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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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    Gold Member

    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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