Is there an idempotent measure on the free LD system? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T16:26:05Z http://mathoverflow.net/feeds/question/58007 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/58007/is-there-an-idempotent-measure-on-the-free-ld-system Is there an idempotent measure on the free LD system? Justin Moore 2011-03-09T22:43:32Z 2012-09-13T18:20:17Z <p>This is follow up question to MO question <a href="http://mathoverflow.net/questions/57903" rel="nofollow">"Idempotent measures on the free binary system?"</a> Let $(A,*)$ be the free binary operation on one generator which satisfies the left self distributive law: $a * (b * c) = (a * b)* (a* c)$. Is there a finitely additive probability $\mu$ on $A$ such that $\mu * \mu = \mu$?</p> <p>Here $\mu * \nu$ is defined by $$\mu * \nu (X) = \int \int \mathbf{1}_{*^{-1}(X)} (x,y)\ d\mu (x) \ d \nu (y).$$ Note: because of the asymmetry of the left self distributive law, the above order of integration is important (it is possible to show that there is an idempotent with the opposite order of integration). I conjecture a positive answer (this would follow from a positive answer for the free binary system on one generator) and this question may be easier to resolve than the other.</p> http://mathoverflow.net/questions/58007/is-there-an-idempotent-measure-on-the-free-ld-system/107112#107112 Answer by Justin Moore for Is there an idempotent measure on the free LD system? Justin Moore 2012-09-13T18:20:17Z 2012-09-13T18:20:17Z <p>The answer is yes and follows from the positive answer to <a href="http://mathoverflow.net/questions/57903/idempotent-measures-on-the-free-binary-system/107111#107111" rel="nofollow">MO question 57903</a>.</p>