Exterior Powers, Orientation and Hodge Duality: Examples and Counterexamples

By [L1], the listed families are exactly the increasing-index wedges of the standard bases, so they are bases of the corresponding exterior powers.

By [L2], the dimensions are the binomial counts: $1,2,1$ for ${\mathbb{R}}^2$; $1,3,3,1$ for ${\mathbb{R}}^3$; and $1,4,6,4,1$ for ${\mathbb{R}}^4$, matching the sizes of the listed families.

By [L3], every degree above the top vanishes, so no further nonzero pieces appear.

Steps 1.1, 1.2 and 1.3 verify all displayed bases, dimensions, and vanishing statements.

The pair $(e_1,2e_1)$ is dependent and its wedge is $e_1\wedge (2e_1)=2(e_1\wedge e_1)=0$, the zero case of [L1].

The pair $(e_1,e_2)$ is independent and its wedge $e_1\wedge e_2$ is a basis vector of ${\mathrm{\Lambda }}^2{\mathbb{R}}^3$, hence nonzero, the nonzero case of [L1].

For $u=(1,2,0)$ and $v=(2,4,0)$, the relation $v=2u$ gives $u\wedge v=2(u\wedge u)=0$, matching the dependence.

For $u=(1,0,0)$ and $v=(0,1,0)$, one has $u\wedge v=e_1\wedge e_2$, which step 1.2 shows is nonzero, matching the independence.

In all four samples the wedge is zero exactly for the dependent pairs, in agreement with [L1].

The matrix $A$ of $T$ in the standard basis is diagonal.

By [L2], the entry of ${\mathrm{\Lambda }}^{k}T$ at $(J,I)$ is the determinant of the $k\times k$ submatrix with rows $J$ and columns $I$; for the diagonal matrix of step 1.1 this is $0$ when $J\mathrm{\ne }I$ and ${\lambda }_I$ when $J=I$.

By [L1], the values of ${\mathrm{\Lambda }}^{k}T$ on the wedge basis $e_I$ determine the whole map, so ${\mathrm{\Lambda }}^{k}T(e_I)={\lambda }_I{e}_I$ and the map is diagonal with the displayed entries; the $n=3$ case reads $\mathrm{diag}(6,10,15)$ on ${\mathrm{\Lambda }}^2{\mathbb{R}}^3$.

Steps 2.1 and 3.1 verify the diagonal form of every exterior power and the concrete example.

Direct computation of the images of the three basis wedges gives ${\mathrm{\Lambda }}^{2}T(e_1\wedge e_2)=(e_1+2e_3)\wedge (2e_1+e_2)=e_1\wedge e_2-4e_1\wedge e_3-2e_2\wedge e_3$, ${\mathrm{\Lambda }}^{2}T(e_1\wedge e_3)=(e_1+2e_3)\wedge (3e_2+e_3)=3e_1\wedge e_2+e_1\wedge e_3-6e_2\wedge e_3$, and ${\mathrm{\Lambda }}^{2}T(e_2\wedge e_3)=(2e_1+e_2)\wedge (3e_2+e_3)=6e_1\wedge e_2+2e_1\wedge e_3+e_2\wedge e_3$, so the columns of the matrix are $(1,-4,-2)$, $(3,1,-6)$, $(6,2,1)$.

By [L1], the expected entries are the nine $2\times 2$ minors: $\det A_{12,12}=1$, $\det A_{13,12}=-4$, $\det A_{23,12}=-2$, $\det A_{12,13}=3$, $\det A_{13,13}=1$, $\det A_{23,13}=-6$, $\det A_{12,23}=6$, $\det A_{13,23}=2$, $\det A_{23,23}=1$.

The nine coordinates of step 1.1 coincide entry by entry with the nine minors of step 1.2, so the displayed matrix equals the signed minor matrix.

By [L1], $\mathrm{\parallel }u\wedge v{\mathrm{\parallel }}^2=\det \left(\begin{array}{cc}⟨u,u⟩& ⟨u,v⟩\\ ⟨v,u⟩& ⟨v,v⟩\end{array}\right)=\det \left(\begin{array}{cc}5& 5\\ 5& 10\end{array}\right)=25$, so the parallelogram area is $\sqrt{25}=5$.

