10
$\begingroup$

Consider the unit ball in $\mathbb R^2$ endowed with the Poincaré metric:

$$ds^2=\frac{4(dx^2+dy^2)}{(1-x^2-y^2)^2}.$$

I want to compute the Gaussian curvature and find that it is $-1$. Given that the curvature is the same at every point (the space is homogeneous and isotropic), it suffices to compute the curvature at $0$.

By definition, $$K=\frac{Rm(\partial_1, \partial_2, \partial_2, \partial_1)}{|\partial_1|^2 |\partial_2|^2}=\frac{Rm(\partial_1, \partial_2, \partial_2, \partial_1)}{16}$$ at the point $(0,0)$, as $\langle \partial_i, \partial_i\rangle = 4$ at $(0,0)$.

Computing $R(\partial_1, \partial_2)\partial_2$ and taking the $\partial_1$ component, I get

$$\partial_1 \Gamma_{22}^1 - \partial_2 \Gamma_{12}^1.$$ There are also a bunch of products of Christoffel symbols, but since the Christoffel symbols are sums of first derivatives of $f$, and all first derivatives of $f$ are $0$ at $(0,0)$, we can ignore these.

I compute $$\Gamma^1_{22}=\frac{-f_x}{2f}.$$ $$\Gamma^1_{12}=\frac{f_y}{2f}.$$ $$\partial_1 \Gamma^1_{22}=-\frac{2f f_{xx} - 2f_xf_x}{4f^2}.$$ $$\partial_1 \Gamma^1_{12}=\frac{2f f_{yy} - 2f_yf_y}{4f^2}.$$

Ignoring the first derivatives again and using symmetry, this leads us to consider $$-\frac{f_{xx}}{f}.$$ This is -4 at 0 (-16/4), and dividing by the factor of 16 noted above gives $-1/4$. We are off by a factor of $4$.

Where have I gone wrong?

edit: I note it really is a factor of $4$, not another computational error, because doing the calculations with

$$ds^2=\frac{4R^4(dx^2+dy^2)}{(R^2-x^2-y^2)^2}$$

yields $-1/(4R^2)$, which is off by $1/4$ from the desired $-1/R^2$.

Edit: Let me see if I can adapt @mauddib's answer to my own way of thinking.

The error is that the above actually computes $R_{122}^1$/16. (I can't get the LaTeX right, but the raised index should be in the last place.) We must lower the index to get the desired tensor. $$R_{1221}=g_{m1}R^1_{221} =g_{11} \left(-\frac{f_{xx}}{f}\right) = -f_{xx}.$$

Now this is $-16$, so dividing by $16$ gets the right answer. Wonderful.

$\endgroup$
2
  • 2
    $\begingroup$ Polar coordinates would likely give easier calculations. $\endgroup$
    – A.Γ.
    Aug 6, 2015 at 0:29
  • $\begingroup$ Polar coordinates in the disk model, cartesian coordinates in the upper half plane model. $\endgroup$
    – Neal
    Aug 6, 2015 at 2:32

1 Answer 1

3
$\begingroup$

Attempt #2: I think the issue here is that we need to transform from the $(1, 3)$ curvature tensor to the $(0, 4)$ one. The formula for the components of the $(1, 3)$ version are given by: $$R^l_{ijk} = \partial_i\Gamma^l_{jk} - \partial_j\Gamma^l_{ik} + \sum_r(\Gamma^r_{jk}\Gamma^l_{ir} - \Gamma^r_{ik}\Gamma^l_{jr})$$ This means that the $(1, 3)$ curvature can be written as $$R_{ijk} = \sum_l \Gamma^l_{ijk} \partial_i$$ We must lower an index to produce the $(0, 4)$ tensor: $$R_{ijkm} = g(\sum_l \Gamma^l_{ijk} \partial_i, \partial_m) = \sum_l \Gamma^l_{ijk} g(\partial_i, \partial_m)$$ In your example, the vectors $\partial_i$ are orthogonal and in particular you calculated: $$R_{1221} = \Gamma^1_{122} g(\partial_1, \partial_1) = -4 \cdot 4 = -16$$ Now from your formula: $$K=\frac{Rm(\partial_1, \partial_2, \partial_2, \partial_1)}{|\partial_1|^2 |\partial_2|^2}=\frac{Rm(\partial_1, \partial_2, \partial_2, \partial_1)}{16}$$ we see that curvature is $-1$.

$\endgroup$
3
  • $\begingroup$ That would certainly account for the factor of 4! $\endgroup$
    – Potato
    Aug 6, 2015 at 2:36
  • $\begingroup$ @Potato Indeed :) $\endgroup$
    – muaddib
    Aug 6, 2015 at 2:45
  • $\begingroup$ Many thanks! $\bf{}$ $\endgroup$
    – Potato
    Aug 6, 2015 at 2:53

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .