Equivalent of unitary matrix with "spectrum" lying on the 2-sphere

Question

The set of $n\times n $ unitary matrices has eigenvalues $\{\lambda_i\}_{i=1}^n$ which lie on the unit circle (or 1-sphere) $\mathbb{S}^1$.

Is there an "object" A (with appropriate algebra) in which a notion of an adjoint $A^*$ and left/right eigenvalues $Av = \lambda v$, $Av = v\lambda'$ make sense, such that if $A$ satisfies the identity: $$ AA^* = I $$ then either the left- or right-eigenvalues of $A$ lie on a manifold which is geometrically equivalent to the 2-sphere?

In searching for this, I have come across quaternionic matrices and their eigenvalues. But, as I understand, the unit quaternions lie on the hypersphere $\mathbb{S}^3$, so even if the above identity holds, it's 1 dimension too high. The only other thing I can think of is to define some notion of a spectrum of a rank-3 tensor.

Answer 1

If you want your eigenvalues to be usable for anything you probably want to be able to multiply them (e.g. so that we can have $A^n v = \lambda^n v$, or so that we can calculate the eigenvalues of tensor products). In that case the answer is no, because $S^2$ does not admit a topological group structure.

For example, using the Lefschetz fixed point theorem you can show that any compact connected triangulable space admitting a topological group structure must have Euler characteristic zero, but $\chi(S^2) = 2$. This argument shows more generally that if $S^n$ has a topological group structure then $n$ must be odd, which can also be proven using cohomology as seen here.

In fact it's possible to show that $S^n$ has a topological group structure only for $n = 1, 3$. This is closely related to the classification of real division algebras.