commutative diagram with Yoneda pairing, Weil pairing and edge morphism

Why does the following diagram commute?$\require{AMScd}$ \begin{CD} H^0(X,\mathscr{A}) \times \mathrm{Ext}^2_X(\mathscr{A},\mu_{\ell^n}) @>>> H^2(X,\mu_{\ell^n}) \\ @VVV @| \\ H^1(X,\mathscr{A}[\ell^n]) \times H^1(X,\mathscr{E}xt_X^1(\mathscr{A},\mu_{\ell^n})) @>>> H^2(X,\mu_{\ell^n}). \end{CD} ($\mathscr{A}/X$ is an Abelian scheme.) The vertical homomorphism $\delta$ on the left is induced by the Kummer sequence $$0 \to \mathscr{A}[\ell^n] \to \mathscr{A} \to \mathscr{A} \to 0.$$

The upper row is the Yoneda pairing.

The lower row is induced by the Weil pairing $$\mathscr{A}[\ell^n] \times \mathscr{A}^\vee[\ell^n] \to \mu_{\ell^n}$$ using $\mathscr{E}xt^1(\mathscr{A},\mu_{\ell^n}) = \mathscr{H}om(\mathscr{A}[\ell^n],\mu_{\ell^n}) = \mathscr{A}^\vee[\ell^n]$ by the long exact Ext sequence for the Kummer sequence $0 \to \mathscr{A}[\ell^n] \to \mathscr{A} \to \mathscr{A} \to 0$: $$0 = \mathscr{H}om_X(\mathscr{A},\mu_{\ell^n}) \to \mathscr{H}om_X(\mathscr{A}[\ell^n],\mu_{\ell^n}) \to \mathscr{E}xt^1_X(\mathscr{A},\mu_{\ell^n}) \stackrel{0}{\to} \mathscr{E}xt^1_X(\mathscr{A},\mu_{\ell^n}),$$ the latter transition map being $0$ since $\ell^n$ kills $\mu_{\ell^n}$. (Edit 2: Or, better, $$0 = \mathscr{H}om_X(\mathscr{A},\mathbf{G}_m) \to \mathscr{E}xt^1_X(\mathscr{A},\mu_{\ell^n}) \to \mathscr{E}xt^1_X(\mathscr{A},\mathbf{G}_m) \stackrel{\ell^n}{\to} \mathscr{E}xt^1_X(\mathscr{A},\mathbf{G}_m)$$ and using $\mathscr{E}xt^1_X(\mathscr{A},\mathbf{G}_m) = \mathscr{A}^\vee$, as pointed out in the comments.)

Finally, for the middle vertical arrow: The local-to-global Ext spectral sequence $H^p(X,\mathscr{E}xt_X^q(\mathscr{A},\mu_{\ell^n})) \Rightarrow \mathrm{Ext}_X^{p+q}(\mathscr{A},\mu_{\ell^n})$ gives us an injection (an edge morphism, see also Natural morphism appearing in Grothendieck spectral sequence) $$H^1(X,\mathscr{E}xt_X^1(\mathscr{A},\mu_{\ell^n})) \hookrightarrow \mathrm{Ext}_X^2(\mathscr{A},\mu_{\ell^n})$$ since $\mathscr{H}om_X(\mathscr{A},\mu_{\ell^n}) = 0$.

Edit: Differently stated, why does \begin{CD} \mathrm{Ext}^2_X(\mathscr{A},\mu_{\ell^n}) @>>> \mathrm{Hom}(H^0(X,\mathscr{A}), H^2(X,\mu_{\ell^n})) \\ @AAA @A\delta^*AA \\ H^1(X,\mathscr{E}xt_X^1(\mathscr{A},\mu_{\ell^n})) @>>> \mathrm{Hom}(H^1(X,\mathscr{A}[\ell^n]), H^2(X,\mu_{\ell^n})) \end{CD} commute? (the left vertical arrow a edge map of the spectral sequence and the right vertical arrow $\delta^*$ with $\delta$ the connecting map of the Kummer sequence)

• If you write everything out with cocycles (eg wrt a fppf hypercovering), then you should be able to decide whether it is true or not by a computation. Did you try? Also, a reference for a good number of commutative diagrams involving the dual abelian scheme A^t, identifying it with Ext^1(A, G_m), and identifying A^t[l] with Hom(A[l], G_m), double duality, etc is Oort's LNM 15. – answer_bot Mar 7 '14 at 14:06
• The description of the bottom row is "cheating" a bit, since in reality one should say we use the isomorphism $\mathscr{Ext}^1(A,\mu_{\ell^n}) \simeq \mathscr{Ext}^1(A,\mathbf{G}_m)[\ell^n] = A^{\vee}[\ell^n]$ and the pairing between torsion-levels in $A$ and $A^{\vee}$. (Edit: answer_bot wrote a related comment at the same time.) – user76758 Mar 7 '14 at 14:08
• The diagrams appear all jumbled to me. Do other people have the same issue? – Ricardo Andrade Mar 9 '14 at 17:15
• I fixed the second diagram, hopefully. Dear @Timo Keller, please check it is as you intended. Regarding the first diagram in the question, I cannot tell what you were trying to write. Please fix it. – Ricardo Andrade Mar 9 '14 at 17:27
• Thank you very much. I tried to fix the first diagram accordingly, but on the left, there should by two vertical arrows pointing down and up, respectively. – TKe Mar 10 '14 at 15:02

1. Show that (in the lower diagram) $H^1(X,\mathscr{H}om(\mathscr{A}[\ell^n],\mu_{\ell^n}))$ embeds via the first edge homomorphism into $\mathrm{Ext}^1(\mathscr{A}[\ell^n],\mu_{\ell^n})$ (by the local-to-global Ext spectral sequence) and that the lower horizontal and left vertical arrows factor through this.
2. Now both paths map $e = (0 \to \mu_{\ell^n} \to \mathscr{F} \to \mathscr{A}[\ell^n] \to 0) \in \mathrm{Ext}^1(\mathscr{A}[\ell^n],\mu_{\ell^n})$ to $H^0(X,\mathscr{A}) \stackrel{\delta_\mathrm{Kummer}}{\to} H^1(X,\mathscr{A}[\ell^n]) \stackrel{\delta_e}{\to} H^2(X,\mu_{\ell^n})$.