Toric structure on the complex projective plane

Feb 2026

CP2 blog

Summary:

A point in the complex projective plane can be described by a triple of complex numbers, each described by a magnitude and a phase. Forgetting the phase defines a map from $\mathbb{CP}^2$ to a triangle, whose fibers are tori. Here is a collection of visualizations of this toric structure.

UNDER CONSTRUCITON

Consider this distorted hexagonal tiling of a triangle:

A distorted hexagonal tiling illustrating the topology of $\mathbb{CP}^2$

Imagine identifying the opposite edges of each hexagon, to form a torus. This picture shows a family of 2 dimensional tori parametrized by a triangle, which traces out a four dimensional manifold. However, as we near the edges of the triangle, the tiles degenerate into lines. The associated tori degenerate into circles. On the corners, the tiles (and thus the tori) degenerate into points. All together, we get a topologically nontrivial four manifold without boundary. This is the complex projective plane $\mathbb{CP}^2$.

This decomposition of $\mathbb{CP}^2$ into families of tori is called a toric structure. We should think of this like a four-dimensional manifold of revolution, where we revolve around two separate axes. In this post, I’d like to explain the toric structure on $\mathbb{CP}^2$ and how I made this image.

Toric structure on $\mathbb{CP}^2$

The toric structure on $\mathbb{CP}^2$ is defined by isolating the phase and magnitude of the coordinates of a point in projective space. In particular, we will define a map from $\mathbb{CP}^2$ to a triangle $\Delta$ using only the magnitude. The fiber over each point will be a torus of possible phases. As a four dimensional manifold, $\mathbb{CP}^2$ can be represented as a parameter family of 2-dimensional tori parametrized by a triangle.

Let’s start with $\CC^2$, which is built out of tori in a simpler manner. A point in $\CC^2$ is a pair of complex numbers $(z_1,z_2)$. Consider the map $\mu: \CC^2 \to \RR_{\geq 0}^2$ defined by taking the magnitude squared of each number

\(\mu (z_0,z_1) = (\vert z_0 \vert^2, \vert z_1 \vert^2)\) This maps $\CC^2$ to the positive quadrent of $\RR^2$. The fibers of $\mu$ are pairs of complex numbers with perscribed magnitude, which are defined by a pair of phases. The fibers are genercially 2 dimensional tori, called the Clifford tori:

However, some of the fibers are lower dimensional tori. If $r_1 = 0$, then $z_1=0$ and the fiber $\mu\inv(0,r_2^2) = (0,r_2 e^{i \theta_2})$ is a circle. Likewise, if $r_2=0$, the fiber is a circle, now in the $\theta_1$ direction instead of the $\theta_2$ direction. Finally, if $r_1=r_2=0$, then the fiber is simply the origin, $\mu\inv(0,0)= 0$. The structure of the fibers is summarized in the figure below.

$\CC^2$ as a torus fibration over the positive quadrent

To illustrate the degeneration of the tori, it is helpful to unwrap them into rectangles, with the opposite sides identified. As the torus moves to the y-axis, it degenerates, the rectangle collapses onto a vertical line. At the x-axis, the rectangle collapses to a horizontal line. At the origin, the rectangle degenerates to a point. This is encoded in the following checkerboard, which

Let $[z_0:z_1:z_2]$ be a point in $\mathbb{CP}^2$ in projective coordinates. Take the norm squared of each coordinate to get the point $(\vert z_0\vert ^2,\vert z_1\vert ^2,\vert z_2\vert ^2)$, which lives in $\RR_{\geq 0}^3$, the positive orthant of $\RR^3$ (map 1 in the figure below). The projective coordinates are only well defined up to scaling, so we need to identify points $(x,y,z) \sim (\lambda x, \lambda y, \lambda z)$ for $(x,y,z) \in \RR_{\geq 0}^3$ and $\lambda \in \RR_{>0}$. It is most convenient to chose a representative, by scaling the $(x,y,z)$ to like on the plane $x+y+z=1$ (map 3 in the figure below). Define the “Toric moment map” $\mu$ by

The image of $\mu$ is the intersection of $\RR^3_{\geq 0}$ with a plane, which is a triangle. Choosing coordinates on the plane, we get a map $\mu:\mathbb{CP}^2 \to \RR^2$ with image any fixed triangle $\Delta$(map 4 in the figure below). Succinctly, $\mu([z_0,z_1,z_2])$ is the point $(\vert z_0\vert ^2,\vert z_1\vert ^2,\vert z_2\vert ^2)$ in barycentric coordinates of $\Delta$.

The maps in the definition of the toric moment map $\mu$.

For a point $\vec{x} = (r_0^2,r_1^2,r_2^2)$ in the interior of $\Delta$, the preimage $\mu\inv(\vec{x})$ consists of the points \([r_0e^{i \theta_0} : r_1 e^{i \theta_1}: r_2 e^{i \theta_2}] \in \mathbb{CP}^2\) We have three parameters $(\theta_0,\theta_1,\theta_2) \in U(1)^3$. But we must divide out by the scaling action on $\CC^3$:

Therefore, the pre-image $\mu^{-1}(p)$ is the quotient of $U(1)^3$ by the diagonal subtorus $U(1) \subset U(1)^3$. The result is a two dimensional torus $T^2$.

