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user108998
user108998

I don't believe this is true. Consider $D_{\mathbb{A}_{\mathbb{C}}^{1}}:=R$$D_{\mathbb{A}_{\mathbb{C}}^{1}}=:R$. Then we have $HH^{1}(R)=0$ by a standard computation. In fact $HH^{*}(D_{X})\cong H^{*}_{dR}(X)$ holds more generally. Now $Out(R)=Aut(R)$ as $R^{*}=\mathbb{C}^{*}$ is central. $Aut(R)$ includes all the transformations $$x\mapsto x+f(\partial), \partial\mapsto \partial,$$ where $f$ is any polynomial in one variable.

I don't believe this is true. Consider $D_{\mathbb{A}_{\mathbb{C}}^{1}}:=R$. Then we have $HH^{1}(R)=0$ by a standard computation. In fact $HH^{*}(D_{X})\cong H^{*}_{dR}(X)$ holds more generally. Now $Out(R)=Aut(R)$ as $R^{*}=\mathbb{C}^{*}$ is central. $Aut(R)$ includes all the transformations $$x\mapsto x+f(\partial), \partial\mapsto \partial,$$ where $f$ is any polynomial in one variable.

I don't believe this is true. Consider $D_{\mathbb{A}_{\mathbb{C}}^{1}}=:R$. Then we have $HH^{1}(R)=0$ by a standard computation. In fact $HH^{*}(D_{X})\cong H^{*}_{dR}(X)$ holds more generally. Now $Out(R)=Aut(R)$ as $R^{*}=\mathbb{C}^{*}$ is central. $Aut(R)$ includes all the transformations $$x\mapsto x+f(\partial), \partial\mapsto \partial,$$ where $f$ is any polynomial in one variable.

Source Link
user108998
user108998

I don't believe this is true. Consider $D_{\mathbb{A}_{\mathbb{C}}^{1}}:=R$. Then we have $HH^{1}(R)=0$ by a standard computation. In fact $HH^{*}(D_{X})\cong H^{*}_{dR}(X)$ holds more generally. Now $Out(R)=Aut(R)$ as $R^{*}=\mathbb{C}^{*}$ is central. $Aut(R)$ includes all the transformations $$x\mapsto x+f(\partial), \partial\mapsto \partial,$$ where $f$ is any polynomial in one variable.