Codifferential of 1-form associated to a Killing field vanishes?

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In summary, if a Killing field \xi is given on a manifold (M, g), it can be associated with a 1-form \eta through the metric. To show that the codifferential of \eta is zero, it suffices to consider the case of Kähler-Einstein manifolds, where the 1-form can be decomposed into three parts: dh, d^{c}f, and \eta^{H}. Since Killing fields are holomorphic, the codifferential vanishes as a result of the Dolbeaut decomposition. This also holds for the usual Laplace operator on Kähler manifolds.
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dongo
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Suppose you are given a Killing field [itex]\xi[/itex] on a manifold [itex] (M, g) [/itex]. We can associate to this vector field a 1-form via the metric. Denote this 1-form by [itex]\eta[/itex]. Show that the codifferential of the 1-form vanishes, i.e., [itex]\delta\eta = 0.[/itex]



As far as I know, a Killing field is a vector field on a manifold whose local flow consists of local isometries. The more pragmatic to check wether [itex]\eta[/itex] is Killing or not is to calculate the Live derivate of the metric of the manifold with respect to the vector field in question. It has to be zero, i.e. [itex]L_{\eta}(g) = 0[/itex]. The codifferential of the usual differential operator is defined as the formally adjoint operator to the differential, where one can explicitly calculate it to be [itex]\delta = (-1)^{nk}\ast d \ast = -\sum_{i=1}^{n}\iota_{e_{i}}\nabla_{e_i}[/itex]. Here, [itex] n = dim(M) [/itex] and the [itex] e_{i}[/itex] form an orthonormal basis with respect to the metric [itex]g[/itex]. The codifferential acts on the space of k + 1 -forms and sends them to k - forms (on M), i.e., [itex]\delta : \Omega^{k+1} \rightarrow \Omega^{k}.



Since the question arose in a book on Kähler geometry, more precisely in asection where holomorphic vector field on Kähler-Einstein manifolds, I assume, it suffices to treat this case. However, in the general case, we know that the 1-form, associated to a real holomorphic vector field, can be (some sort of) hodge decomposed in three parts, namely [itex] /ix = dh + d^{c}f + /ix^{H}[/itex] where the first term on the right-hand side involves the usual differential. the second is the twisted differential, i.e., [itex]d^{c}=[itex]\sum_{i=1}^{n}Je_{i}\wedge \nable_{e_{i}}[/itex], and the last component is the harmonic part. Since we deal with a Kähler manifolds, and we know, that Killing fields are (real) holomorphic, we can use that the Laplace-operator on M is equal to: [itex]\Delta = 2\Delta^{\bar{\partial}}[/itex]. Note that this is only the case on Kähler-manifolds. And this is basically the point from which on I ahve no clue any longer. I'd appreciate hints rather solutions as I would ultimately like to come (with a little help of course) to a solution on my own.

By the way, this is my first post, so if you have any recommendations concerning typesetting etc. you're welcome to utter them. Especially TEX help would be valuable because I have almost no experinece when it comes to that.


dongo
 
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No need to reply, got a hang on it. We know that Killing fields are holomorphic, and we know that when it comes to the Dolbeaut decomposition, a (p,0) form is holomorphic if and only if it is Dolbeaut-harmonic. But this actually implies that the codifferential vanishes. (This is a complex analogon to the Hodge-decomposition). As we have on Kähle manifolds that being Dolbeaut-harmonic implies being harmonic w.r.t to the usual Laplace operator, we have the solution (at least for Kähler manifolds, but I am satisfied with that ;) )
 

Related to Codifferential of 1-form associated to a Killing field vanishes?

1. What is a Killing field?

A Killing field is a vector field in a Riemannian manifold that preserves the metric tensor, meaning that the Lie derivative of the metric tensor with respect to the vector field is equal to zero. In simpler terms, a Killing field is a vector field that generates an isometry, or a transformation that preserves distances and angles.

2. What is a 1-form associated to a Killing field?

A 1-form is a linear map that assigns a real number to each tangent vector at a point on a manifold. A 1-form associated to a Killing field is a special type of 1-form that is defined by the inner product of the vector field with the metric tensor. This 1-form is often used in the study of Riemannian manifolds and their geometric properties.

3. How is the codifferential of a 1-form associated to a Killing field calculated?

The codifferential of a 1-form associated to a Killing field is calculated by taking the exterior derivative of the 1-form and then taking the dual of the result. In other words, it is the adjoint of the exterior derivative operator applied to the 1-form. This can also be expressed as the negative of the Lie derivative of the 1-form with respect to the Killing field.

4. What does it mean for the codifferential of a 1-form associated to a Killing field to vanish?

If the codifferential of a 1-form associated to a Killing field vanishes, it means that the 1-form and the Killing field are orthogonal to each other at every point on the manifold. This can also be interpreted as the dual of the exterior derivative of the 1-form being equal to zero, or the Lie derivative of the 1-form with respect to the Killing field being equal to zero.

5. What are the implications of the codifferential of a 1-form associated to a Killing field vanishing?

The vanishing of the codifferential of a 1-form associated to a Killing field has important implications in the study of Riemannian manifolds. It means that the 1-form and the Killing field are mutually orthogonal, which can provide useful information about the geometric properties of the manifold. It also implies that the manifold may have additional symmetries or isometries that can be studied and utilized in various applications.

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