Let $T^n$ be the $n$-dimensional torus and $g$ be a Riemannian metric on $T^n$. Let $\tilde g$ be the induced metric on the universal covering; using suitable coordinates, $\tilde g$ is therefore a $\mathbb{Z}^n$-periodic metric on $\mathbb{R}^n$ (I shall conflate the lattice $\mathbb{Z}^n$ with the fundamental group of $T^n$ in the sequel).

Let $d$ be the distance induced by $\tilde g$ and $t:\mathbb{Z}^n\to (0,+\infty)$ be the function defined by $t(\gamma)=d(0,\gamma(0))$. Recall that the stable norm $\Vert\cdot\Vert_S$ is a norm on $\mathbb{R}^n$ defined by the property that for any $\gamma\in\mathbb{Z}^n$, $$ \Vert\gamma\Vert_S = \lim_k \frac{t(\gamma^k)}{k} $$

Question: is it true that for all $\gamma\in\mathbb{Z}^n$, we have $$t(\gamma)\le \Vert \gamma \Vert_S+2\mathrm{diam}(g)?$$

If not, does some similar control hold? The formula could depend on $n$ but not on the metric $g$.

A pointer to good literature on this kind of metric Riemannian geometry would already be much appreciated.

Important edit: if needed, I am ok with the additional, very strong assumption that the sectional curvature of $g$ is bounded above by some positive $\varepsilon=\varepsilon(n)$. The formula for a lower bound on $\Vert\cdot\Vert_S$ can depend upon $\varepsilon$ (explicitely).

As for motivation, I need this kind of control for a project of showing some constraints on Riemannian metrics on the torus $T^n$ by using quantitative versions of Milnor's argument in his paper on the growth of fundamental groups and volume of Riemannian manifolds.

Happy ending edit: while I was not able to complete the argument proposed in the accepted answer, with Stéphane Sabourau we managed to get a good enough bound, by passing to a finite cover not increasing diameter too much, see https://arxiv.org/abs/1707.07876

  • $\begingroup$ Could you please more explain on your terminology? Is not the stable norm a norm on real homology or cohomology? Is this some thing different from the stable norm which you refer to? what do you mean by $\gamma(0)$? Is it the lifting of $\gamma$ to the universal covering space?In this case is my edit to $\gamma(1)$ correct? I apologize if my questions are elementary. $\endgroup$ Commented Nov 17, 2014 at 17:07
  • $\begingroup$ There are several things called stable norm, but I precisely defined what I meant by this term here. $\gamma(0)$ is the image of the point $0\in\mathbb{R}^n$ under the action of $\gamma$ (I could have written $\gamma$ instead, but it is looks natural to me this way). When you have such questions, please ask them and wait an answer before editing. $\endgroup$ Commented Nov 17, 2014 at 21:46
  • $\begingroup$ I apologize for unnecessary edit. $\endgroup$ Commented Nov 18, 2014 at 11:58

2 Answers 2


If you allow an additive term $C(n)diam(g)$ rather than $2diam(g)$, then yes, the statement is true. In the paper D.Burago, "Periodic metrics", Adv. Soviet Math. 9, (1992), 205-210, he proves that for every periodic metric on $\mathbb R^n$ there is a constant $C$ such that $$ | d(x,y)-\|x-y\|_{st} | \le C $$ for all $x,y\in\mathbb R^2$. In this non-invariant formulation, $C$ depends not only on the metric but also from a particular choice of coordinates used to identify the torus with the standard $T^n$. However, if $x$ and $y$ are from one orbit of $\mathbb Z^n$, then going through the proof shows that one can take $C$ of the form $C(n)diam(g)$.

Unfortunately the paper is published in an obscure place and I can no longer find it on Google Books. But the proof is rather short and I think I can write down all details here if needed.

Added later. I'm adding a proof outline upon a request from comments.

Fix $p\in \mathbb R^n$ and $v\in\mathbb Z^n$. Define $f:\mathbb Z_+\to\mathbb R$ by $f(m)=d(p,p+mv)$. Note that $ \|v\|_{st} = \lim_{k\to\infty} f(k)/k$. The function $f$ satisfies the following properties:

(1) $f(m_1+m_2)\le f(m_1)+f(m_2)$

(2) $f(2m)\ge 2f(m)-C$,

where $C$ depends only on the metric (and in fact equals $C(n)diam(g)$). These properties immediately imply that $k\|v\|_{st}\le f(k)\le k\|v\|_{st}+C$ with the same $C$. For $k=1$, this gives us the desired estimate on $d(p,p+v)=f(1)$.

The property (1) is just the triangle inequality. The property (2) is based on the following topological lemma:

Lemma. Let $s:[0,1]\to\mathbb R^n$ be a continuous path. Then there is a collection of disjoint intervals $[a_i,b_i]\subset[0,1]$, $i=1,2,\dots,k\le (n+1)/2$, such that $$ \sum (s(b_i)-s(a_i)) = \frac{s(1)-s(0)}2 . $$

Let me derive property (2) from the lemma. Apply the lemma to a shortest path $s$ connecting $p$ to $p+2mv$. This gives us a collection of at most $n/2$ vectors $v_i=s(b_i)-s(a_i)$ in $\mathbb R^n$ whose sum equals $mv$. Let $p_0=p,p_1,\dots,p_k$ be points such that $p_i-p_{i-1}=v_i$. Then $p_k=p+mv$, and each pair of points $p_{i-1},p_i$ is a (possibly non-integer) parallel translation of $s(a_i),s(b_i)$. Due to periodicity, this implies that $d(p_{i-1},p_i)\le d(s(a_i),s(b_i))+C_1$ for some constant $C_1$ depending on $g$. I will address this dependence later. Therefore $f(m)=d(p,p_k)$ is bounded by a constant $C_1n$ plus the length of the parts of $s$ covered by the intervals $[a_i,b_i]$.

