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Choose $r$ of the $n$ basis vectors and require their images in the quotient to be a basis. Normalize those columns to the identity; the remaining $r(n-r)$ entries are arbitrary affine coordinates. Between column sets $I,J$, the overlap is the principal open subset where $\det A_J$ is a unit, and the new normalized matrix is $A_J^{-1}A$. These matrix transformations satisfy the cocycle identity. Locally every surjective matrix has an invertible maximal minor, so the charts cover all rank-$r$ locally free quotients. The normalized matrices glue the universal quotient and give the natural identification with its functor of points.

// Target: ringed-space.bigb

The standard basis of the vector space ${\mathbb{R}}^n$ consists of the vectors $e_i$ whose $i$th coordinate is one and whose other coordinates are zero. Every vector $x$ has the unique expression $x={\sum }_i{x}_i{e}_i$ in this basis.

A standard Young tableau of shape $\lambda ⊢n$ is a Young tableau that uses $1,\dots ,n$ exactly once and strictly increases left to right along every row and top to bottom down every column. Formally, a bijection $T:Y(\lambda )\to \{1,\dots ,n\}$ is standard whenwhenever the indicated cells belong to the Young diagram. We write $\mathrm{SYT}(\lambda )$ for the set of such fillings and $f^{\lambda }=|\mathrm{SYT}(\lambda )|$ for its cardinality.

For $(3,2)$, the five standard Young tableaux are listed below, with each pair giving the top row followed by the bottom row:There are no others: the upper-left entry must be $1$, and choosing the two bottom-row entries fixes both increasing row orders. Of the six choices of two entries from $\{2,3,4,5\}$, only bottom row $(2,3)$ fails, because its $3$ would lie below $4$. Hence $f^{(3,2)}=5$ by direct enumeration, independently of the hook-length formula.

The cells occupied by $1,\dots ,r$ form a Young diagram: every cell above or to the left of an occupied cell has a smaller label. Consequently a standard Young tableau is equivalently a path from the empty shape to $\lambda$ in the Young branching graph, adjoining one addable node of a Young diagram at each step. For $T_2$, the successive shapes areConversely, number the cell added at step $r$ by $r$ to recover the standard Young tableau. This path interpretation also makes a standard Young tableau a linear extension of a partially ordered set defined by the row and column requirements; see Stanley's Algebraic Combinatorics, Chapter 8, on tableaux and paths in Young's lattice.

The usefulness goes beyond counting: the standard polytabloid basis of a Specht module $S^{\lambda }$ over the complex numbers is indexed by $\mathrm{SYT}(\lambda )$, so ${\dim }_{\mathbb{C}}{S}^{\lambda }=f^{\lambda }$. Each of the five pictured standard Young tableaux therefore labels one basis element of the irreducible representation $S^{(3,2)}$ of the symmetric group $S_5$, whose dimension over the complex numbers is five. The filled diagrams index the basis; the actual vectors are polytabloids. The hook-length formula computes this dimension without listing all the standard Young tableaux.

If $X$ is simply connected and $h_{\ast }:{\pi }_{\ast }(X)\otimes \mathbb{Q}\to {\tilde{H}}_{\ast }(X;\mathbb{Q})$ is onto, choose a homogeneous homology basis and representing sphere maps into $X_{\mathbb{Q}}$. Their wedge sum gives a rational homology equivalence, hence a rational homotopy equivalence by the rational Whitehead theorem. All positive cup products therefore vanish. In a minimal model, the induced map from cohomology to the indecomposable quotient is injective; replacing a set of generators by cocycle representatives produces a quasi-isomorphism to the square-zero cohomology algebra, proving formality of a topological space.

Let $S:W\times W\to W$ be a symmetric bilinear map on a finite-dimensional real vector space. If $S(v,v)$ is parallel to $v$ for every $v$, then it has the displayed form for one covector $V$. Write $S(e_i,e_i)={\alpha }_i{e}_i$ in a basis. Applying the condition to $e_i+e_j$ and $e_i-e_j$ excludes every component outside their span and gives $S(e_i,e_j)=({\alpha }_j{e}_i+{\alpha }_i{e}_j)/2$. Set $V(e_i)={\alpha }_i/2$ and extend by bilinearity. Dimension one follows by the same definition. The trace identity $S^a{}_{ac}=(n+1)V_c$ proves uniqueness without depending on the chosen basis.

A bilinear form $B$ is symmetric when $B(v,w)=B(w,v)$ for every pair of vectors $v,w$. Equivalently, its matrix of a bilinear form in any basis is a symmetric matrix, since $A_{ij}=B(e_i,e_j)=B(e_j,e_i)=A_{ji}$. Symmetry makes the relation $B(v,w)=0$ reciprocal, giving the bilinear orthogonal complement a geometric interpretation.

