6
$\begingroup$

What exactly would we get by calculating the cross product of vectors in $R^n, n>3$

using the formula $\vec a \times\vec b=(||\vec a||||\vec b||\sin\theta)\vec n$

$\vec n$ being a vector normal to the two 3D vectors $\vec a$ and $\vec b$

For a set of n-1 vectors that are n dimensional, would there be a generalization of the cross product?

$\endgroup$
  • $\begingroup$ How do you define your "cross product" for $n>3$? $\endgroup$ – user127.0.0.1 Feb 12 '14 at 17:38
  • 1
    $\begingroup$ Well, some people refer to a multi-product creature like $v_1\times\dots\times v_{n-1}\in\Bbb R^n$ as a cross-product. What do you intend as your definition? $\endgroup$ – Ted Shifrin Feb 12 '14 at 17:42
  • 1
    $\begingroup$ Seven-dimensional_cross_product $\endgroup$ – superAnnoyingUser Feb 12 '14 at 17:43
  • 1
    $\begingroup$ If you want to use the definition above, which normal you will consider, there are many!? $\endgroup$ – Semsem Feb 12 '14 at 17:53
  • 2
    $\begingroup$ We often use wedge product or exterior product, as a generalization of cross product in higher dimension. You can easily find the examples in many differential geometry or Manifold textbook. $\endgroup$ – phy_math Aug 24 '14 at 12:17
4
$\begingroup$

Why not use the matrix notation: $$\vec a\times \vec b=\left\vert\begin{array}{ccc}\vec i&\vec j&\vec k\\a_1&a_2&a_3\\b_1&b_2&b_3\end{array}\right\vert.$$ Now in $\mathbb{R}^4$ you have $$\vec a\times \vec b\times \vec c=\left\vert\begin{array}{cccc}\vec i&\ \vec j&\vec k&\vec l\\a_1&a_2&a_3&a_4\\b_1&b_2&b_3&b_4\\c_1&c_2&c_3&c_4\end{array}\right\vert,$$ where $\vec i,\vec j,\vec k,\vec l$ are the basis vectors for $\mathbb{R}^4$.

This is easily expanded to any dimension $n$.

In practice I wouldn't use the (possibly confusing) notation $$\vec a\times \vec b\times \vec c,$$ I would write something like $$\text{cross}(\vec a,\vec b,\vec c).$$

$\endgroup$
10
$\begingroup$

Jeff's answer is good, but I feel that it hides what is really going on a little bit.

Given $n-1$ vectors $v_1,v_2,...,v_{n-1}$ in $\mathbb{R}^n$, you can form a linear map

$$L: \mathbb{R}^n \to \mathbb{R}$$ by the rule

$$L(w) = \det(v_1,v_2, \ldots ,v_{n-1},w)$$

In other words, $L$ is the linear map you get by partially applying the determinant. It is linear because the determinant is linear in each slot, in particular the last one.

Any linear map $\mathbb{R}^n \to \mathbb{R}$ is just a row vector, so its transpose is a column vector which represents the linear map. In other words, there is a vector $\operatorname{Cross}(v_1,v_2, \ldots ,v_{n-1}) \in \mathbb{R}^n$ such that

$$\operatorname{Cross}(v_1,v_2, \ldots ,v_n) \cdot w = L(w) = \det(v_1,v_2, \ldots ,v_{n-1},w)$$

Now all the properties of the cross product flow from this one equation:

  1. $\operatorname{Cross}(v_1,v_2, \ldots ,v_{n-1}) \perp v_i$ for all $i = 1,2, \ldots ,n-1$ because $\operatorname{Cross}(v_1,v_2, \ldots ,v_{n-1}) \cdot v_i = \det(v_1,v_2,...,v_{n-1},v_i) = 0$, because the determinant is alternating.

  2. Let $C = \operatorname{Cross}(v_1,v_2, \ldots ,v_{n-1})$.
    Since $C$ is perpendicular to all of the $v_i$, the volume of the $n$-dimensional parallelepiped $P_n$ spanned by $v_1,v_2, \ldots ,v_{n-1},C$ is just the length of $C$ times the $n-1$-dimensional volume of the parallelepiped $P_{n-1}$ spanned by $v_1,v_2, \ldots , v_{n-1}$. But we can also think of the volume of $P_n$ as being given by the determinant of $(v_1,v_2, \ldots ,v_n,C)$, which in turn is just $C \cdot C$ by the defining property of the cross product. So we have $$C \cdot C = |C| \operatorname{Vol}(P_{n-1})$$ $$|C|^2 = |C| \operatorname{Vol}(P_{n-1})$$ $$|C| = \operatorname{Vol}(P_{n-1})$$ So just like in the 3-dimensional case, the length of the cross product is the $n-1$-dimensional volume of the parallelepiped spanned by the vectors going into the cross product.

  3. $C$ is placed in the orientation so that $\det(v_1,v_2, \ldots ,v_{n-1},C)$ is positive, because that is $C \cdot C$ which must be positive.

We see that the $n$-dimensional cross product enjoys all of the features you know and love about the cross product in $\mathbb{R}^3$: It is perpendicular to all $n-1$ vectors, its length is the volume spanned by those vectors, and $(v_1,v_2, \ldots ,v_{n-1},C)$ is positively oriented.

This is not just abstract nonsense. Try actually computing some cross products using this rule. All you need to get the row vector for $L$ is to put the basis vectors into the defining equation for $L$! This is what is behind this "take a determinant with vectors in some of the slots" business. The way I have told the story, the vectors would be more naturally viewed as column vectors, with $i,j,k,$ etc. occurring as the last column. Of course, this computes the same thing as Jeff mentioned above.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.