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Start with a finite-dimensional algebra $A$ over a topologically nice field $k$ (which means essentially the reals or the complexes). Can one construct a nontrivial measure on $A \setminus \{0\}$ that is invariant under the internal multiplication? If not, does restricting $A$ to a semigroup algebra or group algebra make this possible? (In general, elements of $A$ do not have inverses, so we can't just use the 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.

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

Dale
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  • This is the question https://math.stackexchange.com/questions/4954575/extended-haar-measure-on-algebras-and-projective-spaces with the outright wrong parts removed. – Dale Aug 11 '24 at 17:52
  • If $A$ is finite-dimensional over $\mathbb R$ or $\mathbb C$, then $A$ is locally compact, so $A \setminus {0}$ is locally compact. Then you can go ahead with Qiaouchu's answer. – GEdgar Aug 12 '24 at 00:04

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This question still does not really make sense. $A \setminus \{ 0 \}$ is not invariant under multiplication by $a \in A$ unless $a$ is invertible (because in a finite-dimensional algebra, every element is either invertible or a zero divisor). So

  • If you only care about multiplication by invertible $a \in A$ you can just use Haar measure on $A^{\times}$, or rather the pushforward of Haar measure. So you just assign zero measure to all zero divisors.
  • If you want to multiply by all nonzero $a \in A$, then you have to work with $A$, not $A \setminus \{ 0 \}$. But all the non-invertible nonzero elements are zero divisors, so invariance under both $a \in A$ and $b \in A$ such that $ab = 0$ forces the measure of every non-empty subset to be the measure of $\{ 0 \}$. So the only measure which is invariant in this sense is a Dirac measure concentrated at $\{ 0 \}$.

It sounds like you want this measure in order to do something; you'd get better answers if you just explained what that something is.

Qiaochu Yuan
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  • I wasn't aware of "in a finite-dimensional algebra, every element is either invertible or a zero divisor", and the construction I'm looking for clearly doesn't work if there are zero divisors. But if every element is invertible, you are doing something very close to making a skew field extension of the reals, and there aren't many of those. So there's nothing interesting that comes out of this inquiry. – Dale Aug 20 '24 at 20:20