TITLE: Distribution relation in the Euler system of Heegner points
QUESTION [13 upvotes]: I am trying to understand the details behind the so-called "distribution relations" between Heegner points on the modular curve $X_0(N)$, as given (for instance) in Gross's paper Kolyvagin's work on modular elliptic curves, [Proposition 3.7, (i)]. More precisely, the relation between Hecke and Galois actions on CM-points is still not clear for me. First of all let me recall in details the  general settings, which the familiar reader can skip and go directly to the question.
Settings:
Let $N$ be a positive integer, let $K$ be an imaginary quadratic field of discriminant $D<0$ such that every rational prime divisor of $N$ splits in $K$ (the so-called Heegner hypothesis). The Heegner hypothesis implies that one can find a (non-unique) ideal $\mathcal{N}$ of $\mathcal{O}_K$ such that $\mathcal{O}_N/\mathcal{N}\simeq \mathbb{Z}/N\mathbb{Z}$.
Edit: As noticed by Olivier I also have to assume $D<-4$, to ensure that $\mathcal{O}_K^{\times}=\{\pm 1\}$.
For a general $n$, denote by $\mathcal{O}_n=\mathbb{Z}+n\mathcal{O}_K\subset\mathcal{O}_K$ "the" order of $K$ of conductor $n$, and let $Pic(\mathcal{O}_n)$ be the ideal class group of $\mathcal{O}_n$ (which is isomorphic to $I(n)/P(n)$, with $I(n)$ the group of ideals in $\mathcal{O}_K$ that are prime to $n$; and $P(n)$ the subgroup of principal ideals in $\mathcal{O}_K$ generated by an element congruent mod $n\mathcal{O}_K$ to some $r\in\mathbb{Z}$, $(r,n)=1$).
Class-field theory provides us with an abelian extension $K_n$ of $K$ such that $\mathrm{Gal}(K_n/K)\simeq Pic(\mathcal{O}_n)$.
Let now $n$ be an integer prime to $ND$. Setting $\mathcal{N}_n:=\mathcal{N}\cap\mathcal{O}_n$, we get that $\mathcal{O}_n/\mathcal{N}_n\simeq \mathcal{O}_K/\mathcal{N}\simeq \mathbb{Z}/N\mathbb{Z}$. One can thus define a point $x_n\in X_0(N)$ by setting
$$x_n=[\mathbb{C}/\mathcal{O}_n\rightarrow \mathbb{C}/\mathcal{N}_n^{-1}]$$
(here $[E\rightarrow E']$ denotes the isomorphism class of the pair $(E,E')$ of elliptic curves, with $E\rightarrow E'$ a cyclic $N$-isogeny).
If $l$ is a rational prime not dividing $ND$ nor $n$, we can mimic the previous discussion and set 
$$x_{nl}=[\mathbb{C}/\mathcal{O}_{nl}\rightarrow \mathbb{C}/\mathcal{N}_{nl}^{-1}]$$ 
The theory of complex multiplication ensures that $x_n\in X_0(N)(K_n)$ (resp. $x_{nl}\in X_0(N)(K_{nl})$ ) 
Here comes my question: assuming $l$ is inert in $K/\mathbb{Q}$, why do we have
$$\mathrm{Tr}_{K_{nl}/K_n}x_{nl}:=\sum_{\sigma\in\mathrm{Gal}(K_{nl}/K_n)}\sigma x_{nl} = T_l x_n$$ 
as divisors on $X_0(N)$ ?
What I "understood" is that:

For $\sigma\in \mathrm{Gal}(K_{nl}/K)$, the action of $\sigma$ on $x_{nl}$ is given by
$\sigma x_{nl}=[\mathbb{C}/\mathfrak{a}_{\sigma}^{-1}\rightarrow \mathbb{C}/\mathfrak{a}_{\sigma}^{-1}\mathcal{N}_{nl}^{-1}]$
where $\mathfrak{a}_{\sigma}$ is any (proper) ideal of $\mathcal{O}_{nl}$ such that $[\mathfrak{a}_{\sigma}]\in Pic(\mathcal{O}_{nl})$ corresponds to $\sigma$ via the isomorphism $\mathrm{Gal}(K_{nl}/K)\simeq Pic(\mathcal{O}_{nl})$.
Thus the sum $\mathrm{Tr}_{K_{nl}/K_n}x_{nl}$ rewrites as 

$$\sum_{\sigma\in\mathrm{Gal}(K_{nl}/K_n)}[\mathbb{C}/\mathfrak{a}_{\sigma}^{-1}\rightarrow \mathbb{C}/\mathfrak{a}_{\sigma}^{-1}\mathcal{N}_{nl}^{-1}]$$
Here the condition $\sigma\in\mathrm{Gal}(K_{nl}/K_n)$ implies, I think, that $\mathfrak{a}_{\sigma}\mathcal{O}_n$ is a principal $\mathcal{O}_n$-ideal.
Here $\mathrm{Gal}(K_{nl}/K_n)\simeq \mathbb{F}_{l^2}^{\times}/\mathbb{F}_{l}^{\times}$, so the sum has $l+1$ terms.

The action of the Hecke operator $T_l$ ($l$ not dividing $N$) on divisors of the modular curve $X_0(N)$ can be described (at least in characteristic $0$) as
$$T_l [E\xrightarrow{\phi} E']=\sum_{C\subset E[l], \#C=l} [E/C\rightarrow E'/\phi(C)]$$
(there are $l+1$ such $C$)
In my situation, this can be rewritten (following Gross, Heegner points on $X_0(N)$, §6) as
$$T_l x_n=\sum_{\mathfrak{b}\subset\mathcal{O}_n\text{lattice of index } l}[\mathbb{C}/\mathfrak{b}\rightarrow \mathbb{C}/\mathfrak{b}(\mathcal{N}_n\cap End(\mathfrak{b}))^{-1}] $$

What I don't understand is why the exactly the same terms should appear in both sums. I get that $(id)x_{nl}=x_{nl}$ appears in $T_l x_n$, as $\mathcal{O}_{nl}$ is a sub-lattice of order $l$ in $\mathcal{O}_n$ with $End(\mathcal{O}_{nl})=\mathcal{O}_{nl}$, but I don't see why $\mathfrak{a}_{\sigma}^{-1}$ is a sublattice of order $l$ in $\mathcal{O}_n$  if $\sigma$ fixes $K_n$.
I think this involves properties about fractional ideals of orders which I don't  quite understand. 
I thank everyone taking the time to read this question and trying to provide me with any help !

REPLY [3 votes]: Thank you Olivier for this great answer, I wish I could be one of your students ;-) !
Though I still not quite fully understand the adelic setting, here is some "classical" explanation I found, using the theory of ideals in orders of imaginary quadratic fields. I guess this is supposed to be a very well-known result but, since I haven't found any complete proof of it, I'll write down what I hope to be a detailed one in case some people are interested. My main reference is D.A. Cox 's "Primes of the form $x^2+ny^2$", chapters 7 and 8.
First, slightly modify the notations and set for any $f\in\mathbb{Z}$:

$I_K(f)$: group of ideals of $\mathcal{O}_K$ which are prime to $f$ (i.e., whose norm is prime to $f$)
$P_{K,\mathbb{Z}}(f)$: subgroup of $I_K(f)$ made of principal ideals of the form $\alpha\mathcal{O}_K$, with $\alpha$ congruent to some $a$ mod $f\mathcal{O}_K$, with $a\in\mathbb{Z}$ and $(a,f)=1$.

