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It is a consequence of the generalized Sato-Tate conjecture, that given a non-CM elliptic curve over $\mathbb Q$, any element of $Gal(\bar{\mathbb Q}/\mathbb Q)$, and a real number $\in[−1,1]$, one can construct an ultrafilter on the primes such that Frobenius converges to that element and the angle of Frobenius converges to that real number.

Thus any attempt to answer this question must somehow make use of the transcendentals. I have no idea how one might do that.

EDIT: By ACL's answer earlierto my question, but this comment the nonstandard angle of Frobenius is a better one.)totally independent from all first-order statements about the nonstandard elliptic curve.

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The trace of Frobenius of an elliptic curve and its the number of points over the finite field it is over are both integers, $a_p$ and $p$, satisfying the relation $a_p^2\leq 4p$.

It seems to me that a nonstandard angle should come from is a pair consequence of nonstandard numbers, $a_u$ and $u$, satisfying $a_u^2\leq 4u$. Then with the natural map from nonstandard integers to nonstantard realsgeneralized Sato-Tate conjecture, $a_u/2\sqrt{u}$ would be a nonstandard real number in the interval $[-1,1]$, thus that given a real number, and the $\cos$ of some real number.

In particular you're working over an ultrafilter non-CM elliptic curve over the set of primes, so you know what you want $u$ to be already. So the question is for ways to find nonstandard numbers $a_u$ satisfying this inequality.

The Tate module allows you to compute\mathbb Q$, from an any element of$\sigma \in Gal(\bar{\mathbb Q}/\mathbb Q)$, an element of and a real number$\hat{\mathbb Z}$, \in[−1,1]$, one can construct an ultrafilter on the trace, primes such that if $\sigma=Frob_p$, is equal Frobenius converges to $a_p$ everywhere except $\mathbb Z_p$. Since $\hat{\mathbb Z}$ is a compactification that element and the angle of an integers, there is a map Frobenius converges to it from any notion of nonstandard integersthat real number.

Thus any notion that sends automorphisms of $\bar{\mathbb Q}$, or $\mathbb C$, attempt to angles or traces of Frobenius, should probably form a commutative diagram with the trace answer this question must somehow make use of the Tate module and that maptranscendentals.

The reason this cannot serve as a definition on its own is I have no idea how one might do thatas far as .

(I know this map would not be injectivehad a different answer earlier, nor would the traces need to satisfy the relevant inequality.but this comment is a better one.)

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The trace of Frobenius of an elliptic curve and its the number of points over the finite field it is over are both integers, $a_p$ and $p$, satisfying the relation $a_p^2\leq 4p$. It seems to me that a nonstandard angle should come from a pair of nonstandard numbers, $a_u$ and $u$, satisfying $a_u^2\leq 4u$. Then with the natural map from nonstandard integers to nonstantard reals, $a_u/2\sqrt{u}$ would be a nonstandard real number in the interval $[-1,1]$, thus a real number, and the $\cos$ of some real number.

In particular you're working over an ultrafilter over the set of primes, so you know what you want $u$ to be already. So the question is for ways to find nonstandard numbers $a_u$ satisfying this inequality.

Do

The Tate module allows you want different ways to computethe same number , from an elliptic curveelement of $\sigma \in Gal(\bar{\mathbb Q}/\mathbb Q)$, or entirely different ways to compute an element of $\hat{\mathbb Z}$, the number? Do you want trace, that if $\sigma=Frob_p$, is equal to compute $a_p$ everywhere except $\mathbb Z_p$. Since $\hat{\mathbb Z}$ is a compactification of an integers, there is a map to it from an elliptic curve over an ultra-finite field?

I'm not sure how an automorphism any notion of the complex numbers relates nonstandard integers. Thus any notion that sends automorphisms of $\bar{\mathbb Q}$, or $\mathbb C$, to angles or traces of Frobenius, should probably form a commutative diagram with the trace of the Tate module and that map.

The reason this cannot serve as a definition on its own is that as far as I know this map would not be injective, nor would the traces need to satisfy the relevant inequality.

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