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Define $g$ to be the metric on $R^3$ with components $$g_{11}=g_{22}=g_{33}=(x^1)^2+1, g_{ij}=0 \mbox{ for } i\neq j.$$ What are the geometric implications of imposing this metric on $R^3?$ Is this the same with the Euclidean space?

More generally, let $f:R^n \to (0, \infty)$. Then $g$ with components $$g_{ij}=\delta_{ij}f$$ is a metric on $R^n$, right? How is this different from the Euclidean space?

Thanks much, in advance.

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Why don't you try to compute its curvature? You'll find it's not flat, unless f is constant, so it's not the "same" metric. Try looking up "conformal" changes of metrics. –  Spiro Karigiannis Sep 17 '12 at 12:12
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Things get longer and longer as $|x^1|$ gets bigger. The general form you produced is a metric conformal to the Euclidean metric. This in general preserves angles but stretches or shrinks lengths. –  timur Sep 17 '12 at 12:13
    
As a first guess I would've thought it had something to do with the induced metric on the graph $\Gamma_f\subset\mathbb{R}^{n+1}$, but a short computation shows it's not so. –  Qfwfq Sep 17 '12 at 12:24
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@Spiro: Actually, you have to be careful. There are several nonconstant functions $f$ on $\mathbb{R}^n$ with the property that $f\ g_0$ is flat. The point is that, if $T:\mathbb{R}^n\to \mathbb{R}^n$ is a conformal transformation, then $T^*(g_0) = f\ g_0$ for some $f$. In general, since the (local) conformal transformations are not all affine in $\mathbb{R}^n$, there will be $T$ that yield nonconstant $f$. Roughly speaking, the set of such $f$'s is equivalent to the group of conformal transformations modulo the group of Euclidean isometries. –  Robert Bryant Sep 17 '12 at 14:50
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Although this might not be a homework problem, it looks to me like wonderful exercise for a student in differential geometry to study this metric using the concepts learned in class or from a textbook. I encourage the original poster to explore this further on her or his own instead of relying experts to explain it all. –  Deane Yang Sep 17 '12 at 18:00
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Note that the $(x_1,x_2)$-plane is totally geodesic and the lines parallel to $x_1$-axis are geodesics inside. These lines spread apart when you go to infinity. Therefore your $(\mathbb R^n,g)$ is not isometric to the Euclidean space.

About the second question. As it was noticed in the comments, you need first calculate the curvature tensor. (Since your metric is conformal, it is determined by Ricci tensor.)

If the curvature is not zero then it is not flat.

If the curvature is zero, your space admits an open isometric immersion in the Euclidean space. So if you know little more, say $f(x)>\tfrac1{|x|^2}$ then it has to be isometric to the Euclidean space.

A bad example is $f=e^{x_1}$ for $\mathbb R^2$. (At the moment I do not see such examples for $\mathbb R^3$.)

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