3
$\begingroup$

Hello, I have a question... I think it is worthy for MO.

Let $f: M \to S_{n+1}$ a spacelike immersion in the de Sitter-space with $S_{n+1}:=\lbrace X \in \mathbb{R}_{2}^{n+2}: \left\langle X, X \right\rangle=1\rbrace$.

Here is

$\mathbb{R}_{2}^{n+2}=(\mathbb{R}^{n+2}, \left\langle \cdot, \cdot \right\rangle)$

and

$\left\langle X, Y \right\rangle:=X_{1}Y_{1}+\ldots+X_{n}Y_{n}-X_{n+1}Y_{n+1}-X_{n+2}Y_{n+2}.$

We now consider the Gauss-map $G: M \to Gr_{n}^{+}(2, n+2)$ into Grassmannian manifold of all oriented spacelike n-planes in $\mathbb{R}_{2}^{n+2}$, where

$p \mapsto T_{p}M \in Gr_{n}^{+}(2, n+2)$.

Now denote with $g_{ij}$ the first and with $b_{ij}$ the second fundamentalform of $f$. Now the question ist:

What is the first and second fundamentalform of $G$?

I mean I have to calculate

$\tilde{g}_{\alpha\beta}\frac{\partial G^{\alpha}}{\partial x^{i}}\frac{\partial G^{\beta}}{\partial x^{j}}$.

But what is the metric $\tilde{g}$ in local coordinates of $Gr_{n}^{+}(2, n+2)$ and how can I calculate $\frac{\partial G^{\alpha}}{\partial x^{i}}$?

In the case $n=2$ it is easy, because $Gr_{2}^{+}(2, 4)$ is isometric to $\mathbb{H}^{2} \times \mathbb{H}^{2}$. But how does it work in the general case?

Many greetings Wolfgang

$\endgroup$
5
  • $\begingroup$ nobody an idea? $\endgroup$
    – Wolfgangm
    Dec 5, 2011 at 15:24
  • $\begingroup$ @Wolfganm: Well, some of us do know the answer and also know that it can be found in most good differential geometry books. I'm a little concerned that you have not carefully written out the second fundamental form $II_f$ of the map $f$. You've just written $b_{ij}$, but, of course, the second fundamental form is vector valued since you are in codimension $2$. If you aren't aware of that, then you probably should be looking for a good introductory book on submanifold theory. If you are already aware of that and just want a hint as to what the formula looks like, I can give you one. $\endgroup$ Dec 5, 2011 at 18:16
  • $\begingroup$ Thanks Robert. I think I want a hint as to what the formula looks like. Can one written the second fundamentalform in dependence of the first and second fundamentalform $g_{ij}$ and $h_{ij}$ of $F$? $\endgroup$
    – Wolfgangm
    Dec 7, 2011 at 8:05
  • $\begingroup$ Or can you give me a book, where I can read something about? $\endgroup$
    – Wolfgangm
    Dec 8, 2011 at 14:41
  • $\begingroup$ @Wolfgangm: There are several books that treat pseudo-Riemannian geometry and submanifold theory. I don't know the best one to recommend for you because I don't know what you know. I can give you an answer to the above question, except that I now realize that there is something wrong with your question: It is not possible to have an $n$-dimensional spacelike submanifold of $S_{n+1}$ because the induced metric on this submanifold has type $(n{-}1,2)$. Did you mean to have $S_{n+1}$ be defined by $\langle x,x\rangle=-1$ (instead of $+1$)? With the minus, it would have type $(n,1)$. $\endgroup$ Dec 10, 2011 at 14:33

1 Answer 1

6
$\begingroup$

Note that $S_{n+1}$ is to be defined by $\langle x,x\rangle = -1$ (instead of $+1$, as in the OP). The answers are computed by means of the structure equations and are as follows:

