I'm reading on Haar measure and we know that every locally compact group admits a Haar measure, is the same true for semigroups? if not, is there a class of semigroups that admits a Haar measure?
Thank you for the help.
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1$\begingroup$ "admit a Haar measure" has to be defined for a semigroup before the question makes any sense. You have to redefine the invariance and there's no immediate unique way to do so. For instance, writing $L_g(h)=gh$, one way to define invariance is to assume that $L_g$ preserves $\mu$ for all $g$, i.e., $(L_g)^*\mu=\mu$, i.e., $\mu((L_g)^{-1}B)=\mu(B)$ for every Borel subset. Then, one should think about what it means on finite semigroups before going into more fancy semigroups... $\endgroup$– YCorCommented Mar 6, 2019 at 17:32
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4 Answers
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Not every locally compact semigroup admits a (locally finite) left-invariant measure. In fact, this has nothing to do with any sort of analytic technicalities and already fails for finite semigroups. For example, consider $S=\{a,b\}$ with $ab=a^2=a$ and $ba=b^2=b$. Then $aS=\{a\}$ and $bS=\{b\}$, so no (finite) measure on $S$ can have $\mu(S)=\mu(aS)=\mu(bS)$.
answered Sep 3, 2014 at 19:11
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If memory serves a locally compact semigroup admits a left invariant Borel measure iff it has a minimal ideal which is a direct product of a group and a right zero semigroup.
The measure is supported on the minimal ideal.
Update:
The result I was trying to recall is from http://www.ams.org/journals/proc/1966-017-02/S0002-9939-1966-0188341-7/S0002-9939-1966-0188341-7.pdf.
A locally finite Borel measure $\mu$ on a locally compact inverse semigroup $S$ is said to be left invariant if $\mu(s^{-1}B)=\mu(B)$ for all Borel sets $B$ where $s^{-1}B=\{x\in S\mid sx\in B\}$.
It is proved in the above paper that if $S$ is locally compact and left translations are proper (i.e. $s^{-1}K$ is compact when $K$ is compact), then $S$ has a left invariant measure iff it has a unique minimal left ideal. This is the same as saying it has a minimal ideal which is a direct product of a group and a right zero semigroup. You can take as the measure the product of Haar measure on the semigroup with your favorite locally finite Borel measure on the right zero semigroup.
answered Sep 3, 2014 at 23:17
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$\begingroup$ What about the positive reals with usual addition and usual topology? $\endgroup$ Commented Sep 3, 2014 at 23:50
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$\begingroup$ Maybe this is the result for compact semigroups. Also invariant measure for a semigroup means the measure of the inverse image of a left translation is the same. $\endgroup$ Commented Sep 4, 2014 at 0:38
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$\begingroup$ @YemonChoi, your example is not a counterexample because preimages of Borel sets can change measure. $\endgroup$ Commented Sep 4, 2014 at 1:21
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1$\begingroup$ That said: for an analyst, the benefit of Haar measure on a LC group is that one can define the L^1 convolution algebra. Now this also works for $R_+$, and $L^1(R_+)$ is a natural object, so perhaps the OP would also be interested in semigroups with measures satisfying the "translates have the same measure" property? $\endgroup$ Commented Sep 4, 2014 at 2:56
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1$\begingroup$ @YemonChoi, people have also studied your variant of an invariant measure. I seem to recall that under minor hypotheses such a measure of global support exists when precisely if you embed in a group, but I am not sure. $\endgroup$ Commented Sep 4, 2014 at 10:56
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NOTATION Let $\ (G\ \cdot)\ $ be a semigroup. Let $\ c\in G.\ $ Then let $\ m_c : G\rightarrow G\ $ be defined by: $\ m_c(x) = x\cdot c\ $ for every $\ x\in G$.
Let $\ (G\ \cdot)\ $ be a semigroup. Let $\ \mu\ $ be a $\sigma$-measure in $G$ (with its respective $\sigma$-algebra). Measure $\mu$ is called liberal $\ \Leftarrow:\Rightarrow\ $ for every measurable set $\ A\subseteq G\ $ such that $\ m_c^{-1}(m_c(A))\ = A\ $, and for every $\ c\in G,\ $ there exists a measurable $\ B\subseteq G\ $ such that $\ m_c(A)\subseteq B\ $ and $\ \mu(A)\ge\mu(B)$.
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The natural generalization of Haar measure onto semigroups doesn't feel harmonious (as even the material already provided in this thread seems to indicate). Thus let me introduce two other notions: strict and liberal. When a semigroup is actually a group then its liberal measures (henceforth the strict too) are invariant, as seen below. By invariance I mean a one-sided invariance (I don't know which side though since I am ambidextrous :-)
NOTATION Let $\ (G\ \cdot)\ $ be a semigroup. Let $\ c\in G.\ $ Then let $\ m_c : G\rightarrow G\ $ be defined by: $\ m_c(x) = x\cdot c\ $ for every $\ x\in G$.
Let $\ (G\ \cdot)\ $ be a semigroup. Let $\ \mu\ $ be a $\sigma$-measure in $G$ (with its respective $\sigma$-algebra). Measure $\mu$ is called liberal $\ \Leftarrow:\Rightarrow\ $ for every measurable set $\ A\subseteq G\ $ such that $\ m_c^{-1}(m_c(A))\ = A\ $, and for every $\ c\in G,\ $ there exists a measurable $\ B\subseteq G\ $ such that $\ m_c(A)\subseteq B\ $ and $\ \mu(A)\ge\mu(B)$.
Also, measure $\ \mu\ $ is called strict $\ \Leftarrow:\Rightarrow\ $ for every measurable $\ A\subseteq G\ $ there exists a measurable $\ B\subseteq G\ $ such that $m_c(A)\subseteq B\ $ and $\ \mu(A)\ge \mu(B)$.
Thus every strict measure is liberal is strict (while there are simple examples of non-strict liberal measures). Nevertheless, even in the liberal case, they are all invariant in the case of any group. Next, in the case of finite semigroups, cardinality is a strict measure. It'd be awesome (:-) to connect the general liberal and strict measures with the structure of finite semigroups. And in the case of (infinite) topological semigroups we would get a lot to think about it seems.
answered Sep 4, 2014 at 4:10
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$\begingroup$ measure $μ$ is called strict, if for every measurable $A⊆G$ there exists a measurable $B⊆G$ such that $mc (A)⊆B$ and $ μ(A)≥μ(B)$? for any $c$? Or for some $c$? $\endgroup$ Commented May 18, 2015 at 12:19
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Existence of left invariant measures on a semigroup $S$ with definitions of $m(A)=m(xA)$ or $m({y:y \ \text{belongs to}\ xA})$ does not mean that support $m$ is a right group because it could be embedded in a right group, i.e. the direct product of a semigroup embeddable in a group and right nulls semigroup. But even the later description is not equivalent to existence of a left invariant measure on the left cancellable semigroup because it requires Malcev conditions.
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$\begingroup$ I find this a bit abstruse. Why not add more details, and clarify the post? In particular, your post repeats much of what has already been said in other answers. $\endgroup$– Alex M.Commented Mar 6, 2019 at 18:35