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A multilinear map is antisymmetric when swapping two arguments negates its value. Over a field of characteristic different from $2$, this is equivalent to being an alternating multilinear map: repeated arguments give $T=-T$, hence $T=0$; conversely expand $T(\dots ,u+v,\dots ,u+v,\dots )=0$ to get the sign change. In characteristic $2$, the repeated-argument condition is stronger than the sign-change condition alone.

Over a field of characteristic zero, an associative Lie product is a bilinear map that is simultaneously an associative algebra product and a Lie bracket. Such a product has every triple product zero. To prove this, write multiplication by juxtaposition. By associativity and antisymmetry of a Lie bracket,since moving $z$ left through $y$ and then through $x$ introduces two minus signs. Applying antisymmetry of a Lie bracket instead to the two elements $xy$ and $z$ gives $(xy)z=-z(xy)$. Thus $xyz=-xyz$, and $A^3=0$ in characteristic zero.

Conversely, an alternating multilinear map defining a product whose triple products all vanish is associative, because both associations give zero, and satisfies the Jacobi identity, because each nested product gives zero. A nonzero example has basis $p,q,z$, products $pq=z=-qp$, and all other products of basis vectors zero. It is the Heisenberg Lie algebra with its Lie bracket used as the associative algebra product. Its image $A^2=kz$ annihilates $A$ on both sides, and it has no identity element.

The determinant assigns a scalar to a square matrix. For a linear map on the real plane, it measures signed area scaling: its absolute value tells how much area grows, and its sign records whether the orientation of a vector space is preserved or reversed. In three dimensions, the same idea measures signed volume. Over a field, a square matrix is an invertible matrix exactly when its determinant is nonzero. These are two views of one phenomenon: losing an independent direction collapses full-dimensional volume and makes the original input impossible to recover uniquely.^[ref]

Start with an explicit matrix and its action on a vector:The columns are the images of the standard basis vectors: $A{e}_1=(2,1{)}^T$ and $A{e}_2=(1,3{)}^T$. Consequently the unit square with corners $(0,0),(1,0),(1,1),(0,1)$ becomes the parallelogram with corners $(0,0),(2,1),(3,4),(1,3)$. Its area is five. Every small patch of this square undergoes the same linear map, so the factor describes the whole deformation, rather than just one specially chosen boundary. The matrix mixes horizontal and vertical components, but its determinant compresses that combined effect into one area factor.^[ref]

Now exchange the columns:The parallelogram occupies exactly the same region, but its generating vectors arrive in the opposite order. Moving from the first column to the second now turns clockwise rather than counterclockwise. Ordinary area remains five; the oriented volume in two dimensions changes sign. A negative determinant therefore signals reversed orientation of a vector space, not a negative amount of material. Keeping the sign makes successive reversals calculable: two reversals restore the original orientation of a vector space.

Compare a singular matrix:Its columns are linearly dependent: the second is twice the first. The unit square collapses onto the segment from $(0,0)$ to $(3,6)$, which has zero area despite nonzero length. Since $C(-2,1{)}^T=0$, the kernel of a linear map contains a nonzero vector; inputs differing by that vector have identical outputs. The collapsed picture explains why an inverse matrix cannot exist.

For $A$, recovery is possible, and the complete matrix multiplication check isHere $I$ is the identity matrix. The inverse matrix restores the original square, shrinking its area by $1/5$. Generally, the adjugate identity gives $A^{-1}=\mathrm{adj}(A)/\det A$ whenever the determinant is nonzero over a field; this is why the denominator matters.^[ref]

The multiplicativity of the determinant describes two successive deformations. Matrix multiplication means that $AB$ applies $B$ first, then $A$. TakeThe first linear map doubles area; the second multiplies the resulting area by five. Thus the unit square passes through areas $1,2,10$, with the last parallelogram bounded by $(0,0),(2,1),(4,7),(2,6)$. Direct calculation agrees:This explains physically why area factors multiply instead of adding. The general identity holds for square matrices of the same size, including singular matrices. If either stage loses a direction, later processing cannot recover the discarded information.^[ref]

The same interpretation works in three dimensions without drawing a new picture. The diagonal matrixsends the unit cube to a rectangular box with side lengths two, three, and one. Its ordinary volume is six. The minus sign comes from reversing the third direction, so the oriented volume is minus six. Replacing the last diagonal entry by zero flattens the cube into a planar region: its area may be positive, but its three-dimensional volume is zero. This distinction matters whenever a linear map changes shape: the determinant measures size in the full input dimension, rather than the size of whichever lower-dimensional object remains.