Over the real numbers, set $q(v)=B(v,v)$. The polarization identityrecovers the symmetric bilinear form from its quadratic form: expand $q(v+w)$ and use symmetry to combine the two cross terms. For $q(x,y)=x^2-y^2$, take $v=(2,1)$ and $w=(1,3)$. Then $q(v+w)=-7$, $q(v)=3$, and $q(w)=-8$, so $B(v,w)=(-7-3+8)/2=-1=2\cdot 1-1\cdot 3$. The same formula works whenever the characteristic is not two.^[ref]

Three real diagonal matrices show how much variety symmetry allows. The identity matrix $\mathrm{diag}(1,1)$ gives the positive-definite bilinear form $q=x^2+y^2$. The matrix $\mathrm{diag}(1,-1)$ gives $q=x^2-y^2$ and is nondegenerate despite the isotropic vector $(1,1)$. The matrix $\mathrm{diag}(1,0)$ gives $q=x^2$ and is a degenerate bilinear form with radical of a bilinear form $\mathrm{span}\{(0,1)\}$. Their level sets make the distinction visible: a circle, two branches of a hyperbola, and two parallel lines for $q=1$.

The inertia of a bilinear form records the numbers of positive, negative, and zero directions after matrix congruence diagonalization. By Sylvester's law of inertia, these counts survive change of basis; the three examples have inertia of a bilinear form $(2,0,0)$, $(1,1,0)$, and $(1,0,1)$. Historically, Sylvester stated this invariance in an 1852 paper on reducing real quadratic forms to sums of positive and negative squares. Ghys and Ranicki reproduce his statement and distinguish it from his subsequent 1853 proof.^[ref]

The Fundamental theorem of symmetric polynomials gives freely generated elementary functions, so sending each $e_r$ to $h_r$ defines an algebra homomorphism. For $E(t)=\sum e_r{t}^r$ and $H(t)=\sum h_r{t}^r$, $E(-t)H(t)=1$. Applying the homomorphism shows $\omega (H(t))=E(t)$, hence ${\omega }^2=1$. Taking logarithms of $H$ gives the displayed action on power sums. It preserves the Hall inner product of symmetric functions because each power-sum basis vector acquires only a sign.

A symplectic basis organizes a nondegenerate alternating bilinear form into independent pairs. Let $(V,\omega )$ be a finite-dimensional vector space over a field $K$, with $\omega$ a nondegenerate bilinear form satisfying $\omega (v,v)=0$ for every vector $v$. Its dimension is even, say $2n$, and a symplectic basis is an ordered basis $\mathcal{B}=(e_1,\dots ,e_n,f_1,\dots ,f_n)$ such thatHere ${\delta }_{ij}$ is the Kronecker delta. Each $(e_i,f_i)$ is a hyperbolic pair; different pairs have zero mutual pairing. These conventions apply in every characteristic. In characteristic two, $-1=1$, but the condition $\omega (v,v)=0$ is still essential: a symmetric bilinear form need not be an alternating bilinear form. The existence theorem and the characteristic caveat are given in Bilinear Forms by Keith Conrad (Keith Conrad, Bilinear Forms, §1, Theorem 1.6; §5, Theorems 5.1 and 5.4 and Definition 5.3).

The benefit is a common coordinate language for every symplectic vector space of the same finite dimension over the same field. The span of all $e_i$, and separately of all $f_i$, is a Lagrangian subspace; each paired span is a symplectic subspace. In Hamiltonian mechanics, the corresponding pairing of position and momentum coordinates describes phase space and canonical transformations. On a symplectic manifold, a symplectic basis describes one tangent space; Darboux coordinates additionally describe a whole local neighborhood.^[ref]

A systematic encoder retains the message symbols unchanged in designated coordinates of its codeword and calculates the remaining check symbols. For a linear code, choosing an invertible set of $k$ columns of a generator matrix and changing its row basis gives this form.

A coordinate chart $x$ trivializes the tangent bundle by expressing $v\in T_{p}M$ in the coordinate basis. For $y=f(x)$, the chain rule gives the displayed transition. Smoothness and fibrewise linear invertibility of the Jacobian matrix construct the natural smooth vector bundle structure. The chain rule also supplies the transition cocycle.

A tangent vector is a linear map on germs of smooth functions satisfying $D(ab)=a(x)D(b)+b(x)D(a)$. In a manifold chart, first-order expansion shows $Dh={\sum }_{i}D(u^i){\partial }_{i}h(x)$. Thus the coordinate derivations form a basis, and the dimension equals the manifold dimension. Composition of germs with a smooth map defines its differential of a smooth map without choosing charts.

If $(e_1,\dots ,e_m)$ and $(f_1,\dots ,f_n)$ are bases of finite-dimensional vector spaces $V,W$ over $F$, then the $mn$ products $e_i\otimes f_j$ form a basis of $V{\otimes }_{F}W$. There is one independent coordinate for every pair of indices.^[ref]

For a proof outline, bilinearity expands each pure tensor asso these products form a spanning set. For linear independence, take the coordinate functionals $e^k,f^{\ell }$ from the two dual bases. The bilinear map $(v,w)\mapsto e^k(v)f^{\ell }(w)$ induces a linear functional ${\lambda }_{k\ell }$ on the tensor product, satisfying ${\lambda }_{k\ell }(e_i\otimes f_j)={\delta }_{ki}{\delta }_{\ell j}$. Applying it to a proposed zero linear combination selects just $c_{k\ell }$, forcing every coefficient to vanish. Thus dimensions multiply, rather than add. This finite-dimensional count includes a zero factor, whose tensor product has dimension zero.

