Posted by: cnpde | September 18, 2012

## Basics of Riemann Geometry 3

I will devote this article to Cartan’s package.

Assume $u$ is a vector field on $\mathcal{U}\subset M$. The flow of $u$ is denoted as $\varphi_t$. Suppose $\varphi_t: \mathcal{U} \rightarrow \mathcal{W}$ is diffeomorphism, and its inverse is $\varphi_{-t}$ with push-forward $\varphi_{-t*}: T\mathcal{W}\rightarrow T\mathcal{U}$.

Definition1: For vector field $u$ with flow $\varphi_t$, the Lie derivative $\mathfrak{L}_u: T_s^r\rightarrow T_s^r$ of a tensor is defined as $(\mathfrak{L}_u \Phi)(x)=\lim\limits_{t\rightarrow 0}\frac{1}{t}[\varphi_{-t*}\Phi(x)-\Phi(x)]$.

Definition2: For a vector $u$, define interior product of tensor $\Phi$ with vector $u$ as $i_u: T_r(V)\rightarrow T_{r-1}(V): \Phi\mapsto i_u\Phi$, satisfying

$i_u\Phi(u_1,..., u_{r-1})=\Phi(u, u_1,...,u_{r-1})$.

Cartan’s package is

$[d,d]=0, \quad [i_X,i_Y]=0$

$[d,\mathfrak{L}_X]=0,\quad [\mathfrak{L}_X,i_Y]=i_{[X,Y]}$

$[\mathfrak{L}_X,\mathfrak{L}_Y]=\mathfrak{L}_{[X,Y]}, \quad [d,i_X]=\mathfrak{L}_X$.

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1) Assume $\Phi\in T_s^r$

$\mathfrak{L}_u(\Phi(w_1,...,w_r,v_1,...,v_s))$

$=(\mathfrak{L}_u\Phi)(w_1,...,w_r,v_1,...,v_s)$

$+\sum\limits_{i=1}^r\Phi(w_1,...,\mathfrak{L}_uw_i,...w_r,v_1,...,v_s)$

$+\sum\limits_{i=1}^r\Phi(w_1,...,w_r,v_1,...\mathfrak{L}_uv_i,...,v_s)$

Proof: $LHS=\lim\limits_{t\rightarrow 0}\frac1t[\Phi(\varphi_t(x))(\varphi^*_tw_1(\varphi_t(x)),..., \varphi_t^*w_r(\varphi_t(x)),\varphi_{t*}u_1(\varphi_t(x)),...,$

$\varphi_{t*}u_s(\varphi_t(x))) -\Phi(x)(w_1(x),...,w_r(x),u_1(x),...,u_s(x))]$
$=\lim\limits_{t\rightarrow 0}\frac1t[\Phi(\varphi_t(x))(\varphi^*_tw_1(\varphi_t(x)),..., \varphi_t^*w_r(\varphi_t(x)),\varphi_{t*}u_1(\varphi_t(x)),...,$

$\varphi_{t*}u_s(\varphi_t(x))) -\Phi(\varphi_t(x))(w_1(\varphi_t(x)),..., w_r(\varphi_t(x)),u_1(\varphi_t(x)),...,u_s(\varphi_t(x))]$
$+\lim\limits_{t\rightarrow 0}\frac1t[\Phi(\varphi_t(x))(w_1(\varphi_t(x)),..., w_r(\varphi_t(x)),u_1(\varphi_t(x)),...,u_s(\varphi_t(x))$
$-\Phi(x)(w_1(x),...,w_r(x),u_1(x),...,u_s(x))]$

$=\sum\limits_{i=1}^r \Phi(w_1,...,\mathfrak{L}_uw_i,...,w_r,u_1,...,u_s)$

$+\sum\limits_{i=1}^s \Phi(w_1,...,w_r,u_1,...,\mathfrak{L}_uu_i,...,u_s)$

$+\mathfrak{L}\Phi(w_1(x),...,w_r(x),u_1(x),...,u_s(x)).$

2) $[\mathfrak{L}_u,\mathfrak{L}_v]=\mathfrak{L}_{[u,v]}$

Proof by induction:

$[L_u,L_v]f=L_uL_vf-L_vL_uf=uvf-vuf=L_{[u,v]}f$.

Suppose it holds for $\Phi$, then we can prove it also holds for $df\wedge \Phi.$

3) $[d,d]=0.$

Proof: $[d,d]\Phi=2d^2\Phi=0.$

4)$[d,L_u]=0.$

Proof: $dL_u\Phi=d\lim\frac{\Phi(\varphi_t(x))-\Phi(x)}{t}=\lim\frac{d\Phi(\varphi_t(x))-d\Phi(x)}{t}=L_ud\Phi.$

5) $[i_u,i_v]=0.$

Proof: $i_ui_v\Phi(w_3,...,w_s)+i_vi_u\Phi(w_3,...,w_s)$

$=\Phi(v,u,w_3,...,w_s)+\Phi(u,v,w_3,...,w_s)=0,$

due to skey-symmetry.

6) $[L_u,i_v]=i_{[u,v]}.$

Proof: $[L_u,i_v]\Phi=L_ui_v\Phi-i_vL_u\Phi$

$=\lim\limits_{t\rightarrow 0}\frac{\Phi(\varphi_t(x)(v(\varphi_t(x)),u_2(x),...,u_r(x))-\Phi(x)(v(x),u_2(x),...,u_r(x)}{t}$

$-\lim\limits_{t\rightarrow 0}\frac{\Phi(\varphi_t(x)(v(x),u_2(x),...,u_r(x))-\Phi(x)(v(x),u_2(x),...,u_r(x)}{t}$

$=\lim\limits_{t\rightarrow 0}\frac{\Phi(\varphi_t(x)(v(\varphi_t(x)),u_2(x),...,u_r(x))-\Phi(\varphi_t(x)(v(x),u_2(x),...,u_r(x))}{t}$

$=\Phi(x)(L_uv,u_2,...,u_r)$

$i_{[u,v]}\Phi.$

7) $[d,i_u]=L_u.$

Proof by induction: $[d,i_u]f=i_udf=uf=L_u.$

Supppose it holds for $\Phi$, prove it for $df\wedge \Phi.$

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When we look back, we may find it has many terms on RHS because the manifold is not flat, i.e. the frame is changing with manifold. So when we take derivative, we not only need to differentiate the component of the tensor fields, which is a function of manifold, but also need to differentiate the frame, which also rely on the space point. In the end, we get several terms when we calculate the derivative.

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