# Generalized Cross Product

I know that the cross product can be generalized as $$\text{cross}(x_0,...,x_{n-1})=\det\begin{vmatrix}&x_0&\\&x_1&\\&\vdots&\\e_1&\cdots&e_n\end{vmatrix}$$ where $e_i$ is the $i$'th standard unit vector. We have $n-1$ vectors in $n$-dimensional Euclidean Space, so there is a one-dimensional orthogonal complement to that set (if they are independent) and the cross product above gives a vector in that subspace.

I also know that the "area"/"n-volume" of an n-parallelopiped spanned by the vectors $v_1,...,v_n$ is given by $$\sqrt{\det A^TA}$$ where $A=\begin{bmatrix}v_0&\cdots&v_n\end{bmatrix}$. In three dimensions this reduces to $$\sqrt{\det\begin{bmatrix}a_0&a_1&a_2\\b_0&b_1&b_2\end{bmatrix}\begin{bmatrix}a_0&b_0\\a_1&b_1\\a_2&b_2\end{bmatrix}}=\sqrt{||a||^2||b||^2-(a\cdot b)^2}=||a\times b||$$ I am wondering if it is true in general that, taking the cross product as defined above, $$||\text{cross}(x_0,...,x_{n-1})||=\sqrt{\det A^TA}\;\;\;\;\;\; A=\begin{bmatrix}x_0&\cdots&x_{n-1}\end{bmatrix}$$

The algebra seems horrific but I can't find any nice way to prove (or disprove) it.

• Which side of the parallelopiped are you measuring the area of in general? Since I thought the volume formula was $\det (v_0, v_1, v_2)$ in 3D. – BananaCats Category Theory App Apr 14 '14 at 2:50
• @EnjoysMath I am not measuring the area of a side, I am taking the "n-volume" of the whole parallelopiped. In 3D I found the area of a parallelogram. – user142299 Apr 14 '14 at 2:52
• – user137794 Apr 14 '14 at 2:53
• @user137794 Thank you. In the accepted answer to this: math.stackexchange.com/questions/185991/… I am wondering how the author knows that "Then the magnitude of the cross product of n-1 vectors is the volume of the higher-dimensional parallelogram that they determine." I do not understand his notation or terminology because I have not studied those subjects yet. – user142299 Apr 14 '14 at 2:56
• @user142299 I believe he is essentially giving a definition to the cross product that allows it to be generalized in any dimensions. In $\mathbb{R}^n$, $n-1$ independent vectors for an $(n-1)$-dimensional parallelogram. (In 3D, two vectors form a 2D parallelogram). You can define the cross product of $n-1$ vectors as equal to a vector which is normal to all the vectors, and has a magnitude of that $(n-1)$-dimensional parallelogram they make. – user137794 Apr 14 '14 at 3:07

If $x_1,\dotsc,x_{n-1} \in \mathbb{R}^n$, one defines $x_1 \times \cdots \times x_{n-1} \in \mathbb{R}^n$ to be the unique vector such that $$\forall y \in \mathbb{R}^n, \quad \langle x_1 \times \cdots \times x_{n-1},y \rangle = \operatorname{det}(x_1,\dotsc,x_{n-1},y),$$ where the determinant is being viewed as a function of the rows or columns of the usual matrix argument, i.e., as the unique antisymmetric $n$-form $\operatorname{det} : \mathbb{R}^n \times \cdots \times \mathbb{R}^n \to \mathbb{R}$ such that $\det(e_1,\dotsc,e_n) = 1$ for $\{e_k\}$ the standard ordered basis of $\mathbb{R}^n$.

