# Lie derivative is zero $\Leftrightarrow$ "$\phi_t$ is a symmetry transformation for $T$ $\,\,\forall t\in\mathbb{R}$."

Let $T$ be a smooth tensor field on a manifold $\mathcal{M}$. Let $\phi\in\mathcal{M}^\mathcal{M}$ be a diffeomorphism.

• Question 1: I'm wondering if "$\phi$ is a symmetry transformation for $T$" iff $\forall p\in\mathcal{M}$, $\left.T\right|_p = \phi^\ast\left(\left.T\right|_{\phi^{-1}\left(p\right)}\right)$, where $\phi^\ast$ is the push forward of $\phi$ and this equation is an equation of two tensors in $\mathcal{T}_p\left(k,\,l\right)$ (the space of all tensors of type $\left(k,\,l\right)$ at $p\in\mathcal{M}$. Is this right?

Next, I've defined the Lie derivative as a map from smooth tensor fields to smooth tensor fields.

If:

1. $p\in\mathcal{M}$,
2. $T$ is a smooth tensor field on $\mathcal{M}$ of type $\left(k,\,l\right)$,
3. $v$ is a smooth vector field on $\mathcal{M}$,
4. $\phi_t$ is the one-parameter group of diffeomorphisms that $v$ naturally defines,
5. $\left\{\mu_i\right\}_{i=1}^k$ are dual vectors in $V_p^\ast$ (the cotangent space at $p\in\mathcal{M}$) and $\left\{t_j\right\}_{j=1}^l$ are vectors in $V_p$ (the tangent space at $p\in\mathcal{M}$),

then the Lie derivative of $T$ along $v$ is a tensor defined by the following action on the tuple of dual vectors / vectors $\left(\mu_1,\,\dots,\,\mu_k;\,t_1,\,\dots,\,t_l\right)$: $$\left(\left.\mathcal{L_v}\left(T\right)\right|_p\right)\left(\mu_1,\,\dots,\,\mu_k;\,t_1,\,\dots,\,t_l\right) := \left.\left[\frac{d}{dt}\,\left(\left(\phi_{-t}\right)^\ast\left(\left.T\right|_{\phi_t\left(p\right)}\right)\right)\left(\mu_1,\,\dots,\,\mu_k;\,t_1,\,\dots,\,t_l\right)\right]\right|_{t=0}$$

• Question 2: Is this definition completely correct or am I missing / distorting some details?

Now for the main point of this post:

Claim: $\left(\left.\mathcal{L_v}\left(T\right)\right|_p\right)=0\forall p\in\mathcal{M}\Leftrightarrow$ "$\phi_t$ is a symmetry transformation for $T$ $\,\,\forall t\in\mathbb{R}$."

The $\Longleftarrow$ direction of this proof seems easy to me and follows directly from the definitions.

• Question 3: How do you prove the $\Longrightarrow$ direction?

3: let us take first the case of a $(0,0)$ tensor, i.e. a function $f:M\to\mathbb R$. In this case we are given that $v(p)f=0$ for all $p\in M$ and are asked to show that for every $p\in M$, $g(t):=f(\phi_t(p))$ is constant. Now calculate that $g'(t)=v(\phi_t(p))f=0$, hence $g$ is constant.
(Note that the calculation uses the group property of $\phi_t$, i.e. $\phi_{t+s}=\phi_t\circ\phi_s$).