Suppose $(M,g)$ and $(N,g')$ are smooth Riemannian manifolds and $J^r(M,N)$ is the smooth manifold of $r$-jets $j^r_xf$ of smooth maps $f:M\to N$.

Is there an 'induced' Riemannian metric $g''$ on $J^r(M,N)$?

Of course the term 'induced' is conceptual vague here...

That's because, I don't want to restrict the question in a particular direction. However $J^r(M,N)$ gets other things induced from $M$ and $N$. For example atlases and chart transitions of $M$ and $N$ induces an atlas and chart transitions of $J^r(M,N)$.

In general and for any $m \in \mathbb{N}$, the jet $J^r(\cdot,\cdot)$ is a functor $J^r(\cdot,\cdot): \mathbf{M}_m \times \mathbf{M}\to \mathbf{M}$ from the product of the category of smooth $m$-dimensional manifolds and local diffeomorphisms with the category of smooth manifolds and smooth maps into the latter.

So 'maybe' with induced matric I mean, if $J^r(\cdot,\cdot)$ is still a functor from the product of the category of smooth $m$-dimensional riemannian manifolds with local (isometric) diffeomorphisms with the category of smooth riemannian manifolds and (ismometric) smooth maps into the latter.

I used parantheses around the term 'isometric' here because, the more general situation is of course the preferred one.


1 Answer 1


There are, of course, several different functorially induced metrics on $J^r(M,N)$ when $M$ and $N$ are endowed with given Riemannian metrics.

For example, $J^0(M,N)=M\times N$ and one can just take the product metric.

Meanwhile $J^1(M,N)$ can be regarded as a vector bundle over $M\times N$ with fiber $T^\ast_xM\otimes T_yN$ over $(x,y)\in M\times N$. The metrics on $M$ and $N$ induce an inner product on the bundle $T^\ast M\otimes TN$ (since you have innter products on each factor bundle separately), and the Levi-Civita connections on the two factors can be used to define a connection on this bundle. This information is enough to specify a metric on $J^1(M,N)$.

Added at the request of the OP: Given two bundles $E$ and $F$ over a base that are endowed with inner products and connections, the bundle $E\otimes F$ inherits both an inner product and a connection in the usual way: The inner product is the one for which the square length of $e\otimes f$ is just $\langle e,e\rangle_E\ \langle f,f\rangle_F$, and the connection is the tensor product connection. (You should look this up if you don't know it; I won't spell it out here.) Once you have a bundle $E$ with inner product $\langle,\rangle$ and connection $\nabla$ over a Riemannian manifold $M$, there is a unique metric on the total space $E$ that makes the horizontal space provided by the connection perpendicular to the fibers of $E\to M$, makes the projection to $M$ a Riemannian submersion, and restricts to each fiber to be the given inner product.

One can continue on this this way or just use an induction based on a certain natural inclusion of $J^{r+1}(M,N)$ into $J^1\bigl(M,J^r(M,N)\bigr)$ to finish the construction.

This construction is one choice among several possible ones, which I won't spell out here.

  • $\begingroup$ Thanks. Can you give references for a more in-deep look at the constructions you mentioned? What do you mean by "[..] induce a metric on this bundle [...]". Which bundle? $J^1(M,N)$? Then how can this be done? --You said there are many ways, but given at least one example would be required to qualify as an answer. $\endgroup$
    – Nevermind
    Mar 15, 2013 at 18:25
  • $\begingroup$ You use 1.) the identification $J^1(M,N) \simeq TN \otimes T^*M$ 2.) If $g$ is a metri on $M$,then there are induced metrics on $TM$ and $T^*M$ 3.) There is a metric on $TN \otimes T^*M? Right? -- If yes, the missing link for me is a reference to the construction in 3. $\endgroup$
    – Nevermind
    Mar 15, 2013 at 18:32
  • $\begingroup$ I would also be very interested to see explicitly how this construction can be extended beyond $J^1$, or a reference, since this question is essentially a more abstract version of a question I posed earlier. $\endgroup$ Mar 15, 2013 at 22:04
  • $\begingroup$ Robert, this is not an answer, because you don't give a proof or a reference for a proof. Bad style, not to reply $\endgroup$
    – Nevermind
    Mar 17, 2013 at 18:47
  • 2
    $\begingroup$ @Nevermind: I think that I gave a fairly complete description of a metric $g''$ on $J^1(M,N)$ that is canonically associated with a pair $(g,g')$ of metrics on $M$ and $N$ respectively. I just didn't write out the explicit formulae. I could do this, of course, if you really need it, but there's no guarantee, as far as I can see, that you would be satisfied with that since you might not like my notation or my method. I don't have a reference to give you because I don't know where it is in the literature, but you shouldn't need to look it up anywhere if you can follow my description above. $\endgroup$ Mar 17, 2013 at 19:35

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