# Equivalence of the following two terms

#### Mharris

##### New member
how is \$j\beta =\beta j\$ is equivalent to \$\beta^*+\beta\$ being multiple of identity?

A vector field $X$ is said to be conformal if $L_Xj=0$ where j is the almost complex structure. The conformality condition is equivalent to $j\beta =\beta j$. Where $\beta : ker(\theta) \mapsto ker(\theta)$ such that $\beta(u)= \nabla_uX$ and $\theta$ is contact form such that $\theta(X)=1$. \\
my question is how can i see that $j\beta =\beta j$ is equivalent to $\beta^*+\beta$ being a multiple of the identity? \\
What are the conditions so that i can see $j=\beta-\beta^*$?

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#### Euge

##### MHB Global Moderator
Staff member
Welcome, Mharris !

Is your vector field defined on a Kählerian manifold?

#### Mharris

##### New member
. X is the Reeb vextor field on a contact manifold.

#### Mharris

##### New member
Can someone give me some hint or idea please

#### Euge

##### MHB Global Moderator
Staff member
Re: how is \$j\beta =\beta j\$ is equivalent to \$\beta^*+\beta\$ being multiple of identity?

You should clearly define or describe all terms (e.g. $\theta$, $\beta^*$, etc.) in order to receive assistance. Problems of contact geometry require careful analysis.

#### Mharris

##### New member
Re: how is \$j\beta =\beta j\$ is equivalent to \$\beta^*+\beta\$ being multiple of identity?

You should clearly define or describe all terms (e.g. $\theta$, $\beta^*$, etc.) in order to receive assistance. Problems of contact geometry require careful analysis.
$\theta$ is the contact form defined as $\theta=g(X,-)$. Where $g$ is the Riemanian metric on the manifold M and X is the Reeb vector field. $\beta^*$ is the adjoint of the map $\beta : ker(\theta) \mapsto ker(\theta)$ defined as $\beta(u)= \nabla_uX$ and $g(\beta(Y),Z)=g(\beta^*(Z),Y)$.

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#### Mharris

##### New member
MHarris, please do not bump the thread. First suppose the contact manifold has dimension 3. In this case, the hyperplane field $\operatorname{ker}(\theta)$ is spanned by two vector fields, the commutator of which is not in the hyperplane. Consider the action of $\beta$ and $\beta^*$ on the two basis elements and see what you get. Also, why does $\theta = g(X,\cdot)$? For $X$ to be a Reeb vector field, it simply must satisfy the equations $d\theta(X,\cdot) = 0$ and $\theta(X) = 1$. In local coordinates, you can express $\theta = dz + x\, dy$.