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A fiber bundle whose fibers are affine spaces and whose transition functions are affine. Differences of points in one fiber lie in a corresponding model vector bundle, but there need not be a preferred zero in each fiber.

An affine combination is a weighted point $P={\sum }_{i=0}^k{\lambda }_i{P}_i$ with ${\sum }_{i=0}^k{\lambda }_i=1$. The coefficients can be negative. The sum is meaningful in an affine space without selecting a preferred origin: relative to any point $O$, it means $P=O+{\sum }_i{\lambda }_i(P_i-O)$. Affine maps preserve these weighted points. ^[ref]

For an original example in the Euclidean plane, take $A=(0,0)$, $B=(2,0)$ and $C=(0,2)$. Then $P=\frac{1}{4}A+\frac{1}{2}B+\frac{1}{4}C=(1,\frac{1}{2})$ lies inside their triangle. Its coefficients are nonnegative, making it a convex combination. In contrast, $R=-\frac{1}{2}A+\frac{3}{2}B=(3,0)$ is beyond $B$: it is an affine combination but not a convex combination of these three points. Negative weights allow extrapolation, rather than only interpolation inside the convex hull.

For the affine map $T(x,y)=(2x+y+1,y-2)$, the same weights produce $T(P)=(\frac{7}{2},-\frac{3}{2})$ from $T(A),T(B),T(C)$, and produce $T(R)=(7,-2)$ from $T(A),T(B)$. Thus one can compute a weighted point before transforming the vertices or afterwards, obtaining exactly the same point.

This is useful whenever an object is built from weighted control points: its construction commutes with an affine map, so transforming the control points transforms the constructed object as well. The sum of the weights is the essential condition; no assumption about lengths or angles is needed.

The geometry of affine spaces, studying incidence and parallelism without choosing an origin or a metric.

Points $p_0,\dots ,p_k$ in an affine space are affinely independent if the only affine dependence displayed above has all coefficients zero, using any choice of origin to read the vector equation. Since the coefficients sum to zero, replacing the origin leaves that equation unchanged. Substituting ${\lambda }_0=-{\sum }_{i=1}^k{\lambda }_i$ shows that this condition is equivalent to linear independence of $p_1-p_0,\dots ,p_k-p_0$. It also says that an affine combination of the points has unique coefficients. In ${\mathbb{R}}^n$, at most $n+1$ points can be affinely independent; $n+1$ such points span a full-dimensional simplex.

The affine line over a field $k$ is ${\mathbb{A}}_k^1=\mathrm{Spec}k[t]$.

The one-dimensional scheme has an affine space of rational points over its base field; its scheme structure also records non-rational points and local rings.

The affine plane over a field $k$ is the affine scheme $\mathrm{Spec}k[x,y]$.

The two-dimensional scheme has an affine space of rational points over its base field; its scheme structure also records non-rational points and local rings.

The difference of two connections on a vector bundle is a smooth one-form with values in the endomorphism bundle: the Leibniz derivative terms cancel, leaving pointwise linearity in both the tangent vector and section value. Conversely every such form added to a connection gives another connection. Thus the collection is an affine space with this model vector space. Choosing ${\nabla }^0$ supplies a noncanonical origin; there is no distinguished zero connection.

Affine three-space is $\mathrm{Spec}k[x_1,x_2,x_3]$.

The three-dimensional scheme has an affine space of rational points over its base field; its scheme structure also records non-rational points and local rings.

The graded Leibniz rule terms of two connections cancel, making their difference $C^{\infty }$-linear in the section. It is therefore an element of $A^1(\mathrm{End}E)$. Conversely adding any such form to a connection gives another connection, making the space of connections an affine space.

The dimension vector of a finite-dimensional quiver representation is $({\dim }_k{X}_i{)}_{i\in Q_0}$. Direct sums add dimension vectors. Fixing the vector gives an affine space of possible arrow matrices, the quiver representation space.

