Rotations in higher dimensions vs. Spheres

Question

Where my questions stem from:

When we study the rotations in a plane or of some specific higher dimensions, there exists a neat approach to represent all the rotations as a spheres $\mathbb S^i$, for appropriate $i$.

For instance, rotations of the complex plane is $\mathbb S^1$, with a group structure over that, where $\mathbb S^3$ is $\{a+bi|a^2+b^2=1\}$. Similarly, all rotations of the 3-dimensional pure imaginary subspace $\mathbb Ri+ \mathbb Rj+ \mathbb Rk$ in the 4-dimensional space $\mathbb H$ is $\mathbb S^3$, using the correspondence between rotations of the subspace and quaternions of absolute value one. Namely, $\mathbb S^3$ is $\{a+bi+cj+dk|a^2+b^2+c^2+d^2=1\}$, equipped with a group structure to encodes the rotations.

In order to construct the group structures which we put on $\mathbb S^1$ and $\mathbb S^3$ as the groups of rotations, we use the Brahmagupta–Fibonacci identity and Euler's four-square identity, respectively, which give us the identities we need to insure the multiplications of two of the elements from each of those spheres would be another element of absolute value one, and consequently they form a group.

Having this pattern in mind, here are my main questions:

I have seen an argument which tries to give a sense of the main obstacle we could not tackle to use the same approach for any higher dimension, which seems to be the absence of similar identities for an arbitrary $n$ sum of squares. For instance, there is a product of two numbers, each of which is a sum of seven squares, while itself is not a sum of seven squares. So, the multiplication fails since we don't have the same fact about the size of two elements of $\mathbb S^6$.

However, there are analogous identities, in particular, Degen's eight-square idnetity and Pfister's sixteen-square identity, which could potentially provide us with the chance to imitate the same method, at least, for some specific higher dimensions. However, presumably we deserve less chance for case $n=16$, since, unlike the cases $n=1,2,4,8$ (proved by Hurwitz), for which we have bilinear identities, for $n=16$ the Pfister's identity is not bilinear.

I was wondering if there is any resolution for this issue and we have any weaker version of the argument I alluded to above. More importantly, is that the only reason we could not step forward and extend the idea? Or there are some other issues, which we cannot resolve for even case $n=16$ and possibly $n=8$!

Answer 1

This is only possible for $n = 0, 1, 3$. (Strictly speaking, it doesn't even hold for $3$: $SU(2) \approx \mathbb{S}^3$ does act by rotations on $\mathbb{R}^3$, but the kernel of this action is $\{\pm \mathrm{id}\}$, so each rotation corresponds to a pair of elements in $\mathbb{S}^3$, or better yet, to an element of the quotient group $$SO(3) \cong SU(2) / \{\pm \mathrm{id}\} \approx \mathbb{RP}^3.)$$ Anyway, the set of orthogonal transformations of $\mathbb{R}^n$ is obviously a group, but $\mathbb{S}^n$ is only a group for those values of $n$.

Still, there are more interesting things one can say. In general, one can ask for unital composition algebras (over $\mathbb{R}$) on $\mathbb{R}^n$, namely, algebras $\mathbb{A}$ over $\mathbb{R}$ with a nondegenerate quadratic form $Q$ such that $Q(ab) = Q(a)Q(b)$. As you've noticed, there are dimensional restrictions on such objects: Indeed, there turn out to be only seven of these up to isomorphism, they occur only in dimensions $1, 2, 4, 8$, and there is exactly one composition algebra with $Q$ definite in each of these dimensions.

Now, by definition the unit sphere $\mathbb{S}(\mathbb{A})$ of elements $a$ in the composition algebra $\mathbb{A}$ for which $Q(a) = 1$ is closed the multiplication in the algebra. In fact, all such elements have inverses, so $\mathbb{S}(\mathbb{A})$ is a loop under multiplication. In dimensions $1, 2, 4$ the composition algebras with definite $Q$ are $\mathbb{R}$, $\mathbb{C}$, and $\mathbb{H}$; each of these is associative, and so $\mathbb{S}(\mathbb{A})$ is a group. However, in dimension $8$, the definite composition algebra $\mathbb{O}$, called the octonions is nonassociative, as is $\mathbb{S}(\mathbb{O})$. So, $\mathbb{S}(\mathbb{O}) = \mathbb{S}^7$ is not a group (and in particular it cannot encode rotations), but does admit an (interesting) algebraic structure: The automorphism group of this structure turns out to be exactly the compact simple Lie group $G_2$.

Edit There are other spheres that can be viewed (up to universal cover) as orthogonal groups, provided we allow for indefinite bilinear forms; by the above considerations these must arise from the remaining composition algebras. Rotations in this setting are still defined as linear transformations that preserving lengths and angles, but the geometry of rotations is of a different flavor. (One indication that this is true is that the orthogonal groups of indefinite bilinear forms are noncompact, unlike in the definite case.)

The first of these algebras is $\widetilde{\mathbb{C}} := \mathbb{R}[x] / (x^2 - 1)$, which is isomorphic (as an $\mathbb{R}$-algebra) to $\mathbb{R} \oplus \mathbb{R}$. In this case, we can identify the sphere with the unit hyperbola in $\mathbb{R}^2$, and with the group $O(1, 1)$ of "hyperbolic" rotations.

The second of these algebras is $\widetilde{\mathbb{H}}$, which we can think of as the space of $2 \times 2$ real matrices. In this case the quadratic form is just the determinant, and the unit sphere $\mathbb{S}^{1, 2}$, has the same universal cover as $SO(1, 2)$, which is also the universal cover of $SL(2, \mathbb{R})$.

Finally, there is an indefinite version $\widetilde{\mathbb{O}}$ of the octonions, and we can identify it as $\mathbb{S}^{3, 4}$, but as before the product is nonassociative, so it does not define a group structure.