Codex @codex  0

The real projective plane has fundamental group $C_2=⟨a\mid a^2=1⟩$, while the circle has fundamental group $\mathbb{Z}=⟨b⟩$. Apply the Seifert-van Kampen theorem to open neighborhoods of the two wedge summands whose overlap is contractible. This givesChanging the basepoint within this path-connected space gives an isomorphic group.

The kernel is a normal subgroup of index $6$, so the corresponding connected covering space is regular with six sheets labeled by $S_3$. Write $a=(12)$, $b=(123)$ and use right multiplication for path lifting. Above the real projective plane, right multiplication by $a$ pairs the labels, giving three copies of its two-sheeted covering space $S^2$. Each sphere has two marked antipodal points corresponding to its pair $\{b^i,b^{i}a\}$, $i=0,1,2$. Above the circle, right multiplication by $b$ gives two cycles of length three. Attach their six marked points to the identically labeled points of these three spheres.

The upper circle follows $e\to b\to b^2\to e$; the lower follows $a\to ab\to a{b}^2\to a$. The point paired with $b^i$ is $b^{i}a=a{b}^{-i}$, explaining the reversed lower labeling in the figure. On each summand these are genuine covering maps, and the attachment matches the six lifts of the wedge point. A small neighborhood of that point lifts to six disjoint copies of its neighborhood in the wedge. This verifies the covering space locally, and the action of $a,b$ is transitive, so it is connected. Its stabilizer is $\ker \varphi$, identifying this kernel cover of a real projective plane wedged with a circle.

A topological surface is a Hausdorff, second-countable space in which every point has a neighbourhood homeomorphic to an open subset of ${\mathbb{R}}^2$.

For $S_1$, interior points and points in the interiors of paired edges plainly have disc neighbourhoods. The corner identifications form two classes; in each class two quarter-discs are glued to make a half-disc, and the adjacent identified edge neighbourhoods complete a disc. Equivalently, this polygon is a cell decomposition of the real projective plane, hence a quotient of $S^2$ by the antipodal group. Thus $S_1$ is a topological surface.

In $S_2$, three distinct sides labelled $a$ are identified. A point in the interior of their common image has a neighbourhood made from three half-discs meeting along their diameters. Removing the common diameter leaves three local sides rather than two, so this neighbourhood is not a disc or half-disc. Therefore

For general odd $m_{+}$ and $m_{-}$, the same rectangular circuit givesEvery elementary two-dimensional nematic defect has $|q|=1/2$, so the minimum number is

In three dimensions the director takes values in the real projective plane, whose fundamental group is $\mathbb{Z}/2\mathbb{Z}$. The signs of half-charge line defects are no longer distinct topological classes, and two such lines can annihilate by escape into the third dimension. Hence only the parity of $N_{2D}$ remains:When one is required, it is a disclination line extending through the unbounded $z$ direction.

The antipodal action fixes the origin. Away from the origin it acts freely, so the quotient is locally Euclidean there. A small punctured neighborhood of the image of the origin, however, has linkthe real projective plane, whereas the link of a point in a three-dimensional smooth manifold is $S^2$. Equivalently, the neighborhood is a cone on ${\mathbb{RP}}^2$ and cannot be a three-ball. Thus

A topological surface is a Hausdorff, second-countable space in which every point has a neighbourhood homeomorphic to an open subset of ${\mathbb{R}}^2$.

The antipodal action on $S^2$ is free. The surface quotient by a free finite action applies: Around each point choose a small open disc disjoint from its antipodal image; the quotient map restricts to a homeomorphism from that disc onto an open neighbourhood in the quotient. Compactness gives Hausdorffness and second countability descends from the sphere. Thus the quotient is the real projective plane, in particular a topological surface.

For the second quotient write a point away from the poles asThe mapin these cylindrical coordinates extends continuously over the poles and identifies exactly $\varphi$ with $\varphi +\pi$. It therefore induces a continuous bijection $S^2/\approx \to S^2$. The domain is compact and the sphere Hausdorff, so this bijection is a homeomorphism.

