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\title{Isogeny invariance of the BSD conjecture}
\author{Akshay Venkatesh \footnote{Notes by Tony Feng}}
\begin{document}
\maketitle
\section{Examples}
The BSD conjecture predicts that for an elliptic curve $E$ over $\Q$ with $E(\Q)$ of rank $r \ge 0$,
\begin{equation}\label{BSD}
\boxed{\frac{L^{(r)}(1, E)}{r!} = \frac{(\prod_p c_p) \Omega_E \cdot \Sha_ER_E}{\#E(\Q)_{\rm{tor}}\#\widehat{E}(\Q)_{\rm{tor}}}}
\end{equation}
where
\begin{itemize}
\item $\widehat{E}$ is the dual elliptic curve (so $\widehat{E} \simeq E$, unlike for higher dimensions in general),
\item $\Omega = \int_{E(\R)} |\omega|$, and $\omega$ is the global section of $\Omega^1_{N(E)/\Z}$ corresponding to a choice of basis
of the $\Z$-line ${\rm{Cot}}_0(N(E))$ for the N\'eron model $N(E)$ of $E$ over $\Z$,
\item $c_p$ is the number of connected components of $N(E)_{\mathbf{F}_p}$ that are geometrically connected over $\mathbf{F}_p$, or equivalently (by Lang's theorem
applied to $N(E)^0_{\mathbf{F}_p}$-torsors) have a rational point, so $c_p$ coincides with the number of $\mathbf{F}_p$-points of
the finite \'etale component group $N(E)_{\mathbf{F}_p}/N(E)^0_{\mathbf{F}_p}$.
\item $R_E$ is a regulator term that equals 1 when $E(\Q)$ is finite.
\end{itemize}
We consider the three elliptic curves over $\Q$ with conductor $11$:
\begin{enumerate}
\item $E_1 : y^2+y=x^3-x^2$, which happens to be $X_1(11)$,
\item $E_2 : y^2+y = x^3-x^2-10x-20$, which happens to be $X_0(11)$,
\item $E_3 : y^2+y = x^3-x^2-7820 x - 263580$.
\end{enumerate}
These are all minimal Weierstrass models, so their smooth loci over $\Z$ are the relative identity components of the N\'eron models.
The evident action of $(\Z/11\Z)^{\times}$ on the fine moduli scheme $X_1(11)$ makes $\{\pm 1\}$ act trivially, and the resulting action of
the cyclic group $C = (\Z/11\Z)^{\times}/\{\pm 1\}$ on $X_1(11)$ leaves invariant the forgetful map
$X_1(11) \rightarrow X_0(11)$. The resulting map $X_1(11)/C \rightarrow X_0(11)$ between smooth projective (geometrically connected)
curves is clearly bijective on $\overline{\Q}$-points away from $j = 0, 1728$, so it is birational and hence an isomorphism. In other words,
we have a natural 5-isogeny $E_1 \rightarrow E_2$.
To understand this isogeny in another way, we consider the moduli-theoretic viewpoint.
By moduli-theoretic considerations, the two geometric cusps on $E_2$ (correaponding to the 11-gon and 1-gon equipped with their unique order-11 ample cyclic subgroups take {\em up to automorphism} of the polygon) are both $\Q$-points, and 5 of geometric cusps on $E_1$ are $\Q$-points
(namely, the ones corresponding to the 11-gon equipped with a generator of its component group $\Z/11\Z$ up to sign).
On $E_1$ with the model above, these are the points $\{(0,0), (0,-1), (1,0), (1,-1), \infty\}$. This exhausts $E_1(\Q)$.
\begin{remark} The fact that $E_1(\Q)$ consists entirely of cusps reflects the fact that no elliptic curves over $\Q$ have a rational $11$-torsion point.