A geometric background for Newtonian mechanics consisting of a four-dimensional affine space, an absolute time covector $t_a$, a rank-three positive spatial inverse metric tensor $h^{ab}$ with $t_a{h}^{ab}=0$, and a flat torsion-free affine connection preserving both. A trajectory parametrized by absolute time has tangent $u$ with $t_a{u}^a=1$. Its absolute acceleration is ${\nabla }_{u}u$; an unforced particle obeys ${\nabla }_{u}u=0$. The structure distinguishes inertial motion without selecting an absolute state of uniform rest. Additional force laws, including a Newtonian gravitational force when wanted, specify a dynamical theory on this background.

A closed half-space consists of a hyperplane and one of its two sides in a real affine space. Replacing the weak inequality by a strict inequality gives an open half-space. Half-spaces are convex sets, and finite intersections of closed half-spaces are linear polyhedra.

A homogeneous linear differential equation is a linear differential equation with identically zero forcing term, $L[y]=0$. On a common interval, its solutions of a differential equation form the vector space $\ker L$, the kernel of a linear map. They include the zero function and are closed under addition and scalar multiplication. For any two solutions of a differential equation $u,v$ and constant scalars $\alpha ,\beta$, linearity gives $L[\alpha u+\beta v]=\alpha 0+\beta 0=0$: this is the superposition principle. The scalars must be constant; variable multipliers introduce extra derivatives.

For example, $y^{\prime }+2y=0$ has general solution $y=C{e}^{-2t}$, where $C$ is any constant real number. A short verification uses the product rule: $(e^{2t}y{)}^{\prime }=e^{2t}(y^{\prime }+2y)=0$, so $e^{2t}y$ is constant. For $y_1=e^{-2t}$ and $y_2=2e^{-2t}$, the sum is $3e^{-2t}$ and also solves the same homogeneous linear differential equation. At $t=1$, their values are approximately $0.13534$, $0.27067$ and $0.40601$. Every curve is a multiple of one exponential function; the solution space of a homogeneous linear differential equation has dimension one. Both $t$ and $y$ in this example and the following diagram are dimensionless mathematical variables.

Now keep the same differential operator $L[y]=y^{\prime }+2y$ but prescribe the forcing term $1$. The inhomogeneous linear differential equation $y^{\prime }+2y=1$ has particular solution $y_p=1/2$ and general solution $y=1/2+C{e}^{-2t}$. The initial conditions $y(0)=0,1,2$ select, respectively,At $t=1$, their values are approximately $0.43233$, $0.56767$ and $0.70300$. The midpoint $\frac{1}{2}(u_0+u_2)=u_1$ solves the same inhomogeneous linear differential equation, because $L[\frac{1}{2}(u_0+u_2)]=\frac{1}{2}(1+1)=1$. The sum $s=u_0+u_2=1+e^{-2t}$ instead satisfies $s^{\prime }+2s=2$, so it fails the equation with forcing term $1$. More generally, a linear combination of solutions of a differential equation for the same fixed nonzero forcing term retains that forcing term precisely when its scalar weights add to one. This is the affine space structure, rather than unrestricted superposition within that fixed family.

On the left, all three curves solve the same homogeneous linear differential equation. On the right, $u_0,u_1,u_2$ share the particular solution $1/2$, and the shaded difference $u_2-1/2$ is the homogeneous solution $\frac{3}{2}{e}^{-2t}$. The dashed sum has twice the forcing term, even though its starting value coincides with that of $u_2$. Adding the two initial conditions also adds their inputs; taking their midpoint averages both. This distinction makes the superposition principle useful for assembling responses to different inputs without mistaking that assembly for a solution of a differential equation with the original input. See MIT's superposition notes, including the particular solution one half.

The points with integer coordinates in a real d-dimensional affine space with a chosen origin and coordinate basis. Translations by integer vectors and dilations by positive integers preserve this set. The Gallai theorem for an integer lattice concerns finite colourings of these points and homothetic copies of finite configurations of them.

