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Let $X$ be an abelian variety defined over a number field $K$. We know that the Neron--Tate height machine associates to a class in the Picard group of $X$ a unique quadratic function which is zero at the identity of $X$. And it is known that modulo torsion this association homomorphism is injective. Consider the homomorphism $$ h:\text{Pic}(X_{\bar K})\otimes_\mathbb{Z}\mathbb{R}\rightarrow \{ \text{quadratic real functions on }X(\bar K)\text{ which vanish at the identity} \}. $$My question is, is this homomorphism surjective (and an isomorphism?)?

I'm asking this out of curiosity mainly. I don't know if this is well-known to the experts (or just another silly question of mine). Thank you in advance!

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    $\begingroup$ Why is the homomorphism injective? $\endgroup$
    – Will Sawin
    Commented Oct 16, 2022 at 22:47
  • $\begingroup$ @Will Sawin: Dear Will, I am not sure if this is correct actually. My question should have two consecutive question marks there. $\endgroup$
    – user unknown
    Commented Oct 17, 2022 at 2:22
  • $\begingroup$ OK, I confirm that after tensoring with $\mathbb{R}$, the map is still injective, which is actually quite easy to show. $\endgroup$
    – user unknown
    Commented Oct 21, 2022 at 2:39

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The source has countable dimension over $\mathbb R$, since $A$ has countably many divisors defined over a finite extension of $K$, while the target, being the space of quadratic functions on a countably-infinite-dimensional vector space, has uncountable dimension over $\mathbb R$, so the map can never be surjective.

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  • $\begingroup$ Sorry I should have been more clear on it. I really mean the map from the infinite dimensional space to the infinite dimensional space. The quadratic function here could contain a linear part, so for example, an algebraically trivial line bundle can give a nonzero linear function. The reference I'm looking at is Lang's Fundamentals of Diophantine geometry, chapter 5. $\endgroup$
    – user unknown
    Commented Oct 17, 2022 at 2:43
  • $\begingroup$ @userunknown See edited version. $\endgroup$
    – Will Sawin
    Commented Oct 17, 2022 at 12:59
  • $\begingroup$ Thanks a lot! I know very little about functional analysis; is there an easy way to think about this fact: "the space of quadratic functions on a countably-infinite-dimensional vector space has uncountable dimension"? Is it also true for linear functions? $\endgroup$
    – user unknown
    Commented Oct 17, 2022 at 17:17
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    $\begingroup$ @userunknown Yes, it's also true for linear functions. Exactly for the reason you say: A linear function on a countable-dimensional vector space is determined by countably many values, i.e. it's the product $\mathbb R^{\mathbb N}$ of countably many copies of $\mathbb R$. But that product does not have countable dimension, i.e. it lacks a countable basis. $\endgroup$
    – Will Sawin
    Commented Oct 17, 2022 at 18:25
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    $\begingroup$ @userunknown If it had a countable basis, it could have only countably many linearly independent vectors (at most $n$ using each set of $n$ basis vectors, and there are countably many sets of $n$ basis vectors for all $n$) but in fact the vectors $(1,t,t^2,\dots)$ are linearly independent for all real $t$. $\endgroup$
    – Will Sawin
    Commented Oct 17, 2022 at 18:28

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