\end{remark}
Using the rational cusp for the 11-gon as the identity for the group law turns $X_0(11)$ (so {\em not} the cusp $\infty$ in the standard
analytic model!) turns it into the elliptic curves $E_2$, and likewise
for $E_1$ using any of the $\Q$-cusps on $X_1(11)$. Hence, the forgetful map $E_1 \rightarrow E_2$ is a 5-isogeny
that carries all 5 rational cusps to the identity; i.e., its kernel is the constant $\Q$-group $\Z/5\Z$. Thus, the dual isogeny has
kernel $\mu_5$.
But $E_2(\Q)$ is also finite with order 5, consisting of the points $$\{(5,5)(5,-6), (16,60), (16,-61), \infty\},$$ so the
quotient of $E_2$ by that $\Q$-subgroup $\Z/5\Z$ is {\em not} $E_1$ (as otherwise composing these
would yield an endomorphism of $E_1$ of degree 25, necessarily with kernel $E_2[5]$ since $E_2$ has non-integral $j$-invariant and hence is not CM,
so then $E_1[5]$ would be an extension of $\Z/5\Z$ by $\Z/5\Z$ as a $\Q$-group; the $\mu_5$-valued Weil pairing on $E_1[5]$ would then give
a contradiction). This quotient of $E_2$ must then be another elliptic curve over $\Q$, so it is $E_3$.
To summarize, we have 5-isogenies
\[
E_1 \rightarrow E_2 \rightarrow E_3.
\]
By design, each has kernel $\Z/5\Z$ as a $\Q$-group.
Since the $L$-function is invariant under isogeny, we have
\[
L(1, E_1) = L(1, E_2) = L(1, E_3) \approx 0.2538 \ldots.
\]
Therefore, the BSD conjecture predicts that the quantity on the right side of \eqref{BSD} is also the same for $E_1, E_2$, and $E_3$. In all
three cases $\Sha = 0$. The regulator $R_E$ is also trivial since the common rank of these $\Q$-isogenous curves is 0.
However, the volume and Tamagawa factors vary, as follows.
\begin{enumerate}
\item The common Galois module $E_i[2]$ is not split over $\R$ (the cubic $4x^3 - 4x^2 + 1$ for $E_1$ has negative discriminant $-44$
and so has one real root), so the 1-dimensional compact commutative Lie groups $E_i(\R)$ are all {\em connected} and hence are circles.
Since the Weierstrass models described above are minimal, and the smooth part of the minimal Weierstrass model coincides
with the relative identity component of the N\'eron model, a N\'eron differential on each $E_i$ is given by ${\rm{d}}x/(2y+1)$. Hence, for $E_1$
the volume term is
\[
\Omega_1 =: \Omega_{E_1} = \int_{E(\R)} \left|\frac{{\rm{d}}x}{2y+1}\right| = 2 \int_{\alpha}^{\infty} \frac{{\rm{d}}x}{\sqrt{4x^3-4x^2+1}}
\]
where $\alpha$ is a real root of $4x^3-4x^2-1$.
This volume turns out to be $25 L(1,E)$, expressing that $\Sha(E_1)=1$, $\# E_1(\Q) = 5$, and
$c_p=1$ for all $p$. The triviality of $c_p$ for $p \ne 11$ is clear by the moduli-theoretic meaning of $E_1$ (or by hand: good reduction away
from 11), and for $p=11$ we note that $E_1$ has {\em split} multiplicative reduction. Consequently, by the theory of Tate curves
and the link between the minimal regular proper model and the N\'eron model for an elliptic curve (the latter being the smooth part of the former)
it follows that $c_{11} = -v_{11}(j) = -1$.
It is a general theorem
that the Jacobian of $X_1(\ell)$ has N\'eron model with connected fiber at $\ell$ for all primes $\ell > 3$, but this requires computing
the minimal regular proper model of $X_1(\ell)$ over $\Z_{(\ell)}$, which is much harder than for $X_0(\ell)$ and is also harder in general than for
the genus-1 case $\ell = 11$.
