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Edit. I have not realised that this question is on Birational automorphisms, and not on automorphisms, so what I wrote below does not really answer the question. But I will leave this for a background to the question.

This answer concerns automorphisms of complex projective manifolds (and more generally Kahler ones).

Statement. If $X$ is Kahler then $\mathbb Z[1/2]$ action on $X$ extends to an action of $\mathbb R$ on $X$ unless it factors through the action of a finite group (thanks to Yves). This follows immediately from Lieberman's-Fujiki theorem (thanks to YangMills for a refernece to Fujiki, see his comment), which I will state now (D. I. Lieberman. Compactness of the Chow scheme: applications to automorphisms and deformations of Kahler manifolds, 1978).

Denote by $Aut_0(X)$ the connected component of identity map in the group of automorphisms of $X$.

Lieberman-Fujiki Theorem. Consider the action of $Aut(X)$ on $H^*(X,\mathbb Z)$,

$\phi: Aut(X)\to GL(H^*(X,\mathbb Z))$.

Then the group $Aut_0(X)$ has finite index in $ker(\phi)$.

Proof of the statement. Clearly if $\mathbb Z[1/2]$ belongs to $Aut(X)$, then a finite index subgroup of it belong to $ker(\phi)$ (indeed $GL(H^*(X,\mathbb Z))$ does not have infinitely divisible elements apart form $Id$). So by Lieberman-Fujiki theorem it belongs to $Aut_0(X)$. But $Aut_0(X)$ is a Lie group. This finishes the proof.

I don't know if the same reasoning can work in the real case (one might first look closer into the proof of Liebermann's result).

Edit. I have not realised that this question is on Birational automorphisms, and not on automorphisms, so what I wrote below does not really answer the question. But I will leave this for a background to the question.

This answer concerns automorphisms of complex projective manifolds (and more generally Kahler ones).

Statement. If $X$ is Kahler then $\mathbb Z[1/2]$ action on $X$ extends to an action of $\mathbb R$ on $X$ unless it factors through the action of a finite group (thanks to Yves). This follows immediately from Lieberman's-Fujiki theorem (thanks to YangMills for a refernece to Fujiki, see his comment), which I will state now (D. I. Lieberman. Compactness of the Chow scheme: applications to automorphisms and deformations of Kahler manifolds, 1978).

Denote by $Aut_0(X)$ the connected component of identity map in the group of automorphisms of $X$.

Lieberman-Fujiki Theorem. Consider the action of $Aut(X)$ on $H^*(X,\mathbb Z)$,

$\phi: Aut(X)\to GL(H^*(X,\mathbb Z))$.

Then the group $Aut_0(X)$ has finite index in $ker(\phi)$.

Proof of the statement. Clearly if $\mathbb Z[1/2]$ belongs to $Aut(X)$, then a finite index subgroup of it belong to $ker(\phi)$ (indeed $GL(H^*(X,\mathbb Z))$ does not have infinitely $2$-divisible elements apart form $Id$). So by Lieberman-Fujiki theorem it belongs to $Aut_0(X)$. But $Aut_0(X)$ is a Lie group. This finishes the proof.

I don't know if the same reasoning can work in the real case (one might first look closer into the proof of Liebermann's result).

Edit. I have not realised that this question is on Birational automorphisms, and not on automorphisms, so what I wrote below does not really answer the question. But I will leave this for a background to the question.

This answer concerns automorphisms of complex projective manifolds (and more generally Kahler ones).

Statement. If $X$ is Kahler then $\mathbb Z[1/2]$ action on $X$ extends to an action of $\mathbb R$ on $X$. This follows immediately from Lieberman's-Fujiki theorem (thanks to YangMills for a refernece to Fujiki, see his comment), which I will state now (D. I. Lieberman. Compactness of the Chow scheme: applications to automorphisms and deformations of Kahler manifolds, 1978).

Denote by $Aut_0(X)$ the connected component of identity map in the group of automorphisms of $X$.

Lieberman-Fujiki Theorem. Consider the action of $Aut(X)$ on $H^*(X,\mathbb Z)$,

$\phi: Aut(X)\to GL(H^*(X,\mathbb Z))$.

Then the group $Aut_0(X)$ has finite index in $ker(\phi)$.

Proof of the statement. Clearly if $\mathbb Z[1/2]$ belongs to $Aut(X)$, then it belong to $ker(\phi)$ (indeed GL(H^*(X,\mathbb Z) does not have infinitely divisible elements apart form $Id$). So by Lieberman-Fujiki theorem it belongs to $Aut_0(X)$. But $Aut_0(X)$ is a Lie group. This finishes the proof.

