Exterior Powers, Orientation and Hodge Duality

Let $F$ be a field and let $V,W$ be $F$-vector spaces (Vector space over a field). For $k\ge 0$, a map

$f:V^k⟶W$

is $k$-linear (multilinear) when it is linear in each of its $k$ arguments separately, with the other arguments held fixed. It is alternating when

$f(v_1,\dots ,v_k)=0\text{whenever }{v}_i=v_j\text{ for some }1\le i<j\le k\mathrm{.}$

For $k=0$, a $0$-linear map is a choice of one element of $W$; for $k=1$, a $1$-linear map is a linear map. Both are alternating vacuously, because there is no pair of arguments to compare.

When $2\mathrm{\ne }0$ in $F$, alternation is equivalent to the sign rule $f(\dots ,v_i,\dots ,v_j,\dots )=-f(\dots ,v_j,\dots ,v_i,\dots )$: expand $f(\dots ,v_i+v_j,\dots ,v_i+v_j,\dots )=0$ and use $2\mathrm{\ne }0$. In characteristic two the sign rule collapses to the tautology $x=-x$ and does not imply alternation; the published witness is Over $\mathbb{Z}\mathrm{/}2$, an antisymmetric bilinear form need not be alternating. This page therefore works directly with the repeated-argument condition, never with a division by $k!$.

Let $V$ be a vector space over a field $F$ and let $k\ge 0$. The $k$-fold tensor power is $V^{\otimes 0}=F$, $V^{\otimes 1}=V$, and for $k\ge 2$ the iterated tensor product of $k$ copies of $V$ under any parenthesization. By Finite iterated tensor products represent multilinear maps independently of parenthesization every parenthesization represents the $k$-linear maps out of $V^k$, and different parenthesizations are joined by the unique isomorphism preserving pure tensors, so $V^{\otimes k}$ is determined up to that isomorphism.

Let $W_k\subseteq V^{\otimes k}$ be the subspace spanned by all pure tensors $v_1\otimes \cdots \otimes v_k$ for which $v_i=v_j$ for some pair $1\le i<j\le k$ (so $W_0=W_1=0$). The $k$th exterior power of $V$ is the quotient vector space

${\mathrm{\Lambda }}^{k}V:=V^{\otimes k}\mathrm{/}{W}_k,$

and the basic wedge map is

$\wedge :V^k⟶{\mathrm{\Lambda }}^{k}V,(v_1,\dots ,v_k)⟼v_1\otimes \cdots \otimes v_k+W_k\mathrm{.}$

The construction never divides by $k!$, so ${\mathrm{\Lambda }}^{k}V$ is defined in every characteristic. Its relation to the Alternating $k$-linear maps maps is the universal property of Exterior powers represent alternating multilinear maps and are unique up to unique isomorphism; the convention fixed here is that a repeated entry makes the wedge zero.

For $v_1,\dots ,v_k\in V$, the image of the basic wedge map of The $k$th exterior power as the tensor-power quotient by repeated-vector relations is written

$v_1\wedge \cdots \wedge v_k:=v_1\otimes \cdots \otimes v_k+W_k\in {\mathrm{\Lambda }}^{k}V,$

and is called the wedge (or exterior product) of the list. A $k$-vector $\alpha \in {\mathrm{\Lambda }}^{k}V$ is decomposable (or pure) when it equals $v_1\wedge \cdots \wedge v_k$ for some list; a general element of ${\mathrm{\Lambda }}^{k}V$ is a finite sum of decomposables, because the pure tensors span $V^{\otimes k}$ and their classes span the quotient.

For $k=0$ the empty wedge is $1\in F={\mathrm{\Lambda }}^{0}V$; for $k=1$ the wedge of a single vector is the vector itself under the identification ${\mathrm{\Lambda }}^{1}V=V$.

The wedge of $v_1,\dots ,v_k$ depends on the order of the list only up to a sign (see Exterior multiplication is well defined, graded, associative, unital, and graded-commutative); the notation $v_1\wedge \cdots \wedge v_k$ records the order.

