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I'm confused about how many different 7D cross products there are. I'm defining a 7D cross product to be any bilinear map $V \times V \to V$ (where $V$ is the inner product space $\mathbb{R}^7$ endowed with the Euclidean inner product) such that for all $a, b \in V$:

  1. $(a \times b) \cdot a = (a \times b) \cdot b = 0$ and
  2. $|a \times b|^2 = |a|^2 |b|^2 - (a \cdot b)^2$.

(I know that other definitions are sometimes used in the literature.)

Massey 1983 (PDF) provides a construction that they claim characterizes the 7D cross product "uniquely up to isomorphism", but I don't understand what they mean by "up to isomorphism."

On the other hand, the Wikipedia page on the 7D cross product claims that there are 480 different multiplication tables for the 7D cross product. But are these actually distinct functions on the abstract (coordinate-free) vector space, or just the same function written with respect to different choices of ordered basis?

The Wikipedia page also claims that "the" cross product (does their use of the word "the" imply that it's unique, or do they really mean "a" cross product?) is only invariant under the 14-dimensional subgroup $G_2$ of the 21-dimensional group $SO(7)$. I don't have a great intuitive understanding of how to think about this result. But since the defining properties of the cross product are clearly $SO(7)$-invariant, the result seems to indicate to me that there is actually a continuous family of 7D cross products isomorphic to the homogeneous space $SO(7)/G_2$. If that's correct, then the space of 7D cross products actually forms a 7-dimensional manifold, and so there are an uncountably infinite number of distinct 7D cross products.

So how many different 7D cross products are there? One? 480? An uncountably infinite number? How should I think about this?

(My guess is that the answer is "an uncountably infinite number", and that (a) there's some subtlety hidden in Massey's phrase "up to isomorphism" which makes the answer greater than one, and (b) the 480 multiplication tables mentioned on Wikipedia are actually just the restriction of the full manifold of cross products to those whose basis vectors are permutations of each other.)

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    $\begingroup$ a long survey by Baez, who is one of the people I would want answering web.archive.org/web/20100707194509/http://math.ucr.edu/home/… also a book on the octonions by Conway, this link is a review by the same Baez. Oh, he pronounces it "Buys," he told me once at a conference $\endgroup$ – Will Jagy Dec 15 '19 at 0:22
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    $\begingroup$ I believe that after settling upon an orthonormal basis $\{e_1,\ldots,e_7\}$ there are $480$ cross products with the feature that $e_\alpha\times e_\beta=\pm e_\gamma$ (when $\alpha\neq\beta$). So you have a lot more freedom to choose that basis first, accounting for the continuous family. $\endgroup$ – alex.jordan Dec 17 '19 at 7:06
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    $\begingroup$ There are uncountably many. If $\times_1$ and $\times_2$ are two 7D cross products with multiplication tables having pointwise mutually orthogonal entries (e.g. the two induced from octonionic multiplications in this Wikipedia page, one shown in the table and the other described under "A variation of this..." after replacing $0$ by $7$), then $(a,b) \mapsto (a \times_1 b) \cos \theta + (a \times_2 b) \sin \theta$ are also cross products for any $\theta$. $\endgroup$ – pregunton Dec 17 '19 at 10:50
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    $\begingroup$ "Unique up to isomorphism" means any two cross products are related by an orthogonal change of coordinates. (If we only consider rotations, then there are probably two "classes" of cross products, corresponding to the two connected components of $O(7)/G_2$ as per orbit-stabilizer theorem.) $\endgroup$ – runway44 Dec 18 '19 at 1:09
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    $\begingroup$ @tparker Why would you think it's obvious? Let's say you've got two cross products $\times_1$ and $\times_2$. How would you use the fact you cited to prove there exists an $R$ such that $R((R^{-1}v)\times_1(R^{-1}w))=v\times_2 w$ for all $v,w\in\Bbb R^7$? If the cross product were just a single vector, then of course there would be a rotation between any two cross products, but a cross products is not just a single vector, it is a vector-valued function of two vector variables! $\endgroup$ – runway44 Dec 19 '19 at 15:50
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My question has gathered several useful comments, which together have given me enough pieces of information that I think I now have the answer. I think this is correct, but please don't hesitate to correct me if I'm making any mistakes:

Throughout this answer, the letter $R$ will denote an element of (the fundamental representation of) $SO(7)$, and $\times_1$ and $\times_2$ will denote 7D cross products. We'll use square brackets to denote the rotated vector $R[a]$ and subscripts to denote any additional quantities that the rotation $R$ might depend on.

