Another question related to the isogeny theorem for elliptic curves

I was reading the following question: About isogeny theorem for elliptic curves and was interested in the following statement at the end of Torsten Ekedahl's answer:

"Note also that the situation is similar (not by chance) to the case of CM-curves. If we look at CM-elliptic curves with a fixed endomorphism ring, then algebraically they can not be put into bijection with the elements of the class group of the endomorphism ring (though they can analytically), you have to fix one elliptic curve to get a bijection."

Could someone please clear up exactly what could be meant by algebraically they cannot be put into bijection...'?

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I would add the words "as of yet". Maybe it is possible, but just not currently known. We want to choose a "correct" identity element. Over $\mathhbb{C}$ this is immediate: $\mathbb{C}/O$, where $O$ is the given order. In some sense, this can also be done p-adic analytically at the primes of multiplicative reduction. I don't know if others would agree, but I think the essence of this problem is the following: given a rational weight two eigenform, how do you algebraically get the j-invariant of the associated optimal curve? As above, currently, we only know how to do this analytically. –  Dror Speiser Oct 9 '12 at 20:26
@Dror:What do you mean by associated optimal curve'? Could you point me to somewhere this is discussed? –  Adam Harris Oct 9 '12 at 20:44
@Dror:Also, could you formalise what problem you have in mind when you state that you could add "as of yet"..? (I am interested in this due to purely model theoretic considerations, and it would be very interesting to know if number theorists have already thought about similar things for different reasons) –  Adam Harris Oct 9 '12 at 20:48
@Adam: this is the name for the quotient $J_1(N)/I_fJ_1(N)$. Also called Weil curve. –  Dror Speiser Oct 9 '12 at 20:49
@Dror:Cheers. Could you give a brief idea of how the Weil curve and the problem of recovering $j(\mathcal{O})$ (amongst the solutions of the class polynomial) algebraically are related? –  Adam Harris Oct 9 '12 at 21:37

I think that maybe what is meant is that there is no functorial way to define the bijection in the category of algebraic geometry. So suppose that we let $\hbox{Ell}(R)$ denote the set of elliptic curves with $\hbox{End}(E)\cong R$, where for simplicity $R$ is the maximal order of an imaginary quadratic field. (The isomorphism with $R$ is over $\overline{\mathbb{Q}}$.) The ideal class group $H_R$ is, as its name proclaims, a group. So it has a preferred element, namely the identity element. But $\hbox{Ell}(R)$ does not have a preferred element in any natural sense. The right way to think of this is that there is a natural action of $H_R$ on $\hbox{Ell}(R)$, and this action is simply transitive. In particular, $H_R$ and $\hbox{Ell}(R)$ have the same number of elements. But if you want to identify them $\hbox{Ell}(R)\leftrightarrow H_R$, you need to choose an element of $\hbox{Ell}(R)$ to be distinguished.

On the other hand, if you fix an embedding $R\subset\mathbb{C}$, then every ideal $\mathfrak{a}$ in $R$ is a lattice in $\mathbb{C}$, so you can identify the ideal class $\overline{\mathfrak{a}}$ with the complex torus $\mathbb{C}/\mathfrak{a}$. This is analytically isomorphic to an elliptic curve $E_{\mathfrak{a}}$ in $\hbox{Ell}(R)$.

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In case the class group of $R$ is odd, perhaps we can recognize the identity element on the elliptic curve side: it should be the one with real j-invariant, right? Corresponding to the unique lattice isomorphic to its complex conjugate. –  Tommaso Centeleghe Oct 10 '12 at 7:43
yes - this was my initial motivation for asking the question, and whether any more can be done when the class number isn't odd –  Adam Harris Oct 10 '12 at 8:00
I have the feeling that in order to set up your bijection one would have to define what is the j-invariant of an element of the ideal class group H_R, and then make elliptic curves and ideal classes correspond though j. However this requires the choice of an orientation of R\otimes\reals, i.e., of an embedding of R into the complex #s. right? –  Tommaso Centeleghe Oct 10 '12 at 12:22
@Tommaso: I'm not sure what you mean. Aren't the solutions of the class polynomial exactly the j-invariants of the corresponding ideals? In which case - why do you have to mention the j-invariant? I'm not sure why you can't just look for a real solution of the class polynomial? –  Adam Harris Oct 10 '12 at 13:07
The problem seems to be that you cannot define j of an ideal class without first regarding it as a lattice in the complex plane C. Right? A posteriori the two embeddings of R in C will tell you that there are two reasonable j-invariants you can attach to each ideal class, they are complex conjugate of each other. Another point about the real j-invariant, is that if you give yourself just the Class polynomial, it is not clear how to get the real root. I think you should first split it in the complex numbers. I find these questions interesting. –  Tommaso Centeleghe Oct 10 '12 at 13:34

The class group of the endomorphism ring $\mathcal O_K$ is defined over $K$. But unless the class group of $\mathcal O_K$ is trivial, none of the CM curves are defined over $K$. Thus there is no bijection defined over $K$.

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@Will:What do you mean by the class group of the endomorphism ring is defined over $K$? –  Adam Harris Oct 9 '12 at 20:40
Also, does this mean there is a bijection defined over $\overline{\mathbb{Q}}$? –  Dror Speiser Oct 9 '12 at 21:16
@Adam: I mean that the elements of the class group are literally certain equivalence classes of certain sets of elements of $K$. @Dror: The sets have the same cardinality, so yes, there is. –  Will Sawin Oct 9 '12 at 23:43

I just want to add to Joe's excellent answer the following much simpler example that shows the same behavior. You might ask: what is the relationship between the set of square roots of -1 in $\bar{\mathbf{Q}}$ and the group $A = \pm 1$? As mere sets, one may say they're in bijection, which is not a very rich statement; it just says there are two of each.

When we say that $\pm i$ and $A$ are not algebraically in bijection, we are saying that there is no bijection between the two sets which is commutes with the action of the Galois group $G_Q$ on the left. On the other hand, just as in the case Joe describes, A acts on $\pm i$ (by multiplication) simply transitively, and this action is compatible with Galois; it satisfies

$a t^\sigma = (at)^\sigma$

for each $t$ in $\pm 1$ and each $\sigma$ in Galois.

We say that $\pm i$ is a torsor for $A$.

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