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Chapter 15: Problem 18

Show that the Jacobi identity can be written $$ \mathcal{L}_{[x, r]} Z=\mathcal{L}_{x} \mathcal{L}_{r} Z-\mathcal{L}_{r} \mathcal{L}_{x} Z $$ and this property extends to all tensors \(T\) :

Short Answer

Expert verified
The Jacobi identity for Lie derivatives is verified through application of the properties of Lie brackets of vector fields and the Leibniz rule for the Lie derivative. This result extends to all tensors due to the derivation property of the Lie derivative on the algebra of tensor fields.

Step by step solution

01

Define tensors

First, identify the tensors in the given equation. In this case, x and r are vector fields and Z is a generic tensor.
02

Apply the Jacobi identity

The Jacobi identity is a property of the Lie bracket of vector fields, and it is commonly given by: \([x, [y, z]] + [z, [x, y]] + [y, [z, x]] = 0\). Based on this, we express the left-hand side in the following way: \(\mathcal{L}_{[x, r]} Z = \mathcal{L}_{x} \mathcal{L}_{r} Z-\mathcal{L}_{r} \mathcal{L}_{x} Z = \mathcal{L}_{x} \mathcal{L}_{r} - \mathcal{L}_{r} \mathcal{L}_{x} Z\).
03

Apply the properties of the Lie derivative

The Lie derivative obeys the Leibniz rule, which means \(\mathcal{L}_{x}(fg) = (\mathcal{L}_{x}f)g + f(\mathcal{L}_{x}g)\), where f and g can be any tensors. Apply this property for each term: \(\mathcal{L}_{[x, r]} = \mathcal{L}_{x} \mathcal{L}_{r} - \mathcal{L}_{r} \mathcal{L}_{x}\). Hence, \(\mathcal{L}_{[x, r]} = 0\), demonstrating The Jacobi identity.
04

Extension to other tensors

The Lie derivative \(\mathcal{L}\) is a derivation on the algebra of tensor fields. It therefore follows that the same properties that hold for Lie derivatives of vector fields also hold for Lie derivatives of arbitrary tensors. Thus, the result given by the proof extends to all tensors.

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Most popular questions from this chapter

Show that the curve $$ 2 x^{2}+2 y^{2}+2 x y=1 $$ can be converted by a rotation of axcs to the standand form for an ellipse $$ x^{\prime 2}+3 y^{2}=1 $$ If \(x^{\prime}=\cos \psi, v^{\prime}=\frac{1}{\sqrt{3}} \sin \psi\) is used as a parametrization of this curve, show that $$ x=\frac{1}{\sqrt{2}}\left(\cos \psi+\frac{1}{\sqrt{3}} \sin \psi\right), \quad y=\frac{1}{\sqrt{2}}\left(-\cos \psi+\frac{1}{\sqrt{3}} \sin \psi\right) $$ Compute the components of the tangent vector $$ X=\frac{\mathrm{d} x}{\mathrm{~d} \psi} \partial_{x}+\frac{\mathrm{d} y}{\mathrm{~d} \psi} \partial_{y} $$ Show that \(X(f)=(2 / \sqrt{3})\left(x^{2}-y^{2}\right)\).

Let \(\mathbb{R}^{\prime}\) be the marifold consisting of \(\mathbb{R}\) with differentiable structure generated by the chart \(\left(R ; y=x^{3}\right)\). Show that the identity map id \(\mathbb{R}: \mathbb{R}^{\prime} \rightarrow \mathbb{R}\) is a differentiable homeomorphism, which is not a diffeomorphism.

Show that the Lie derivative \(\mathcal{L}_{x}\) commules with all operations of contraction \(C_{t}^{y}\) on a tensor field \(T\), $$ \mathcal{L}_{X} C_{J} T=C_{j} \mathcal{L}_{\mathrm{Y}} T $$

Show that the set of real \(m \times m\) matrices \(M(m, n ; \mathbb{R})\) is a manifold of dimension \(m n\). Show that the matrix multiplication map \(M(m, k ; \mathbb{R}) \times M(k, n ; \mathbb{R}) \rightarrow M f(m, n ; \mathbb{R})\) is differentiable.

On the \(n\)-sphere \(S^{n}\) find coordinates corresponding to (i) stereographic projection, (ii) spherical polars.
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