Help understanding the derivation of weingarten formula.

Question

Let $f: M^n \to \tilde{M}^m$ be an isometric immersion of Riemannian manifolds (that is $\langle X,Y \rangle_M =\langle f_*(x) X, f_*(x)Y \rangle_{\tilde{M}}$ with $\langle , \rangle_M$ and $\langle , \rangle_{\tilde{M}}$ be the Riemannian structures of $M$ and $\tilde{M}$ respectively and with $\tilde{\nabla}$ the Levi-Civita connection for $\tilde{M}^m$

For $X,Y \in \mathcal{X} (M)$ we have the decomposition $$\tilde{\nabla}_X f_* Y =(\tilde{\nabla}_X f_* Y )^T +(\tilde{\nabla}_X f_*Y )^{\perp}$$ with respect to the decomposition $$f^* T \tilde{M} =f_*TM \oplus N_fM,$$ where $N_fM$ is the normal bundle of $f$.

Let $a:\mathcal{X} (M) \times \mathcal{X} (M) \to \Gamma (N_fM)$ defined by $$a^f (X,Y) =(\tilde{\nabla }_X f_* Y)^{\perp}$$ be the second fundamental form of $f$.

Let $A_{\xi}$ της $f$ στο $x \in M^n$ the shape operator at $\xi \in N_fM(x)$ defined as $$\langle A_{\xi} X,Y \rangle =\langle a(X,Y),\xi \rangle$$ for all $X,Y \in T_xM$.

We will derive the weingarten formula. For vector fields $X,Y \in \mathcal{X}(M)$ and $\xi \in \Gamma (N_fM)$ we have $$\langle \tilde{\nabla}_X \xi ,f_* Y\rangle =-\langle \xi ,\tilde{\nabla}_X f_*Y \rangle $$ $$ =-\langle \xi , a(X,Y) \rangle$$ $$ = -\langle A_{\xi} X,Y \rangle$$ Therefore the tangential component of $\tilde{\nabla_X} \xi$ is $-f_* A_{\xi} X$.

I don't understand the proof, specifically

1. why is it true that $ -\langle \xi ,\tilde{\nabla}_X f_*Y \rangle=-\langle \xi , a(X,Y) \rangle$, since $a$ is defined to be equal to $(\tilde{\nabla }_X f_* Y)^{\perp}$, not $\tilde{\nabla }_X f_* Y$, there must be some property of inner products that I am missing and
2. why does $\langle \tilde{\nabla}_X \xi ,f_* Y\rangle = -\langle A_{\xi} X, Y \rangle$ imply that the tangential component of $\tilde{\nabla_X} \xi$ is $-f_* A_{\xi} X$, I think this should imply that $\tilde{\nabla}_X \xi = -f_* A_{\xi } X$ by the properties of inner products, but here the author of the proof concludes that $$\tilde{\nabla}_X \xi = -f_* A_{\xi } X +\text{normal component}$$ why is that?

Can you explain?

Answer 1

I believe I have work it out.

For the first part we have $$\langle \xi , \tilde{\nabla}_X f_*Y\rangle$$ but $\xi$ is in the normal bundle, therefore when we take an inner product on the normal bundle we should get the normal component only, hence $$a(X,Y) =(\tilde{\nabla}_X f_* Y)^{\perp}.$$

Similarly for the second part $f_*Y$ lives in the tangent bundle therefore we should only get the tangent component of $\tilde{\nabla}_X \xi$.