Here the ideals are meant to be integral ideals, and the group structure is obtained by identifying $\mathfrak{a}$ and $\mathfrak{b}$ iff there are some $\alpha$, $\beta\in \mathcal{O}_K$ such that $\alpha\mathfrak{a}=\beta\mathfrak{b}$.
The quotient $I_K(f)/P_{K,\mathbb{Z}}(f)$ is what we call a generalized ideal class group, isomorphic to $Pic(\mathcal{O}_f)\simeq \mathrm{Gal}(K_f/K)$ ($K_f/K $ is the ring class field of conductor $f$, the isomorphism being defined by sending any prime-to-$f$ prime ideal of $\mathcal{O}_K$ to its Frobenius element). Here we set $Pic(\mathcal{O}_f)$ to be the group formed by the quotient of proper ideals of $\mathcal{O}_f $ (i.e., ideals $\mathfrak{a}\subset \mathcal{O}_f$ such that $End(\mathfrak{a})=\mathcal{O}_f)$, by principal ideals.
An important fact is that any (proper or not) $\mathcal{O}_f$-ideal is a free  $\mathbb{Z}$-module of rank $2$.
The isomorphism between $I_K(f)/P_{K,\mathbb{Z}}(f)$ and $Pic(\mathcal{O}_f)$ is given by sending
$$\mathcal{O}_K\supset\mathfrak{a} \mapsto \mathfrak{a}\cap\mathcal{O}_f$$
One checks that elements of $P_{K,\mathbb{Z}}(f)$, of the form $\alpha\mathcal{O}_K$ are sent to $\alpha\mathcal{O}_f $ (this follows from $(a,f)=1$, with $a$ "the" integer congruent to $\alpha$ mod $f\mathcal{O}_K$.)
The inverse isomorphism is given by sending
$$\mathcal{O}_f\supset\mathfrak{a}\mapsto \mathfrak{a}\mathcal{O}_K$$
(see Cox's 7.18, 7.20).
Though it's not strictly necessary, I mention that the norm is preserved by these maps. The case$ f=1$ gives us the Hilbert class field $K_1 $ of $K$, such that $I_K/P_K=Pic(\mathcal{O}_K)\simeq \mathrm{Gal}(K_1/K)$. With these identifications, one obtains that the group $\mathrm{Gal}(K_f/K_1)$ corresponds to $I_K(f)\cap P_K/P_{K,\mathbb{Z}}(f)$.
An important fact is that, if $\mathcal{O}_K^{\times}=\{\pm 1\}$, we have the following exact sequence:
$$1\rightarrow (\mathbb{Z}/f\mathbb{Z})^{\times}\rightarrow (\mathcal{O}_K/f\mathcal{O}_K)^{\times}\rightarrow I_K(f)\cap P_K/P_{K,\mathbb{Z}}(f) \rightarrow 1$$
(see Cox, 7.27)
Applying this to $f=n$, $nl$, one gets that
$$\mathrm{Gal}(K_{nl}/K_1)\simeq (\mathcal{O}_K/nl\mathcal{O}_K)^{\times}/(\mathbb{Z}/nl\mathbb{Z})^{\times}$$
$$\mathrm{Gal}(K_{n}/K_1)\simeq (\mathcal{O}_K/n\mathcal{O}_K)^{\times}/(\mathbb{Z}/n\mathbb{Z})^{\times}$$
Therefore $\mathrm{Gal}(K_{nl}/K_n)$ is isomorphic to $$\frac{(\mathcal{O}_K/nl\mathcal{O}_K)^{\times}/(\mathbb{Z}/nl\mathbb{Z})^{\times}}{(\mathcal{O}_K/n\mathcal{O}_K)^{\times}/(\mathbb{Z}/n\mathbb{Z})^{\times}}
$$
$$\simeq (\mathcal{O}_K/l\mathcal{O}_K)^{\times}/(\mathbb{Z}/l\mathbb{Z})^{\times}$$
as $(n,l)=1$. This last group is isomorphic to $\mathbb{F}_{l^2}^{\times}/\mathbb{F}_{l}^{\times}$, for $l$ is inert in $K/\mathbb{Q}$, so has cardinal $l+1$.