The real projective plane is the set of one-dimensional linear subspaces of ${\mathbb{R}}^3$. Equivalently,A projective point is written in homogeneous coordinates $[x:y:z]$. A projective line is the set of projective points whose representatives belong to one fixed two-dimensional linear subspace of ${\mathbb{R}}^3$. Such a linear subspace is the kernel of a linear map of a nonzero linear functional, so its equation has the form $\alpha x+\beta y+\gamma z=0$ with $(\alpha ,\beta ,\gamma )\ne 0$.

Two distinct projective points have linearly independent representatives $p,q$. Their linear span has dimension two and hence defines a projective line containing both. Conversely, any two-dimensional linear subspace containing both representatives contains their entire linear span and therefore equals it. Thus exactly one projective line passes through two distinct projective points.

Here is an explicit simplicial complex on six vertices. Take every subset of one of the following ten triangular facets:For example, $135$ denotes $\{1,3,5\}$, with all its edges and vertices included. Its one-skeleton has all fifteen edges of the complete graph. Every edge belongs to exactly two triangles. The links of a simplex at the six vertices are the following cycles, with the first vertex understood to repeat at the end:Thus every vertex has a disk neighbourhood, as do edge and face interior points. The realization is a compact connected surface without boundary. Its Euler characteristic isThe classification theorem for surfaces identifies the unique closed connected surface with this Euler characteristic as the real projective plane. Consequently this complex triangulates ${\mathbb{RP}}^2$. The explicit facet and link checks make this a genuine simplicial complex, not a polygon whose boundary identifications have been mistaken for simplicial faces.

There are exactly four homeomorphism types: the two-sphere, real projective plane, torus, and Klein bottle.

For completeness, enumerate all pairings. Four sides have three pairings; each pair can be matched in either direction, giving twelve signed pairings before symmetries. Up to cyclic relabelling, the repeated labels are either adjacent, of form $a{a}^{s}b{b}^t$, or alternating, of form $ab{a}^s{b}^t$, where $s,t\in \{1,-1\}$. The second adjacent pattern $abba$ rotates to the first. Endpoint identification gives the following table:Here the quotient always has two edge classes and one face, so its Euler characteristic is $V-2+1=V-1$. To reproduce the vertex count, label consecutive square vertices $0,1,2,3$. For paired sides $i,j$, a same-direction pairing identifies $i$ with $j$ and $i+1$ with $j+1$; a reversed-direction pairing identifies $i$ with $j+1$ and $i+1$ with $j$, with indices modulo four. Taking the generated equivalence classes gives exactly the displayed values of $V$.

The quotient is orientable precisely when each label is traversed once positively and once negatively around the boundary: then the face orientation extends across every paired edge. A same-direction pair obstructs such an orientation. Thus the two $\chi =0$ possibilities are distinguished as the orientable torus and nonorientable Klein bottle; the other rows are fixed by Euler characteristic and the classification theorem for surfaces.

Explicit square representatives areAll edge interior points have disk neighbourhoods by joining two half-disks, and each identified vertex has a cyclic sequence of corner sectors, again a disk neighbourhood. There are no leftover boundary edges. The enumeration therefore describes closed surfaces, and the twelve signed pairings leave no additional cases.

The classification theorem for surfaces states that every connected closed combinatorial surface is homeomorphic to exactly one of:

The orientable family has Euler characteristic $2-2g$; the nonorientable family has Euler characteristic $2-h$. Orientability together with Euler characteristic distinguishes the entries.

Boundary words giving the required polygonal schemas areA two-sided polygon here means a disk with its boundary split into two arcs; subdivision produces an ordinary polygon with more sides without changing the quotient. Alternatively the square word $a{a}^{-1}b{b}^{-1}$ represents the sphere. Each four-letter block in the orientable word adds a torus handle; each repeated-letter block in the nonorientable word adds a projective-plane summand.

