# Embedding manifold equipped with FLRW metric

The Friedmann–Lemaître–Robertson–Walker (FLRW) metric is as follows, in natural units: $$\mathrm{d}s^2=-\mathrm{d}t^2+a(t)^2\left(\frac{\mathrm{d}r^2}{1-\kappa r^2}+r^2\mathrm{d}\theta^2+r^2\sin^2\theta\,\mathrm{d}\phi^2\right)$$ For the sake of my own visualization, I am attempting to embed a slice of this manifold in $$\mathbb{R}^3$$, where $$\theta=\frac{\pi}{2}$$ and $$t=$$ constant. This transforms the metric into $$\mathrm{d}s^2=a(t_0)^2\left(\frac{\mathrm{d}r^2}{1-\kappa r^2}+r^2\mathrm{d}\phi^2\right)$$ Using the pullback, we know that ($$\hat{g}_{\mu\nu}$$ is the Euclidean metric) $$g_{\mu\nu}=(\phi^*\hat{g})_{\mu\nu}=\frac{\partial y^\alpha}{\partial x^\mu}\frac{\partial y^\beta}{\partial x^\nu}\hat{g}_{\alpha\beta}$$ From this, the following system follows, solving for $$y_1$$, $$y_2$$, and $$y_3$$: \begin{align} \frac{a(t_0)^2}{1-\kappa r^2}&=(\partial_r y^1)^2+(\partial_r y^2)^2+(\partial_r y^3)^2 \\ r^2a(t_0)^2&=(\partial_\phi y^1)^2+(\partial_\phi y^2)^2+(\partial_\phi y^3)^2 \end{align} These equations elude me. Is there any way to solve this analytically, or will I have to solve numerically?

• Your metric is missing some squares: $$\frac{dr^2}{1 - k r^2}$$. – Cham Dec 2 '19 at 19:34
• Ah, bless you. It has been edited. – Tesseract Dec 2 '19 at 19:43

You're considering some 3D space, of euclidian metric in cylindrical coordinates: $$\tag{1} ds^2 = dx^2 + dy^2 + dz^2 = dr^2 + r^2 \, d\varphi^2 + dz^2.$$ Now introduce a surface in that space, of height $$z = f(r)$$ (assuming isotropy in the $$x \, y$$ plane). Then $$dz = f^{\prime} \, dr$$ and (1) becomes $$\tag{2} ds^2 = (1 + f^{\prime \, 2}) \, dr^2 + r^2 \, d\varphi^2.$$ You want this metric to coincide with the FLRW metric, in case of $$\theta = \frac{\pi}{2}$$: $$\tag{3} ds^2 = \frac{1}{1 - k r^2} \, dr^2 + r^2 \, d\varphi^2.$$ Thus, you need to impose the following differential equation (assuming $$k = 1$$. It is trivial for $$k = 0$$ and there's no solution for $$k = -1$$ with metric (1)): $$\tag{4} \frac{df}{dr} = \pm \, \frac{r}{\sqrt{1 - r^2}}.$$ This imposes $$f(r) = C \mp \sqrt{1 - r^2}$$. You may choose the negative sign and $$C = 1$$, so $$z(r) = 1 - \sqrt{1 - r^2}$$ for $$0 \le r < 1$$ (so $$z(0) = 0$$ and $$z(1) = 1$$).
Notice that this surface is half a sphere of radius 1 and centered on $$z_c = 1$$, in 3D Euclidian space: $$\tag{5} x^2 + y^2 + (z - 1)^2 = r^2 + (z - 1)^2 = 1.$$ This embedded sphere corresponds to the geometry associated to the space curvature parameter $$k = 1$$.
For $$k = -1$$, you need a pseudo-euclidian metric instead of (1) above.