Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $M=\mathbb R^n$ and define for each $x\in \mathbb R^n$, define $$\langle v,w\rangle_x= \langle v,w\rangle_0$$ where $v,w\in T_x\mathbb R^n\equiv \mathbb R^n\equiv T_0\mathbb R^n$. Hence we see that $M=\mathbb R^n$, admits constant metrics.

Is there any other manifold (other than $\mathbb R^n$) which admits constant metrics...

One guess may be Lie group admit constant metric. Is there any other example.

share|cite|improve this question
LaTeX tip: Use \langle and \rangle for angle brackets. < and > are relation symbols; they are not only different, the spacing is different too. – Harald Hanche-Olsen Feb 11 '13 at 8:02
I'm perhaps a little confused as to what this should mean in general. You got "lucky" here because you can canonically identify the tangent spaces of $\mathbb{R}^n$ with each other as just $\mathbb{R}^n$ making the idea of a "constant metric" sensible. In general though I feel like you need to make identifications first. If you are on a Lie group there is a canonical way to do this via the pushforward of $L_g$ identifying $T_1G$ with $T_gG$, but besides that case the question doesn't make sense to me. – Alex Youcis Feb 11 '13 at 8:02

I'm interpreting your "admits constant metrics" in the following way: Any point $p\in M$ has a neighborhood $U$ with local coordinates $(u_i)_{1\leq i\leq n}$ such that in terms of the $u_i$ the metric tensor $g_{ik}$ is constant. This means that any point $p$ has a neighborhood $U$ which is isometric to a euclidean $n$-ball.

Examples of such manifolds can be produced in the following way: In $\Bbb R^n$, provided with the standard euclidean metric, take $r$ linearly independent vectors $a_1$, $\ldots$, $a_r$ and call two points $x$, $y\in\Bbb R^n$ equivalent if $$y-x=\sum_{k=1}^r j_k \>a_k,\qquad j_k\in{\mathbb Z}\quad (1\leq k\leq r)\ .$$ It's easy to check that $M:=\Bbb R^n{/}\!\sim\ $ has the desired property. When $r=n$ then $M$ is compact and is called an $n$-dimensional flat torus.

share|cite|improve this answer
It may be worth pointing out that with the technical assumption of completeness, all examples which "admit constant metrics" arise in the way you described, possibly with different notion of "equivalent". – Jason DeVito Feb 11 '13 at 17:52

I'm interpreting the question differently: Namely, when can I use precisely what you did to define a metric?

The issue is that you can't always make a well defined identification of the tangent space at one point with the tangent space at another. So, saying $v\in T_p M$ "corresponds" to $v'\in T_q M$ doesn't always make sense.

For example, on $S^2$, given a nonzero tangent vector at the north pole, it should be clear that if this "corresponds" to anything, it should "correspond" to something nonzero. Then, a "correspondance" to every point on $S^2$ is nothing but a vector field on $S^2$ which vanishes nowhere. Unfortunately, the Hairy Ball Theorem states that there is no such vector field.

The class of manifolds you are looking for are the so called Parallelizable Manifolds. These are precisely those manifolds for which the tangent bundle admits a global trivialization. After choosing such a trivialization, choosing any basis for the tangent space at some particular point. Then, via the trivialaztion, this defines a basis at each point of your manifold and this gives you a way of identifying tangent spaces at different points.

I want to stress that this is very noncanonical - you must choose a trivialization and a basis. Different choices will give rise to different Riemannian metrics with different properties.

Parallizable manifolds include $\mathbb{R}^n$ and all Lie groups, as you guessed, but also contain many more manifolds. For example, $S^7$ is not a Lie group, but is parallelizable. Further, the product of any sphere with an odd dimensional sphere is parallelizable.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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