The determinant measures neither trace nor perimeter. For $A$, the trace happens also to be five, but the identity matrix in two dimensions has trace two and determinant one. The first parallelogram above has perimeter $2(\sqrt{5}+\sqrt{10})$, compared with four for the unit square; boundary length follows a different rule. Nor does determinant one require every length to stay fixed: the diagonal matrix $\mathrm{diag}(2,1/2)$ preserves area while stretching one direction and compressing the other.

These properties make the determinant useful when coordinates obscure what a deformation does. A nonzero determinant certifies that linearly independent input directions remain distinguishable; its magnitude counts the area or volume represented by a transformed cell. That second role reappears in the change of variables formula, where the Jacobian determinant corrects the sizes of small cells. Signed volume also separates a deformation that reverses the orientation of a vector space from one that merely stretches it, even when both give identical ordinary volume factors.^[ref]

For general $n$, regard the columns $v_1,\dots ,v_n$ as ordered arguments. The determinant is the unique alternating multilinear map to the scalar field normalized by $\det I=1$. The multilinearity of the determinant means linearity in each column separately, with all others held fixed:Alternation means equal columns give zero; exchanging columns negates the result. Normalization sets the oriented volume of the standard basis to one. These properties also hold in rows. For real three-dimensional columns, their determinant is the scalar triple product, and its absolute value is the enclosed volume. Over other fields, the algebraic definition still works even without a geometric notion of positive volume.^[ref]

Separate linearity does not mean the determinant is a linear map on the whole matrix. Scaling every column of an $n$-by-$n$ matrix by $t$ gives $\det (tA)=t^n\det A$. For our two-dimensional example, $\det (2A)=4\det A=20$: doubling both directions quadruples area. The power records how many independent directions are being scaled together.

The properties suggest practical computations. Adding a multiple of one column to another preserves the determinant: the extra term has repeated columns and vanishes. This is the algebra behind an area-preserving shear mapping. Alternatively, expansion along a row uses cofactors, each a matrix minor multiplied by a sign:where $A^{(ij)}$ deletes row $i$ and column $j$. The Leibniz formula for determinants supplies the general entry formula. Both calculations encode the same geometric invariant; the defining properties explain their structure without requiring memorization of a separate recipe for every size.^[ref]

A determinantal function on $M_n(F)$ is an alternating multilinear map of the $n$ columns, normalized to take value one at the identity matrix. It is uniquely the determinant. Indeed, expand each column in the standard basis. Terms with a repeated basis vector vanish, and all remaining terms differ from the value on the standard basis only by the sign of a permutation. Thus an alternating multilinear map of $n$ columns is determined by its value on the identity matrix.

A differential form is a smooth field of alternating multilinear maps that can be integrated over an appropriately oriented manifold of matching dimension. A one-form measures a directed displacement; a two-form measures an oriented area element. The point is to express what is being accumulated along a curve or over a surface without tying the answer to one set of coordinates. This makes line integrals, surface integrals, and the integral theorems of vector calculus parts of one construction.

For a concrete one-form, take $\alpha =xdy$ in the plane. Travel from $(0,0)$ to $(1,1)$ along the bottom and then the right edge of the unit square. The horizontal leg contributes zero because $dy=0$; the vertical leg has $x=1$ and contributes ${\int }_0^{1}dy=1$. Taking the left and then the top edge instead gives zero: the vertical leg has $x=0$, and the horizontal leg again has $dy=0$. The two paths have the same endpoints but different integrals.

