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Topology Sources about Killing vector fields?

  1. May 28, 2016 #1


    Staff: Mentor

    I'm interested in Killing vector fields and want to ask whether anybody can name me a good textbook or online-source about them, preferably with a general treatment with local coordinates as examples and not at the center of consideration.
  2. jcsd
  3. May 28, 2016 #2

    George Jones

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    Staff Emeritus
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    Starting on page 83, "Differential Geometry and Lie Groups for Physicists", by Fecko, has my favourite treatment of Killing vectors. It has lots of great examples in its short exrecises. I am not sure if has enough coordinate-invariant stuff for your tastes, but it does start with the definition ##\mathcal{L}_\xi g = 0##, where ##\mathcal{L}## is the Lie derivative and the vector field ##\xi## is generated by a one-parameter group of isometries.
  4. May 29, 2016 #3


    Staff: Mentor

    @George Jones Thank you.
  5. Jun 4, 2016 #4
    Peter Petersen ~ Riemannian geometry, curvature chapter (grad texts in math) :

    ** First, we can measure the change in X by asking whether or not X is a gradient field. If iXg = θX is the 1-form dual to X, i.e., (iXg) (Y ) = g (X, Y ), then we know that X is locally the gradient of a function if and only if dθX = 0. In general, the 2-form dθX therefore measures the extend to which X is a gradient field. Second, we can measure how a vector field X changes the metric via the Lie derivative LXg. This is a symmetric (0, 2)-tensor as opposed to the skew-symmetric (0, 2)-tensor dθX. If Ft is the local flow for X, then we see that LXg = 0 if and only if Ft are isometries (see also chapter 7). If this happens then we say that X is a Killing field. Lie derivatives will be used heavily below. The results we use are standard from manifold theory and are all explained in the appendix. In case X = ∇f is a gradient field the expression L∇f g is essentially the Hessian of f. We can prove this in Rn were we already know what the Hessian should be. Let X = ∇f = ai ∂i, ai = ∂if, then LX δijdxi dxj = (LXδij ) + δijLX dxi dxj + δijdxi LX dxj = 0+ δij dLX xi dxj + δijdxi dLX xj = δij dai dxj + δijdxi daj = δij ∂kai dxkdxj + δijdxi ∂kaj dxk = ∂kai dxkdxi + ∂kai dxi dxk = ∂kai + ∂iak dxi dxk = (∂k∂if + ∂i∂kf) dxi dxk = 2(∂i∂kf) dxi dxk = 2Hessf. - From this calculation we can also quickly see what the Killing fields on Rn should be. If X = ai ∂i, then X is a Killing field iff ∂kai + ∂iak = 0. This shows that ∂j∂kai = −∂j∂iak = −∂i∂jak = ∂i∂kaj = ∂k∂iaj = −∂k∂jai = −∂j∂kai . Thus we have ∂j∂kai = 0 and hence ai = αi jxj + βi with the extra conditions that αi j = ∂jai = −∂iaj = −αj i . The angular field ∂θ is therefore a Killing field. This also follows from the fact that the corresponding flow is matrix multiplication by the orthogonal matrix cos (t) − sin (t) sin (t) cos (t) . More generally one can show that the flow of the Killing field X is Ft (x) = exp (At) x + tβ, A = αi j , β = βi . In this way we see that a vector field on Rn is constant iff it is a Killing field that is also a gradient field. **
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