Elements of the group $\mathrm{Gal}(K_{nl}/K_n)$ also corresponds to prime-to-$nl$ principal ideals of the form $\mathfrak{a}=\alpha\mathcal{O}_K$, with $\alpha$ congruent to $a\in\mathbb{Z}$ mod $n\mathcal{O}_K$, $(a,n)=1$. Notice that $\alpha\in\mathcal{O}_{nl}$ iff. it corresponds to the identity of $\mathrm{Gal}(K_{nl}/K_n)$.
Denote by $\mathfrak{a}(\sigma)=\alpha(\sigma)\mathcal{O}_K$ such an ideal corresponding to $\sigma\in\mathrm{Gal}(K_{nl}/K_n)$, and set $\mathfrak{a}_{\sigma}:=\mathfrak{a}({\sigma})\cap\mathcal{O}_{nl}$  the corresponding proper ideal of $\mathcal{O}_{nl}$.
As a consequence, if $\sigma\in\mathrm{Gal}(K_{nl}/K_n)$, then $\mathfrak{a}(\sigma)\cap\mathcal{O}_n=\alpha(\sigma)\mathcal{O}_n$ is principal, thus 
$\mathfrak{a}_{\sigma}\subset\mathcal{O}_{nl}$ satisfies
$$\alpha(\sigma)^{-1}\mathfrak{a}_{\sigma}\subset \mathcal{O}_n.$$
We have 
$$\frac{\mathcal{O}_n}{\alpha(\sigma)^{-1}\mathfrak{a}_{\sigma}}\xrightarrow{\sim}\frac{\mathfrak{a}(\sigma)\cap\mathcal{O}_n}{\mathfrak{a}_{\sigma}}\hookrightarrow \frac{\mathcal{O}_n}{\mathcal{O}_{nl}},$$
the first map being multiplication by $\alpha(\sigma)$.
As $\frac{\mathcal{O}_n}{\mathcal{O}_{nl}}\simeq \mathbb{Z}/l\mathbb{Z}$ and as $\frac{\mathfrak{a}(\sigma)\cap\mathcal{O}_n}{\mathfrak{a}_{\sigma}}$ is non-trivial (if $\sigma$ is not trivial then $\alpha(\sigma$) belongs to $\mathfrak{a}(\sigma)\cap\mathcal{O}_n$ but not to $\mathcal{O}_{nl})$; if $  \sigma$ is trivial then one directly gets $\frac{\mathfrak{a}(\sigma)\cap\mathcal{O}_n}{\mathfrak{a}_{\sigma}}=\frac{\mathcal{O}_{n}}{\mathcal{O}_{nl}})$,
this finally implies that 
$$\frac{\mathcal{O}_n}{\alpha(\sigma)^{-1}\mathfrak{a}_{\sigma}}\simeq \mathbb{Z}/l\mathbb{Z}$$
Combining this with the formula
$$\mathrm{Tr}_{K_{nl}/K_n}(x_{nl})=\sum_{\sigma\in\mathrm{Gal}(K_{nl}/K_n)}[\mathbb{C}/\mathfrak{a}_{\sigma}^{-1}\rightarrow \mathbb{C}/\mathfrak{a}_{\sigma}^{-1}\mathcal{N}_{nl}^{-1}]$$
(see my first post), and the facts that 1- $End(\alpha(\sigma)^{-1}\mathfrak{a}_{\sigma})=End(\mathfrak{a}_{\sigma})=\mathcal{O}_{nl}$, 2- the isomorphism class of an elliptic curve $\mathbb{C}/\Lambda$ is invariant under $\Lambda\mapsto \alpha\Lambda$ and 3- if $\sigma\neq \sigma' $ then $ \mathfrak{a}_{\sigma}$ and $\mathfrak{a}_{\sigma'}$ define different elements in $Pic(\mathcal{O}_{nl})$, one gets that $\mathrm{Tr}_{K_{nl}/K_n}(x_{nl})$ is a sum of $l+1$ distinct elements of the form
$$[\mathbb{C}/\mathfrak{b}\rightarrow \mathbb{C}/\mathfrak{b}(\mathcal{N_n}\cap End(\mathfrak{b}))^{-1}]$$
with $\mathfrak{b}\subset \mathcal{O}_n$ a sub-lattice of index $l$.
This is precisely the equality
$$\mathrm{Tr}_{K_{nl}/K_n}(x_{nl})=T_l(x_n)$$
which I was looking for.