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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.

The dual projective space is $\mathbb{P}(({\mathbb{R}}^3{)}^{\ast })$. A projective line with coefficients $[\alpha :\beta :\gamma ]$ becomes a projective point in this dual plane; a projective point $[a:b:c]$ becomes the dual projective line $a\alpha +b\beta +c\gamma =0$.

The point-line duality reverses incidence:Thus the dual statement is that the dual point $l^{\ast }$ lies on the dual line $P^{\ast }$. A join of projective points becomes an intersection of dual projective lines, and the converse holds too. In particular, collinear projective points correspond to concurrent dual projective lines.

The dual of Desargues's theorem is its converse: if the intersections of the three pairs of corresponding sides are collinear, then the three lines joining corresponding vertices are concurrent. In the same nondegenerate setting, perspectivity from a line therefore implies perspectivity from a point.

The corresponding side equations, computed using projective joins and meets by cross products, areSubtracting the two equations in each row and substituting back gives the three projective pointsFor example, the first pair forces $y=z$ and $x=3z$; the second pair forces $y=0$ and $x=-z$; the last pair forces $x=z=2y$. All three satisfysince $3-4+1=0$, $1-0-1=0$, and $2-4+2=0$. The coefficient vector is nonzero, so this is a projective line. Thus the corresponding-side intersections are collinear, explicitly confirming Desargues's theorem.

The diagram uses the affine chart $x+y+z\ne 0$ with coordinates $u=y/(x+y+z)$ and $v=z/(x+y+z)$. Here the center $O$ is finite, the axis is $u=1/5$, and $Y$ lies on the projective line at infinity. The apparent parallelism of $QR$ and $Q^{\prime }{R}^{\prime }$ records their common vertical point at infinity, which also belongs to the axis.

Conjugate by a projective transformation sending $P$ to $[1:0]$ and $Q$ to $[0:1]$. A matrix representing the conjugated transformation must have zero diagonal, since its first column represents $[0:1]$ and its second represents $[1:0]$. Therefore it has the formIts square is $abI$, a nonzero scalar matrix, which induces the identity on the projective line. Conjugating back provesFor the requested explicit involution, takeMore generally $[x:y]\mapsto [ay:bx]$ works whenever $ab\ne 0$.

The four points form a projective frame: no three are collinear. By the allowed uniqueness theorem and cross-ratio invariance, useIntersecting the indicated projective lines givesA point of $Q_0{Q}_2$ has coordinates $[a:b:a+b]$. Intersecting with $P_0{P}_1$, whose equation is $z=0$, gives $Q_3=[1:-1:0]$. On that line use its first two homogeneous coordinates. The cross-ratio from (a)(ii) isThe characteristic of a field being zero ensures that $Q_3\ne Q_1$. The construction is the diagonal-point property of a complete quadrangle, producing harmonic conjugate points.

A projective conic in ${\mathbb{P}}^2(\mathbb{C})$ is the zero locus of a nonzero homogeneous quadratic polynomial, up to nonzero scaling of that polynomial. Under an invertible linear change of homogeneous coordinates, its symmetric matrix has rank $3$, $2$, or $1$. Over $\mathbb{C}$, the corresponding forms are a nonsingular conic such as $XZ-Y^2=0$, two distinct lines such as $XY=0$, or a double line $X^2=0$. In the last case the repeated factor records multiplicity, even though the underlying point set is one line.

A projective line is the projectivization of a two-dimensional linear subspace of ${\mathbb{C}}^3$, equivalently the zero locus of one nonzero linear form. For two such subspaces $U,W$, the dimension formula for a sum of subspaces gives $\dim (U\cap W)\ge 2+2-3=1$. A nonzero vector in the intersection therefore supplies a common projective point. Distinct lines have exactly one such point.