Consider the complement intervals $[0,a_1]$, $[b_i,a_{i+1}]$, $[b_k,1]$. The corresponding vectors $s(a_1)-s(0)$ etc, also add up to $mv$. Hence the above argument applies to these intervals as well and implies that $f(m)$ is no greater than $C_1n$ plus the remaining part of the length of $s$. One of these two parts is no greater that half of the total length of $s$, hence the result.

The problem is how to control the constant $C_1$ by the diameter only. To do so, one moves the division points $s(a_i)$, $s(b_i)$ to nearby lattice points (losing at most the diameter on each move), then the required parallel translations preserve the distance and the dependence on the metric goes away. But the new collection of vectors no longer sums up to $mv$. So we have to choose those "nearby" lattice points wisely, so that the accumulated error is bounded by the diameter times $C(n)$. This (hopefully) can be done as follows: approximate $s$ by a sequence of lattice points (with distances between neighboring ones at most twice the diameter), then apply the lemma to the Euclidean broken line connecting these points, then move each division point to the next vertex of this broken line. The error in the sum of the resulting vectors has bounded stable norm and this should imply a bound on the metric distance. (I have not worked through this detail yet.)

Proof of lemma. It was discussed in this answer but the Google Book link there died since that (probably some copyright maniac killed it, now it looks as if the volume was never digitized). The proof it the following, Consider the standard unit sphere $S^n$ of points $x\in\mathbb R^{n+1}$ with $\sum x_i^2=1$. To each point $x\in S^n$, associate a partition $0=t_0\le t_1\le\dots t_{n+1}=1$ of $[0,1]$ such that $t_i-t_{i-1}=x_i^2$, and define $f(x)\in\mathbb R^n$ by $$ f(x) = \sum sign(x_i) (s(t_i)-s(t_{i+1}) . $$ The resulting function $f:\mathbb S^n\to\mathbb R^n$ is continuous and odd (i.e. $f(-x)=f(x)$). Therefore $f(x)=0$ for some $x$. Now let $[a_j,b_j]$ be the segments of the partition $\{t_i\}$ associated to this $x$, whose corresponding coordinates $x_i$ are positive. If there are too many of them, take the negative ones instead.

  • $\begingroup$ I confess that I would be very interested to see a proof. $\endgroup$
    – JHM
    Commented Mar 7, 2013 at 18:14
  • $\begingroup$ This is great! I would be very interested in the details. $\endgroup$ Commented Mar 7, 2013 at 19:47
  • $\begingroup$ I wrote a proof but there is a little detail missing: to show that displacement is bounded by, say $1000ndiam$ for vectors with stable norm less than $10ndiam$. It should be doable but I don't see a short argument at the moment. $\endgroup$ Commented Mar 7, 2013 at 23:56
  • $\begingroup$ @Sergei Ivanov: I think there is no need for that: instead of changing $s([a_i,b_i])$ so that its endpoint are on the lattice grid, go from $p_i$ to the closest point in the $\mathbb{Z}^n$ orbit of $s(a_i)$, then follow an integer translate of $s|[a_i,b_i]$, and you end up at a point in the $\mathbb{Z}^n$-orbit of $s(b_i)$ close to $p_{i+1}$. This shows that $d(p_i,p_{i+1})\leqslant d(s(a_i),s(b_i)) + 2 \textrm{diam}(g)$. $\endgroup$ Commented Mar 13, 2013 at 13:52
  • $\begingroup$ @Benoît: This does not work. What is near in terms of the RIemannian distance can be far in terms of coordinates. After a couple of steps you may end up far away from the intended destination. $\endgroup$ Commented Mar 13, 2013 at 20:32

This is true for n=2, but for n>2 one is unlikely to have such inequalities regardless of what the additive constant is. This is due to "tunneling" phenomena. Take a high multiple K of a primitive class, and represent it by an imbedded loop. Then take a thin tubular neighborhood of the loop. Here the metric can be modified in such a way that the 1-systole stays uniformly bounded from below, but the stable 1-systole tends to zero. Such examples are in the literature. This is not quite enough to disprove the inequality as stated above, but a finer analysis of the metrics should show that sufficiently high multiples of the primitive class also stay "long" so long as the multiple is considerably less than K.

  • $\begingroup$ I know your example, but the additive constant here helps a lot: with $2\mathrm{diam}$ you have enough to enter the tunnel and go out of it. It follows that at least if you make the tunnel "linear" (everywhere roughly in the direction of your chosen primitive class, not wandering up and down), in such an example the upper bound I'd like to have does hold. $\endgroup$ Commented Mar 6, 2013 at 18:09
  • $\begingroup$ Moreover it seems to me that making a tunnel with a complicated shape forces you to make the metric very short inside it, possibly helping. $\endgroup$ Commented Mar 6, 2013 at 18:11
  • $\begingroup$ I wish i knew what is the `tunnelling' phenomenon you are both referring to. $\endgroup$
    – JHM
    Commented Mar 7, 2013 at 18:15
  • $\begingroup$ @J. Martel: start from any metric, and along a small tubular neighborhood of a loop, change the metric to make it very small so that the loop becomes short (you can also make the metric very large in a thin layer around the loop so that it "disconnects" it from the rest of the manifold). $\endgroup$ Commented Mar 7, 2013 at 19:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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