2

Consider $M := \mathbb R^3$ as a smooth manifold with a Riemannian metric $g := \sum_{i=1}^3 dx^i\otimes dx^i$, where $(x^1, x^2, x^3)$ is the standard coordinate of $M$. Let $N\subset M$ be a submanifold defined as $S^2 \cap H$, where $H = \{(x^1, x^2, x^3) \mid x^3 > 0\}$. Then $N$ has the coordinate $(x^1,x^2)$.

The problem is to represent the restriction of $g$ to the tangent bundle $TN$ w.r.t. the (global) frame $(dx^1, dx^2)$.

I tried to solve it in the following way: I introduced a coordinate neighborhood $(M \cap H , \psi)$ to $M$, where $\psi:M\cap H \rightarrow(0,\infty)\times(0,\pi)\times(0,\pi)$ is a diffeomorphism defined by $\psi^{-1}(r, \phi, \theta) = (r\cos\phi, r\sin\phi\cos\theta, r\sin\phi\sin\theta)$. Then I write $g$ in terms of $dr$, $d\phi$ and $d\theta$. Since I get $g|_{TN^{\otimes 2}}$ by dropping the terms that involve $dr$ from $g$, I rewrite this in terms of $dx^1, dx^2$ by applying inverse mapping theorem to obtain the result.

While this seems to work in theory, it is a bit cumbersome. I would be most grateful if you could help me solve this problem more easily.

Pteromys
  • 7,220
  • 1
    if $x_3>0$, then you can (globally) parametrize $N$ as $N={(x_1,x_2,\sqrt{1-x_1^2-x_2^2})}$ and compute the restriction of $g$ to $N$, right? – Avitus Jul 17 '13 at 11:05
  • Thank you, but I am totally unfamiliar with these things, so I would appreciate if you could elaborate on that. – Pteromys Jul 17 '13 at 11:09

2 Answers2

2

Let me elaborate on Avitus's comment.

  1. Let $\iota : N \to \mathbb{R}^3$ be the canonical embedding. Then what you want to understand is actually $\iota^\ast g = \delta_{ij} \iota^\ast dx^i \otimes \iota^\ast dx^j$.
  2. As you observed, you actually have a diffeomorphism $\phi : N \to U := \{(y^1,y^2) \in \mathbb{R}^2 \mid (y^1)^2 + (y^2)^2 < 1\}$ given by $$\phi(x^1,x^2,x^3) := (x^1,x^2),$$ so that $T^\ast N$ admits the global frame $$\{\phi^\ast dy^1,\phi^\ast dy^2\}.$$ What you would like to do is express $\iota^\ast g$ in terms of $\phi^\ast dy^1$ and $\phi^\ast dy^2$, which involves expressing $\{\iota^\ast dx^1,\iota^\ast dx^2,\iota^\ast dx^3\}$ in terms of $\{\phi^\ast dy^1,\phi^\ast dy^2\}$.
  3. As Avitus observed, $\phi^{-1} : U \to N$ is given by $$ \phi^{-1}(y^1,y^2) = \left(y^1,y^2,\sqrt{1-(y^1)^2-(y^2)^2}\right). $$ You can readily check, then, that $$ (\iota \circ \phi^{-1})^\ast(dx^1) = dy^1, \quad (\iota \circ \phi^{-1})^\ast(dx^2) = dy^1,\\(\iota \circ \phi^{-1})^\ast(dx^3) = -\frac{y^1}{\sqrt{1-(y^1)^2-(y^2)^2}} dy^1 -\frac{y^2}{\sqrt{1-(y^1)^2-(y^2)^2}} dy^2. $$

Now, observe that by the basic properties of pullbacks, for any $\omega \in \Omega^1(\mathbb{R}^3)$, $$ \phi^\ast\left(\iota \circ \phi^{-1}\right)^\ast(\omega) = \left(\left(\iota \circ \phi^{-1}\right) \circ \phi\right)^\ast(\omega) = \iota^\ast(\omega). $$ What does this imply, then, for $\iota^\ast dx^1$, $\iota^\ast dx^2$, $\iota^\ast dx^3$? What does this, in turn, imply for $\iota^\ast g$?

