Does the Lie derivative of a symmetric $(1,1)$- tensor along a non-vanishing vector field remain symmetric? If the answer is positive how to prove it, else does the following equation hold?

$$g((\mathscr{L}_VT)X,V)\stackrel{?}{=}g(X,(\mathscr{L}_VT)V),\qquad\forall X\in \mathfrak{X}(M).$$

where $(M,g)$ is a Riemannian manifold and $T$ a symmetric $(1,1)$- tensor and $V$ is a non-vanishing vector field.


In general, the Lie derivative of a symmetric $(1,1)$-tensor need not be symmetric. The basic reason is because "symmetry" of a $(1,1)$-tensor really means self-adjointness, which depends on the metric. So unless $V$ is a Killing vector field (meaning $\mathscr L_V g\equiv 0$), there's no reason to expect $\mathscr L_V T$ to be symmetric when $T$ is.

Now the identity you asked about is a little weaker than saying $\mathscr L_V T$ is self-adjoint, because one of the vectors you're applying $\mathscr L_V T$ to is $V$ itself. But even this weaker identity need not hold if $V$ is not a Killing vector field.

Here's a counterexample. In $\mathbb R^2$ with the Euclidean metric, let \begin{align*} T &= \partial_ x \otimes dy + \partial_y \otimes dx,\\ V &= y\partial_y,\\ X &= \partial_x. \end{align*} Then we compute \begin{align*} \mathscr L_V dx &= 0, &\mathscr L_V dy &= dy,\\ \mathscr L_V \partial_x &= 0, & \mathscr L_V \partial_y &= -\partial_y,\\ \mathscr L_V T &= \partial _x \otimes dy - \partial_y \otimes dx, \end{align*} and therefore \begin{align*} g((\mathscr L_V T)X,V) &= g(-\partial_y,y\partial_y) =-y,\\ g(X,(\mathscr L_V T)V) &= g(\partial_x,y\partial_x) =y. \end{align*}

On the other hand, if $V$ is a Killing vector field, then in fact $\mathscr L_V T$ is symmetric whenever $T$ is. Here's a proof. The fact that $V$ is a Killing vector field means $\mathscr L_V \big(g(X,Y)) = g(\mathscr L_V X,Y) + g(X,\mathscr L_V Y)$ for all vector fields $X$ and $Y$. Thus \begin{align*} 0 &= \mathscr L_V \big( g(TX,Y) - g(X,TY)\big)\\ &= g((\mathscr L_V T)X,Y) + g(T(\mathscr L_V X),Y) + g(TX,\mathscr L_V Y)\\ &\quad - g(\mathscr L_V X,TY) - g(X,(\mathscr L_V T)Y) - g(X,T(\mathscr L_V Y))\\ &= g((\mathscr L_V T)X,Y) - g(X,(\mathscr L_V T)Y), \end{align*} where in the second line, I've used $\mathscr L_V(TX) = (\mathscr L_V T)X + T(\mathscr L_V X)$, and in the last line I've canceled four terms using the fact that $T$ is self-adjoint.

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    $\begingroup$ Prof J. Lee. many thanks for your nice answer. and thanks so much for your beautiful book in Riemannian manifolds, I really appreciate it. $\endgroup$ – C.F.G Sep 17 '17 at 5:05

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