Now, suppose that $x_1,\dotsc,x_{n-1} \in \mathbb{R}^n$ are linearly independent, and hence span a hyperplane $H$ ($n-1$-dimensional subspace) in $\mathbb{R}^n$. Then, in particular, $x_1 \times \cdots \times x_{n-1} \neq 0$ is orthogonal to each $x_k$, and hence defines a non-zero normal vector to $H$; write $$x_1 \times \cdots \times x_{n-1} = \|x_1 \times \cdots \times x_{n-1}\|\hat{n}$$ for $\hat{n}$ the corresponding unit normal. Let $y \notin H$. Then $x_1,\dotsc,x_{n-1},y$ are linearly independent and span an $n$-dimensional parallelopiped $P$ with $n$-dimensional volume $$|\operatorname{det}(x_1,\dotsc,x_{n-1},y)| = |\langle x_1 \times \cdots x_{n-1},y\rangle| = \|x_1 \times \cdots \times x_{n-1}\||\langle \hat{n},y\rangle|.$$ Now, with respect to the decomposition $\mathbb{R}^n = H^\perp \oplus H$, let $$T = \begin{pmatrix} I_{H^\perp} & 0 \\ M & I_{H} \end{pmatrix}$$ for $M : H^\perp \to H$ given by $$M(c \hat{n}) = -c \langle \hat{n},y \rangle^{-1} P_H y = -c\langle \hat{n},y\rangle^{-1}(y-\langle\hat{n},y\rangle\hat{n}),$$ where $P_H(v)$ denotes the orthogonal projection of $v$ onto $H$. Then $T(P)$ is a $n$-dimensional parallelepiped with with vertices $Tx_1 = x_1,\dotsc,Tx_{n-1}=x_{n-1}$, and $$Ty = \langle \hat{n},y \rangle \hat{n} = P_{H^\perp} y = y - P_H y,$$ with the same volume as $P$. On the one hand, since $Ty = y - P_H y$ for $P_H y \in H = \{x_1 \times \cdots \times x_{n-1}\}^\perp$, $$\operatorname{Vol}_n(T(P)) = |\operatorname{det}(Tx_1,\dotsc,Tx_{n-1},Ty)|\\ = |\operatorname{det}(x_1,\dotsc,x_{n-1},y-P_H y)|\\ = |\operatorname{det}(x_1,\dotsc,x_{n-1},y)|\\ = \|x_1 \times \cdots \times x_{n-1}\||\langle \hat{n},y\rangle|.$$ On the other hand, since $Ty \in H^\perp$, $T(P)$ is an honest cylinder with height $\|Ty\| = |\langle \hat{n},y\rangle|$ and base the $(n-1)$-dimensional parallelopiped $R$ spanned by $x_1,\dotsc,x_{n-1}$, so that $$\operatorname{Vol}_n(T(P)) = \operatorname{Vol}_{n-1}(R)|\langle \hat{n},y\rangle|.$$ Thus, $$\operatorname{Vol}_{n-1}(R)|\langle \hat{n},y\rangle| = \operatorname{Vol}_n(T(P)) = \|x_1 \times \cdots \times x_{n-1}\||\langle \hat{n},y\rangle|,$$ so that $$\operatorname{Vol}_{n-1}(R)| = \|x_1 \times \cdots \times x_{n-1}\|,$$ as required.

