Extended Haar measure on algebras and projective spaces

Question

Start with a finite-dimensional algebra $A$ over a well-behaved skew field $k$ (which means essentially the reals, the complexes, or the quaternions). We can construct the projective space $Pr(k, n)$ by taking the quotient $(A^n \setminus \{0\})/(A \setminus \{0\})$. [Edited to correct: I wrote this formula incorrectly; what I meant was $(A \setminus \{0\})/(k \setminus {0})$, which is/should be a richer structure on $(k^n \setminus \{0\})/(k \setminus \{0\})$.] $Pr(k, n)$ is well-behaved (locally compact and Hausdorff). The linearity of the internal multiplication of $A$ means that it is consistent with the quotient operation and becomes an internal product on $Pr(k, n)$.

The product on $Pr(k, n)$ is associative, has a unit if $A$ has a unit as an algebra, and is commutative if $A$ is commutative.

The product has inverses if the product on $A$ has inverses. If so, $A$ is a division algebra over the reals (by restricting the scalars $k$) and must be either the reals, the complexes, or the quaternions.

If the product has inverses, then $Pr(k, n)$ is a locally compact topological group, and has a Haar measure.

I want to consider more general cases, including when $A$ is a group algebra $kG$ or a semigroup algebra. Generally, these cases do not have an inverse of the product. In which of these cases does $Pr(k, n)$ have a nontrivial translation-invariant measure, an extended Haar measure?

Argabright (A note on invariant integrals on locally compact semigroups https://www.ams.org/journals/proc/1966-017-02/S0002-9939-1966-0188341-7/S0002-9939-1966-0188341-7.pdf) gives a criterion which looks to my inexperienced eye that it should decide these cases, but I don't know enough group/ring theory to apply it.

Can such a measure be mapped back into a measure on $A$ that is suitably invariant under translation by multiplication?

I expect that this rather natural question has been solved; does anyone have a reference handy to the solution?

Answer 1

The product has inverses if the product on $A$ has inverses.

This is false. There is no group structure on projective spaces over a division algebra; already $\mathbb{CP}^1 \cong S^2$ does not support any topological group structures whatsoever (which has been discussed before). I also do not believe that definition of the projective space over $A$ does the right thing unless $A$ is a division algebra.

You haven't defined what product you have in mind but I assume it's induced from the pointwise product on $A^n$; this does not have inverses for all nonzero elements even if $A$ is a division algebra if $n \ge 2$, since e.g. $(1, 0, 0, \dots )$ is not invertible.

What is still true is that the projective spaces $K \mathbb{P}^n$ where $K = \mathbb{R}, \mathbb{C}, \mathbb{H}$ are homogeneous spaces over compact Lie groups, namely the groups $O(n+1), U(n+1), Sp(n+1)$ respectively. There is a generalization of Haar measure to homogeneous spaces which I believe implies in this case that a Haar measure (as in, a measure invariant under the action of $O(n+1), U(n+1), Sp(n+1)$ respectively) continues to exist; this generalizes the example of spherical measure. But this requires a choice of inner product on the corresponding vector spaces $\mathbb{R}^{n+1}, \mathbb{C}^{n+1}, \mathbb{H}^{n+1}$, and I don't this can be avoided; that is, I don't think there's a measure invariant under the action of $GL_{n+1}(K)$.

I am not going to venture a guess as to what happens more generally. It's not at all clear that there's a good definition of the projective space for, say, an arbitrary f.d. real or complex algebra $A$. It's also not at all clear that there's even a good definition of "an algebra over the quaternions."