The exterior algebra of $V$ is the graded vector space

$\mathrm{\Lambda }V:={⨁}_{k\ge 0}{\mathrm{\Lambda }}^{k}V={\mathrm{\Lambda }}^{0}V\oplus {\mathrm{\Lambda }}^{1}V\oplus {\mathrm{\Lambda }}^{2}V\oplus \cdots ,$

whose homogeneous piece of degree $k$ is the exterior power of The $k$th exterior power as the tensor-power quotient by repeated-vector relations. On decomposables of Decomposable $k$-vectors and the basic wedge product define the wedge product

$(v_1\wedge \cdots \wedge v_k)\wedge (w_1\wedge \cdots \wedge w_{\mathrm{\ell }}):=v_1\wedge \cdots \wedge v_k\wedge w_1\wedge \cdots \wedge w_{\mathrm{\ell }}\in {\mathrm{\Lambda }}^{k+\mathrm{\ell }}V,$

and extend bilinearly to all of ${\mathrm{\Lambda }}^{k}V\times {\mathrm{\Lambda }}^{\mathrm{\ell }}V$. The product of two homogeneous elements is homogeneous of the sum of the degrees, so $\mathrm{\Lambda }V$ is a graded algebra with unit $1\in F={\mathrm{\Lambda }}^{0}V$.

The product is well defined. Each generator $v_1\otimes \cdots \otimes v_k$ of $W_k$ contains a repeated pair, so its concatenation with any pure tensor of degree $\mathrm{\ell }$ lies in $W_{k+\mathrm{\ell }}$. By linearity, concatenation sends $W_k\times V^{\otimes \mathrm{\ell }}$ and $V^{\otimes k}\times W_{\mathrm{\ell }}$ into $W_{k+\mathrm{\ell }}$. It therefore descends to a bilinear map on the two quotient spaces, and that descended map is exactly the displayed wedge product. Associativity and the graded commutation law are proved in Exterior multiplication is well defined, graded, associative, unital, and graded-commutative.

Let $T:V\to W$ be linear. Iterating the induced map of Module homomorphisms induce tensor-product homomorphisms functorially gives the $k$-fold tensor power

$T^{\otimes k}:V^{\otimes k}⟶W^{\otimes k},T^{\otimes k}(v_1\otimes \cdots \otimes v_k)=T(v_1)\otimes \cdots \otimes T(v_k),$

with $T^{\otimes 0}={\mathrm{id}}_F$ and $T^{\otimes 1}=T$. A pure tensor with $v_i=v_j$ maps to the pure tensor with $T(v_i)=T(v_j)$, so $T^{\otimes k}(W_k(V))\subseteq W_k(W)$ for the relation subspaces of The $k$th exterior power as the tensor-power quotient by repeated-vector relations. Hence $T^{\otimes k}$ descends to the quotient, defining the $k$th exterior power

${\mathrm{\Lambda }}^{k}T:{\mathrm{\Lambda }}^{k}V⟶{\mathrm{\Lambda }}^{k}W,{\mathrm{\Lambda }}^{k}T(v_1\wedge \cdots \wedge v_k)=T(v_1)\wedge \cdots \wedge T(v_k)\mathrm{.}$

For $k=0$ this is ${\mathrm{\Lambda }}^{0}T={\mathrm{id}}_F$, and for $k=1$ it is $T$ itself under ${\mathrm{\Lambda }}^{1}V=V$.

Let $V$ be a finite-dimensional real vector space (Finite-dimensional vector space, and its dimension ${\dim }_{F}V$; infinite-dimensional means having no finite basis). Two ordered bases $\mathcal{B},{\mathcal{B}}^{\mathrm{\prime }}$ of $V$ (Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis) are declared equivalent when the determinant of the unique change-of-basis linear isomorphism $P:V\to V$ carrying $\mathcal{B}$ to ${\mathcal{B}}^{\mathrm{\prime }}$ satisfies $\det P>0$, where the determinant is The determinant of an endomorphism of a finite-dimensional vector space: its matrix determinant in an ordered basis in positive dimension, and $1$ on the zero space and positivity is the order of the real field The reals form a totally ordered field.

This is an equivalence relation. The identity change of basis has determinant $1>0$. If $\mathcal{B}\sim {\mathcal{B}}^{\mathrm{\prime }}$ through $P$, then ${\mathcal{B}}^{\mathrm{\prime }}\sim \mathcal{B}$ through $P^{-1}$, and $\det (P^{-1})=(\det P{)}^{-1}>0$. If $\mathcal{B}\sim {\mathcal{B}}^{\mathrm{\prime }}\sim {\mathcal{B}}^{\mathrm{\prime }\mathrm{\prime }}$ through $P,Q$, then $\det (Q\circ P)=\det Q\det P>0$ by For same-sized finite square matrices over a commutative ring, $\det (AB)=\det (A)\det (B)$.