Let's see what $\phi x_1 \times \cdots \times \phi x_n$ is in terms of $x_1 \times \cdots \times x_{n-1}$ for $\phi$ a linear transformation on $\mathbb{R}^n$.
Define a linear map $T : (\mathbb{R}^n)^{\otimes(n-1)} \to (\mathbb{R}^n)^\ast$ by $$T : x_1 \otimes \cdots \otimes x_{n-1} \mapsto \operatorname{det}(x_1,\cdots,x_{n-1},\bullet),$$ so that if $S : \mathbb{R}^n \to (\mathbb{R}^n)^\ast$ is the isomorphism $v \mapsto \langle v,\bullet \rangle$, then $$x_1 \times \cdots \times x_n = (S^{-1}T)(x_1 \otimes \cdots \otimes x_n).$$ Now, since the determinant is antisymmetric, so too is $T$, and hence $T$ descends to a linear map $T : \bigwedge^{n-1} \mathbb{R}^n \to (\mathbb{R}^n)^\ast$, $$x_1 \wedge \cdots \wedge x_{n-1} \mapsto \operatorname{det}(x_1,\cdots,x_{n-1},\bullet);$$ indeed, if $\operatorname{Vol} = e_1 \wedge \cdots \wedge e_n$ for $\{e_k\}$ the standard ordered basis for $\mathbb{R}^n$, then for any $y \in \mathbb{R}^n$, $$\langle x_1 \otimes \cdots \otimes x_{n-1},y \rangle \operatorname{Vol} = \operatorname{det}(x_1,\cdots,x_{n-1},y)\operatorname{Vol} = x_1 \wedge \cdots \wedge x_{n-1} \wedge y,$$ which, in fact, shows that $$x_1 \times \cdots \times x_{n-1} = \ast (x_1 \wedge \cdots \wedge x_{n-1}),$$ where $\ast : \wedge^{n-1} \mathbb{R}^n \to \mathbb{R}^n$ is the relevant Hodge $\ast$-operator. Thus, a cross product is really an $(n-1)$-form in the orientation-dependent disguise given by the Hodge $\ast$-operator; in particular, it will really transform as an $(n-1)$-form, as we'll see now.
Now, let $\phi : \mathbb{R}^n \to \mathbb{R}^n$ be linear. Observe that the adjugate matrix $\operatorname{Adj}(\phi)$ of $\phi$ can be invariantly defined as the unique linear transformation $\operatorname{Adj}(\phi) : \mathbb{R}^n \to \mathbb{R}^n$ such that for any $\omega \in \bigwedge^{n-1} \mathbb{R}^n$ and $y \in \mathbb{R}^n$, $$(\wedge^{n-1})\omega \wedge y = \omega \wedge \operatorname{Adj}(\phi) y,$$ e.g., in our case, $$x_1 \wedge \cdots \wedge x_{n-1} \wedge \operatorname{Adj}(\phi) y = (\wedge^{n-1}\phi)(x_1 \wedge \cdots \wedge x_{n-1}) \wedge y = \phi x_1 \wedge \cdots \wedge \phi x_{n-1} \wedge y,$$ and that, as a matrix, $\operatorname{Adj}(\phi) = \operatorname{Cof}(\phi)^T$, where $\operatorname{Cof}(\phi)$ denotes the cofactor matrix of $\phi$. Then for any $y$, $$\langle \phi x_1 \times \cdots \times \phi x_{n-1},y \rangle \operatorname{Vol} = \operatorname{det}(\phi x_1,\cdots,\phi x_{n-1},y)\operatorname{Vol}\\ = \phi x_1 \wedge \cdots \wedge \phi x_{n-1} \wedge y\\ = (\wedge^{n-1}\phi)(x_1 \wedge \cdots \wedge x_{n-1}) \wedge y\\ = (x_1 \wedge \cdots \wedge x_{n-1}) \wedge \operatorname{Adj}(\phi)y\\ = \langle x_1 \times \cdots \times x_{n-1},\operatorname{Adj}(\phi)y \rangle \operatorname{Vol}\\ = \langle \operatorname{Cof}(\phi)(x_1 \times \cdots \times x_{n-1}),y \rangle \operatorname{Vol},$$ and hence, since $y$ was arbitrary, $$\phi x_1 \times \cdots \times \phi x_{n-1} = \operatorname{Cof}(\phi)(x_1 \times \cdots \times x_{n-1}) = (\ast \circ \wedge^{n-1}\phi \circ \ast^{-1})(x_1 \times \cdots \times x_{n-1}),$$ in terms of the Hodge $\ast$-operation and the invariantly defined $\wedge^{n-1}\phi$.
• Thanks! This is pretty epic and basically what I was looking for... the only thing tripping me up is how to rigorously justify the use of "Cavalieri." Is there any way to establish a connection (perhaps by looking at the special case of $y=\hat{n}$) between the magnitude of the cross product and the area-determinant formula from linear algebra? – user142299 Apr 14 '14 at 3:22
• I sense that sheer awesomeness is happening - unfortunately, you're getting a bit over my head. Could you please clarify some points? 1. If $M(v) = \langle\hat{n},v\rangle\hat{n} -v$ then it would map vectors from $\mathbb{R}^n\to H$ right? And any vector in $H$ itself would be left unchanged? 2. What is the dimension of $M$ as a matrix? I see that $M(x_i)=-x_i$ but I'm not sure where $Tx_i=x_i$ is coming from. I haven't technically studied direct products yet so I'm just struggling a little to grasp everything you wrote... but I'm learning! Thanks in advance! – user142299 Apr 14 '14 at 15:25
• @user142299 There's an error in the definition of $T$; let me correct it and make things clearer. – Branimir Ćaćić Apr 14 '14 at 18:10
• @user142299 The definition of $T$ is now correct. The way it works is that it fixes anything in $H$, and hence each of the $x_k$, but it's rigged to send $y$ to $P_{H^\perp} y = \langle \hat{n},y\rangle \hat{n}$, which is now orthogonal to the hyperplane $H$ containing the $x_k$. Explicitly, if $P_H$ denotes the orthogonal projection onto $H$ and $P_{H^\perp}$ denotes the orthogonal projection onto $H^\perp$, then $$Tv = P_{H^\perp}v + (MP_{H^\perp} v + P_H v),$$ where $$P_{H^\perp} v = \langle \hat{n},v \rangle \hat{n}, \quad P_H v = v - P_{H^\perp} v.$$ – Branimir Ćaćić Apr 14 '14 at 18:25