Branimir Ćaćić
  • 13,684
  • That simplified the calculation. Thank you. How do I know that what I want to understand is $\iota^*g$? – Pteromys Jul 18 '13 at 01:28
  • Because that's actually how you induce a Riemannian metric on a submanifold, by pulling back via the inclusion. If you like, you can view it roughly as restricting $g$ to a section of $T^\ast \mathbb{R}^3 \otimes T^\ast \mathbb{R}^3|_N$, and then reinterpreting the result as a section of $T^\ast N \otimes T^\ast N$, which is what you actually want. – Branimir Ćaćić Jul 18 '13 at 08:10
  • Is the definition of a metric on a submanifold by pulling back equal to the one in my first attempt to solve the problem, i.e., by dropping the terms that contain $dr$? (The problem is that the professor did not define how to induce a metric on a submanifold...) – Pteromys Jul 18 '13 at 08:44
  • What you did is at least a heuristic derivation of exactly the same induced metric $\iota^\ast g$, namely, the round metric on $S^3$, except in spherical coordinates instead of the one you were actually interested in. Rigorously, though, what you should do instead is repeat the procedure of the answer, except replacing the diffeomorphism $\phi : N \to U$ with the coordinate chart $\psi : V \to (0,2\pi) \times (0,\pi)$, where $\psi^{-1}(\phi,\theta) := (\cos \phi \sin \theta, \sin \phi \sin \theta, \cos \theta)$. – Branimir Ćaćić Jul 18 '13 at 09:00
  • That your original derivation with spherical coordinates works out is ultimately because spherical coordinates on $\mathbb{R}^3$ and spherical coordinates on the sphere $S^3$ are compatible with the inclusions $$ (0,2\pi) \times (0,\pi) \hookrightarrow (0,\infty) \times (0,2\pi) \times (0,\pi), \quad (\phi,\theta) \mapsto (1,\phi,\theta),\ S^3 \hookrightarrow \mathbb{R}^3, $$ fitting into a nice commutative diagram. – Branimir Ćaćić Jul 18 '13 at 09:06
  • Correction: It should be $\psi : V \to (0,2\pi) \times (0,\pi/2)$, to get just the upper hemisphere $N$. – Branimir Ćaćić Jul 18 '13 at 09:28
2

Here is a pedestrian approach:

Denote the coordinates in $N$ by $z=(z_1,z_2)$. Then we have a map $$f:\quad N\to {\mathbb R}^3, \qquad (z_1,z_2)\mapsto\left(z_1,z_2,\sqrt{1-z_1^2-z_2^2}\right)\ .$$ One computes $$df(z).e_1=f_{.1}(z)=\bigl(1,0,-z_1/\sqrt{1-z_1^2-z_2^2}\bigr),\quad df(z).e_2=f_{.2}(z)=\bigl(0,1,-z_2/\sqrt{1-z_1^2-z_2^2}\bigr)\ .\tag{1}$$ Now the fundamental form $g$ in ${\mathbb R}^3$ is just the scalar product. The matrix $\bigl[g^*_{ik}(z)\bigr]$ of the pullback $g^*$ on $N$ is therefore given by $$\eqalign{g^*_{ik}(z)&=g^*(z).(e_i,e_k)= df(z).e_i\>\cdot\>df(z).e_k\cr &=f_{.i}(z)\cdot f_{.k}(z)\ .\cr}$$ Now plug in the expressions obtained in $(1)$ and obtain $$\bigl[g^*_{ik}(z)\bigr]=\left[\matrix{{1-z_2^2\over 1-z_1^2-z_2^2} &{z_1z_2\over 1-z_1^2-z_2^2} \cr {z_1z_2\over 1-z_1^2-z_2^2} & {1-z_1^2\over 1-z_1^2-z_2^2} \cr}\right]\ .$$

Christian Blatter
  • 226,825