It is a stupid question i guess but like they say if you ask you are stupid for 5 minutes and if you don't ask you are stupid forever . here is the question given a closed manifold $(M,g)$ and $\alpha$ in $\pi_1(M,p)$ we can define the length of $\alpha$ as the minimum riemannian length of a representative now obviousle length $\alpha^2$ is less or equal then $2\mathrm{length}(\alpha)$ but it seems it is always equal $2\mathrm{length}(\alpha)$ i want to know why ?
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4$\begingroup$ The statement is false. Think projective plane. $\endgroup$– Autumn KentCommented Jun 24, 2011 at 16:01
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$\begingroup$ under what conditions the statement is true ? and can the length of \alpha^2 be less the length of \alpha ? $\endgroup$– unkownCommented Jun 24, 2011 at 16:34
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3$\begingroup$ When $\pi_1$ has an element $\alpha$ of order 2, $\alpha^2$ can have length 0 (that's what R.Kent was suggesting). There are intermediate possibilities too, such as $\alpha$ of order 3 implies $\alpha$ and $\alpha^2$ have the same length; this doesn't happen in two dimensions but does in three (e.g. lens space). $\endgroup$– Noam D. ElkiesCommented Jun 24, 2011 at 17:02
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8$\begingroup$ Generically, a geodesic loop realizing the minimum length does not close up smoothly at $p$, and then the double loop is not a geodesic and hence not a minimizer. So the statement is almost never true unless you minimize in a free homotopy class. $\endgroup$– Sergei IvanovCommented Jun 24, 2011 at 17:40
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$\begingroup$ (sorry, I see I have unwillingly down-voted this question) $\endgroup$– Pietro MajerCommented Jun 27, 2011 at 16:16
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2 Answers
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There are geometric hypotheses that ensure the property you want. For example, suppose that $(M,g)$ has negative curvature. Then every $\alpha \in \pi_1(M)$ is freely homotopic to a unique geodesic representative $\alpha^*$. Usually people write $\ell_g(\alpha)$ for the length of $\alpha^*$. Finally, uniqueness of geodesic representatives implies that $\ell_g(\alpha^k) = k \cdot \ell_g(\alpha)$. This is just the beginning of an important area in Riemannian geometry. (When are geodesic representatives unique? What is the interaction between the metric and the variational properties of geodesics? And in a different direction: How does the fundamental group act on the universal cover? What does the metric tell us about the algebraic topology of $M$ and the universal cover? Etc.)
It is amusing to contemplate all the ways in which the real projective plane, or more generally any closed manifold with finite fundamental group, differs from a negatively curved manifold.
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$\begingroup$ is there is some algebraic properties of the fundamental group that can garuentee the property without talking about curvature say for example if the fundamental group is torsion free $\endgroup$– unkownCommented Jun 24, 2011 at 17:22
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$\begingroup$ From another hand thanks for your answer but it doesn't answer the question because the length i am talking about is for \alpha=[c] in pi_1(M,p) length(\alpha) is the infimum length of loops that are homotpic to c and not freely homotopic $\endgroup$– unkownCommented Jun 24, 2011 at 17:26
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$\begingroup$ I personally seriously doubt that there exists such an algebraic condition, as there are many manifolds with a given fundamental group and you can perturb the metric on a Riemannian manifold in several ways. I would be very surprised if the property you require was stable under those perturbations (except in the case of
$S^1$), see the comment by Sergei Ivanov. Btw, the only examples of manifolds with that property I can think of are tori (with flat metric). And simply connected manifolds. $\endgroup$ Commented Jun 24, 2011 at 18:47 -
$\begingroup$ @unknown - I assumed that free homotopy classes were meant and I explicitly wrote free homotopy in my answer. In the setting of based homotopy classes (as others have pointed out) length is basically never multiplicative with respect to powers... Somethings can still be shown even in the based case. For example, if you look at $\ell_g(\alpha^k)/k$ using based length (where $g$ is negatively curved) then this approaches the free length. $\endgroup$– Sam NeadCommented Jun 24, 2011 at 22:24
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$\begingroup$ @unknown - Regarding your first comment. The Klein bottle has torsion free fundamental group. At the same time there is a based loop whose (based) square has length less than twice the original. @Alessandro - and products of spheres and tori, I guess. Are there other examples? $\endgroup$– Sam NeadCommented Jun 24, 2011 at 22:34
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The following construction shows why algebraic properties will never give you what you're looking for. I hope the idea is clear without much detail. Take any $\alpha \in \pi_1(M, p)$ and let $\gamma$ be a simple curve in the homotopy class of $\alpha^2$. (If dimension of $M$ is at least 3 this is always possible.)
Now, we're going to construct a metric, and we're going to construct it so that $l_g(\alpha^2) < 2 l_g(\alpha)$. First, measure the tangent vectors to $\gamma$ so that the length of $\gamma$ is very small. Then, measure vectors in a tubular neighborhood of $\gamma$ so that any path which "escapes" $\gamma$ incurs a huge length. Then extend the metric to the whole manifold, but make sure that you keep the length of any curve in $\alpha$ of medium size.
In contrast, in Lemma 5 of A lower bound for the diameter of solutions to the Ricci flow with nonzero $H_1(M^n;\mathbb{R})$, the authors (Tom Ilmanen and Dan Knopf) show that if the image of $\alpha$ in $H_1(M, \mathbb{R})$ is nonzero, then $$\liminf_{k \to \infty}\frac{l_g(\alpha^k)}{k}>0$$