A scalar linear differential equation of order $n\ge 1$ has the formHere the coefficients $a_j$ and the forcing term $g$ are prescribed functions of the independent variable $t$. A classical solution of a differential equation is an $n$-times differentiable function satisfying the identity at every point of its interval. The unknown function $y$ and its derivatives enter to the first power, without products between them or coefficients depending on $y$. Equivalently, the differential operator $L$ is a linear map: for constant scalars $\alpha ,\beta$,The forcing term may be zero, giving a homogeneous linear differential equation, or nonzero, giving an inhomogeneous linear differential equation. Linearity of $L$ does not mean that every function graph of a solution of a differential equation is a straight line.

To use the usual order-$n$ initial value problem theory on an interval $I$, assume that $a_n(t)\ne 0$ throughout $I$ and that all $a_j$ and $g$ are continuous functions there. Dividing by $a_n$ normalizes the highest derivative; specifying $y(t_0),y^{\prime }(t_0),\dots ,y^{(n-1)}(t_0)$ at $t_0\in I$ then selects one solution of a differential equation throughout $I$. A zero of $a_n$ needs separate analysis: one cannot divide through it and retain these conclusions. The first-order normalization appears in OpenStax's definition and standard form of linear equations; the corresponding system theorem is in MIT's existence and uniqueness notes for linear systems.

The coefficients need not be linear functions of $t$. For example, $t^2{y}^{\prime }+y=0$ is a linear differential equation on $t>0$, although $t^2$ is quadratic. Its solution of a differential equation with initial condition $y(1)=1$ is $y(t)=e^{1/t-1}$: its derivative is $y^{\prime }=-y/t^2$, so substitution gives $t^2{y}^{\prime }+y=0$, and $y(2)=e^{-1/2}\approx 0.60653$. The zero of the highest coefficient at $t=0$ explains the domain restriction. In contrast, $y^{\prime }+y^2=0$, $y{y}^{\prime }+y=0$ and $y^{\prime }+\sin y=0$ are nonlinear ordinary differential equations because of their dependence on the unknown function. Variable coefficients and nonlinear ordinary differential equations are different issues; see MIT's definitions and examples of first-order linear equations.

The practical benefit of linearity is that it separates the effect of the initial condition from the effect of the forcing term. Once one particular solution $y_p$ satisfies $L[y_p]=g$, every solution of a differential equation isHere $\ker L$ is the kernel of a linear map, consisting of solutions of a differential equation to the associated homogeneous linear differential equation. Indeed, $L[y-y_p]=g-g=0$, and conversely $L[y_p+h]=g+0=g$. Thus the family for a fixed forcing term is an affine space obtained by translating $\ker L$. If $g$ is not identically zero, the zero function is absent, so this family is not a vector space under the usual operations on functions. This short argument explains why a general solution is a particular solution plus a homogeneous solution.

Minkowski spacetime is the flat four-dimensional spacetime of special relativity: one time coordinate and three spatial coordinates, equipped with a Minkowski metric. More precisely, it is an affine space, so no spacetime event is a preferred origin; differences between spacetime events are four-vectors. Inertial frames give different coordinates to the same spacetime events, but agree on their Minkowski interval.

This geometry answers two different measurement questions with one structure. Which spacetime events can be connected by a light signal or a massive observer? The light cone classifies them. How much time does a moving clock record? Its world line and the Minkowski metric determine its proper time. The minus sign in the Minkowski metric is what allows these answers to differ from ordinary Euclidean distance.

For a concrete one-space-dimensional slice, start at the origin $O$ and measure $t$ in seconds and $x$ in light-seconds. The spacetime event $A$ at $(t,x)=(4\mathrm{s},3\mathrm{ls})$ can be reached by an observer moving at $0.75c$, whereas $B=(2\mathrm{s},3\mathrm{ls})$ would require $1.5c$ to be reached directly from $O$. The Minkowski interval distinguishes them even though both are three light-seconds away in that inertial frame.

