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

  1. Jul 28, 2011 #1
    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.

    Best regards,
    dongo
     
  2. jcsd
  3. Jul 28, 2011 #2
    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 ;) )
     
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