Assume $L\ne 0$. The quadratics divisible by $L$ form $L\cdot \mathrm{span}\{x,y\}$, a two-dimensional linear subspace; their projectivization is a projective line. Evaluating at the projective zero $[-\beta :\alpha ]$ of $L$ gives its equationAt the point $[L^2]=[{\alpha }^2:2\alpha \beta :{\beta }^2]$ of $C$, the tangent equation obtained from the gradient of $b^2-4ac$ is $-4{\beta }^{2}a+4\alpha \beta b-4{\alpha }^{2}c=0$. It is exactly the same line, establishing tangency.

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

Two projective triangles are in perspectivity from a line when the three intersections of corresponding sides lie on one projective line, called their axis of perspectivity. In the nondegenerate setting this is dual to perspectivity from a point under point-line duality.

Two projective triangles are in perspectivity from a point when the three projective lines joining corresponding vertices meet in one projective point, called their center of perspectivity.

For any field $k$, degree identifies the Picard group of the projective line with $\mathbb{Z}$, generated by $\mathcal{O}(1)$. A closed point on the affine chart is cut out by an irreducible polynomial $f(t)$, whose divisor is that point minus $\mathrm{deg}(f)$ times infinity. Hence every divisor is linearly equivalent to its degree times infinity. Principal divisors have degree zero, so the degree map is injective as well as surjective. Thus a line bundle of degree $d$ is $\mathcal{O}(d)$.

A degree-two projective plane curve, allowing singular and nonreduced cases. In characteristic different from two, write its equation as $z^{\mathsf{T}}Mz$ with $M$ a symmetric matrix. It is smooth exactly when $M$ has matrix rank three. Over an algebraically closed field, rank two gives two distinct projective lines, while rank one gives a double line.

In the affine coordinate $z\mapsto [z:1]$, the complex projective line consists of the complex plane together with the single extra projective point $[1:0]$, called its point at infinity. It is the only point whose second homogeneous coordinate vanishes. This coordinate choice distinguishes the point; it is not intrinsically special under projective transformations. The map $[u:v]\mapsto [v:u]$ exchanges zero and infinity. The resulting one-point compactification is the Riemann sphere.

A projective conic is the zero locus in a projective plane of a nonzero homogeneous degree-two polynomial. The equation is considered up to multiplication by a nonzero scalar. Over the complex numbers, a symmetric matrix for the quadratic form has rank three, two, or one: these give a nonsingular conic, two distinct projective lines, or one line counted twice. Diagonalization of a quadratic form and rescaling its nonzero diagonal entries prove this classification. The smooth model $XZ-Y^2=0$ is parametrized by the degree-two Veronese map $[s:t]\mapsto [s^2:st:t^2]$.

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 quadric is the zero set of a nonzero homogeneous quadratic polynomial in projective space. Over a field of characteristic different from two, represent it by a symmetric matrix $B$, up to a nonzero scalar. The gradient at a representative $v$ is $2Bv$, so the singular locus is the projectivization of $\ker B$. Thus the quadric is nonsingular exactly when $B$ is invertible. In the complex projective plane, ranks three, two, and one give respectively a smooth projective conic, two distinct projective lines, and a double projective line.

For a smooth algebraic surface defined by a homogeneous polynomial $F$ in projective space of dimension three, its tangent plane at $P$ is the displayed plane. Every projective line on the surface through $P$ lies in this plane.

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.

Three distinct points of multiplicity two use the entire arithmetic genus three of an integral projective curve of degree four. One point blowup at each gives arithmetic genus zero. The normalization of an algebraic curve then has genus zero and is the projective line over an algebraically closed field, by rationality of a smooth projective genus-zero curve. This allows double points with one branch as well as ordinary nodes.

Over an algebraically closed field, choose a point $P$ on the smooth projective curve. The Riemann-Roch theorem gives $\dim L(P)=2$: the complementary canonical divisor term has negative degree. A nonconstant function in $L(P)$ has exactly one simple pole, so it defines a degree-one morphism of algebraic curves to the projective line. A degree-one map between smooth projective curves is an isomorphism. The existence of a rational point is essential over a field which is not algebraically closed.