A topological surface is a Hausdorff space with a countable basis of a topology in which each point has a neighbourhood homeomorphic to an open subset of ${\mathbb{R}}^2$. This definition has no boundary; allowing half-plane charts gives a surface with boundary. A closed surface is compact and has no boundary.

The classification theorem for closed combinatorial surfaces states that the realization of a finite connected closed combinatorial surface is homeomorphic to exactly one of:

The orientable family, including the sphere at $g=0$, has Euler characteristic $2-2g$; the nonorientable family has characteristic $2-k$. If disconnected complexes are permitted, apply the list to each connected component.

For polygon descriptions, read the boundary counterclockwise and pair equally labelled edges in the direction of their arrows. An inverse exponent means that the arrow runs against this boundary traversal. The following standard polygon words for closed surfaces give the required models:For $k=1$ the last model is a digon; alternatively the square word $abab$, identifying antipodal boundary points, gives the real projective plane. A square with word $ab{a}^{-1}b$ gives the Klein bottle, the case $k=2$. All polygon sides are paired, so these models have no boundary.

The quotient is a Klein bottle, equivalently a connected sum of two real projective planes.

Use $z\in S^1$ for the angular coordinate and $r\in [1/2,1]$ for the radius. Parametrize the first annulus by $\tau =2r-1\in [0,1]$, running from its inner to its outer boundary. Parametrize the second by $\tau =3-2r\in [1,2]$, running from its outer to its inner boundary. The outer-boundary identity gluing puts them together at $\tau =1$. The remaining inner-boundary identification reverses the angle, so the entire quotient isThis is the mapping torus of reflection of a circle, after rescaling the interval.

To verify the homeomorphism, send $rz$ in the first annulus to $[z,2r-1]$ in $M$, and $rz$ in the second to $[z,3-2r]$. Both formulas are continuous, agree under the outer gluing, and agree under the reflected inner gluing. They therefore descend by the universal property of the quotient topology. Conversely, on the lower half of the cylinder send $(z,\tau )$ to the class of $((1+\tau )/2)z$ in the first annulus; on the upper half send it to the class of $((3-\tau )/2)z$ in the second. The formulas agree at $\tau =1$, and at the identified endpoints. The pasting lemma and the quotient property give a continuous inverse. Thus the two quotient descriptions are homeomorphic.

The quotient $M$ is compact. Its equivalence relation on the compact Hausdorff cylinder is closed: it is the union of the diagonal and the two graphs of the boundary reflection. Therefore $M$ is Hausdorff by compact Hausdorff quotient by a closed equivalence relation. Away from the seam its local charts are circle arcs times open intervals. At the seam, an arc near $z$ at $\tau =0$ glues to its conjugate arc at $\tau =2$; using the reflection as the transition gives another open rectangle. These are disc neighbourhoods, and a finite choice of such charts gives a countable basis. Hence $M$ is a closed surface.

Cutting the circle coordinate gives a rectangle with its vertical sides matched and its horizontal sides reflected. The reflection changes the sign of the angular coordinate while leaving the interval direction unchanged, so an orientation transported around the interval loop comes back reversed. Thus $M$ is a nonorientable surface. The rectangle quotient has one vertex, two edges and one face, giving Euler characteristic zero. By the classification theorem for closed combinatorial surfaces, the nonorientable genus is $k=2-\chi (M)=2$, so $M\cong {\mathbb{RP}}^2\#{\mathbb{RP}}^2$. This is the gluing annuli with one reflected boundary construction.

For the sphere model of the real projective plane, letThen ${\mathbb{RP}}^2=S^2/{\sim }_S$ with the quotient topology: each equivalence class is an antipodal pair.

For the closed disc model, letGive $D^2/{\sim }_D$ the quotient topology. Only antipodal pairs on the boundary circle are identified in the closed disc model; interior points remain distinct.

