---
title: "Why a small universe draws circles on the sky"
description: "Two copies of the last-scattering sphere cross in a circle. Derive its angular radius, its multiplicity and why it vanishes when the box exceeds the view."
canonical_url: "https://sandboxphysics.com/stories/cosmic-topology/why-a-small-universe-draws-circles/"
language: "en"
page_type: "story"
alternate_languages:
  zh-CN: "https://sandboxphysics.com/zh/stories/cosmic-topology/why-a-small-universe-draws-circles/"
last_modified: "2026-11-12"
---

[The shape of space](https://sandboxphysics.com/stories/cosmic-topology/)

# Why a small universe draws circles on the sky

Sandbox Physics · 2026-11-12

When the three-torus is smaller than the sphere from which the microwave background reaches us, that sphere crosses its own copies. Each crossing is a circle, and every point of the circle is seen twice, in two directions on opposite sides of the sky. The angular radius of the circle follows from one line of geometry, $\cos\alpha = |\mathbf{g}|/2\chi_{\mathrm{rec}}$, and the circles vanish once every copy is more than twice the sphere radius away. This chapter derives that result, counts the circles and states what is left when there are none.

It builds on the [three-torus chapter](https://sandboxphysics.com/stories/cosmic-topology/flat-but-finite-three-torus/): images of the observer sit at $\mathbf{g} = L\mathbf{n}$.

## The sphere we see

Light from the cosmic microwave background left the hot gas of the early universe when it became transparent, and all of it has travelled the same comoving distance to reach us. Comoving distance is distance measured on a grid that expands with the universe, so the expansion does not change it. Call this distance $\chi_{\mathrm{rec}}$. The points that sent the light form a sphere of radius $\chi_{\mathrm{rec}}$ centred on us, the last-scattering surface.

From now on lengths are measured in units of $\chi_{\mathrm{rec}}$, so the sphere has radius 1. For scale, Planck quotes $\chi_{\mathrm{rec}} \simeq 3.1\,H_0^{-1}$, where $H_0^{-1}$ is the Hubble length, the speed of light divided by the present expansion rate.

## Where a sphere meets its copy

In the covering space, our image at $\mathbf{g}$ has its own last-scattering sphere, also of radius 1. A point $\mathbf{x}$ lies on both spheres when

$$
|\mathbf{x}| = 1 \quad \text{and} \quad |\mathbf{x} - \mathbf{g}| = 1 .
$$

Square both conditions and subtract. The terms $|\mathbf{x}|^2$ cancel and what remains is linear:

$$
2\,\mathbf{x}\cdot\mathbf{g} = |\mathbf{g}|^2 \quad\Longrightarrow\quad \mathbf{x}\cdot\hat{\mathbf{g}} = \frac{|\mathbf{g}|}{2} .
$$

So the shared points lie on a plane perpendicular to $\mathbf{g}$, halfway to the copy. A plane cuts a sphere in a circle. Seen from the centre, the circle subtends an angular radius $\alpha$ around the direction $\hat{\mathbf{g}}$, with

$$
\cos\alpha = \frac{|\mathbf{g}|}{2\chi_{\mathrm{rec}}} .
$$

This is equation 8 of Cornish, Spergel and Starkman. A circle exists only when the right-hand side is less than 1, that is, when the copy is closer than twice the sphere radius. At exactly $|\mathbf{g}| = 2$ the spheres touch at a single point.

## One circle, seen in two places

A point $\mathbf{x}$ on the circle is reached by light along the straight path in direction $\hat{\mathbf{x}}$. The same physical point is also the point $\mathbf{x} - \mathbf{g}$ in our own brick, because the two differ by a lattice step. Light from it reaches us along the direction of $\mathbf{x} - \mathbf{g}$. As $\mathbf{x}$ runs around the circle about $\hat{\mathbf{g}}$, the partner $\mathbf{x} - \mathbf{g}$ runs around a circle of the same radius about $-\hat{\mathbf{g}}$.

So the sky shows two circles, centred on opposite directions, carrying the same sequence of temperatures. Searches call these back-to-back circles. If you use one fixed pair of perpendicular directions to label angles on both circles, matching points sit at the same angle. If instead you order the points of each circle the same way as the observer sees it, looking towards its centre, one sequence runs backwards. Planck calls this ordering anti-phased and states that orientable topologies, those that keep left and right hands distinct, give anti-phased matches.

## Counting circles

Each pair of images $\pm\mathbf{g}$ closer than 2 gives one pair of circles. Cornish, Spergel and Starkman worked out the cubic case $L = \chi_{\mathrm{rec}}$. The three nearest shells have $|\mathbf{g}| = 1$, $\sqrt{2}$ and $\sqrt{3}$. They give

$$
\alpha = 60^\circ,\ 45^\circ,\ 30^\circ \quad\text{with}\quad 3,\ 6,\ 4\ \text{pairs},
$$

plus 3 pairs of copies with $|\mathbf{g}| = 2$ that only touch the sphere, $\alpha = 0^\circ$.

Worked example. For $L = 1.2$ the first shell gives $\cos\alpha = 0.6$, so $\alpha = 53.13^\circ$, three pairs centred on the three face directions. The second shell gives $\cos\alpha = 1.2\sqrt{2}/2 \approx 0.849$, so $\alpha \approx 31.95^\circ$, six pairs. The third shell is at $2.08$, out of reach. The sky holds 9 pairs.

