8

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?

xuehy
  • 83
  • $B$ is only defined on the image of $f$ – Thomas May 18 '14 at 15:17
  • I still don't understand. Can you please explain it in detail? $B$ is defined on $\chi^\perp(U)$. So actually $B$ does not belong to $T_p(M)$. How could it be defined on the image of $f$. I think it is defined on $\bar{M}$-image of $f$? – xuehy May 18 '14 at 15:22
  • One sometimes loosely speaks of $B$ being defined on the image of $f$ when it is actually defined on some vector bundle over that space. Think of a surface in $\overline M = \mathbb{R}^3$. $B$ is only defined where both derivative make sense, which is (a bit more precisely than before) on the tangent bundle of $\overline M$ restricted to $f(M)$, and only for vectors tangent to $f(M)$. On that set extensions of vector fields $X$ tangent to $f(M)$ agree (trivially), and only there it is relevant. – Thomas May 18 '14 at 16:00
  • All right. If $B$ is defined on the tangent bundle of $\bar{M}$ restricted to $f(M)$, I think $B$ will always be 0. because the two items in the definition of $B$ should equal, which is $\bar{\nabla}_{\bar{X}}\bar{Y}=\nabla_XY$. In addition, is it right $B\in T_p(\bar{M})$? – xuehy May 18 '14 at 16:22
  • Yes, as the difference of two elements of that space it is in that space. It is, however, always normal to $f(M)$, it is, actually, the normal component of the covariant derivative in the surrounding space. – Thomas May 19 '14 at 05:06

1 Answers1

5

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".

Yuri Vyatkin
  • 11,279