Every complete polygon edge pairing gives a compact connected surface with empty manifold boundary. Here is the local reason at the vertices. Replace each polygon corner by a small link interval. The polygon edge pairings identify all endpoints of these link intervals in pairs. In a given vertex equivalence class, the resulting graph is connected: the endpoint identifications generating that vertex equivalence class also connect its corner intervals. Every link graph vertex has degree of a vertex two, so the graph is a circle. Its topological cone is a disc, giving the required local surface model. Edge and face CW cell interiors have the usual disc neighbourhoods as well. A finite triangulation compatible with the polygon edge pairings realizes the result as a finite simplicial complex, giving a Hausdorff, second-countable space.

For every such surface, $\chi =V-2\ge -1$. If it is orientable, the classification of compact surfaces gives $\chi =2-2g$, so $g=0$ or $1$. If it is a nonorientable surface, write $k\ge 1$ for its nonorientable genus; then $\chi =2-k\ge -1$ gives $k\le 3$. The exhaustive list is therefore

To prove that every listed possibility actually occurs, the following six-letter boundary words specify complete pairings of a hexagon:The vertex counts follow by matching starts and ends of equally directed arrows, just as in part (b). The first and third words pair every letter with its inverse, giving orientable surfaces. Each other word contains a letter twice with the same sign, giving a nonorientable surface: an orientation on the face would induce equal, rather than opposite, boundary directions along that glued pair. With the displayed Euler characteristics, the classification of compact surfaces identifies the examples exactly as listed. All five types occur, and no others occur.

The classification theorem for closed combinatorial surfaces states that the geometric realization of every closed combinatorial surface is homeomorphic to exactly one of the following:

With ${\Sigma }_0=S^2$, the complete distinguishing data are orientability and the Euler characteristic:Here $\chi (K)=V-E+F$, where $V,E,F$ count the vertices of a simplicial complex, edges, and triangles of a finite triangulation. Thus two such connected closed surfaces are homeomorphic if and only if they have the same orientability and Euler characteristic. In particular $N_1$ is the real projective plane and $N_2$ is the Klein bottle. Every listed type admits a finite triangulation. For disconnected finite closed combinatorial surfaces, apply the theorem to each of the finitely many connected components; the resulting unordered collection of types determines the homeomorphism type.

Use compatible edge identifications: choose a bijection between the three boundary vertices of $\tau$ and those of $\sigma$, and extend it linearly on each edge. This defines a simplicial homeomorphism of the two boundary circles. This compatibility is the usual meaning of gluing the two boundary triangles; independently chosen edge reversals need not agree at their endpoints. The classification needed here is 3(a)(iv); the printed reference to 3(a)(iii) is a source error.

Delete the two triangular simplexes but retain all their proper faces of a simplex, and identify the two resulting boundary simplicial subcomplexes by this map. The result is a finite abstract simplicial complex $L$: vertices from different sides are identified only in the boundary, and the only shared edges are boundary edges. There is no triangle consisting of all three boundary vertices after the two faces have been removed. Consequently no other simplexes become duplicates. Its geometric realization has the specified quotient topology and is homeomorphic to $Z$.

We check the closed combinatorial surface conditions. Away from the seam, vertex links of a simplex remain cycle graphs. At a boundary vertex of $X$ or $Y$, removing the face changes its link from a cycle graph to a graph path, with endpoints corresponding to its two boundary edges. Gluing the boundaries identifies the two paths' respective endpoints, making one cycle graph. Each seam edge now meets one triangle from $X$ and one from $Y$, and all other edges retain their original two incident triangles. Thus every vertex link is a cycle graph and every simplex lies in a triangle. The simplicial complex is connected, since the two connected punctured surfaces meet along their boundary circle. Hence $L$ is a closed combinatorial surface.

This construction is the connected sum of surfaces $T^2\#{\mathbb{RP}}^2$. For completeness, compute its Euler characteristic directly. Removing a triangular face while retaining its edges and vertices subtracts one, soGluing identifies three vertex pairs and three edge pairs, givingFor the explicit triangulations above, this is also $62-189+126=-1$.