Let the first and second fundamental forms of $f$ be given by $I_f = g_{ij}\ dx^idx^j$ and $I\!I_f = h_{ij}\ dx^idx^j$ in some local coordinates. For the Gauss map $G:M\to \text{Gr}_n^+(2,n{+}2)$, one then has $$ I_G = \left(g_{ij} + g^{kl}h_{ik}h_{jl}\right)\ dx^idx^j. $$

To interpret the second fundamental form, one first remembers that $\text{Gr}_n^+(2,n{+}2)$ has the structure of a Kähler manifold (it is the dual of the Hermitian symmetric space $\text{SO}(n{+}2)/\text{SO}(2)\text{SO}(n)$). Then one observes that $G$ immerses $M$ into $\text{Gr}_n^+(2,n{+}2)$ as a Lagrangian submanifold (with respect to the Kähler form), so that the normal bundle of the immersion $G$ into $\text{Gr}_n^+(2,n{+}2)$ is naturally isomorphic to the tangent bundle of $M$. This means that the second fundamental form, which is usually thought of as a quadratic form with values in the normal bundle, can be regarded as a cubic form. Because the immersion is Lagrangian, this cubic form is symmetric. Calculation with the structure equations then shows that one has $$ I\!I_G = \nabla^{I_f}\left(I\!I_f\right). $$ I.e., the second fundamental form of the Gauss map $G$ is the covariant derivative of the second fundamental form of $f$ with respect to the Levi-Civita connection of the first fundamental form of $f$. (Note that the fact that this covariant derivative is indeed symmetric is simply the Codazzi equation for the immersion $f$.)

$\endgroup$
6
  • $\begingroup$ Oh yes, sry. I mean $S_{n+1}:=\lbrace X \in \mathbb{R}_{2}^{n+2}: \left\langle X, X \right\rangle=-1\rbrace$. Sry. Greetings! $\endgroup$
    – Wolfgangm
    Dec 11, 2011 at 20:02
  • $\begingroup$ Ah, very cool! But, if you calculate the first fundamentalform you need $\mathcal{G}_{i}^{\alpha}$. Right? And for the second fundamental form too!? Thanks! $\endgroup$
    – Wolfgangm
    Dec 12, 2011 at 14:55
  • 3
    $\begingroup$ @Wolfgangm: Actually, if you do the computation via the moving frame (as I did), you don't need to know those particular quantities; you get the answers without having to write out the natural metric on $\text{Gr}^+(n,n{+}2)$. Also, I should point out that I'm regarding $\text{Gr}^+(n,n{+}2)$ as an abstract manifold, not as embedded into some vector space (similar to the way you are thinking of $S_{n+1}$ as an abstract manifold when you write the second fundamental form of $f$ as a quadratic form, rather than as a quadratic form with values in the normal bundle in $\mathbb{R}^{n+2}_2$. $\endgroup$ Dec 12, 2011 at 15:57
  • $\begingroup$ Thanks for your efforts. I have two much questions: 1.) Can one seen that $\mathcal{G}$ is Lagrangian immersion in local coordinates? This follows from the Kahler property, too? 2.) It is very interesting that I do not need the metric or $\mathcal{G}_{i}^{\alpha}$ to calculate the fundamentalforms. But how do you mean "via the moving frame"? Sry for this stupid question. But I do not know. Greetings! $\endgroup$
    – Wolfgangm
    Dec 13, 2011 at 9:40
  • $\begingroup$ @Wolfgangm: Unfortunately, I don't have the time (and it's probably not appropriate anyway) to explain the method of the moving frame here. Assuming the necessary background in Lie groups and differential forms, there are many sources for the reader who wants to learn it, though. Monographs of Chern and his mathematical relatives and descendants cover this very well. Look for authors Chern, Griffiths, Jensen, Green, Sharpe, Spivak, and Sternberg, as well as more recent ones such as Ivey and Landsberg's "Cartan for Beginners..." Of course, there are also Élie Cartan's great works on MoMF. $\endgroup$ Dec 13, 2011 at 15:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.