6
$\begingroup$

Let $M, N$ be Riemannian manifolds and $f: M \to N$ be a smooth map (I'm actually only considering diffeomorphisms (flows) $\Phi^t: M \to M$, but just for the sake of generality).

The first derivative of $f$ can be understood as its tangent map $T f: T M \to T N$. Higher derivatives can abstractly be viewed as maps between higher order tangent bundles.

I want to make estimates on the (operator norm) size of these higher derivatives. In the higher order tangent spaces (see also the recent question In how many ways can an iterated tangent bundle (T^k)M be viewed as a fibre bundle over (T^(k-1))M?) I'd have to use induced metrics, which I don't readily know how to work with, and besides, I think these would include the base, lower order derivatives as well.

I would prefer to keep things defined on the tangent/tensor bundle, in a similar way as taking covariant derivatives for vector fields, but I don't know how to do this for for maps $f: M \to N$.

So my question roughly is: are there natural/practical representations of norms of higher order derivatives of maps between manifolds?

One thing I did come up with is representing $f$ in normal coordinates, as these are the most canonical charts and then use the norms in the tangent spaces at the argument and image points $x$ and $y = f(x)$.

(The basis for this question is that I want to obtain a Gronwall-like growth estimate for the higher derivatives of a flow $\Phi^t$ in terms of the exponential growth of its tangent flow $D \Phi^t$.)

Improve this question
$\endgroup$

1 Answer 1

Reset to default
7
$\begingroup$

You could indeed pull back each tensor bundle of $N$ via the map $f$ to $M$ and use the naturally induced metric and connection to define norms of higher covariant derivatives of $f$. You can do this for every order greater than or equal to $1$. Depending on your needs, this might work fine.

If you need something that will work under weaker a priori assumptions, then you might need to embed $N$ isometrically into a higher dimensional Euclidean space (via Nash's theorem) and treat $f$ as a vector-valued function. This allows you to work with maps $f$ that are not necessarily continuous and define its weak derivatives. This is often used in the study of harmonic maps.

Improve this answer
$\endgroup$
6
  • $\begingroup$ Thanks for these suggestions! I'm trying to understand whether they are useful for my problem. For the pullback construction, I understand that one should interpret $D f$ as a vector bundle map $T M \to f^\star(T N)$. But then the induced connection on $f^\star(T N)$ involves the pushforward along $f$, right? This would seem to introduce additional terms that I think I don't want. Unfortunately, I think Nash doesn't help me. I'm working on noncompact spaces and require uniform estimates (by bounded geometry assumptions). A Nash embedding doesn't control curvature in the normal directions. $\endgroup$
    – Jaap Eldering
    Commented Jan 13, 2011 at 10:35
  • $\begingroup$ The norm of $Df$ is well-defined, since it does not use the induced connection at all. The norm of $D^2f$ does, but that shouldn't be a problem. In general, the norm of $D^kf$ does depend on $D^jf$ for $j < k$, but that shouldn't be a serious issue. $\endgroup$
    – Deane Yang
    Commented Jan 13, 2011 at 17:39
  • $\begingroup$ I don't quite understand why Nash would not work for you, but I don't know the details. If you are not varying the Riemannian metric on $N$, then the isometric embedding is fixed. So all it does is add lots of constants to your estimates. The curvature in the normal directions should not need to be controlled, since they never change in what you do. $\endgroup$
    – Deane Yang
    Commented Jan 13, 2011 at 17:40
  • $\begingroup$ Indeed I agree about the dependence of $D^k f$ on $j < k$ terms, but the weighted degree of all $j_i$ terms should be at most $k$, as in en.wikipedia.org/wiki/Faà_di_Bruno's_formula. I think that here I get one term $D f$ too many due to the pushforward along $f$ in the induced connection. This would prevent me from getting the right (inductively obtained) exponential growth behaviour $D^k \Phi^t \le C_k e^{k \rho t}$ when $\D \Phi^t \le C e^{\rho t}$. $\endgroup$
    – Jaap Eldering
    Commented Jan 14, 2011 at 11:19
  • $\begingroup$ If I'd have $M = N = \mathbb{R}$ flat manifolds and $f(x) = x$, but a Nash embedding of $N$ into $\mathbb{R}^n$ that is strongly curved, then the representation of $D^k f$ would pick up these curvature terms, which can globally be unbounded for embeddings of noncompact manifolds. I don't immediately see how to correct for this. $\endgroup$
    – Jaap Eldering
    Commented Jan 14, 2011 at 11:29

You must log in to answer this question.

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