\item For $E_2$, we have $\Omega_2 = \Omega_1/5$. The other factors must change to compensate, and it turns out that the change is $c_{11}=5$. This comes from the fact that $v_{11}(j(E_2))=-5$ and $E_2$ has {\em split} multiplicative reduction (since $E_1$ has that property); in contrast, a quadratic twist $E'_2$ of $E_2$ by a character that is unramified but nontrivial locally at 11
would have $c'_{11} = 1$ even though $v_{11}(j') = -5$ too.
\item For $E_3$, we have $\Omega_3 = \Omega_2/5$. Here the changes relative to $E_2$ are that $c_{11}=1$ (because the reduction type is split multiplicative
and $v_{11}(j) = -1$) but $E_3(\Q) = 0$.
\end{enumerate}
We have seen that the variation in the $j$-invariants, coupled with the theory of Tate curves and minimal regular proper models,
explains the variation of the Tamagawa factors.
Let's explain why the volume terms are changing.
In all three cases, we have seen that $E_i(\R)$ is a circle. The induced maps $E_i(\R) \rightarrow E_{i+1}(\R)$
are therefore finite-degree homomorphisms from the circle {\em onto} itself as a Lie group, of degree equal to the size of the kernel.
But we rigged both isogenies over $\Q$ to have kernel $\Z/5\Z$, so on $\R$-points the kernel has order 5; i.e., it is a degree-5 map
between Lie groups. Hence, the effect on periods is
precisely division by 5 (from the degree of the map) {\em provided} that a N\'eron differential pulls back to a N\'eron differential under
each map $N(E_i) \rightarrow N(E_{i+1})$. In other words, we have to prove
that this $\Z$-homomorphism between smooth $\Z$-groups is \'etale.
Over $\Z[1/11]$ the map $N(E_i) \rightarrow N(E_{i+1})$ between abelian schemes
must be finite flat, and its generic fiber is a constant $\Q$-group of order 5 by design. But away from 5 this
kernel must then be finite \'etale, hence the same constant group. By Raynaud's work on finite flat group schemes,
the kernel must be constant over $\Z_{(5)}$ as well (or more concretely, the points in the kernel are clearly
distinct modulo 5 for each isogeny), so overall we have the \'etaleness away from 11.
It remains to study the situation at 11.
If you look at the points in $E_2(\Q)$ aside from the identity, they all have the same reduction at 11, namely $(5,5)$. This is the {\em singularity}
in the mod-11 fiber of the minimal Weierstrass model, so over $\Z_{(11)}$ these points have reduction in the N\'eron model that lie in {\em non-identity}
components of the mod-11 fiber. Consequently, we see that the quasi-finite flat schematic closure in $N(E_2)$ of the kernel of $E_2 \rightarrow E_3$
has mod-11 fiber that is also a constant group consisting of 5 distinct points. Since $N(E_2)_{\mathbf{F}_{11}}$ has 5 connected components
and $N(E_3)_{\mathbf{F}_{11}}$ is connected, we conclude that $N(E_2) \rightarrow N(E_3)$ is actually surjective even on mod-11 fibers
and its kernel is the constant group $\Z/5\Z$ over $\Z$. Hence, this map between N\'eron models is the quotient by that constant group,
and in particular it is an \'etale morphism as desired.
%Why doesn't the period increase from $E_2$ to $E_3$? The key difference is that the kernel of the isogeny is defined over $\Q$, but the individual torsion points in it are %not; they are only defined over an extension.
\subsection{Computing $\Sha$}
Finally, let's discuss computing $\Sha$, focusing on $\Sha(E)[2]$ for $E$ one of the curves discussed above.
This is a subgroup of $H^1(G_{\Q, S}, E[2])$, where $S = \{2,11\}$ (the ramified places for $E[2]$).
We will compute this ambient degree-1 cohomology group, and find that it is 2-dimensional over $\F_2$
(and then when further local conditions are imposed to get $\Sha(E)[2]$ one gets 0).
Since 5-isogenies induce isomorphisms on 2-torsion, the problem is literally the same for each $E$.
Abstractly $M := E[2] \simeq (\Z/2\Z)^2 \simeq \{(u,v,w) \in (\Z/2\Z)^3 \colon u+v+w = 0 \}$ on which the Galois action is given by a map $G_S \rightarrow S_3$ corresponding to the cubic equation defining $E[2]$. This is the standard permutation representation, and the splitting field of $E_i[2]$ is an $S_3$-extension of $\Q$.
\begin{remark}
Tate's global Euler characteristic for Galois cohomology (as will be discussed in Jeremy's talk) says
\[
\frac{\# H^1(G_S, M)}{\# H^0(G_S, M) \cdot \# H^2(G_S, M)} = \frac{\# M}{\# M^{\Gal(\CC/\R)}}.
\]
Since our cubic has only one real root, we have $\# M^{\Gal(\CC/\R)} = 2$. Also, $\# M = \# E[2] = 4$. We also know $\# H^0(G_S, M) = 1$ because $E(\Q)$
has no non-trivial $2$-torsion points, so $\# H^1(G_S, M)/\# H^2(G_S, M) = 2$.
We'll show $\# H^1(G_S, M) = 4$ (so $\#H^2(G_S,M) = 2$).
\end{remark}
There is only one way to approach the $H^1$-computation, which is to \emph{pass to the splitting field} (which lies inside $\Q_S$).
Let $K$ be the cubic extension of $\Q$ obtained by adjoining a root $\alpha$ of $4x^3-4x+1$, and $L$ its Galois closure, so $L$ has a unique quadratic extension $\Q(\sqrt{-11})$.
\[
\xymatrix{
& L \ar@{-}[dl]_2 \ar@{-}[dr]^3 \ar@{-}[dd]_{S_3} & \\
K \ar@{-}[dr]_3 & & \Q(\sqrt{-11}) \ar@{-}[dl]^2 \\
& \Q
}
\]
Obviously ${\rm{Gal}}(\Q_S/L)$ acts trivially on $E[2]$. There is a spectral sequence
$$H^p(G_{L/\Q},H^q(\Q_S/L, M)) \implies H^{p+q}(G_{\Q,S}, M)$$ whose $E_2$ page is
\[
\xymatrix{
H^0(G_{L/\Q}, H^2(\Q_S/L, E[2]))& \ldots \\
H^0(G_{L/\Q}, H^1(\Q_S/L, E[2])) \ar[drr] & H^1(G_{L/\Q}, H^1(\Q_S/L, E[2])) & \ldots \\
H^0(G_{L/\Q}, H^0(\Q_S/L, E[2])) & H^1(G_{L/\Q}, H^0(\Q_S/L, E[2])) & \ldots \\
}
\]
It turns out that everything beyond the left column vanishes, but this is not obvious, so the spectral sequence degenerates at this page.
The reason is that on a $p$-torsion module Galois cohomology injects into that of
a $p$-Sylow subgroup, and $E[2]$ happens to be a free module over the group algebra on a 2-Sylow of this $S_3$.
