Shape Operator $S_n (x) = - (\bar{\nabla}_xN)^t$ This is from page 128 from Da Carmo's Riemannian Geometry.

let $p \in M$, $x \in T_p M$ and $\eta \in (T_p M)^\perp.$ Let $N$ be a local extension of $n$ normal to $M$. Then $$S_n (x) = - (\bar{\nabla}_xN)^t$$

His proof goes like this

Let $y \in T_pM$ and $X,Y$ be local extensions of $x,y$ which are tangent to $M$. Then $(N,Y) = 0$, therefore.
$(S_n(X),y) = (B(X,Y)(p),N) = (\bar{\nabla}_X Y - \bar{\nabla}Y,N)(p) = (\bar{\nabla}_X Y, N)(p) = -(Y,\bar{\nabla}_XN)(p) = (-\bar{\nabla}_x N,y)$
for all $y \in T_pM$.

I don't understand the last equality $-(Y,\bar{\nabla}_XN)(p) = (-\bar{\nabla}_x N,y)$

*

*Is he evaluating at $p$ so we are back to the vectors $x$ and $y$? Then why do we still have $N$ and not $n$?


*How does this show the result? The claim is that it is equal to the tangent of the derivative. It is important that $\nabla_XY = (\bar{\nabla}_{\bar{X}}\bar{Y})^t$


*Elementary question, but I thought covariant derivative is always in the direction of a vector field not a single vector? So what is meant by $\nabla_x$?
 A: *

*Yes, he's evaluating $-\langle Y, \overline{\nabla}_{X} N \rangle$ at $p$. This equals $-\langle Y(p), (\overline{\nabla}_{X} N)(p) \rangle = -\langle y, (\overline{\nabla}_{X} N)(p) \rangle $. Now, remember that given any affine connection $\nabla$, $(\nabla_X Y)(p)$ depends only on $X(p)$ and on the values of $Y$ along a curve $\alpha$ such that $\alpha(0) = p$ and $\alpha'(0) = X(p) = x$. So it makes perfect sense to let $\nabla_x Y$ (which is a tangent vector in $T_p M$) be shorthand for $(\nabla_X Y)(p)$, which is what he's doing. However, $(\nabla_X Y)(p)$ depends on $X(p)$ and on the values of $Y$ along a tangent curve which satifies what I wrote before. It does not depend only on $X(p)$ and $Y(p)$. That's why writing $\nabla_x y$ wouldn't make sense, and that's why we still have $N$ instead of $n$.


*We just have to write a little more. Notice that if you write $\overline{\nabla}_X Y = \left(\overline{\nabla}_X Y  \right)^{\perp} +\left(\overline{\nabla}_X Y  \right)^{T}  $, he actually proved that:
$$\langle S_{\eta}(x), y \rangle = \left\langle - \left(\overline{\nabla}_X Y  \right)^{T}, y  \right\rangle \text{ for all $x, y \in T_p M$ }$$
And since by the properties of a Riemannian metric a tangent vector is uniquely determined by it's inner products with the whole tangent space (actually just a basis would be enough), we get the desired result.


*You're right, it's an abuse of notation. But given what I mentioned in 1., this is a completely harmless abuse.

