# Lie derivative on a riemannian manifold

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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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. – Ted Shifrin Feb 7 '14 at 19:23
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. – levap Feb 7 '14 at 19:23
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 – Sameh Shenawy Feb 7 '14 at 19:28
@levap do you mean Lie derivatives of functions and vector fields only or generally, it doesn't depend on the metric. Many thanks – Sameh Shenawy Feb 7 '14 at 19:32
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. – Ted Shifrin Feb 7 '14 at 19:34

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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Notable appreciated answer, so it depends on the connection even it does not in my three cases – Sameh Shenawy Feb 7 '14 at 22:22
@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. – Yuri Vyatkin Feb 7 '14 at 23:07
accepted @yuriVyatkin, thanks – Sameh Shenawy Feb 7 '14 at 23:09
Is there is a method to find it out without connections – Sameh Shenawy Feb 7 '14 at 23:15
Sure, you can use the Lie brackets instead, similar to the identities in your question. This would show the independence even more strikingly. – Yuri Vyatkin Feb 7 '14 at 23:18