Let $M$ be a Riemannian manifold with Levi-Civita connection $ \nabla$. Let $S$ be a differentiable manifold and $ \varphi : S \to M $ be a $C^{\infty}$ immersion. Let
$$ D : TS \times \{ \text{vector fields along } \varphi \} \to TM $$ s.t.
-
$ (v,X) \mapsto D_v(X) \in T_{\varphi(\pi(v))} M $
-
$D_{\alpha v_1 + \beta v_2}(X) = \alpha D_{v_1}(X) + \beta D_{v_2}(X)$
- $D_v( X+Y) = D_v(X)+D_v(Y)$ and $ D_v(fX) = f(\pi(v))D_v(X) + v(f)X_{\pi(v)}$
- $D_v (\varphi^* (X) ) = \nabla_{d\varphi_{\pi(v)}(v)} (X) $ for any $X$ vector field in $M$.
where $ \pi : TM \to M$ is the natural projection $ v \in T_pM \mapsto p $ and $\varphi^*(X) = X \circ \varphi $
I want to prove
$$ D_X( d\varphi(Y)) – D_Y( d\varphi(X)) = d\varphi ( [X,Y] )$$
Being $d\varphi(X)$ and $d \varphi(Y)$ vector fields along $\varphi$ I can write
$$ d\varphi(X) = U^i \partial_i \quad \quad d\varphi(Y) = V^j \partial_j $$
where $\partial_i \bigr |_p = \frac{\partial}{\partial x^i} \bigr |_{\varphi(p)}$ for each $p \in S$ and $ \{ x^1, \dots x^n \} $ are coordinates near $\varphi(p)$ in $M$. By linearity and using Leibniz rule the LHS becomes
$$ V^jD_X(\partial_j) + X(V^j) \partial_j – U^iD_Y(\partial_i) – Y(U^i) \partial_i =$$
$$ = V^j U^i \nabla_{\frac{\partial}{\partial x^i}} (\frac{\partial}{\partial x^j}) + X(V^j) \partial_j – V^j U^i \nabla_{\frac{\partial}{\partial x^j}} (\frac{\partial}{\partial x^i})- Y(U^i) \partial_i$$
$$= X(V^j) \partial_j – Y(U^i) \partial_i$$
The RHS is, for any $p \in S$ and $f \in C^{\infty}_{\varphi(p)}M$
$$(d\varphi ( [X,Y] ))_p (f) = d\varphi_p ( [X,Y]_p) (f) = [X,Y]_p (f \circ \varphi) $$
$$= X_p(Y(f \circ \varphi)) – Y_p(X(f \circ \varphi)) = X_p(d\varphi(Y)(f)) – Y_p(d\varphi(X)(f))$$
$$= X_p( V^j \frac{\partial f}{\partial x^j} \bigr |_{\varphi( ) } ) – Y_p( U^i \frac{\partial f}{\partial x^i} \bigr |_{\varphi( ) } ) $$ $$= X_p(V^j)\frac{\partial f}{\partial x^j} \bigr |_{\varphi(p) } – Y_p(U^i)\frac{\partial f}{\partial x^i} \bigr |_{\varphi(p) } + V_p^jX_p ( \frac{\partial f}{\partial x^j} \bigr |_{\varphi( ) }) – U_p^iY_p ( \frac{\partial f}{\partial x^i} \bigr |_{\varphi( ) }) $$
$$= LHS + V_p^jX_p ( \frac{\partial f}{\partial x^j} \bigr |_{\varphi( ) }) – U_p^iY_p ( \frac{\partial f}{\partial x^i} \bigr |_{\varphi( ) }) $$
Where is my mistake?
Best Answer
There is no mistake. Actually the remaining term is zero:
We have
\begin{align}V_p^jX_p ( \frac{\partial f}{\partial x^j} \bigr |_{\varphi( ) })=V_p^jX_p ( \frac{\partial f }{\partial x^j} \circ\varphi) =V_p^j(d\varphi (X))_p ( \frac{\partial f}{\partial x^j} ) =V_p^j U^i_p\frac{\partial }{\partial x^i}\bigr |_{\varphi(p) } ( \frac{\partial f}{\partial x^j} )\end{align}
Similar
$$U_p^iY_p ( \frac{\partial f}{\partial x^i} \bigr |_{\varphi( ) }) =U_p^i V^j_p\frac{\partial }{\partial x^j}\bigr |_{\varphi(p) } ( \frac{\partial f }{\partial x^i} )$$
so the remaining term is given by
$$V^j_pU^i_p\left(\left[ \frac{\partial }{\partial x^i}, \frac{\partial }{\partial x^j}\right]_{\varphi(p)}(f)\right)=0$$
since on canonical tangent vectors given by a coordinate chart the Lie bracket vanishes.