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This claim comes from Henri Darmon, and I would believe it, but can't exactly prove it. Let $\mathcal{F}$ denote the space of $\mathbb{C}$-valued functions on $\mathbb{P}_1(\mathbb{Q})$, and let $\mathcal{M}$ denote the space of $\mathbb{C}$-valued modular symbols. Let $d:\mathcal{F}\to\mathcal{M}$ by $f\mapsto df$ where $(df)\{x\to y\} = f(y) - f(x)$.

The claim is that $d$ is surjective. I see that the right hand side indeed gives a complex valued modular symbol. Is there a very easy way to see that all complex valued modular symbols arise in such a way? Is it obvious?

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up vote 2 down vote accepted

I figured it out (with a little help from a friend). For those of you who are interested:

Let $m\in\mathcal{M}$ be a modular symbol and define $f(y) = m\{1\to y\}$. Then $f$ is a complex valued function on $\mathbb{P}_1(\mathbb{Q})$, hence $f\in\mathcal{F}$. Moreover we have, by transitivity of modular symbols, that $$m\{x\to y\} = m\{x\to 1\} + m\{1\to y\}.$$ Then, since modular symbols are antisymmetric we have $$m\{x\to y\} = -m\{1\to x\} + m\{1\to y\} = f(y) - f(x),$$ so indeed the map $d$ is surjective.

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Nice -- this argument works over an arbitrary base ring as well (it's not specific to $\mathbb{C}$-valued symbols). – David Loeffler Aug 28 '12 at 9:07

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