Now replace $\alpha$ with the exact differential $\beta =d(xy)=ydx+xdy$. Both paths give $1$, because integrating an exact differential along a curve gives the endpoint difference of its potential. Here that difference is $xy{|}_{(1,1)}-xy{|}_{(0,0)}=1$. This is the fundamental theorem of calculus applied to a function along a parametric curve, and is the basic reason conservative vector fields have path-independent work.

The exterior derivative records the local obstruction in the first example: $d(xdy)=dx\wedge dy$. The Generalized Stokes theorem turns the difference between the two path integrals into the integral of this two-form over the square, whose oriented area is $1$. Interior edges cancel when adjoining pieces are added; only the outer boundary survives. That cancellation connects a local calculation with a global measurement.

More precisely, a degree-$k$ differential form assigns to each point $p$ a multilinear scalar-valued function of $k$ tangent vectors at $p$. Interchanging two arguments changes its sign. A one-form is therefore a covector field, while a two-form vanishes on a pair of parallel vectors. Degree zero simply means a smooth function. The space ${\Omega }^k(M)$ is a vector space; the wedge product of differential forms combines degrees, and the exterior derivative raises degree by one.

The usefulness goes beyond bookkeeping. Pullbacks of a differential form supply the right transformation law automatically under a coordinate change. The Generalized Stokes theorem unifies the fundamental theorem of calculus, Green theorem, Stokes theorem, and divergence theorem. Meanwhile, the distinction between closed differential forms and exact differential forms detects holes through de Rham cohomology. Reyer Sjamaar develops these connections in sections 5.4 and 9.3 of his university notes. ^[ref].

Consider lines $r=a+sm$ and $r=b+tn$ with nonzero direction vectors. If they intersect, then $a-b=tn-sm$, so the necessary coplanarity condition is $[a-b,m,n]=0$. For nonparallel directions it is also sufficient, since the orthogonal complement of $m\times n$ is the linear span of $m,n$. For parallel directions the condition alone is insufficient: distinct parallel lines also satisfy it.

Assume intersection and $D=[a,m,n]\ne 0$. Take the scalar triple products $[a,b,n]$ and $[a,b,m]$ of $b=a+sm-tn$. The scalar triple product is an alternating multilinear map, giving respectively $sD$ and $tD$, proving the title formulas. The nonzero denominator fixes a convenient determinant chart; intersections with $D=0$ can still exist and require a different auxiliary vector or another method.

For a Lie algebra $\mathfrak{g}$ over a field $k$ and a Lie algebra representation $V$, set $C^n(\mathfrak{g},V)={\mathrm{Hom}}_k({\Lambda }^n\mathfrak{g},V)$. These alternating multilinear maps form a cochain complex with differentialThe hat means that the argument is omitted. The Jacobi identity and the Lie algebra representation identity give $d^2=0$. The Lie algebra cohomology is the quotient vector space $H^n(\mathfrak{g},V)=\ker (d:C^n\to C^{n+1})/\mathrm{im}(d:C^{n-1}\to C^n)$. In degree zero it consists of the invariant vectors of a Lie algebra representation; with a trivial Lie algebra representation as coefficients, $H^1(\mathfrak{g},k)={\mathrm{Hom}}_k(\mathfrak{g}/[\mathfrak{g},\mathfrak{g}],k)$.^[ref]

For trivial real coefficients, $H^2(\mathfrak{g},\mathbb{R})$ classifies central extensions of a Lie algebra by a fixed one-dimensional center, up to equivalences inducing the identity on the quotient and center. If $\mathfrak{g}$ is abelian and two-dimensional, every alternating bilinear form is a Lie algebra two-cocycle and every coboundary is zero; a nonzero such form produces the Heisenberg Lie algebra.

