4
$\begingroup$

I am studying Riemannian Geometry from the book by M.P. do Carmo and I am trying to get a complete picture of Connections by working out some examples and in particular I to calculate for the class of sub-manifolds given as a regular level set of a regular value, I'm trying to calculate it for that case. As a corollary, I wish to see the case of sphere, or some other simple manifold.

If $f:\Bbb{R}^n \to \Bbb{R}^k$ where $ (k<n)$ and $p\in\Bbb{R}^k$ be a regular value, then $M = f^{-1}(p)$ will be a regular submanifold of $\Bbb{R}^n$ of dimension $n-k$. Now how to give a connection on $M$?

Any hint will be appreciated. It will be really helpful if one can give a reference to a book, online material or online lecture notes in which similar calculations are done or where sufficient hints are provided.

$\endgroup$
  • 1
    $\begingroup$ math.stackexchange.com/questions/155173/… this might be a helpful link for you. $\endgroup$ – Anubhav Mukherjee Nov 24 '16 at 16:04
  • 1
    $\begingroup$ @Anubhav Thanks! That seems like a good approach to give a metric on any embedded manifold but if a function is given and we explicitly have a regular submanifold of domain (through the regular level set), then if I'm not mistaken, this would be hard to do via the given method, wouldn't it? Any alternative? $\endgroup$ – glip-glop Nov 24 '16 at 17:25
3
+50
$\begingroup$

I assume that by "connection" you just mean "covariant derivative". If this is the case, then you can basically proceed as in the case of hypersurfaces and define the Levi-Civita connection on the submanifold via the orthogonal projection of the "ordinary derivative" to the tangent spaces of $M$. Unfortunately, this is not partiuclarly well adapted to the presentation of a submanifold as the pre-image of a regular value, but it can be formulated in these terms.

In the setup you have, you can view a vector field $\eta$ on $M$ as a smooth function $\eta:M\to\mathbb R^n$ such that for each $x\in M$, the value $\eta(x)$ lies in $T_xM$, i.e. that $df(x)(\eta(x))=0$. Now you can simply differentiate $\xi$ as a function to $\mathbb R^n$ in directions tangent to $M$. Most easily, you can represent a tangent vector to $M$ in $x$ as $c'(0)$ for a smooth curve $c:(-\epsilon,\epsilon)\to M$ such that $c(0)=x$. The derivative of $\eta$ in direction of this tangent vector then is given by $\frac{d}{dt}|_{t=0}\eta(c(t))\in\mathbb R^n$. If you have two vector fields $\xi$ and $\eta$ on $M$, you can use this construction to define a function $\xi\cdot\eta:M\to\mathbb R^n$ whose value in $x$ is the derivative of $\eta$ in $x$ in direction $\xi(x)$. (Alternatively, you can also extend $\eta$ to a smooth function defined on an open neighborhood of $M$ in $\mathbb R^n$ and then use the standard derivative for $\mathbb R^n$-valued functions.)

In general, the value $(\xi\cdot\eta)(x)$ does not lie in the tangent space $T_xM$ and to obtain the value of the covariant derivative $\nabla_\xi\eta$ in the point $x$, one has the project this orthogonally into $T_xM$. So far nothing was specific to the way the submanifold is presented, but one can compute this orthogonal projection using the given presentation. This is particularly easy in the case $k=1$. Here you can use the gradient of $f$ as a normal vector to the tangent space, let me denote it by $grd(f)$ to avoid confusion. In terms of this, you then get $$ \nabla_\xi\eta(x)=\xi\cdot\eta(x)-\frac{\langle \xi\cdot\eta(x),grd(f)(x)\rangle}{\langle grd(f)(x),grd(f)(x)\rangle}grd(f)(x). $$ Alternatively, you can take the normalized gradient $\mathfrak{n}(x):=\tfrac1{\sqrt{\langle grd(f)(x),grd(f)(x)\rangle}}grd(f)(x)$ to write things as $\nabla_\xi\eta(x)=\xi\cdot\eta(x)-\langle\xi\cdot\eta(x),\mathfrak{n}(x)\rangle\mathfrak{n}(x)$.

In this language, things generalize to higher codimensions. Take the components $f_1,\dots,f_k$ of $f$ and form their gradients $grd(f_i)$ for $i=1,\dots,k$. Then apply Gram-Schmidt in each point to change this into a family $\mathfrak{n}_i$ of functions $\mathbb R^n\to \mathbb R^k$ whose values in each point are orthonormal. (It is easy to see that the resulting functions are smooth.) Then you can write the covariant derivative as $$ \nabla_\xi\eta(x)=\xi\cdot\eta(x)-\sum_i\langle\xi\cdot\eta(x),\mathfrak{n}_i(x)\rangle\mathfrak{n}_i(x). $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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