By [L1], $\mathrm{\parallel }u\wedge v\wedge w{\mathrm{\parallel }}^2=\det \left(\begin{array}{ccc}2& 0& 1\\ 0& 4& 2\\ 1& 2& 2\end{array}\right)=2(8-4)+(0-4)=4$, so the parallelotope volume is $\sqrt{4}=2$.

By [L2], the unit volume forms have norm $1$, so the norms computed in steps 1.1 and 1.2 are precisely the unoriented area and volume; the orientation data would only attach a sign, not a size.

Steps 1.1, 1.2 and 2.1 recover area and volume from wedges and Gram determinants.

In ${\mathbb{R}}^2$, the complementary basis and permutation signs of [L1] give $\star 1=e_1\wedge e_2$, $\star e_1=e_2$, $\star e_2=-e_1$, and $\star (e_1\wedge e_2)=1$.

In ${\mathbb{R}}^3$, the same formula gives the displayed list: for example $\star e_2=e_3\wedge e_1$ because the permutation $(2,3,1)\mapsto (1,2,3)$ has sign $+1$, and $\star (e_1\wedge e_2)=e_3$ because $(1,2,3)$ is already ordered.

In ${\mathbb{R}}^4$, the formula gives the displayed values, e.g. $\star e_2=-e_1\wedge e_3\wedge e_4$ because $(2,1,3,4)$ has sign $-1$, and $\star (e_1\wedge e_3)=-e_2\wedge e_4$ because $(1,3,2,4)$ has sign $-1$.

Square checks with [L2]: in ${\mathbb{R}}^2$, $\star \star e_1=\star e_2=-e_1=(-1{)}^{1\cdot 1}{e}_1$; in ${\mathbb{R}}^3$, $\star \star e_1=\star (e_2\wedge e_3)=e_1=(-1{)}^{1\cdot 2}{e}_1$; in ${\mathbb{R}}^4$, $\star \star (e_1\wedge e_2)=\star (e_3\wedge e_4)=e_1\wedge e_2=(-1{)}^{2\cdot 2}(e_1\wedge e_2)$.

In each dimension, $\star$ permutes the orthonormal wedge basis up to sign, so it preserves the Gram norm, as [L2] states.

Steps 1.1 through 2.2 verify the displayed stars, the square signs, and the isometry in all three dimensions.

Expanding the wedge gives $u\wedge v=(e_1+2e_3)\wedge (3e_2+e_3)=3e_1\wedge e_2+e_1\wedge e_3+6e_3\wedge e_2=3e_1\wedge e_2+e_1\wedge e_3-6e_2\wedge e_3$.

The coordinate cross product is $u\times v=(u_2{v}_3-u_3{v}_2,u_3{v}_1-u_1{v}_3,u_1{v}_2-u_2{v}_1)=(-6,-1,3)$.

Applying the star to step 1.1, with $\star (e_1\wedge e_2)=e_3$, $\star (e_1\wedge e_3)=-e_2$, $\star (e_2\wedge e_3)=e_1$, gives $\star (u\wedge v)=3e_3-e_2-6e_1=(-6,-1,3)$.

Steps 2.1 and 1.2 produce the same vector, in agreement with the identity of [L1].

The change of basis from $(e_1,e_2,e_3)$ to $(e_2,e_1,e_3)$ is a transposition with determinant $-1$, so by [L1] the two bases represent the two opposite orientation classes.

By [L4], the unit volume forms are ${\omega }_{+}=e_1\wedge e_2\wedge e_3$ and ${\omega }_{-}=e_2\wedge e_1\wedge e_3=-{\omega }_{+}$.

By the characterizing relation of [L2] and its uniqueness clause, $\alpha \wedge {\star }_{-}\beta =⟨\alpha ,\beta ⟩{\omega }_{-}=-⟨\alpha ,\beta ⟩{\omega }_{+}=\alpha \wedge (-{\star }_{+}\beta )$ for all $\alpha$, so ${\star }_{-}=-{\star }_{+}$ on every degree.

In particular ${\star }_{+}(e_1\wedge e_2)=e_3$ by [L2], and step 2.1 gives ${\star }_{-}(e_1\wedge e_2)=-e_3$; the metric used in both computations is the same standard inner product, and [L3] records that each star is an isometry for it.

Steps 1.1 through 3.1 show that reversing the orientation negates the Hodge star while the metric stays fixed.