Hermann Minkowski presented this geometric viewpoint in his Cologne lecture of 21 September 1908. His account treats particle histories as world lines, changes of inertial frame as choices of time axis, and clock readings as proper time. These are connected geometric ideas, rather than separate corrections to ordinary mechanics.^[ref]

The algebraic set defined by the annihilator of a module of the associated graded module for a good filtration for the order filtration is the order characteristic variety. It lies in affine space with coordinates $x_1,\dots ,x_n,{\xi }_1,\dots ,{\xi }_n$ and is invariant under scaling the ${\xi }_i$. Its dimension is the dimension of a Weyl algebra module for the order filtration. The order filtration of a Weyl algebra assigns degrees zero to the $x_i$ and one to the ${\xi }_i$; a characteristic variety formed with the Bernstein filtration need not be the same algebraic set.

The point assigned all-zero coordinates in a chosen Cartesian coordinate system. Choosing an origin identifies position displacements with vectors; a bare affine space has no distinguished origin.

We work over an algebraically closed field $k$, so the assertion concerns a geometric projective line. Let $V=k^4$ and let $W$ be the vector space of homogeneous cubic forms on $V$. It has dimension $\left(\genfrac{}{}{0ex}{}{6}{3}\right)=20$, so the parameter space of cubic equations up to nonzero scalar is $\mathbb{P}(W)={\mathbb{P}}^{19}$. The Grassmannian $G=\mathrm{Gr}(2,V)$ parametrizes lines in ${\mathbb{P}}^3$. Its standard graph charts are four-dimensional affine spaces, so its dimension is four. The Plücker embedding realizes it as a closed quadric in ${\mathbb{P}}^5$, proving projectivity. The general linear group ${\mathrm{GL}}_4$ acts transitively on two-planes and is an irreducible open subset of the affine space of matrices; its orbit map proves that $G$ is irreducible.

Consider the incidence varietyRestriction of a cubic to a fixed line is surjective onto the four-dimensional space of binary cubic forms: after choosing coordinates with that line given by $x_2=x_3=0$, extend any binary cubic using $x_0,x_1$ alone. Thus its kernel has dimension sixteen. On each Grassmannian chart these restriction maps depend algebraically on the line parameters and have constant rank four. Their kernels form a rank-sixteen vector bundle, and $\mathcal{I}$ is its projective bundle, with fibre ${\mathbb{P}}^{15}$. In particular, $\mathcal{I}$ is irreducible of dimension $4+15=19$. It is closed in the product, since restriction vanishes precisely when four polynomial coefficients vanish.

The projection $\pi :\mathcal{I}\to \mathbb{P}(W)$ is a projective morphism, hence its image is closed. It remains to prove that this image has dimension nineteen. Equal dimensions alone would not suffice: we must control a fibre locally.

TakeIn the open Grassmannian chart around ${\ell }_0$, every line is uniquely a graphRestriction of $F_0$ to that line isThe four monomials are linearly independent over every field. Hence the only line in this chart lying on $F_0$ has $a_0=a_1=b_0=b_1=0$. The fibre ${\pi }^{-1}([F_0])$ therefore has local dimension zero at $({\ell }_0,[F_0])$. We do not need $F_0$ to define a smooth surface, nor do we claim its entire fibre is finite.

The fiber dimension theorem, in its local dimension inequality, givesThus the closed image has dimension at least nineteen. A proper closed subset of the irreducible ${\mathbb{P}}^{19}$ has smaller dimension, so $\pi (\mathcal{I})={\mathbb{P}}^{19}$. Every cubic equation consequently has a line in its zero set. In particular, every nonsingular cubic surface contains a line. The proof in fact establishes existence for arbitrary cubic surfaces over an algebraically closed field; it does not assert a rational line over an arbitrary nonclosed ground field.

For the degree-ten partition action of S6, take unordered partitions of six letters into two triples. There are $\left(\genfrac{}{}{0ex}{}{6}{3}\right)/2=10$. Fix $A\mid A^c$. Its stabilizer is $(S_3\times S_3)⋊S_2=S_3\wr S_2$, permuting each triple and interchanging them. The subgroup generated by a 3-cycle in each triple has order nine, so is a Sylow subgroup of $S_6$. Any normalizer preserves its two three-point orbits; conversely all the indicated within-triple permutations and the swap normalize it. This proves the asserted normalizer exactly.