The vector-length multiset of a rank-two Euclidean lattice determines it up to an orthogonal map. Let $a$ be the shortest nonzero length. Removing two copies of each positive integer multiple of $a$ removes one primitive lattice line; the shortest remaining length $b$ is the shortest independent-vector length. Those two vectors form a lattice basis: otherwise a nonzero point in the centred basis parallelogram would be an independent lattice vector of length less than $b$. Lattice-point density determines the covolume $A$. The basis Gram matrix then has diagonal entries $a^2,b^2$ and absolute off-diagonal entry $\sqrt{a^2{b}^2-A^2}$. This proves two-dimensional spectral rigidity for a flat torus through its dual lattice.

The two-factor tensor rank is the smallest number of pure tensors whose sum equals $t$, with zero allowed as the empty sum. For two finite-dimensional vector spaces over the same field, it equals the matrix rank of the tensor-product coefficient matrix. A sum of $r$ outer products has column space contained in the span of their $r$ first factors, so its matrix rank is at most $r$. Conversely, choose a basis $u_1,\dots ,u_r$ for the column space, express each column in that basis, and collect the coefficients into row vectors $w_k^T$. This yields $C={\sum }_{k=1}^r{u}_k{w}_k^T$, hence a sum of $r$ pure tensors.

The pure tensor $(2,1)\otimes (-1,3)$ therefore has two-factor tensor rank one, whereas $e_1\otimes f_1+e_2\otimes f_2$ has two-factor tensor rank two. Both have two factors: the number of factors and the minimum number of product summands are different counts. Over the real numbers, the singular value decomposition supplies a useful decomposition $C={\sum }_k{\sigma }_k{u}_k{v}_k^T$, with one nonzero term per matrix rank. The two-factor rank equality does not extend to three or more factors by merely choosing one matrix layout.

Let nonnegative partition-normalized B-splines $N_i$ form a basis, with ${\sum }_i{N}_i\le 1$ on the integration interval and $d_i=\int N_i>0$. The row-normalized mixed Gram matrix is $G_{ij}=\int N_i{N}_j/d_i$. The normal equations for the orthogonal projection onto their linear span become $Ga=b$, with $b_i=\int f{N}_i/d_i$ and $\Vert b{\Vert }_{{\ell }^{\infty }}\le \Vert f{\Vert }_{\infty }$. The subpartition of unity for B-splines gives $\Vert {\sum }_i{a}_i{N}_i{\Vert }_{\infty }\le \Vert a{\Vert }_{{\ell }^{\infty }}$ and proves the bound. The ordinary symmetric Gram matrix omits this essential row normalization. The projection maps continuous functions into the spline space, not onto all continuous functions.

For a bilinear map $B:V\times W\to U$, there is exactly one linear map $L:V{\otimes }_{F}W\to U$ satisfying $L(v\otimes w)=B(v,w)$. Equivalently, $B=L\circ \tau$, where $\tau (v,w)=v\otimes w$. The universal property of a tensor product concerns the vector space together with this distinguished bilinear map.^[ref]

Here is a complete calculation over the real numbers. Define the bilinear mapIts values on pairs of basis vectors determine $L$ on the four tensor-product basis elements:Consequently, for $t={\sum }_{i,j=1}^2{c}_{ij}{e}_i\otimes f_j$,For $v=(2,1)$ and $w=(-1,3)$, the direct calculation gives $B(v,w)=(-2+12+3,-1-6)=(13,-7)$. Using the expansion of $v\otimes w$ instead gives $L(v\otimes w)=-2(1,0)+6(2,-1)-(0,1)+3(1,0)=(13,-7)$. The linear map also handles a nonseparable tensor such as $t=e_1\otimes f_1+e_2\otimes f_2$: $L(t)=(2,0)$, even though no single pair has $v\otimes w=t$.

A short proof uses a tensor-product basis: assign $L(e_i\otimes f_j)=B(e_i,f_j)$ and extend by linearity. Expanding both inputs verifies $L(v\otimes w)=B(v,w)$ by bilinearity; the assigned values on a basis force uniqueness. More generally, every multilinear map $F:V_1\times \cdots \times V_r\to U$ factors uniquely through a linear map from $V_1\otimes \cdots \otimes V_r$. This lets formulas involving several inputs define linear maps without choosing coordinates; tensor contraction uses the evaluation pairing between a vector and a covector.

For an unramified extension of local fields $L/K$, an integral element whose residue generates $k_L/k_K$ generates $L/K$: its degree is at least the residue degree $f=[L:K]$. The reduced powers $1,\bar{\alpha },\dots ,{\bar{\alpha }}^{f-1}$ are a basis of the residue field extension. If $x=\sum c_j{\alpha }^j$, factor out a coefficient of least valuation. Reduction of the remaining sum is nonzero by that residue basis, so $v_L(x)={\min }_j{v}_K(c_j)$. Consequently $x$ is integral exactly when all $c_j$ are integral.