In an affine slice of $\mathbb{CP}^2$, the fiber over $p$ is a Clifford torus $\CC^2$, defined by fixing the norm of each coordinate

When $p$ moves to the edge of the triangle, the torus degenerates. We can see this in an affine slice. The coordinates of the affine slice $\CC^2$ provide a natural choice of triangle, with vertices ${(0,0), (0,1), (1,0)}$. The affine slice covers the complement of the hypotenuse. When point $p$ lies on the $y$-axis, any point in the preimage has $r_1=0$. One circle $r_1 e^{i\theta_1}$ degenerates to a point, so the preimage is the remaining circle $r_2 e^{i\theta_2}$. Likewise, when $r_2=0$, the circle the premiere is the circle $r_1 e^{i \theta_1}$. As $r_1=r_2=0$, the preimage is the point $0$.

Choosing different affine charts, we perform a similar analysis for all edges of the triangle. The preimage above any interior point of $\Delta$ is a 2-torus, the preimage of any point on the interior of an edge of $\Delta$ is a circle, and the pre-image of a vertex of $\Delta$ is a point.

Cartoon of the fibers of $\mu : \mathbb{CP}^2 \to \Delta$.

The three sides of the triangle degenerate the central torus into three different quotient tori. This is best understood by thinking of the torus as a square with edges identified. The vertical edge shrinks the $(1,0)$ subtorus, meaning we lose the $x$-direction. The horizontal edge degenerates the $(0,1)$ torus, losing the $y$ direction. The diagonal degnerates the $(1,1)$ subtorus which is generated by a line with slope 1.

Cartoon of the toric fibration on $\mathbb{CP}^2$, with tori rendered as squares with their edges identified.

Here is a turtle shell on this triangle, showing how the unfolded tori degenerate as they move to the boundary.

Tori living on a triangle

Two figures showing how the tori on a triangle degenerate at the edges, to form $\mathbb{CP}^2$. On the left, a the unfolded tori form a checkerboard pattern. On the right, every tile is filled in with a gap between adjacent tiles.

The hexagonal turtle shell at the beginning of the section is mathematically the same picture as this. But, we apply an affine transform to make the triangle equilateral, and we render the Voronoi tiling by hexagons instead of the parallelogram tiling. This illustrates the underlying symmetry exchanging the three edges.

Notice that the tile edges form lines which pass to the verticies of the triangle. This looks like

How I made the picture

Though I draw the tori as tiles, I think of them as lattices. A lattice $\Lambda \subset \RR^2$ defines a torus $\RR^2 / \Lambda$. The toric structure on $\mathbb{CP}^2$ induces a family of tori on the triangle $\Delta$, which I think of as a spatially-varying lattice $\Lambda(x)$ depending on $x\in \Delta$. We would like to draw $\Lambda(x)$ as a single distorted lattice $\tilde{\Lambda}$. That is, $\tilde{\Lambda}$ is a map from the standard lattice $\mathbb{Z}^2$ to the triangle $\Delta$ which, zooming in near $x\in \Delta$ is nearly periodic. Under a magnifying glass near $x$, $\tilde{\Lambda}$ looks like a rotation or scaling of $\Lambda(x)$. Here is the deformed lattice $\tilde{\Lambda}$ corresponding to the $\CC\PP^2$ pictures in right triangles. Explicitly, for the original lattice $\mathbb{Z}^2 = (n,m)$, the deformed lattice is given in barycentric coordinates as $[e^n,e^m,1]$. In this section, I’ll explain where this grid came from.

the deformed grid $[e^n,e^m,1]$ in barycentric coordinates. This was used to create the checkerboard patterns above.

One way to produce such a $\tilde{\Lambda}$ is to find a function $F:\RR^2 \to \Delta$, then define $\tilde{\Lambda}$ as the pushforward of a scaled standard lattice $\tilde{\Lambda} = F(\lambda\mathbb{Z}^2)$. As $\lambda$ becomes smaller, $\tilde{\Lambda}$ becomes finer, and becomes nearly periodic as you zoom in near $x$. More precisely, if $F=(u,v)$ with $u,v:\RR^2 \to \RR$, then near $x$, $F(\lambda \mathbb{Z}^2)$ looks like the lattice spanned by $\langle\nabla u(x), \nabla v(x)\rangle$. To approximate our family of lattices $\Lambda(x)$ with a deformed lattice, we must find functions $u$ and $v$ whose derivatives behave correctly.

Here’s how we want our lattices to degenerate as we move to the edges of the triangle

As a lattice degenerates, some of the rows coalesce into solid lines. The three types of degeneration in $\mathbb{CP}^2$ occur when the rows coalesce in the three directions of the hyperbolic lattice.

For $\tilde{\Lambda}$ to degenerate as $x$ nears the boundary, the inverse function $F\inv: \Delta \to \RR^2$ must have its derivative go to infinity. We make one more simplification, and define the map $F\inv$ as the gradient of a real valued function

We require $g$ to be a convex function, so that $F\inv: \Delta \to \RR^2$ is one-to-one. The deformed lattice locally defines the lattice spanned by the columns of $\text{Hess} g \inv$. The hessian of $g$ must diverge at the boundaries of $\Delta$ for the lattice to degenerate properly. To generate my deformed lattice, I used the following function. Representing a point in $\Delta$ in barycentric coordinates as $(a,b,c)$ with $a+b+c=1$, I define

The graph of this function looks like a pillowcase above the triangular domain.

Graph of $g$ plotted above a right triangle.

At the edges of the triangle, the gradient of $g$ diverges. This ensures that $F = \nabla g : \Delta \to \RR^2$ is 1-1 and onto, so has an inverse.