I don't know if the same reasoning can work in the real case (one might first look closer into the proof of Liebermann's result).

Edit. I have not realised that this question is on Birational automorphisms, and not on automorphisms, so what I wrote below does not really answer the question. But I will leave this for a background to the question.

This answer concerns automorphisms of complex projective manifolds (and more generally Kahler ones).

Statement. If $X$ is Kahler then $\mathbb Z[1/2]$ action on $X$ extends to an action of $\mathbb R$ on $X$ unless it factors through the action of a finite group (thanks to Yves). This follows immediately from Lieberman's-Fujiki theorem (thanks to YangMills for a refernece to Fujiki, see his comment), which I will state now (D. I. Lieberman. Compactness of the Chow scheme: applications to automorphisms and deformations of Kahler manifolds, 1978).

Denote by $Aut_0(X)$ the connected component of identity map in the group of automorphisms of $X$.

Lieberman-Fujiki Theorem. Consider the action of $Aut(X)$ on $H^*(X,\mathbb Z)$,

$\phi: Aut(X)\to GL(H^*(X,\mathbb Z))$.

Then the group $Aut_0(X)$ has finite index in $ker(\phi)$.

Proof of the statement. Clearly if $\mathbb Z[1/2]$ belongs to $Aut(X)$, then a finite index subgroup of it belong to $ker(\phi)$ (indeed $GL(H^*(X,\mathbb Z))$ does not have infinitely divisible elements apart form $Id$). So by Lieberman-Fujiki theorem it belongs to $Aut_0(X)$. But $Aut_0(X)$ is a Lie group. This finishes the proof.

I don't know if the same reasoning can work in the real case (one might first look closer into the proof of Liebermann's result).

Edit. I have not realised that this question is on Birational morphisms, and not on automorphisms, so what I wrote below does not really answer the question. But I will leave this for a background to the question.

This answer concerns automorphisms of complex projective manifolds (and more generally Kahler ones).

Statement. If $X$ is Kahler then $\mathbb Z[1/2]$ action on $X$ extends to an action of $\mathbb R$ on $X$. This follows immediately from Lieberman's theorem, which I will state now (D. I. Lieberman. Compactness of the Chow scheme: applications to automorphisms and deformations of Kahler manifolds, 1978).

Denote by $Aut_0(X)$ the connected component of identity map in the group of automorphisms of $X$.

Lieberman's Theorem. Consider the action of $Aut(X)$ on $H^*(X,\mathbb Z)$,

$\phi: Aut(X)\to GL(H^*(X,\mathbb Z))$.

Then the group $Aut_0(X)$ has finite index in $ker(\phi)$.

Proof of the statement. Clearly if $\mathbb Z[1/2]$ belongs to $Aut(X)$, then it belong to $ker(\phi)$. So by Lieberman's theorem it belongs to $Aut_0(X)$. But $Aut_0(X)$ is a Lie group. This finishes the proof.

I don't know if the same reasoning can work in the real case (one might first look closer into the proof of Liebermann's result).

Edit. I have not realised that this question is on Birational automorphisms, and not on automorphisms, so what I wrote below does not really answer the question. But I will leave this for a background to the question.

This answer concerns automorphisms of complex projective manifolds (and more generally Kahler ones).

Statement. If $X$ is Kahler then $\mathbb Z[1/2]$ action on $X$ extends to an action of $\mathbb R$ on $X$. This follows immediately from Lieberman's-Fujiki theorem (thanks to YangMills for a refernece to Fujiki, see his comment), which I will state now (D. I. Lieberman. Compactness of the Chow scheme: applications to automorphisms and deformations of Kahler manifolds, 1978).

Denote by $Aut_0(X)$ the connected component of identity map in the group of automorphisms of $X$.

Lieberman-Fujiki Theorem. Consider the action of $Aut(X)$ on $H^*(X,\mathbb Z)$,

$\phi: Aut(X)\to GL(H^*(X,\mathbb Z))$.

Then the group $Aut_0(X)$ has finite index in $ker(\phi)$.

Proof of the statement. Clearly if $\mathbb Z[1/2]$ belongs to $Aut(X)$, then it belong to $ker(\phi)$ (indeed GL(H^*(X,\mathbb Z) does not have infinitely divisible elements apart form $Id$). So by Lieberman-Fujiki theorem it belongs to $Aut_0(X)$. But $Aut_0(X)$ is a Lie group. This finishes the proof.

I don't know if the same reasoning can work in the real case (one might first look closer into the proof of Liebermann's result).