Therefore, what we want is $$H^1(\Q_S/L, E[2])^{G_{L/\Q}} = (E[2] \otimes H^1(\Q_S/L, \Z/2\Z))^{G_{L/\Q} \simeq S_3}.$$
Now, $S_3$ has two irreducible representations in characteristic $2$: the trivial representation $T$ and a two-dimensional representation $U$. (In characteristic $2$, the sign representation collapses to the trivial one). It is clear that $E[2] = U$ as an $S_3$-module, and
since $U$ is self-dual we can identify the functor $(U \otimes (\cdot))^{S_3}$ with ${\rm{Hom}}_{S_3}(U, \cdot)$. This is an {\em exact}
functor on $\F_2[S_3]$-modules. Indeed, if $N$ is any $\F_2[S_3]$-module then
$${\rm{Ext}}^i(U, N) = H^i(S_3, U^{\ast} \otimes_{\F_2} N) = H^i(S_3, U \otimes_{\F_2} N)$$
and we claim that this vanishes for $i > 0$. It suffices to check vanishing for the analogous cohomology of a 2-Sylow, over which $U$ is
free over the group algebra. In general if $G$ is a finite group and $V$ is a $k[G]$-module for a ring $k$ then we claim that
$k[G] \otimes_k V$ is an induced module (so has vanishing higher cohomology); this is due to the classical observation
that for the underlying $k$-module $V_0$ with trivial action we have a $G$-module isomorphism $k[G] \otimes_k V \simeq k[G] \otimes_k V_0$
via $g \otimes v \mapsto g \otimes (g.v)$.
The upshot is that to compute the size of $(U \otimes H^1(\Q_S/L, \Z/2\Z))^{S_3}$, it suffices to find a filtration of
$H^1(\Q_S/L, \Z/2\Z)$ as an $S_3$-module and compute the sizes for $S_3$-invariants of $U$ tensored against each successive
quotient in the filtration.
The ring $\Cal{O}_{L,S}$ of $S$-integers of $L$ has trivial class group, so the $S$-integral Kummer sequence gives
$$H^1(\Q_S/L, \Z/2\Z) = H^1(\Q_S/L, \mu_2) = \Cal{O}_{L,S}^{\times}/(\Cal{O}_{L,S}^{\times})^2$$
as a $G_{L/\Q}$-module. The unit group is
\[
\Cal{O}_{L,S}^{\times} = \langle \pi_2, \pi_{11}^a, \pi_{11}^b, \pi_{11}^c \rangle \Cal{O}_L^{\times},
\]
where $\pi_{11}^{a,b,c}$ are respective generators for the three primes lying over $11$ (and $\pi_2$ is a generator for the unique prime over 2).
Thus, we have an exact sequence of $G_{L/\Q}$-modules
$$1 \rightarrow \Cal{O}_L^{\times}/(\Cal{O}_L^{\times})^2 \rightarrow \Cal{O}_{L,S}^{\times}/(\Cal{O}_{L,S}^{\times})^2 \rightarrow
T \oplus T \oplus U \rightarrow 0.$$
Tensoring against $U$ and taking invariants, the right term gives
$$(U \oplus U \oplus (U \otimes U))^{S_3} = (U \otimes U)^{S_3} = (U^{\ast} \otimes U)^{S_3} = {\rm{End}}_{S_3}(U) = \F_2.$$
Thus, it remains to show that $(U \otimes (\Cal{O}_L^{\times}/(\Cal{O}^{\times}_L)^2))^{S_3}$ is 1-dimensional over $\F_2$.
Consider the filtration
$$1 \rightarrow \langle -1 \rangle \rightarrow \Cal{O}_L^{\times}/(\Cal{O}_L^{\times})^2 \rightarrow
\Cal{O}_L^{\times}/(\langle -1 \rangle \cdot (\Cal{O}_L^{\times})^2) \rightarrow 1.$$
The quotient $\Cal{O}_L^{\times}/\langle -1 \rangle$ is free of rank $2$, and by computing a basis one sees by inspection
that the action of $G_{L/\Q}$ gives the standard representation of $S_3$ on $\Z^2$. Hence,
$\Cal{O}_L^{\times}/(\Cal{O}_L^{\times})^2$ is an extension of $U$ by $T$, so
applying the exact functor $(U \otimes (\cdot))^{S_3}$ gives a further contribution of
$(U \otimes U)^{S_3}$ that we have already seen is 1-dimensional
Hence, ${\rm{H}}^1(G_{\Q,S}, E[2])$ is 2-dimensional as claimed.
\end{document}