I am studying Riemannian geometry using Do Carmo's book. I am learning about isometric immersions right now, and I got stuck with the following claim about Codazzi's equation.
Let $f:M^n \to \overline M ^{n+m} $ be an immersion, with the metric of $M$ being the one iduced from $\overline M$.
In this situation we can write $T_p \overline M = T_p M \oplus (T_p M)^{\bot}$.
We can also split the connection $\overline\nabla$ of $\overline M$ to its tangential and normal components.
This gives rise to the normal connection, defined on the normal bundle by $$\nabla^{\bot}_X\eta = \text{the normal component of } \overline\nabla_X\eta$$ where $X\in \mathcal X(M)$ and $\eta \in \mathcal X(M)^{\bot}$.
Let $X,Y,Z \in \mathcal X(M)$ and $\eta \in \mathcal X(M)^{\bot}$. Codazzi's equation is $$ <\overline R(X,Y)Z, \eta> = (\overline\nabla_Y B)(X, Z, \eta) - (\overline\nabla_X B)(Y, Z, \eta)$$ where $B(X, Y, \eta) = <B(X, Y), \eta>$ and $B(X, Y)$ is the normal component of $\overline\nabla_X Y$.
Now, if $\overline M$ has constant sectional curvature, the curvature tensor is 0, so Codazzi's equation can be rewritten as $$(\overline\nabla_Y B)(X, Z, \eta) = (\overline\nabla_X B)(Y, Z, \eta)$$
Up until here I understood everything. Now the author claims that if, in addition $M$ is a hypersurface (i.e. the codimension of the immersion is 1) then $\nabla^{\bot}_X\eta = 0$. This again allows to again rewrite the equation in another form.
My question is, why is that if $\overline M$ has constant sectional curvature and the codimension of $M$ in $\overline M$ is 1 then $\nabla^{\bot}_X\eta = 0$?
This is easy to see if $\eta$ has length $1$, for then, for $X$ tangent to $M$, $$X\langle \eta, \eta\rangle = 0 = 2\langle \eta ,\bar{\nabla}_X \eta\rangle$$
(Note that this reasoning, in fact, only works in codimension $1$).
I don't think this is true in general, though I may have forgotten something during the years.