Return to Answer

Fix a Riemann metric metric $g$ on the manifold. Denote by $\omega$ the $1$-form dual to $X$ defined by

$$\omega(Y)= g(X,Y), $$

for any vector field $Y$. The $1$-form

$$\alpha :=\frac{1}{|X|^2_g}\omega =\frac{1}{g(X,X)}\omega $$

will do the trick.

Mea Culpa I missed the word closed in the question. Obviously the $\alpha$ above need not be closed.

If $\alpha$ is closed and $\alpha(X)=i_X\alpha=1$ ($i_X$ denoting the contraction with $X$), then the identity

$$L_X=di_X+i_Xd $$

implies that $L_X\alpha=0$. Conversely, if $\alpha$ is a closed $1$-form such that $L_X\alpha=0$ then the above identity implies that $i_X\alpha$ is constant. Thus you need to find closed forms which are invariant under the flow generated by $X$ and such that $\alpha(X)$ is nonzero at one point of the manifold.

Over each contractible open subset $U\subset M$ a closed form $\alpha$ is exact so that

$$\alpha|_U= d f_U $$.

The condition $\alpha(X)=1$ implies that

$$ X\cdot f_U = df_U(X) =1. $$.

Thus $f_U$ increases linearly along the orbits of $X$ in $U$. For example, if $X$ has a periodic orbit contained in a simply connected set your question does not have solution.

Fix a Riemann metric metric $g$ on the manifold. Denote by $\omega$ the $1$-form dual to $X$ defined by

$$\omega(Y)= g(X,Y), $$

for any vector field $Y$. The $1$-form

$$\alpha :=\frac{1}{|X|^2_g}\omega =\frac{1}{g(X,X)}\omega $$

will do the trick.

Mea Culpa I missed the word closed in the question. Obviously the $\alpha$ above need not be closed.

If $\alpha$ is closed and $\alpha(X)=i_X\alpha=1$ ($i_X$ denoting the contraction with $X$), then the identity

$$L_X=di_X+i_Xd $$

implies that $L_X\alpha=0$. Conversely, if $\alpha$ is a closed $1$-form such that $L_X\alpha=0$ then the above identity implies that $i_X\alpha$ is constant. Thus you need to find closed forms which are invariant under the flow generated by $X$ and such that $\alpha(X)$ is nonzero at one point of the manifold.

Over each contractible open subset $U\subset M$ a closed form $\alpha$ is exact so that

$$\alpha|_U= d f_U $$.

The condition $\alpha(X)=1$ implies that

$$ X\cdot f_U = df_U(X) =1. $$.

Thus $f_U$ increases linearly along the orbits of $X$ in $U$. For example, if $X$ has a periodic orbit contained in a simply connected set your question does not have solution.

For example, if $X$ is the generator of the usual action of $S^1$ on $S^3$, then you cannot find a closed $1$-form on $S^3$ such that $\alpha(X)=1$.

Fix a Riemann metric metric $g$ on the manifold. Denote by $\omega$ the $1$-form dual to $X$ defined by

$$\omega(Y)= g(X,Y), $$

for any vector field $Y$. The $1$-form

$$\alpha :=\frac{1}{|X|^2_g}\omega =\frac{1}{g(X,X)}\omega $$

will do the trick.

Mea Culpa I missed the word closed in the question. Obviously the $\alpha$ above need not be closed.

If $\alpha$ is closed and $\alpha(X)=i_X\alpha=1$ ($i_X$ denoting the contraction with $X$), then the identity

$$L_X=di_X+i_Xd $$

implies that $L_X\alpha=0$. Conversely, if $\alpha$ is a closed $1$-form such that $L_X\alpha=0$ then the above identity implies that $i_X\alpha$ is constant. Thus you need to find closed forms which are invariant under the flow generated by $X$ and such that $\alpha(X)$ is nonzero at one point of the manifold.

Over each contractible open subset $U\subset M$ a closed form $\alpha$ is exact so that

$$\alpha|_U= d f_U $$.

The condition $L_X\alpha=0$ implies that

$$ X\cdot f_U = df_U(X) =c $$.

If $c>0$ then $f_U$ increases linearly along the orbits of $X$ in $U$. For example if $X$ has a periodic orbit contained in a simply connected set your question does not have solution.

Fix a Riemann metric metric $g$ on the manifold. Denote by $\omega$ the $1$-form dual to $X$ defined by

$$\omega(Y)= g(X,Y), $$

for any vector field $Y$. The $1$-form

$$\alpha :=\frac{1}{|X|^2_g}\omega =\frac{1}{g(X,X)}\omega $$

will do the trick.

Mea Culpa I missed the word closed in the question. Obviously the $\alpha$ above need not be closed.

If $\alpha$ is closed and $\alpha(X)=i_X\alpha=1$ ($i_X$ denoting the contraction with $X$), then the identity

$$L_X=di_X+i_Xd $$

implies that $L_X\alpha=0$. Conversely, if $\alpha$ is a closed $1$-form such that $L_X\alpha=0$ then the above identity implies that $i_X\alpha$ is constant. Thus you need to find closed forms which are invariant under the flow generated by $X$ and such that $\alpha(X)$ is nonzero at one point of the manifold.

Over each contractible open subset $U\subset M$ a closed form $\alpha$ is exact so that

$$\alpha|_U= d f_U $$.

The condition $\alpha(X)=1$ implies that

$$ X\cdot f_U = df_U(X) =1. $$.

Thus $f_U$ increases linearly along the orbits of $X$ in $U$. For example, if $X$ has a periodic orbit contained in a simply connected set your question does not have solution.

Fix a Riemann metric metric $g$ on the manifold. Denote by $\omega$ the $1$-form dual to $X$ defined by

$$\omega(Y)= g(X,Y), $$

for any vector field $Y$. The $1$-form

$$\alpha :=\frac{1}{|X|^2_g}\omega =\frac{1}{g(X,X)}\omega $$

will do the trick.

Fix a Riemann metric metric $g$ on the manifold. Denote by $\omega$ the $1$-form dual to $X$ defined by

$$\omega(Y)= g(X,Y), $$

for any vector field $Y$. The $1$-form

$$\alpha :=\frac{1}{|X|^2_g}\omega =\frac{1}{g(X,X)}\omega $$

will do the trick.

Mea Culpa I missed the word closed in the question. Obviously the $\alpha$ above need not be closed.

If $\alpha$ is closed and $\alpha(X)=i_X\alpha=1$ ($i_X$ denoting the contraction with $X$), then the identity

$$L_X=di_X+i_Xd $$

implies that $L_X\alpha=0$. Conversely, if $\alpha$ is a closed $1$-form such that $L_X\alpha=0$ then the above identity implies that $i_X\alpha$ is constant. Thus you need to find closed forms which are invariant under the flow generated by $X$ and such that $\alpha(X)$ is nonzero at one point of the manifold.

Over each contractible open subset $U\subset M$ a closed form $\alpha$ is exact so that

$$\alpha|_U= d f_U $$.

The condition $L_X\alpha=0$ implies that

$$ X\cdot f_U = df_U(X) =c $$.

If $c>0$ then $f_U$ increases linearly along the orbits of $X$ in $U$. For example if $X$ has a periodic orbit contained in a simply connected set your question does not have solution.