Ricci equation for the normal connection

Apply [F1] first to ${\bar{\mathrm{\nabla }}}_Y\nu =-S_{\nu }Y+{\mathrm{\nabla }}_Y^{\perp }\nu$. The normal component after differentiating in direction $X$ is $({\bar{\mathrm{\nabla }}}_X{\bar{\mathrm{\nabla }}}_Y\nu {)}^{\perp }=-\mathrm{I}\mathrm{I}(X,S_{\nu }Y)+{\mathrm{\nabla }}_X^{\perp }{\mathrm{\nabla }}_Y^{\perp }\nu \mathrm{.}$ The analogous formula holds with $X,Y$ interchanged, and $({\bar{\mathrm{\nabla }}}_{[X,Y]}\nu {)}^{\perp }={\mathrm{\nabla }}_{[X,Y]}^{\perp }\nu$.

Form the bracket-corrected ambient curvature and take its normal component. By [F4], the three normal-connection terms combine to $R^{\perp }(X,Y)\nu$, leaving $(\bar{R}(X,Y)\nu {)}^{\perp }=R^{\perp }(X,Y)\nu -\mathrm{I}\mathrm{I}(X,S_{\nu }Y)+\mathrm{I}\mathrm{I}(Y,S_{\nu }X)\mathrm{.}$

Pair step 2.1 with $\mu$ and use [F1]: the two correction terms become $-g(S_{\mu }X,S_{\nu }Y)+g(S_{\mu }Y,S_{\nu }X)$. Self-adjointness rewrites their sum as $-g(S_{\nu }{S}_{\mu }X,Y)+g(S_{\mu }{S}_{\nu }X,Y)=-g([S_{\nu },S_{\mu }]X,Y),$ which proves the stated equation.

The equation is vacuous on the empty submanifold. If tangent or normal rank is zero every term vanishes; in tangent rank one the curvature pair and the commutator vanish. The same calculation applies for a normal line bundle and at boundary points. Positive definiteness supplies the orthogonal splitting and self-adjointness. The stated ${\mathrm{A}\mathrm{C}}_{\omega }$ is inherited through [F1]–[F3], and no new selection occurs.