1
$\begingroup$

Also asked here: https://math.stackexchange.com/questions/1725491/riemannian-metric-on-a-level-set-of-a-smooth-function-on-a-manifold

Let $(M,g)$ be a finite or infinite dimensional Riemannian manifold. Let $G:M\to \mathbb{R}^d$ be a smooth map. Here $d < dim(M)$ if $M$ is finite dimensional, otherwise there's no restriction on $d$. Let $L:=G^{-1} \{0\} $. Is there a way we can express the restricted Riemannian metric on $L$ by explicitly using $G$?

So, let $q\in L, v,w\in T_q{L}$. How can we define the restricted inner product $g_q(v,w)$ on the submanifold $L$ using $G$, as explicitly as possible?

Thank you!

Improve this question
$\endgroup$
5
  • $\begingroup$ You might try using a moving frame of orthonormal vector fields, with the first so many of them tangent to the level sets of $G$, and the rest perpendicular. $\endgroup$
    – Ben McKay
    Commented Apr 3, 2016 at 7:50
  • 2
    $\begingroup$ Or use the implicit function theorem to write out coordinates $(x,y)$ in which your map is $G(x,y)=x$. Do you know if $G'$ has full rank? $\endgroup$
    – Ben McKay
    Commented Apr 3, 2016 at 7:52
  • $\begingroup$ @BenMcKay Yes assume I know that $DG$ has full rank at all points, so we cna apply the implicit function theorem, so we can express the level set locally as a graph. $\endgroup$
    – Learning math
    Commented Apr 3, 2016 at 9:44
  • 1
    $\begingroup$ Maybe you’re ultimately interested in more than just the metric? What’s more interesting is how to compute the second fundamental form and Riemann curvature of the submanifold using $G$. $\endgroup$
    – Deane Yang
    Commented Jan 8, 2018 at 4:49
  • $\begingroup$ If you assume that the metric on $M$ is known, then the metric on $L$ is, for each $x \in L$, the restriction of the metric on $T_xM$ to the subspace $T_xL$. So the only issue is identifying the subspace $T_xL$. This, however, is the space of vectors $v \in T_xM$ such that $\langle dG(x),v\rangle = 0$. Another way to say this is that it is the subspace of $T_xM$ that is orthogonal to $\nabla G(x)$. This, I believe, is as explicit as you can get with the metric on $L$. $\endgroup$
    – Deane Yang
    Commented Jan 8, 2018 at 20:56

1 Answer 1

Reset to default
3
$\begingroup$

First you need your level set $L \subset M$ to be a smooth manifold. For example you can require $G$ to be a submersion or that all points of $L$ are regular points for the map $G$.

Then, for all $q \in L$, $T_q L = \mathrm{ker} d_q G \subset T_q M$, where $d_q G : T_q M \to T_0\mathbb{R} = \mathbb{R}$.

The Riemannian structure $h$ on $L$ is just the restriction to the submanifold:

$$h_q(v,w) = g_q(v,w), \qquad \forall q \in L, \, \forall v,w \in T_q L \subset T_q M.$$

Remark: you cannot express $h$ using $G$ in a direct way (or at least, there is no explicit formula which gives $h$ in terms of $g$ and $G$). Indeed the metric $h$ is intrinsically defined by the restriction of $g$ to the submanifold $L$, and there are infinitely many ways to describe a fixed submanifold $L$ as the level set of a regular function.

As Ben suggested, choose a neighborhood $U$ of $q \in L$ and submanifold coordinates $(x,y) \in \mathbb{R}^{n-d} \times \mathbb{R}^d$ on $U$ such that $G(x,y) = y$. In this way:

$$ L \cap U \simeq \{ (x,0) \mid x \in \mathbb{R}^{n-d}\}. $$

Then the symmetric matrix representing $h$ on $L$ is just the $n-d\times n-d$ upper left block representing $g$, restricted to $L\cap U$:

$$h_x = \sum_{i,j=1}^{n-d} g_{ij}(x,0) \,dx^i \otimes dx^j. $$

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .