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Killing vector field is often referred as "generator" of infinitesimal isometry. However from what I understood this would mean that the exponential map of these generators would be the relevant transformation. $$x^\mu\to x^\mu + \epsilon K^\mu$$ The above transformation leaves the metric invariant. Now the field $K$ is the defined as generator of isometry here. But from what I learnt in Lie Algebra is that the symmetry transformations form a group and the generators of those transformations are defined via exponential map. If Killing vector fields are indeed generator in group theoretic sense then how do we perform those infinitesimal transformation?

$$g_{\mu\nu}\to g_{\mu\nu} +\delta g_{\mu\nu}$$

with

$$\delta g_{\mu\nu} = \epsilon\mathcal{L}_{K}g_{\mu\nu}$$

where $\mathcal{L}_{K}$ is Lie derivative with respect to killing field $K$, and then it seems to sugges the exponential map for the transformation should look like this

$$g_{\mu\nu}=e^{\epsilon\mathcal{L}_{K}}g_{\mu\nu}$$

I know, since $\mathcal{L}_{K}g_{\mu\nu}=0$, this is just identity map but would it be well defined if considered in this manner? Since according this post Lie derivatives form the representation of Lie algebra obeyed by the vector fields. So is the action of killing field on other tensors via lie derivative? Like could we do something of this manner here like this? \begin{align} x^\mu & \to e^{\epsilon K}x^\mu\\ g_{\mu\nu}&\to e^{\epsilon\mathcal{L}_{K}}g_{\mu\nu}. \end{align}

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  • $\begingroup$ can you define what is $\mathcal{L}_K$ and also explain why the variation of the metric should be proportional to the metric itself? $\endgroup$
    – Rho Phi
    Dec 17, 2023 at 21:02
  • $\begingroup$ Lie derivative of the metric tensor $\endgroup$ Dec 18, 2023 at 2:53
  • $\begingroup$ A generator is a basis vector of your Lie algebra.Depending on how you formalise your Lie algebra a basis vector could be a vector field or a tangent vector at identity tangent space. There are other meanings but in your context it means this. Exponential of generators give u the connected component to identity of the Lie group. I am not sure about how to answer the rest and what the question is $\endgroup$
    – Rescy_
    Dec 18, 2023 at 8:03
  • $\begingroup$ I wanna ask how would the exponential map of killing fields act on vectors and tensors in general. Would it happen via lie derivative? If so then it makes sense to consider them generator in group theoretic sense. $\endgroup$ Dec 18, 2023 at 8:32
  • $\begingroup$ @ChandraPrakash An isometry by definition should leave the metric invariant, shouldn't it? $\endgroup$
    – Sidd
    Dec 21, 2023 at 6:10

1 Answer 1

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  1. The set of vector fields endowed with the Lie bracket of vector fields form a Lie algebra.

  2. If a vector field heuristically corresponds to an infinitesimal change, then a flow corresponds to a finite change.

    In particular, a vector field generates a flow, cf. OP's title question.

  3. Similarly, the notion of Lie derivative wrt. a vector field of various tensors gets replaced by pushforwards and pullbacks by the flow.

References:

  1. M. Nakahara, Geometry, Topology and Physics, 1989; section 5.3.

  2. M. Nakahara, Geometry, Topology and Physics, 2003; section 5.3.

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  • $\begingroup$ I think, i am looking for something along the lines of "While Lie algebras often act on vector spaces through representations, not exclusively via matrices, their interaction with manifolds involves vector fields and their flows rather than direct application on coordinates. These flows essentially capture the action of the Lie algebra on the underlying vector space associated with manifold points." $\endgroup$ Dec 19, 2023 at 14:53

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