Worked example. For $L = 1.93$ only the first shell reaches: $\cos\alpha = 0.965$, $\alpha \approx 15.2^\circ$. Three small pairs of circles remain, one along each box edge. This is close to the smallest circle Planck could detect, as the paper guide explains.

## When the box is bigger than the view

Every image of the observer lies at least $L$ away. If $L$ exceeds 2, no copy of the last-scattering sphere meets ours, and the sky contains no matched circles at all. Each patch of microwave sky was emitted by a different place, exactly as in infinite flat space.

Then the circle test returns to its unidentifiable limit. A finite box that is larger than the view and an infinite flat space give the same answer to it: no pairs. Whether other statistics can still find such a box is a separate research question. The COMPACT preprint proposes correlations between different spherical harmonic coefficients for shapes that circle searches cannot test.

There is a second limit worth stating once. In the plain cubic three-torus every point is equivalent, so it does not matter where the observer sits. In tori whose faces are glued with a rotation, the distance to the nearest image depends on position. The COMPACT preprint shows that the shortest loop around space can then avoid us. Our cubic box is the simplest teaching case, not a claim about the shape of the real universe.

## Lab tasks

1. Predict the angular radius of the largest circle for $L = 1.6$. Then set the box edge to $1.6$ and read the largest circle radius.
2. Drag the box corner from $1.9$ to $2.1$ and watch the circles on the sky map shrink and disappear. Note the edge length at which the last pair vanishes.
3. Compare the circle profiles in the data row: the box-axis pair should overlap, the unrelated pair should not.
4. Explain why the circles are centred on opposite points of the sky rather than at arbitrary separations.

## Exercises

1. Show that a circle of angular radius $\alpha$ has an actual radius $\sin\alpha$ in units of $\chi_{\mathrm{rec}}$, and evaluate it for $L = 1.2$, first shell.
2. For what range of $L$ does the cubic torus show exactly three pairs of circles?
3. A circle search finds no pairs with radius larger than $25^\circ$. Assuming a cubic three-torus, what lower limit on $L$ in units of $\chi_{\mathrm{rec}}$ does that imply?
4. Why does the derivation not depend on which brick the observer is in?

Answers

1. The circle lies in the plane $\mathbf{x}\cdot\hat{\mathbf{g}} = \cos\alpha$ on a unit sphere, so its radius is $\sqrt{1 - \cos^2\alpha} = \sin\alpha$. For $L = 1.2$, $\cos\alpha = 0.6$ and the radius is $0.8$.
2. Three pairs means the first shell reaches but the second does not: $L$ below 2 and $L\sqrt{2}$ at least 2, so $\sqrt{2} \le L$ with $L$ below 2, about $1.414$ to $2$.
3. The largest circle comes from the first shell: $\cos\alpha = L/2$. No circle above $25^\circ$ means $L/2 \ge \cos 25^\circ \approx 0.906$, so $L \ge 1.81$. Smaller circles are still allowed by that search alone.
4. The lattice is the same seen from any brick, because every image is related to every other by a lattice step. Only the difference $\mathbf{g}$ enters.

For the first task: $\cos\alpha = 0.8$, so $\alpha \approx 36.9^\circ$.

Sources: [Cornish, Spergel and Starkman 1998, equation 8 and the cubic multiplicities](https://arxiv.org/abs/astro-ph/9801212v1), [Planck 2015 results XVIII, section 4.2 on back-to-back and anti-phased circles](https://arxiv.org/abs/1502.01593v2), [COMPACT collaboration preprint on observer position and twisted tori](https://arxiv.org/abs/2211.02603v4).

[Predict the circle radius for a box edge of 1.6, then check it in the Lab.](https://sandboxphysics.com/labs/flat-torus-matched-circles/)

## Sources

1. [Cornish, Spergel and Starkman 1998, equation 8 and the cubic multiplicities](https://arxiv.org/abs/astro-ph/9801212v1)
2. [Planck 2015 results XVIII, section 4.2 on back-to-back and anti-phased circles](https://arxiv.org/abs/1502.01593v2)
3. [COMPACT collaboration preprint on observer position and twisted tori](https://arxiv.org/abs/2211.02603v4)

## Keep reading

The shape of space · Reading path

### From a flat box to matched circles and Planck limits

1. [Article **Can a flat universe be finite?**](https://sandboxphysics.com/stories/cosmic-topology/can-a-flat-universe-be-finite/)
2. [Article **How the three-torus makes flat space finite**](https://sandboxphysics.com/stories/cosmic-topology/flat-but-finite-three-torus/)
3. [You are here **Why a small universe draws circles on the sky**](https://sandboxphysics.com/stories/cosmic-topology/why-a-small-universe-draws-circles/)
4. [Article **Telling a real circle match from a lucky one**](https://sandboxphysics.com/stories/cosmic-topology/telling-a-real-circle-match-from-a-lucky-one/)
5. [Article **Recompute matched circles in a toy three-torus**](https://sandboxphysics.com/stories/cosmic-topology/recompute-matched-circles-in-a-toy-torus/)
6. [Article **How to read Planck's cosmic topology limits**](https://sandboxphysics.com/stories/cosmic-topology/reading-planck-topology-limits/)