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Equivalence of the following two terms

Mharris

New member
Jan 8, 2018
5
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^*$?
 
Last edited:

Euge

MHB Global Moderator
Staff member
Jun 20, 2014
1,890
Welcome, Mharris ! (Wave)

Is your vector field defined on a Kählerian manifold?
 

Mharris

New member
Jan 8, 2018
5
. X is the Reeb vextor field on a contact manifold.
 

Mharris

New member
Jan 8, 2018
5
Can someone give me some hint or idea please
 

Euge

MHB Global Moderator
Staff member
Jun 20, 2014
1,890
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
Jan 8, 2018
5
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
Jan 8, 2018
5
Any help or hint please
 

Euge

MHB Global Moderator
Staff member
Jun 20, 2014
1,890
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 $\theta = dz + x\, dy$.

Sorry I won’t have time to go into great detail. I’m very busy this week.