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Suppose we have a Riemannian manifold $(M,g,\nabla)$ with Levi-civita connection $\nabla$. We define a new symmetric non-metric connection $\bar\nabla$ on $M$. Then the Lie derivative of functions and vector fields are related as follows
$$\bar{L}_Xf=X(f)={L}_Xf \\ \bar{L}_XY=[X,Y]={L}_XY \\ (\bar{L}_Xg)(U,V)=X(g(U,V)-g([X,U],V)-g(U,[X,V])=(L_Xg)(U,V)$$ Is this true? Does it make sense? Thanks in advance.

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    $\begingroup$ The Lie derivative depends only on the differentiable structure on $M$ and has nothing to do with a connection. It is defined using only flows of vector fields. $\endgroup$ Feb 7, 2014 at 19:23
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    $\begingroup$ The Lie derivative does not depend on a choice of connection or metric or any structure on $M$ other than the structure of $M$ as a smooth manifold.. Given this, it is not completely clear what you are asking. $\endgroup$
    – levap
    Feb 7, 2014 at 19:23
  • $\begingroup$ Many thanks, I was asking about the term $X(g(U,V))$ in the last step. I think it depends on the the connection (metric or non-metric) @TedShifrin $\endgroup$
    – Semsem
    Feb 7, 2014 at 19:28
  • $\begingroup$ @levap do you mean Lie derivatives of functions and vector fields only or generally, it doesn't depend on the metric. Many thanks $\endgroup$
    – Semsem
    Feb 7, 2014 at 19:32
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    $\begingroup$ No, Semsem. You started with the Riemannian metric. You're just differentiating $g(U,V)$ in the direction of $X$. No connection involved here. You could say further things about this if you had a metric-compatible connection. $\endgroup$ Feb 7, 2014 at 19:34

2 Answers 2

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The Lie derivative $L_X T$ of any tensor $T$ along any vector field $X$ is defined directly and only from the underlying manifold structure. If, moreover, the manifold has a symmetric connection, (symmetric means zero torsion), then it is possible to express $\mathcal{L}_X T$ using this connection. Of course, the result is the same no matter which connection is used, (as long as it is symmetric).

Please note, accordingly, there is no meaning in writing $L_X$ and $\bar{L}_X$. This is wrong. The Lie derivative comes first, and then the expression in terms of a connection, not the other way around.

The Lie derivative has two (equivalent) definitions. A dynamical one and an algebraic one. It is very important to appreciate that these reflect two equally useful intuitions. I will not get into this here, (feel free to ask), but I will consider expressing the Lie derivative in terms of a symmetric connection.

If $\nabla$ is a symmetric connection, then for any vector fields $X,Y$

$$ L_XY = [X,Y] = \nabla_XY-\nabla_YX $$ where the first equality is by definition, but the second equality means $\nabla$ is symmetric.

For $(0,s)$-tensor $T$, (think of the metric if you would like), $L_XT$ is also a $(0,s)$-tensor, (one says that $L_X$ is type preserving), and by definition

$$ L_XT(Y_1,\ldots,Y_s) = X(T(Y_1,\ldots,Y_s)) - \sum_i T(\ldots,[X,Y_i],\ldots) $$

The first term is

$$ X(T(Y_1,\ldots,Y_s)) = (\nabla_X T)(Y_1,\ldots,Y_s) + \sum_i T(\ldots,\nabla_XY_i,\ldots) $$

By replacing in the definition of $L_XT$ and using the fact that $\nabla$ is symmetric, this yields the expression

$$ L_XT(Y_1,\ldots,Y_s) = \nabla_X T(Y_1,\ldots,Y_s) + \sum_i T(\ldots,\nabla_{Y_i}X,\ldots) $$

Please note this is not the definition of $L_XT$ but only a formula, using the connection $\nabla$. In other words, $L_XT$ does not depend on $\nabla$ but the righ-hand side does.

If $T$ is parallel, (this means $\nabla_X T = 0$ for any $X$),

$$ L_XT(Y_1,\ldots,Y_s) = \sum_i T(\ldots,\nabla_{Y_i}X,\ldots) $$

You may apply this last formula to the Lie derivative of the Riemannian metric $T=g$, and use the Levi-Civita connection as $\nabla$, to get the ''elasticity tensor'' $L_Xg$ and understand the definition of Killing vector fields, etc.

-- Salem

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  • $\begingroup$ Thank you for writing this up here! (+1) $\endgroup$ Sep 16, 2018 at 1:18
  • $\begingroup$ This is useful! $\endgroup$
    – Guy The
    May 1, 2020 at 3:53
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The Lie derivative $L_X T$ of a tensor $T$ in the direction of a vector field $X$ can be expressed in terms of any connection $\nabla$ on you manifold $M$.

For torsion free (= symmetric) connections $\nabla$ the expression is particularly simple: $$ L_X \, T_{a_1 \dots a_p}{}^{b_1 \dots b_q} = X^c \nabla_c T_{a_1 \dots a_p}{}^{b_1 \dots b_q} + \sum_{i=1}^p T_{a_1 \dots c \dots a_p}{}^{b_1 \dots b_q} \nabla_{a_j} X^c - \sum_{j=1}^p T_{a_1 \dots a_p}{}^{b_1 \dots c \dots b_q} \nabla_{c} X^{b_j} $$

(see R.Wald, General Relativity, p.441)

If connection $\nabla'$ is not torsion free, one can make a torsion free connection $\nabla$, write down the above expression, and plug in the expression for $\nabla$ in terms of $\nabla'$ and its torsion to obtain a general formula.

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  • $\begingroup$ Notable appreciated answer, so it depends on the connection even it does not in my three cases $\endgroup$
    – Semsem
    Feb 7, 2014 at 22:22
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    $\begingroup$ @Semsem Not at all, the Lie derivative does not depend on any connection, because it is defined on any smooth manifold without any additional structure. This is the expression for the Lie derivative that does depend on the connection provided such a connection (an additiona structure) is given. $\endgroup$ Feb 7, 2014 at 23:07
  • $\begingroup$ accepted @yuriVyatkin, thanks $\endgroup$
    – Semsem
    Feb 7, 2014 at 23:09
  • $\begingroup$ Is there is a method to find it out without connections $\endgroup$
    – Semsem
    Feb 7, 2014 at 23:15
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    $\begingroup$ Sure, you can use the Lie brackets instead, similar to the identities in your question. This would show the independence even more strikingly. $\endgroup$ Feb 7, 2014 at 23:18

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