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Let us consider the Cartesian product $X^r$, where $X$ is a smooth projective variety. There is a subgroup $Aut_{\Delta}(X^r)\subset Aut(X^r)$ of automorphisms of $X^r$ mapping the smalla $k$-dimensional diagonal in $\Delta\subset X^r$$X^r$ isomorphically to itselfa $k$-dimensional diagonal. In particular any automorphism in $Aut_{\Delta}(X^r)$ preserves the smallest diagonal $\Delta_{1,...,r}\subset X^r$.

If we consider the diagonal action of $Aut(X)$ on $X^r$ clearly we have $S_r\times Aut(X)\subseteq Aut_{\Delta}(X^r)$, where $S_r$ is the symmetric group. Under which hypothesis on $X$ (perhaps $X$ of general type) do we have $S_r\times Aut(X)= Aut_{\Delta}(X^r)$?

Let us consider the Cartesian product $X^r$, where $X$ is a smooth projective variety. There is a subgroup $Aut_{\Delta}(X^r)\subset Aut(X^r)$ of automorphisms of $X^r$ mapping the small diagonal $\Delta\subset X^r$ isomorphically to itself.

If we consider the diagonal action of $Aut(X)$ on $X^r$ clearly we have $S_r\times Aut(X)\subseteq Aut_{\Delta}(X^r)$, where $S_r$ is the symmetric group. Under which hypothesis on $X$ (perhaps $X$ of general type) do we have $S_r\times Aut(X)= Aut_{\Delta}(X^r)$?

Let us consider the Cartesian product $X^r$, where $X$ is a smooth projective variety. There is a subgroup $Aut_{\Delta}(X^r)\subset Aut(X^r)$ of automorphisms of $X^r$ mapping a $k$-dimensional diagonal in $X^r$ isomorphically to a $k$-dimensional diagonal. In particular any automorphism in $Aut_{\Delta}(X^r)$ preserves the smallest diagonal $\Delta_{1,...,r}\subset X^r$.

If we consider the diagonal action of $Aut(X)$ on $X^r$ clearly we have $S_r\times Aut(X)\subseteq Aut_{\Delta}(X^r)$, where $S_r$ is the symmetric group. Under which hypothesis on $X$ (perhaps $X$ of general type) do we have $S_r\times Aut(X)= Aut_{\Delta}(X^r)$?

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user56259
user56259

Automorphisms of Cartesian products

Let us consider the Cartesian product $X^r$, where $X$ is a smooth projective variety. There is a subgroup $Aut_{\Delta}(X^r)\subset Aut(X^r)$ of automorphisms of $X^r$ mapping the small diagonal $\Delta\subset X^r$ isomorphically to itself.

If we consider the diagonal action of $Aut(X)$ on $X^r$ clearly we have $S_r\times Aut(X)\subseteq Aut_{\Delta}(X^r)$, where $S_r$ is the symmetric group. Under which hypothesis on $X$ (perhaps $X$ of general type) do we have $S_r\times Aut(X)= Aut_{\Delta}(X^r)$?