# Henri

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## Registered User

 Name Henri Member for 3 years Seen 2 days ago Website Location Paris Age 25
Phd Student in Complex Geometry
 May3 revised Hyperbolic Riemann Surfaceadded 91 characters in body May3 answered Hyperbolic Riemann Surface Apr27 awarded ● Yearling Mar20 comment Extension of pluriharmonic functionsYes you are right of course; I just had in mind the case where $M$ is a domain in $\mathbb C^n$. My point was just that the assumptions on $\Omega$ should not be too restrictive like pseudoconvexity e.g.; but on $M surely! Mar20 comment Extension of pluriharmonic functionsNo, not even. You just need$\Omega$relatively compact, and$M\setminus \overline{\Omega}$connected. (btw, you should have defined$u$on$M\setminus \overline{\Omega}$only if you want to make sense of pluriharmonicity) Mar19 comment Extension of pluriharmonic functionsOh, I just realized that I misunderstood your question; I thought that$u$was only defined on$\Omega$. In fact, if$M\setminus \Omega$has$b_1=0$, then every pluriharmonic function on$M\setminus \Omega$is the real part of a holomorphic function; so$u=Re(f)$for$f \in \mathcal O(M\setminus \Omega)$. Now, if$M\setminus \Omega$is connected, Hartog's extension theorem tells you that$f$extends (uniquely) to$M$, and then$Re(f)$extends$u$has a pluriharmonic function. Mar19 comment Extension of pluriharmonic functionsYes; if you want consider$\Omega = B(0, 1/2)$; by uniqueness of the pluriharmonic extension (ph functions are real analytic),$u$won't extend to$\mathbb C^2$either. Mar19 comment Extension of pluriharmonic functionsHow about$\Omega=B(0,1)\subset \mathbb C^2$and$u(z,w)=Re(e^{1/(z-1)})$? (cf remark of Alexandre) Feb27 awarded ● Civic Duty Jan31 comment When does$Aut(X)=Bir(X)$hold?Yes, Y. Consider the two projection$p_1$,$p_2$from the graph to$X$and$Y$. We know that the exceptional locus of$p_1$contains rational curves$C \subset X\times Y$. By assumption,$p_2$must contract$C$, so that$C$is contracted by$p_1$and$p_2$, which is absurd. Jan31 answered When does$Aut(X)=Bir(X)$hold? Nov21 revised restriction of sheafadded 276 characters in body Nov21 comment restriction of sheafIn particular, as$X$is smooth hence normal, the desired property holds as soon as$U$has codimension at least$2$. Nov21 comment restriction of sheafThanks, you are perfectly right! (I had in mind a morphism with connected fibers) So here here one can just say that$i_*i^*F= i_* \mathcal O_X \otimes F\$. Nov21 answered restriction of sheaf