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In the this paper (https://arxiv.org/abs/1203.4392), on page 2 it is said that

Since DVCS amplitude is symmetric under $s \leftrightarrow u$ channel crossing, the CFFs and $^{S}C^i_{\cdots}$ coefficients have definete symmetry properties under $\xi$ reflexion

My problem is that I don't see $u$ channel in DVCS, just the $t$ one due to the $e^- e^- \gamma$ vertex in diagram (a) in Fig. 1. What am i missing and what is the relation among Mandelstam and $\xi$?

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Note that the transformation $\xi\rightarrow-\xi$ corresponds to a change of $q_{in}^2\leftrightarrow q_{out}^2$ (Eq. (4)), which means that $p_1\leftrightarrow -p_3$ (Eq. (1), the minus sign is needed to have momentum conservation), which changes $(p_1+p_2)^2\leftrightarrow (p_2-p_3)^2$ in other words $s\leftrightarrow u$.

Also remember that $s+t+u=0$ meaning that if something depends on $t$ only it means it depends on $-(s+u)$.

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  • $\begingroup$ A change in the sign of $\xi$ is not given by $q_{in}^2 \leftrightarrow q_{out}^2$ bc the $\eta$ dependence compensate that exchange, so $\xi$ remains the same $\endgroup$
    – Vicky
    Commented Jul 6, 2020 at 4:18
  • $\begingroup$ $\eta$ doesn't change sign, both the numerator and denominator change sing under the exchange $p_1\leftrightarrow -p_3$ ($q_{in}\leftrightarrow -q_{out}$. $\endgroup$
    – Diógenes Figueroa
    Commented Jul 6, 2020 at 12:03

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