Wedge of vectors and wedge of forms

Question

Consider the $\mathbb{R}^3$ vectors expressed on terms of the canonical basis $$X=\sum_{i=1}^3x_ie_i,\,Y=\sum_{i=1}^3 y_ie_i,$$ so the wedge product of vectors is $$X\wedge Y=(x_2y_3-x_3y_2)e_1-(x_1y_3-x_3y_1)e_2+(x_1y_2-x_2y_1)e_3.$$ I was wondering if there's any relantio between this wedge and the usual wedge of forms, so i associate a $1$-form to each of these vectors, on the basis $\{de_i\}$ of $\Omega^1(\mathbb{R}^3),$ namely $$X=\sum_{i=1}^3 x_ide_i,\,Y=\sum_{i=1}^3 y_ide_i,$$ so the wedge as forms is $$X\wedge Y=(x_1y_2-x_2y_1)de_1\wedge de_e+(x_1y_3-x_3y_1)de_1\wedge de_3+(x_2y_3-x_3y_2)de_2\wedge de_3.$$ Clearly there is much in common between both expressions, but i cannot longer make a formal statement. I thought about $e_i\mapsto de_j\wedge de_k$ if $i,j,k=1,2,3$ as some sort of maps, also the minus is missing when talking about $e_2$ mapped to $e_1\wedge e_3$. Any thoughts about this?

Answer 1

$ \newcommand\Ext{\mathop{\textstyle\bigwedge}} \newcommand\K{\mathbb K} \newcommand\R{\mathbb R} \newcommand\ExtPow[1]{\mathop{\textstyle\bigwedge^{\mkern-1mu#1}}} \newcommand\form[1]{\langle#1\rangle} \newcommand\dd{\mathrm d} $You have (partly) discovered the Poincaré isomorphism (which is also intimately related to the Hodge star). One reference is Werner Greub's Multilinear Algebra.

It should be noted that a "differential form" is properly a type of field over a manifold. This field structure is irrelevant for this discussion; in the following, any time I say "differential form" what I really mean is more like "differential form at some fixed point on a manifold".

Let $V$ be a finite dimensional vector space over an arbitrary field $\K$. Your "wedge product of vectors" is poor terminology, and it is better to just call it the cross product and reserve the "wedge product" for the exterior algebra $\Ext V$. This is exactly the analog of differential forms $\Ext V^*$ but built up using vectors from $V$ instead of covectors from $V^*$. More concretely, if $e_1,\dotsc,e_n$ is a basis for $V$ then $$ \{e_{i_1}\wedge\dotsb\wedge e_{i_k} \;:\; 1 \leq i_1 < \dotsb < i_k \leq n,\: 0\leq k \leq n \}. $$ is a basis for $\Ext V$, where $k = 0$ corresponds to the scalar $1$. More abstractly, $\Ext V$ can be characterized as the "freest" associative algebra subject to the relations $v\wedge v = 0$ for all $v \in V$. The exterior algebra is naturally a direct sum of exterior powers $$ \Ext V = \ExtPow0V\oplus\ExtPow1V\oplus\dotsb\oplus\ExtPow nV $$ where $\ExtPow0V = \K$, $\ExtPow1V = V$, and more generally $$ \ExtPow kV = \left\{\sum_{i=1}^lv_{i1}\wedge\dotsb\wedge v_{ik} \;:\; v_{11},\dotsc,v_{lk} \in V,\: l \geq 0\right\}. $$

Nonzero simple $k$-vectors $X = v_1\wedge\dotsb\wedge v_k$ can be identified with a $k$-subspaces of $V$: $$ [X] = \{v \in V \;:\; v\wedge X = 0\} = \begin{cases} \mathrm{span}\{v_1,\dotsc,v_k\} &\text{if }X \ne 0, \\ V &\text{otherwise}. \end{cases} $$ (This can be proved using basic properties of the wedge product.) This gives us a geometric characterization of the wedge product: for any simple $X, Y \in \Ext V$ we have $$ [X\wedge Y] = \begin{cases} [X]\oplus[Y] &\text{if }X\wedge Y \ne 0, \\ V &\text{otherwise}. \end{cases} $$ A vector $v \in V$ can be identified with the line $\K v$; dually, a covector $\omega \in V^*$ can be identified with the hyperplane $\ker\omega$. Just as subspaces of $V$ can be built up by spanning lines, they can also be "built down" by intersecting hyperplanes.

$\Ext V$ is the algebra of spanning lines, and $\Ext V^*$ is the algebra of intersecting hyperplanes. Just like we associated $X$ with $[X]$ we can naturally associate $\Omega = \omega_1\wedge\dotsb\wedge\omega_k$ with the $(n-k)$-subspace $$ [\Omega]^* = \begin{cases} \ker(\omega_1)\cap\dotsb\cap\ker(\omega_k) &\text{if }\Omega \ne 0, \\ V &\text{otherwise}, \end{cases} \tag{$*$} $$ and we get a geometric characterization of the "dual wedge product" where for any simple $\Omega, \Phi \in \Ext V^*$ $$ [\Omega\wedge\Phi]^* = \begin{cases} [\Omega]^*\cap[\Phi]^* &\text{if }\Omega\wedge\Phi \ne 0, \\ V &\text{otherwise}. \end{cases} $$ (This idea can be used to prove ($*$) by induction.) This sets up a geometric correspondence between $\ExtPow kV$ and $\ExtPow{n-k}V^*$; for every nonzero simple $X \in \ExtPow kV$ there is a nonzero simple $\Omega \in \ExtPow{n-k}V^*$ such that $$ [X] = [\Omega]^* $$ and this $\Omega$ is unique up to multiplication by a nonzero scalar.

The Poincaré isomorphisms are the realization of this correspondence within $\Ext V$ and $\Ext V^*$. Let $X \mapsto \form{X}_n$ denote the projection of $X \in \Ext V$ onto its $n$-vector part. Because there is exactly one $n$-dimensional subspace of $V$, the space $\ExtPow nV$ is one-dimensional; so for fixed nonzero $I \in \ExtPow nV$ and every $J \in \ExtPow nV$ there is a unique scalar $J/I$ such that $J = (J/I)I$. In this way, we have a bilinear form on $\Ext V$: $$ (X, Y) \mapsto \form{X\wedge Y}_n/I. $$ This form can be proved to be nondegenerate, and so it furnishes an isomorphism $\Ext V \cong (\Ext V)^*$ given by $$ X \mapsto \form{X\wedge{-}}_n/I. $$ Finally, we need to connect $(\Ext V)^*$ with $\Ext V^*$. The exterior powers can be characterized as the sources of alternating multilinear maps; any alternating $k$-linear map $V^k \to W$ extends uniquely to a linear map $\ExtPow kV \to W$. It is in this sense that a differential $k$-form is usually identified with (if not defined by) an element of $(\ExtPow kV)^*$. In other words, we have linear isomorphisms $\ExtPow kV^* \cong (\ExtPow kV)^*$ which build into an isomorphism $\Ext V^* \cong (\Ext V)^*$ given by the natural pairing $\Ext V^*\times\Ext V \to \K$ defined by $$ \form{\omega_1\wedge\dotsb\wedge\omega_k,\; v_1\wedge\dotsb\wedge v_l} = \delta_{kl}\det\Bigl(\omega_i(v_j)\Bigr)_{i,j=1}^k. $$ This pairing tells you how to apply $\Omega \in \ExtPow kV$ to vectors $v_1,\dotsc,v_k$ in order to get a scalar: $$ (v_1,\dotsc,v_k) \mapsto \form{\Omega,\, v_1\wedge\dotsb\wedge v_k}. $$

Finally, for every nonzero $I \in \ExtPow nV$ we have a linear isomorphism $$ \natural : \Ext V \cong (\Ext V)^* \cong \Ext V^*, $$ which, following these isomorphisms, is defined by the equation $$ \form{\natural X, Y}I = \form{X\wedge Y}_n $$ for all $X, Y \in \Ext V$. This is the (left) Poincaré isomorphism associated with $I$.

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As a concrete example, let $V = \R^3$ and let $e_1,e_2,e_3$ be the standard basis defining coordinates $(x,y,z) \mapsto xe_1 + ye_2 + ze_3$. Then for any $v, w \in V$ consider the defining equation of the Poincaré isomorphism associated with $e_1\wedge e_2\wedge e_3$, $$ \form{\natural e_1, v\wedge w}e_1\wedge e_2\wedge e_3 = e_1\wedge v\wedge w. $$ Choosing $v = e_2$ and $w = e_3$, this is saying that applying the differential form $\natural e_1$ to $e_2\wedge e_3$ should yield 1. A possible candidate is $\dd y\wedge\dd z$ since $$ \form{\dd y\wedge\dd z, e_2\wedge e_3} = \dd y(e_2)\dd z(e_3) - \dd y(e_3)\dd z(e_2) = 1. $$ Similarly, we see that $\dd y\wedge\dd z$ evaluates appropriately to zero on $e_3\wedge e_1$ and $e_1\wedge e_2$; thus $\natural e_1 = \dd y\wedge\dd z$. Indeed, $$ [e_1] = \R e_1 = \mathrm{span}\{e_1,e_3\}\cap\mathrm{span}\{e_1,e_2\} = \ker(\dd y)\cap\ker(\dd z) = [\dd y\wedge\dd z]^*. $$

Your observation of the identifications $$ \dd x\wedge\dd y \leftrightarrow e_3,\quad \dd z\wedge\dd y \leftrightarrow e_2,\quad \dd y\wedge\dd z \leftrightarrow e_1 $$ is exactly this Poincaré isomorphism associated with $e_1\wedge e_2\wedge e_3$.

To get an actual cross product out of this requires identifying $V$ with $V^*$, in other words a metric. The metric defined by $$ \flat : e_1 \mapsto \dd x, e_2\mapsto \dd y, e_3 \mapsto \dd z $$ is precisely the standard inner product. The map $\flat$ is, geometrically, a specification of orthogonalization: we are declaring that $$ [e_1]^\perp = [\dd x]^*,\quad [e_2]^\perp = [\dd y]^*,\quad [e_3]^\perp = [\dd z]^*. $$ The cross product of vectors $v, w$ is now $$ v\times w = \natural^{-1}(\flat v\wedge\flat w). $$ The musical isomorphism $\flat$ extends naturally to an algebra homomorphism $\Ext V \to \Ext V^*$; the Hodge star is the composition $\star = \natural^{-1}\circ\flat$ and hence $$ v\times w = \natural^{-1}\flat( v\wedge w) = \star(v\wedge w). $$ Geometrically, $\flat$ takes a subspace $[v_1\wedge\dotsb\wedge v_k]$ as represented in $\Ext V$ to its orthogonal complement $[\omega_1\wedge\dotsb\wedge\omega_k]^*$ as represented in $\Ext V^*$; the Poincaré isomorphism $\natural^{-1}$ then takes this to its representation $[w_1\wedge\dotsb\wedge w_{n-k}]$ in $\Ext V$. Thus $\star$ is the map on $\Ext V$ taking subspaces to their orthogonal complements.

Answer 2

I am not sure where the terminology "wedge product of vectors" comes from. I would view this terminology as rather questionable since (in the way you state it) this is a purely 3-dimensional phenomenon, which there is a meaningful "wedge product of vectors in any dimension".

In principle, for any vector space $V$ and any $k\geq 2$, you can define the $k$th exterior power $\bigwedge^kV$ via a universal construction using tensor products. This comes with a map $V^k\to \bigwedge^kV$ which is written as $(v_1,\dots,v_k)\mapsto v_1\wedge\dots\wedge v_k$ and is $k$-linear and alternating. One then proves that if $a_1,\dots, a_n$ is an ordered basis for $V$, then the elements $a_{i_1}\wedge\dots\wedge a_{i_k}$ with $1\leq i_1<\dots<i_k\leq n$ form a basis for $\bigwedge^kV$. In particular, if you do this for $V=\mathbb R^3$ and $k=2$, you get the space $\bigwedge^2\mathbb R^3$, for which the three vectors $e_1\wedge e_2$, $e_1\wedge e_3$, and $e_2\wedge e_3$ are a basis. For this construction and vectors $X$ and $Y$ as in your question you get (using the fact that the wedge product is bilinear and alternating $$X\wedge Y=(x_1y_2-x_2y_1)e_1\wedge e_2+(x_1y_3-x_3y_1)e_1\wedge e_3+(x_2y_3-x_3y_2)e_2\wedge e_3.$$ So this fits with what you have for forms.

Since $\bigwedge^2\mathbb R^3$ is three-dimensional it is a natural idea to identify it with $\mathbb R^3$ in some way and this is done in order to get the formula for what you call the wedge product of the vectors. The conceptual way to do this is as follows: Consider the map $(X,Y,Z)\mapsto\det(X,Y,Z)$. For fixed $Z\in\mathbb R^3$, this is bilinear and alternating in $X$ and $Y$ and this implies that it only depends on $X\wedge Y\in\bigwedge^2\mathbb R^3$. It then turns out that you get a unique vector $\Phi(X\wedge Y)\in\mathbb R^3$ such that $ \det(X,Y,Z)=\langle\Phi(X\wedge Y),Z\rangle$ where $\langle\ ,\ \rangle$ denotes the standard inner product on $\mathbb R^3$ and this defines a linear isomorphism $\Phi: \bigwedge^2\mathbb R^3\to\mathbb R^3$. By definition, this linear isomorphism maps $e_1\wedge e_2$ to $e_3$, $e_1\wedge e_3$ to $-e_2$, and $e_2\wedge e_3$ to $e_1$, which exactly leads to the formula for the wedge product of vectors you start from.

Connecting this to differential forms needs invoking duality (which is hidden in the inner product in the above approach) and this generalizes to the relation between curl and the exterior derivative and (as stated in the comment by @peek-a-boo) to the Hodge-* operation and musical isomorphisms in Riemannian geometry.