{div}\) so \(M \otimes \ZZ_l \hookrightarrow M^\text{div} \otimes \ZZ_l\) so we may assume \(M = M^\text{div}\text{.}\)

Let \(f_1,\ldots, f_r\) be a \(\ZZ\)-basis for \(M\) and suppose that \(\sum a_iT_l(f_i)=0\) for some \(a_i \in \ZZ_l\) not all 0. We can assume not all \(a_i\) are divisible by \(l\text{.}\) Choose \(a_i' \in \ZZ\) s.t. \(a_i'=a_i \pmod{l}\)

\begin{equation*} f = \sum a_i' f_i \in \End(A) \end{equation*}

we then have

\begin{equation*} f = \sum a_i' T_lf_i \end{equation*}

is 0 on the first coordinate of \(T_l\text{.}\) So \(A\lb l \rb \subseteq \ker f\) so there exists \(g\) with \(f= lg\) \(f\in M\) implies \(g\in M^\text{div} = M\) so \(g = \sum b_i f_i\) and \(f = \sum lb_i f = \sum a_i f_i\) hence \(l\mid a_i\) for all \(i\) a contradiction. So \(\End(A)\otimes \ZZ_l \hookrightarrow \End(T_l(A))\text{.}\)

Therefore

\begin{equation*} \Hom(A,B) \otimes \ZZ_l \hookrightarrow \Hom(T_l(A), T_l(B)) \end{equation*}
\begin{equation*} \rank_\ZZ \Hom(A,B) \le 4 \dim A\dim B\text{.} \end{equation*}