riemannian length of an element of the fundamental group of a manifold  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 ? 
 A: 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.  
A: 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
$$\liminf_{k \to \infty}\frac{l_g(\alpha^k)}{k}>0$$
