Haar Measure of a Topological Ring
A topological ring is a (not necessarily unital) ring $(R,+,\cdot)$ equipped with a topology $\mathcal{T}$ such that, with respect to $\mathcal{T}$, both $(R,+)$ is a topological group and $\cdot:R\times R\to R$ is a continuous map. A left Haar measure on $R$ is a (nonnegative) measure $\lambda$ on $R$ with respect to the Borel $\sigma$-algebra of $R$ such that there is a multiplicative map $l:R\to[0,\infty]$ called a left multiplier for which
- $l(x\cdot y)=l(x)\,l(y)$ for all $x,y\in R$,
- $\lambda(x\cdot S)=l(x)\,\lambda(S)$ for all $x\in R$ and for any Borel measurable subset $S\subseteq R$ (with respect to $\mathcal{T}$), and
- $\lambda$ is a Haar measure for the topological group $(R,+)$.
(We interpret $0\cdot \infty$ and $\infty\cdot 0$ as $0$.)
An example is the ring $\text{Mat}_{n\times n}(\mathbb{R})$ of $n$-by-$n$ matrices over $\mathbb{R}$ under the topology inherited from $\mathbb{R}^{n\times n}\cong\text{Mat}_{n\times n}(\mathbb{R})$ (here, $\cong$ means "is homeomorphic to"). If $\textbf{A}=\left[a_{i,j}\right]_{i,j=1,2,\ldots,n}$, then we take $\text{d}\lambda(\textbf{A})$ to be $\prod_{i=1}^n\,\prod_{j=1}^n\,\text{d}a_{i,j}$ (i.e., $\lambda$ is the Lebesgue measure of $\text{Mat}_{n\times n}(\mathbb{R})\cong \mathbb{R}^{n\times n}$). Then, $\lambda$ is a left Haar measure of $\text{Mat}_{n\times n}(\mathbb{R})$ with respect to the left multiplier $l(\textbf{X}):=\big|\det(\textbf{X})\big|^n$ for all $\textbf{X}\in\text{Mat}_{n\times n}(\mathbb{R})$.
Another example is $(\mathbb{C},+,\cdot)$. We take $\lambda$ to be such that $\text{d}\lambda(z):=\text{d}x\,\text{d}y$ for $z=x+\text{i}y$. Then, a left multiplier is $l(w):=|w|^2$ for each $w\in\mathbb{C}$. Similarly to the case of the ring of real $n$-by-$n$ matrices, $\text{Mat}_{n\times n}(\mathbb{C})$ has a left Haar measure $\lambda$ with respect to $l(\textbf{X}):=\big|\det(\textbf{X})\big|^{2n}$ for all $\textbf{X}\in\text{Mat}_{n\times n}(\mathbb{C})$, i.e., $\text{d}\lambda(\textbf{A}):=\prod_{i=1}^n\,\prod_{j=1}^n\,\left(\text{d}x_{i,j}\,\text{d}y_{i,j}\right)$ for $\textbf{A}=\left[x_{i,j}+\text{i}y_{i,j}\right]_{i,j=1,2,\ldots,n}$.
The third example here is $\mathbb{R}[T]/\left(T^n\,\mathbb{R}[T]\right)$ equipped with the topology inherited from $\mathbb{R}^n$. For $f(T)=\left(a_0+a_1T+\ldots+a_{n-1}T^{n-1}\right)+\left(T^n\,\mathbb{R}[T]\right)$, we take $\text{d}\lambda(f):=\prod_{i=0}^{n-1}\,\text{d}a_i$. Then, a left multiplier is $l(g):=\left|b_0\right|^n$, for $g(T)=\left(b_0+b_1T+\ldots+b_{n-1}T^{n-1}\right)+\left(T^n\,\mathbb{R}[T]\right)$.
Note that we can also similarly define the notion of right Haar measures for rings and their corresponding right multipliers (which are usually denoted by $r$ in this thread). If a ring has both left and right Haar measures, then (due to commutativity of addition) the two Haar measures coincide (up to scalar multiple) and we can omit the adjectives left and right, and simply call them Haar measures (or two-sided Haar measures to be precise). The adjectives "left" and "right" only indicate which kind of multipliers the ring has. A ring with a Haar measure is unimodular if the left multiplier coincides with the right multiplier; otherwise, we call the ring non-unimodular.
