# What does this metric on R^3 geometrically mean?

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?

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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
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
@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
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

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.
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 $\mathbb R^3$.)