For any different partition $C\mid C^c$, $|A\cap C|$ is one or two after excluding the identical partition. Interchanging $C$ and $C^c$ identifies these two possibilities. The stabilizer of $A\mid A^c$ is transitive on each fixed intersection pattern, hence on all other nine partitions. The degree-ten action is two-transitive. It is also faithful: two distinct partitions have a common refinement with two singleton cells and two cells of size two. Every unordered pair of letters occurs as those singleton cells. A permutation fixing every partition therefore fixes every pair and consequently every letter.

For the subgroup factorization from coset transitivity, use cosets $Hg$ (right cosets) and right multiplication by $K$. The orbit of $H$ is precisely $\{Hk:k\in K\}$. This is every coset if and only if every $g$ can be written $hk$, provingA natural $S_5$ fixing one letter is transitive on the ten partitions: the triple containing the fixed letter is determined by any two of the other five. Taking $H=S_3\wr S_2$, $K=S_5$ gives $S_6=HK$; inversion gives the requested order as well,

Now let $B$ be the binary nondegenerate alternating bilinear form. Its binary quadratic refinements form an affine space: the difference of two such forms is linear, so for a fixed $Q_0$ every one is uniquelyIn a symplectic basis each refinement isA two-dimensional summand with $(a_i,b_i)\ne (1,1)$ admits a basis of singular vectors paired to one and is hyperbolic; the $(1,1)$ summand is anisotropic. Two planes spanned by anisotropic vectors for a form together become two hyperbolic planes of a quadratic form: with anisotropic basis pairs $(e_1,f_1),(e_2,f_2)$, the singular vectors $e_1+e_2,f_1+e_2$ pair to one; their orthogonal complement for a sesquilinear form has singular paired basis $e_1+f_1+f_2,e_1+f_1+e_2+f_2$. Consequently every refinement is equivalent to either $m$ hyperbolic planes of a quadratic form or $(m-1)$ hyperbolic planes of a quadratic form and one anisotropic plane (all three nonzero vectors are anisotropic vectors for a form).

These two types are different. Indeed the character sum of each plane is $+2$ or $-2$, soThus there are exactly two symplectic orbits $Q^{+}(B),Q^{-}(B)$, the signs denoting $\mathrm{Arf}=0,1$. Counting the three coefficient pairs with product zero and the one with product one givesThe same character sum gives the corresponding numbers of singular vectors, including zero, in a fixed form of either type.

Fix a form $Q$ of type $ϵ$. Completing the square by $x\mapsto x+v$ showsThe stabilizer of $Q$ is exactly $O(V,Q)$. It sends the parameter $v$ by its natural linear action or inverse action according to the chosen left/right convention. Other forms of the same type correspond to nonzero singular vectors $v$, and all forms of opposite type correspond to nonsingular vectors $v$. For two nonzero vectors of the same $Q$ value, the map between their spans is an isometry of a space with a form. Witt's lemma extends it, so the orthogonal stabilizer is transitive on each of these two vector sets. This proves the two-transitive binary quadratic-form actions: the symplectic group over a finite field is transitive on each type and its point stabilizer is transitive on all other members of that type.

In particular $O_{2m}^{-}(2)$ is transitive on the plus orbit. Identify that orbit with the cosets (right cosets) of a plus stabilizer, using $Q^g(x)=Q(gx)$ as a right action. The coset criterion then provesBoth genuine two-transitive actions require $m\ge 2$. At $m=1$ the plus orbit has three forms and the minus orbit is a singleton; transitivity and the factorization remain true, while two-transitive action on the singleton is only vacuous if one's convention permits degree one. No special simplicity assertion is needed for this argument. At $m=2$ the orbit sizes are ten and six, linking these actions with $S{p}_4(2)\cong S_6$ and the partition action above.