An orientation of $V$ is an equivalence class of ordered bases under this relation; a basis in the chosen class is positively oriented for that orientation. When $\dim V\ge 1$ there are exactly two orientations: fixing one ordered basis ${\mathcal{B}}_0$, every other basis $\mathcal{B}$ has $\det P>0$ or $\det P<0$ by the trichotomy of The reals form a totally ordered field. If $\dim V=1$, replacing the sole basis vector of ${\mathcal{B}}_0$ by its negative produces a basis with determinant $-1$; if $\dim V\ge 2$, interchanging two entries of ${\mathcal{B}}_0$ does the same. Thus both classes occur. When $\dim V=0$ the only ordered basis is the empty one and there is exactly one orientation.

Orientations depend only on the real vector-space structure; no inner product or basis preference enters the definition.

Let $V$ be a vector space over a field $F$ and $k\ge 0$. The basic wedge map $\wedge :V^k\to {\mathrm{\Lambda }}^{k}V$, $(v_1,\dots ,v_k)\mapsto v_1\wedge \cdots \wedge v_k$, is $k$-linear and alternating. For every $F$-vector space $W$ and every alternating $k$-linear map $f:V^k\to W$ of Alternating $k$-linear maps, there is a unique linear map $\bar{f}:{\mathrm{\Lambda }}^{k}V\to W$ with

$\bar{f}(v_1\wedge \cdots \wedge v_k)=f(v_1,\dots ,v_k)\mathrm{.}$

Moreover, if $U$ is a vector space and ${\wedge }^{\mathrm{\prime }}:V^k\to U$ is a $k$-linear alternating map with the same property, then there is a unique linear isomorphism $u:{\mathrm{\Lambda }}^{k}V\to U$ with $u\circ \wedge ={\wedge }^{\mathrm{\prime }}$.

For every $k\ge 0$, the map

$(v_1,\dots ,v_k)⟼v_1\wedge \cdots \wedge v_k$

from $V^k$ to ${\mathrm{\Lambda }}^{k}V$ is $k$-linear and alternating, in the sense of Alternating $k$-linear maps.

Let $V$ have the ordered basis $(e_1,\dots ,e_n)$ over $F$ and let $0\le k\le n$. For each $k$-element subset $I=\{i_1<\cdots <i_k\}\subseteq \{1,\dots ,n\}$ write

$e_I:=e_{i_1}\wedge \cdots \wedge e_{i_k}\mathrm{.}$

Then the family $(e_I{)}_I$ indexed by the $k$-element subsets of $\{1,\dots ,n\}$ is a basis of ${\mathrm{\Lambda }}^{k}V$.

If $V$ is finite-dimensional over $F$ with $\dim V=n$, then for every $0\le k\le n$,

$\dim {\mathrm{\Lambda }}^{k}V=\left(\genfrac{}{}{0px}{}{n}{k}\right)\mathrm{.}$

If $V$ is finite-dimensional with $\dim V=n$ and $k>n$, then ${\mathrm{\Lambda }}^{k}V=0$.

Let $V$ be a finite-dimensional vector space over $F$. For $v_1,\dots ,v_k\in V$,

$v_1\wedge \cdots \wedge v_k\mathrm{\ne }0⟺v_1,\dots ,v_k\text{ are linearly independent}\mathrm{.}$

The wedge product of The graded exterior algebra $\mathrm{\Lambda }V$ is well defined and bilinear in each homogeneous pair of degrees. It is associative, has unit $1\in F={\mathrm{\Lambda }}^{0}V$, is graded in the sense that ${\mathrm{\Lambda }}^{k}V\wedge {\mathrm{\Lambda }}^{\mathrm{\ell }}V\subseteq {\mathrm{\Lambda }}^{k+\mathrm{\ell }}V$, and is graded-commutative:

$\alpha \wedge \beta =(-1{)}^{k\mathrm{\ell }}\beta \wedge \alpha (\alpha \in {\mathrm{\Lambda }}^{k}V,\beta \in {\mathrm{\Lambda }}^{\mathrm{\ell }}V)\mathrm{.}$

All of this holds over every field, including characteristic two.