If $\alpha =u\wedge v$ were decomposable, then by [L2], $\alpha \wedge \alpha =u\wedge v\wedge u\wedge v=-u\wedge u\wedge v\wedge v=0$.

For the displayed $\alpha$, compute $\alpha \wedge \alpha =e_1\wedge e_2\wedge e_3\wedge e_4+e_3\wedge e_4\wedge e_1\wedge e_2=2e_1\wedge e_2\wedge e_3\wedge e_4$, which is the nonzero basis vector of ${\mathrm{\Lambda }}^4{\mathbb{R}}^4$ by [L1].

Steps 1.1 and 1.2 contradict each other, so the bivector $\alpha$ is not decomposable.

By [L1], ${\mathrm{\Lambda }}^{k}V$ is a quotient of $V^{\otimes k}$, and the structure map $\pi$ is a surjection with nonzero kernel: for $\dim V\ge 2$ and $k\ge 2$, extend a nonzero vector $v$ to a basis $(v,e_2,\dots ,e_n)$; then the pure tensor $v\otimes v\otimes e_3\otimes \cdots \otimes e_k\in V^{\otimes k}$ (with the tail omitted when $k=2$) is nonzero yet $\pi$ of it is $v\wedge v\wedge e_3\wedge \cdots \wedge e_k=0$ because the first two entries repeat, so the canonical construction presents ${\mathrm{\Lambda }}^{k}V$ as a quotient, not a subspace.

For the formula-defined antisymmetrization, [L2] gives

Over a field whose characteristic divides $k!$, step 1.2 gives $\pi \circ \mathrm{Alt}=k!\cdot \mathrm{id}=0$, while ${\mathrm{\Lambda }}^{k}V\mathrm{\ne }0$ by the hypothesis $\dim V\ge k$; a section must satisfy $\pi \circ s=\mathrm{id}$, so $\mathrm{Alt}$ is not a section over such a field. The concrete witness is $F={\mathbb{F}}_2$, $V={\mathbb{F}}_2^2$, $k=2$: $\mathrm{Alt}(e_1\wedge e_2)=e_1\otimes e_2+e_2\otimes e_1$ and $\pi$ of that is $2e_1\wedge e_2=0$.

Step 2.1 gives a field $F$, a vector space $V$, and a degree $k$ for which the displayed formula is not a section of $\pi$, so the universal claim "for every field" is false.

By [L1], the same standard inner product is compatible with both the orientation represented by $(e_1,e_2,e_3)$ and the opposite one represented by $(e_2,e_1,e_3)$; neither is preferred by the metric.

By [L2], these are genuinely different orientations: the change-of-basis determinant is $-1$, so the two bases lie in opposite classes.

Since one inner product is compatible with two different orientations, the inner product does not determine an orientation.

The two inner products $g$ and $2g$ are different: $⟨e_1,e_1{⟩}_g=1$ but $⟨e_1,e_1{⟩}_{2g}=2$. However, orientation classes of ordered bases are defined purely by determinants of change-of-basis maps, with no metric input, so both metrics sit over the same orientation class of $(e_1,e_2)$.

By [L1], the unit volume form for $(\text{orientation},g)$ is $e_1\wedge e_2$, while for $(\text{orientation},2g)$ it is $(e_1\mathrm{/}\sqrt{2})\wedge (e_2\mathrm{/}\sqrt{2})=\frac{1}{2}{e}_1\wedge e_2\mathrm{\ne }{e}_1\wedge e_2$.

By [L2], the two metrics also induce different Gram pairings on the exterior powers: norms are rescaled, so the metric data is genuinely different even though the orientation is the same.

Steps 1.1, 1.2 and 2.1 exhibit two different inner products over one fixed orientation, refuting the claim.

By [L1], the definition of the Hodge star consumes a metric and an orientation as input, so the star is data attached to the pair, not to the bare vector space.

By [L3], the same underlying real vector space ${\mathbb{R}}^3$ with the same metric but the two opposite orientations has two different stars, ${\star }_{+}=-{\star }_{-}$.

By the uniqueness clause of [L2], the operator is genuinely tied to the chosen data: the characterizing relation forces the sign change of step 1.2, so no single star is attached to the vector-space structure alone.

Steps 1.1 through 2.1 refute the claimed sufficiency of the vector-space structure.