Local Ext for reflexive sheaves on surfaces Let $X$ be a normal Gorenstein complex surface with $H^i(X,\mathcal{O}_X)=0$ for $i>0$ and $F$ be a rank one reflexive sheaf on $X$. I'm trying to find some ways to determine local Ext $\mathcal{E}xt^i_X(F,F)$.
For $i=0$, by the normality of $F$, I think we have $\mathcal{H}om_X(F,F)=\mathcal{O}_X$. Is there any similar result for $i>0$ (e.g. $\mathcal{E}xt^i_X(F,F)=0$ for $i>0$)?
The only thing now I can show is that $\mathcal{E}xt^i_X(F,F)$ is supported on points for $i>0$.
 A: Firstly, note that formation of $\mathscr Hom_{\mathcal O_X}(\mathscr F,\mathscr F)$ is local, so we don't need global assumptions such as $H^i(X,\mathcal O_X) = 0$ for $i > 0$.
Secondly, I agree that the natural map $\mathcal O_X \to \mathscr Hom_{\mathcal O_X}(\mathscr F,\mathscr F)$ is an isomorphism whenever $\mathscr F$ is a rank $1$ reflexive sheaf on a normal (integral) Noetherian scheme $X$. Indeed, there exists an open $j \colon U \hookrightarrow X$ with $\operatorname{codim}(X\setminus U,X) \geq 2$ such that $j^*\mathscr F$ is locally free of rank $1$ and the unit $\mathscr F \to j_*j^*\mathscr F$ of the adjunction $j^* \dashv j_*$ is an isomorphism; see [Tag 0AY6]. Then the natural map
$$\mathcal O_U \to \mathscr Hom_{\mathcal O_U}(j^*\mathscr F,j^*\mathscr F) = j^*\mathscr Hom_{\mathcal O_X}(\mathscr F,\mathscr F)$$
is an isomorphism, so the claim follows from [Tags 0AY4 and 0EBJ].
But the higher vanishing does not hold:
Example. Let $k$ be a field, and let $R = k[x,y,z]/(z^2-xy)$ be the quadratic cone. Let $I = (x,z)$ be the ideal of the $y$-axis. We get a surjection $R^2 \to I$ by $(a,b) \mapsto ax-bz$, and write $K \subseteq R^2$ for the kernel. Then the second projection $\pi \colon K \to R$ is injective with image $I$: if $(a,0) \in K$, then $a = 0$ since $R$ is a domain, showing injectivity. The image of $\pi \colon K \to R$ is those $r \in R$ such that $rz \in (x)$, i.e. the annihilator of $z$ in $R/(x) \cong k[x,y,z]/(x,z^2)$. This is the ideal $I = (x,z)$. (Concretely, $K$ is generated by $(z,x)$ and $(y,z)$.) Thus we get a short exact sequence
$$0 \to I \to R^2 \to I \to 0,$$
which shows that $I$ is reflexive [Tag 0AV2] (see also [Tag 0EBM]). Applying $\operatorname{Hom}_R(-,I)$ gives an exact sequence
$$0 \to \operatorname{Hom}_R(I,I) \to \operatorname{Hom}_R(R^2,I) \to \operatorname{Hom}_R(I,I) \to \operatorname{Ext}^1_R(I,I) \to 0.$$
By the above, the first terms read
$$0 \to R \to I^2 \to R.$$
The final map cannot be surjective, for then $I$ would be projective: if $M$ is any $R$-module, then the short exact sequence $0 \to R \to I^2 \to R \to 0$ would give
$$0 = \operatorname{Ext}^i_R(R,M) \to \operatorname{Ext}^i_R(I^2,M) \to \operatorname{Ext}^i_R(R,M) = 0,$$
showing that $\operatorname{Ext}^i_R(I,M)^2 = 0$ for $i > 0$. This is absurd since $I$ is not a principal ideal. We conclude that $\operatorname{Ext}^1_R(I,I)$ cannot be zero. $\square$