For linear maps $T:V\to W$ and $S:W\to U$ and every $k\ge 0$, the induced maps of The induced map ${\mathrm{\Lambda }}^{k}T$ on exterior powers satisfy

${\mathrm{\Lambda }}^k({\mathrm{id}}_V)={\mathrm{id}}_{{\mathrm{\Lambda }}^{k}V},{\mathrm{\Lambda }}^k(S\circ T)={\mathrm{\Lambda }}^{k}S\circ {\mathrm{\Lambda }}^{k}T\mathrm{.}$

Thus $V\mapsto {\mathrm{\Lambda }}^{k}V$, $T\mapsto {\mathrm{\Lambda }}^{k}T$ is a functor from $F$-vector spaces to $F$-vector spaces, for each fixed $k$.

Let $V,W$ be finite-dimensional with ordered bases $(e_1,\dots ,e_n)$ and $(f_1,\dots ,f_m)$, let $T:V\to W$ be linear with matrix $A=(a_{ij})$, so $T(e_j)={\sum }_i{a}_{ij}{f}_i$, and let $1\le k\le \min (n,m)$. In the wedge bases $(e_I)$ and $(f_J)$ of Increasing-index wedges of a basis form a basis of ${\mathrm{\Lambda }}^{k}V$, the matrix of ${\mathrm{\Lambda }}^{k}T$ has entries

$[{\mathrm{\Lambda }}^{k}T{]}_{J,I}=\det A_{J,I},$

where $A_{J,I}$ is the $k\times k$ submatrix of $A$ with rows $J=\{j_1<\cdots <j_k\}$ and columns $I=\{i_1<\cdots <i_k\}$.

Let $T:V\to V$ be an endomorphism of a finite-dimensional vector space of dimension $n\ge 1$. Then ${\mathrm{\Lambda }}^{n}V$ is one-dimensional with basis $e_1\wedge \cdots \wedge e_n$ for any ordered basis $(e_1,\dots ,e_n)$, and

${\mathrm{\Lambda }}^{n}T=\det (T)\cdot {\mathrm{id}}_{{\mathrm{\Lambda }}^{n}V}\mathrm{.}$

For $n=0$, ${\mathrm{\Lambda }}^{0}T={\mathrm{id}}_F=\det (T)\cdot {\mathrm{id}}_F$ by convention.

For endomorphisms $S,T$ of an $n$-dimensional vector space $V$ with $n\ge 1$,

$\det (S\circ T)=\det (S)\det (T)\mathrm{.}$

Let $T:V\to V$ be a linear isomorphism of an oriented real vector space with $\dim V=n\ge 1$. If $\det T>0$, then $T$ carries every positively oriented basis to a positively oriented basis; if $\det T<0$, then $T$ carries every positively oriented basis to a negatively oriented one. Thus the sign of $\det T$ is exactly whether $T$ preserves or reverses the orientation.

Let $V$ be a finite-dimensional real inner product space (Real and complex inner product spaces, with the inner product linear in the first argument) and $k\ge 0$. On decomposable wedges of The $k$th exterior power as the tensor-power quotient by repeated-vector relations put

$⟨v_1\wedge \cdots \wedge v_k,w_1\wedge \cdots \wedge w_k⟩:=\det (⟨v_i,w_j⟩{)}_{i,j\le k},$

the determinant (For $n\ge 1$, the determinant over a commutative ring by the Leibniz formula, and $\mathrm{\mid }\det A\mathrm{\mid }$ for a real matrix) of the $k\times k$ matrix of pairings. For $k=0$ both sides are the empty determinant, which is $1$.

For each fixed list $(w_1,\dots ,w_k)$, the assignment $(v_1,\dots ,v_k)\mapsto \det (⟨v_i,w_j⟩)$ is $k$-linear (each entry is linear in the corresponding $v_i$, and the determinant is multilinear in its rows) and alternating (if $v_i=v_{i^{\mathrm{\prime }}}$, two rows of the matrix are equal, so the determinant vanishes). It therefore descends to a linear functional on ${\mathrm{\Lambda }}^{k}V$ in the first slot, and symmetrically in the second slot. The resulting bilinear pairing is the Gram inner product on ${\mathrm{\Lambda }}^{k}V$; that it is an inner product in the sense of Real and complex inner product spaces, with the inner product linear in the first argument is proved in The Gram formula gives a well-defined positive-definite inner product on exterior powers, and $\mathrm{\parallel }{v}_1\wedge \cdots \wedge v_k{\mathrm{\parallel }}^2$ is the Gram determinant ↗, which also discharges the well-definedness obligation recorded above.