Question about the second fundamental form I am studying Riemannian geometry and have a question understanding something.
I use Do Carmo's book.
In the book, a vector field is defined for isometric immersions:
for an immersion $$ f:M\rightarrow\bar{M} $$
a local vector field is defined as
$$ 
B(X,Y) = \bar{\nabla}_{\bar{X}}\bar{Y}-\nabla_xY
$$
It is said that $B(X,Y)$ does not depend on the extensions $\bar{X}$,$\bar{Y}$
.
If $\bar{X}_1$ is another extension of $X$,
We will have
$$
( \bar{\nabla}_{\bar{X}}\bar{Y}-\nabla_xY )
- \bar{\nabla}_{\bar{X}_1}\bar{Y}-\nabla_xY = \bar{\nabla}_{\bar{X}-\bar{X}_1}\bar{Y}
$$
In the book, it is indicated that because $\bar{X}-\bar{X}_1 = 0$ on $M$, so the above difference vanishes on $M$.
But why?
In my opinion, $B(X,Y)$ is defined on $\bar{M}$, so the difference vanishes on $M$ does not mean anything.
Intuitively, we should have $\bar{X} = X + X^\perp$, where $X^\perp$ is normal to $X$. The choice of $X^\perp$ should be arbitrary.
Thus $B(X,Y)$ will not be unique. 
Can someone help me to understand it?
 A: This is a very good question that deserves a thorough answer. The OP has spotted a murky wording in the celebrated textbook of M.P. do Carmo "Riemannian geometry" on page 126 where is is said that $B(X,Y) = \overline{\nabla}_{\overline{X}} \overline{Y} - \nabla_X Y$ 

is a local vector field on $\overline{M}$ normal to $M$.

I guess that this causes problems in understanding because a local vector field on $\overline{M}$ is supposed to be defined on an open subset of $\overline{M}$, but $B(X,Y)$ is well defined only at each point $p \in M$, and as a whole $B$ is a bi-linear operator on $TM$ with values in $(T_p M)^{\perp}$.
Maybe, it would be better to say, that at each point $p$ and for any two tangent to $M$ vector fields $X$ and $Y$ we can find local extensions $\overline{X}$ and $\overline{Y}$ of the fields $X$ and $Y$ respectively, and an extension $\overline{\nabla_X Y}$ of the tangent to $M$ vector field $\nabla_X Y$, and then form a vector field $\overline{B}(\overline{X},\overline{Y},\overline{\nabla_X Y})$ on an open subset $\overline{U}$ of $\overline{M}$ containing the point $p$ defined by the equation
$$
\overline{B}(\overline{X},\overline{Y},\overline{\nabla_X Y}) := \overline{\nabla}_{\overline{X}} \overline{Y} - \overline{\nabla_X Y}
$$
Of course, $\overline{B}(\overline{X},\overline{Y},\overline{\nabla_X Y})$ heavily depends on extensions, but only away from the manifold $M$.

Fact. At each point $p \in M$  $$ B(X,Y)|_p := \overline{B}(\overline{X},\overline{Y},\overline{\nabla_X Y})|_p $$  does not depend on a
  choice of extensions for the tangent to $M$ vector fields $X$, $Y$ and
  $\nabla_X Y$.

With each choice of tangent to $M$ vector fields $X$ and $Y$ we obtain $B(X,Y)$ as a vector field  along $M$ (or, along the immersion $f$, if you like).
Not a local vector field on $\overline{M}$...

Edit. A friend of mine just has pointed to me that OP also misinterprets the notion of extension. 

Definition. A vector field $\overline{X}$ on $\overline{M}$ is an extension of a vector field $X$ on $M$ if and only if $\overline{X} = X$ at all points of $M$.

In other words, the normal component $(\overline{X})^{\perp}$ vanishes along $M$. 
Yet another way to say that, the choices of $X^{\perp}$ are not arbitrary along $M$, in fact $X^{\perp}$ must vanish for $\overline{X}$ to be an extension of $X$.
This is "why".