Since the definition of the cross product fixes its magnitude, we obviously have that $|a \times_1 b| \equiv |a \times_2 b|$ and so $$ a \times_2 b \equiv R_{\times_1, \times_2, a, b} [a \times_1 b]. $$ (I.e. the map is norm-preserving but not a priori linear in $a$ or $b$.)

However, it turns out that any two 7D cross products are related by a single rigid rotation of the entire vector space: that is, for any two cross products $\times_1$ and $\times_2$, we have $$ a \times_2 b \equiv \pm R_{\times_1, \times_2} \left[ R_{\times_1, \times_2}^{-1}[a] \times_1 R_{\times_1, \times_2}^{-1}[b] \right]. $$ (We can drop the "$\pm$" if we allow $R$ to be a general element of $O(7)$ instead of $SO(7)$.) This nontrivial result is what Massey means when he says that the 7D cross product is "unique up to isomorphism"; by "isomorphism" he means a single rigid rotation of the entire vector space.

Now let's think about this from a different perspective. Above, we started with two arbitrary fixed cross products $\times_1$ and $\times_2$ and considered a rotation $R_{\times_1, \times_2}$ that related them. Now let's instead start with a particular cross product $\times_1$ and imagine a generic "rotated version" $$ f_{\times_1,R}(a, b) := R\left[ R^{-1}[a] \times_1 R^{-1}[b] \right], \qquad R \in SO(7). $$ We can show that $f_{\times_1, R}$ is also a cross product $\times_2$ (with the same orientation as $\times_1$ if $R \in SO(7)$). This is in some sense the converse of the previous result; the previous result states that any two cross products are rotations of each other, and this result states that the rotation of any cross product is another cross product.

In 3D, it turns out that there is a single cross product $\times$ (up to orientation reversal) and $f_{R,\times} \equiv \times$ for all $R \in SO(3)$. But this is not true in 7D. In 7D it turns out that for any fixed cross product $\times_1$, the claim $f_{\times_1, R} \equiv \times_1$ is only true for a set of rotations $R$ that forms a proper subgroup of $SO(7)$ that is isomorphic to the exceptional Lie group $G_2$. (As Jason DeVito explains in the comments, the exact subgroup of $SO(7)$ depends on $\times_1$, but it is always isomorphic to $G_2$.)

The set of distinct 7D cross products is therefore $X := SO(7)/G_2$. (More precisely, this is the set of transformations that maps an arbitrary initial cross product into any other cross product.) Since $SO(n)$ is always a simple Lie group, $G_2$ is not a normal subgroup of $SO(7)$. So $X$ does not have the full structure of a Lie group, but only of a homogeneous space. In fact, it turns out that this homogeneous space $SO(7)/G_2$ is isometric to the real projective space $\mathbb{R}P^7$.

Now let's fix an orthonormal basis of $\mathbb{R}^7$ and consider the set of cross products that map basis vectors to basis vectors. One way to do this is to choose an arbitrary such cross product $\times_1$ and then consider the cosets $R\, G_2 \in X$ that transform $\times_1$ to another cross product with the same property of mapping basis vectors to basis vectors. We find that this is a finite subset of $X \cong \mathbb{R}P^7$ with 480 elements. (But the other elements of $X$ are still perfectly valid cross products.)

TLDR: There are an uncountably infinite number of distinct 7D cross products, and they form a Reimannian manifold isometric to the real projective space $\mathbb{R}P^7$.

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    $\begingroup$ Very nice! I just have two small comments. First, I do think that the subgroup isomorphic to $G_2$ can change for different choices of $\times_1$. In more detail, if $G_2$ denotes the subgroup you get for the "standard" $\times$ and $R\in SO(7)$, then the "new" $G_2$ you get using $R$ to rotate $\times $ is $R^{-1} G_2 R$. Since, as you said, $G_2$ is not a normal subgroup of $SO(7)$, this gives you different copies of $G_2$ as you vary $\times_1$. Second, the homogeneous space $SO(7)/G_2$ has a nice name - it's just $\mathbb{R}P^7$. $\endgroup$ – Jason DeVito Dec 20 '19 at 14:49
  • $\begingroup$ @JasonDeVito Thanks, edited! I did not know that $SO(7)/G_2 \cong \mathbb{R}P^7$ - if I could ask a naive question, is that "$\cong$" just a diffeomorphism between smooth manifolds, or an isometry between Riemannian manifolds? $\endgroup$ – tparker Dec 21 '19 at 15:51
  • $\begingroup$ Picking any homogeneous metric on $SO(7)/G_2$, you can scale it to get something isometric to a round $\mathbb{R}P^7$. $\endgroup$ – Jason DeVito Dec 21 '19 at 20:53

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