Since the general linear Lie algebra ${\mathfrak{gl}}_1(k)$ has zero Lie bracket, identify it with $kz$. A central extension of a Lie algebra is a short exact sequenceof Lie algebra homomorphisms with $\iota (kz)$ contained in the center of a Lie algebra $Z(\tilde{\mathfrak{g}})$. Thus $\pi$ is surjective, its kernel is precisely the indicated line, and that line commutes with all of $\tilde{\mathfrak{g}}$.

Choose a linear map $s:\mathfrak{g}\to \tilde{\mathfrak{g}}$ with $\pi s=\mathrm{id}$. Such a section exists by lifting a basis. Identify $\tilde{\mathfrak{g}}$ with the direct sum of vector spaces $\mathfrak{g}\oplus kz$ by $(x,tz)\mapsto s(x)+t\iota (z)$. The difference $[s(x),s(y)]-s([x,y])$ belongs to the kernel of $\pi$, so there is a unique bilinear map $\alpha :\mathfrak{g}\times \mathfrak{g}\to k$ such thatIt is an alternating bilinear form: $\alpha (x,x)=0$, equivalently $\alpha (x,y)=-\alpha (y,x)$ in characteristic zero. Taking the central component of the Jacobi identity gives the Lie algebra two-cocycle conditionConversely, any alternating bilinear form satisfying this identity defines a Lie algebra byIndeed its Lie bracket is an alternating multilinear map and a bilinear map, and the two components of its Jacobi identity are respectively the Jacobi identity in $\mathfrak{g}$ and the displayed Lie algebra two-cocycle condition. The line $kz$ is central, and projection to $\mathfrak{g}$ gives the required central extension of a Lie algebra.

For later use, replacing $s$ by $s^{\prime }(x)=s(x)+t(x)\iota (z)$, where $t\in {\mathfrak{g}}^{\ast }$ is in the dual space, changes the Lie algebra two-cocycle toThus changes of section add a Lie algebra two-coboundary, up to the conventional sign. A section that is a Lie algebra homomorphism has ${\alpha }^{\prime }=0$ and gives a direct sum of Lie algebras.

Let $A$ be a three-dimensional vector space with basis $p,q,z$. Define a bilinear map $\bullet :A\times A\to A$ bywith all other products of basis vectors zero. Equivalently,Every product belongs to $kz$, and every product with an element of $kz$ is zero. Hence both $(u\bullet v)\bullet w$ and $u\bullet (v\bullet w)$ vanish for every $u,v,w\in A$. This proves associativity, so $A$ is an associative algebra. The product is an alternating multilinear map, and each nested product in the Jacobi identity is zero, so the same product is a Lie bracket. It is nonzero because $p\bullet q=z\ne 0$.

There is no multiplicative identity element: any candidate $1_A$ would satisfy $1_A\bullet z=0$, whereas an identity element would require $1_A\bullet z=z$. ThusThis is an associative Lie product; its underlying Lie algebra is the Heisenberg Lie algebra.

The determinant is an alternating multilinear map of its rows, so exchanging two rows gives the negative of the original determinant: $\det A\mapsto -\det A$. This also holds when the determinant is zero.

Interchanging two rows reverses the sign of an alternating multilinear map, soIn the Leibniz formula for determinants, the same result follows by composing every permutation with the corresponding transposition, which reverses its sign.

Write $B=[b_1,\dots ,b_n]$. Since $AB=[A{b}_1,\dots ,A{b}_n]$ and $b\mapsto Ab$ is a linear map, the multilinearity of the determinant gives, for any column position $j$,If $b_i=b_j$ for $i\ne j$, the corresponding columns $A{b}_i,A{b}_j$ also agree, so $D(B)=0$. Thus $D$ is an alternating multilinear map of the columns of $B$, even when $A$ is not invertible.

We have $D(I_n)=\det A$. Define a function of the columns of $B$ byIt is an alternating multilinear map, because both summands are, and $F(I_n)=\det A+1-\det A=1$. Hence $F$ is a normalized determinantal function. By the asserted uniqueness of the determinant, $F(B)=\det B$ for every $B$. Rearranging yieldsThis argument also covers $\det A=0$ without dividing by it.