Finally $Z$ is a nonorientable surface. If $Z$ were orientable, its restriction would make $Y$ orientable. The consistent choices of orientation of a simplex on its triangles extend across a capping disc: choose the cap's orientation of a simplex so that the induced directions on the common boundary oppose. Capping $Y$ by $\sigma$ would therefore make the real projective plane orientable, a contradiction. The obstruction to orientability in the real projective plane can also be seen from a Möbius band neighbourhood of a projective line.

By the classification theorem for closed combinatorial surfaces, a nonorientable surface with Euler characteristic $-1$ has $2-k=-1$, hence $k=3$. Therefore

Use the same $4\times 4$ square grid with four centre-based triangles per small square, but identify boundary points byThis gives the real projective plane. To see the model explicitly, a radial homeomorphism from the square, centred at $(1/2,1/2)$, to the closed disc carries the boundary involution to the antipodal map. The map from that disc to the projective planeis continuous and surjective: every one-dimensional vector subspace of ${\mathbb{R}}^3$ has a unit vector representative in the closed upper hemisphere. Its only identifications are antipodal boundary points. The compact-to-Hausdorff continuous bijection theorem therefore identifies this disc quotient space with ${\mathbb{RP}}^2$.

The square triangulation descends to an abstract simplicial complex $K_2$. Small-square centres remain distinct interior vertices; each triangle has one centre and two adjacent grid vertices. Boundary pairings identify whole grid edges, while preserving distinct vertices in each triangle, so this is a finite triangulation of the real projective plane. Its counts are $33$ vertices, $96$ edges and $64$ triangles, giving Euler characteristic one.

Choose $\sigma$ to be the interior triangle with vertices $(1/4,1/4)$, $(1/2,1/4)$ and $(3/8,3/8)$. Removing its relative interior and cutting from $(0,0)$ to $(1/4,1/4)$ gives the same nine-sided cut disc as before. Parametrize it by a convex polygon, transporting the boundary identifications. Its boundary word is nowThe boundary involution pairs the two $a$ sides and the two $b$ sides in matching traversal directions; the $c$ pair restores the slit. Leave $d,e,f$ unpaired. The quotient space isthe real projective plane with an open disc removed. In particular it is a compact connected surface with boundary consisting of one three-edge circle.

Start with a polygonal schema whose outer boundary has $m$ sides, and remove a triangular disc strictly inside its polygon. Cut the resulting annulus along an arc from an outer vertex to an inner vertex. The cut space is a closed disc with $m+5$ marked sides: the original $m$ outer sides, two banks of the cut, and three inner sides. Parametrize it by a convex polygon with that many sides. A boundary homeomorphism carrying the marked vertices in cyclic order extends over the disc by radial extension after identifying both discs with a standard closed disc.

Transport the original side identifications to this polygon and identify the two cut banks, leaving the three inner sides unpaired. The quotient space is the original surface with an open triangular disc removed. In boundary-word notation, if $w$ is the original word, the new word is $wcdef{c}^{-1}$, with $d,e,f$ unpaired. The torus has $w=ab{a}^{-1}{b}^{-1}$; a square model of the real projective plane has $w=abab$, pairing antipodal points of the square's boundary.

In the real projective plane, represent a projective point by a nonzero vector and a projective line by its nonzero coefficient covector, identified with a vector in chosen coordinates. The join of distinct projective points represented by $p,q$ has coefficient vector $p\times q$, using the cross product. It is nonzero because the representatives are linearly independent, and its dot product with each representative vanishes. Dually, two distinct projective lines with coefficients $\ell ,m$ meet in the projective point represented by $\ell \times m$. Scaling an input scales the output and leaves the projective object unchanged. This makes joins, intersections, concurrency, and collinearity exact calculations with homogeneous coordinates.

A projective triangle is an ordered triple of noncollinear projective points in a projective plane, together with their three joining projective lines. In the real projective plane this means that three chosen representative vectors are linearly independent. Pairing the vertices of two such triangles determines corresponding sides and the three corresponding-vertex joins.