Then, $\text{Mat}_{n\times n}(\mathbb{R})$, $\mathbb{C}$ (as well as $\text{Mat}_{n\times n}(\mathbb{C})$), $\mathbb{R}[T]/\left(T^n\,\mathbb{R}[T]\right)$ (as well as $\mathbb{C}[T]/\left(T^n\,\mathbb{C}[T]\right)$), the ring $\mathbb{Z}_{p}$ of $p$-adic integers (see Crostul's comment below), and the ring of formal power series $\mathbb{F}_p[\![ T]\!]$ (see Jyrki Lahtonen's comment below) are examples of unimodular rings. Of course, many of these rings are commutative rings, and commutative rings with Haar measures are necessarily unimodular.
An example of non-unimodular rings is as follows. Consider $\mathbb{R}\times\mathbb{R}$ (some people may denote it by the semidirect sum $\mathbb{R}\,{\supset\!\!\!\!\!\!+}\,\mathbb{R}$) with the usual entry-wise addition, but with the twisted multiplication $(u,v)\cdot(x,y):=(ux,uy+v)$. Then, with the Haar measure $\text{d}\lambda(a,b):=\text{d}a\,\text{d}b$, we notice that a left multiplier of $\mathbb{R}^2$ is $l(u,v):=|u|^2$ for every $u,v\in\mathbb{R}$, whereas a right multiplier is $r(u,v):=|u|$ for every $u,v\in\mathbb{R}$. The ring $\text{Mat}_{n\times n}(\mathbb{R})\times\mathbb{R}^n$ (also denoted by $\text{Mat}_{n\times n}(\mathbb{R})\,{\supset\!\!\!\!\!\!+}\,\mathbb{R}^n$) and the ring $\text{Mat}_{n\times n}(\mathbb{C})\times\mathbb{C}^n$ (also denoted by $\text{Mat}_{n\times n}(\mathbb{C})\,{\supset\!\!\!\!\!\!+}\,\mathbb{C}^n$), with the usual entry-wise addition and with the twisted multiplication $(\mathbf{A},\mathbf{u})\cdot(\mathbf{B},\mathbf{v}):=(\mathbf{AB},\mathbf{Av}+\mathbf{u})$, are also non-unimodular for the same reason.
Has there been any study on this sort of concepts? I'm sure that I am not the first who thought about the notion of Haar measures for rings. Are there any criteria for a ring to be unimodular? Are there rings with Haar measures of a given characteristic $k>0$ which are non-unimodular, or simple, or at least non-commutative? Are there rings with left Haar measures, but without right Haar measures? (Unlike groups, I expect to find rings with only one kind of Haar measures.) The cardinality of each example of rings with Haar measures I have so far is that of the continuum $\mathfrak{c}=2^{\aleph_0}$. Do there exist rings with Haar measures of higher cardinalities? Please let me know if you have or know of any references.
On the other hand, there may be some pathology with this concept of Haar measures for rings, and because of that, nobody introduces this concept. If that is the case, would you please let me know why?
EDIT I: I removed a wrong example from the question. It turns out that, with respect to the discrete topology, $(\mathbb{Z},+,\cdot)$ does not have a Haar measure.
EDIT II: I think I may need another compatibility condition, as stated in menag's reply below. That is, I may need to enforce the following condition (which I shall name it left compatibility): for any Borel measurable subset $S\subseteq R$ and for every $x\in R$, $x\cdot S$ is also Borel measurable. For right Haar measures, we may similarly define right compatibility. Alternatively, I may need to modify Condition 2 to $\lambda(x\cdot S)=l(x)\,\lambda(S)$ for each Borel measurable subset $S\subseteq R$ and for any $x\in R$ such that $x\cdot S$ is Borel measurable (and likewise for right Haar measures). I am in favor of the former modification (i.e., the left or right compatibility condition) and shall assume it in all instances.
EDIT III: I may be wrong about this claim: "An abelian group has at most one Haar measure (up to scalar multiple)." (See the boldfaced portion of my question.) Without the locally compact Hausdorff assumption, there may be an abelian group with two essentially different Haar measures. If the underlying additive group of a ring is isomorphic to such an abelian group, then the ring may have a left Haar measure and a right Haar measure which are not proportional. Maybe somebody who knows better about topological groups can give me some insight.
To not load the question with too much text, I decided to add my discoveries as an answer, though very lacking it may be. However, these results are simple and have no references. I would greatly appreciate if you find any mistake, if you can add anything to this, or if you make some comments. My answer here is still loaded with many questions.
Proposition: For a fixed integer $k>0$, there exists a commutative topological ring of characteristic $k>0$ with a Haar measure.
Let $R_1,R_2,\ldots,R_m$ be topological rings with Haar measures $\lambda_1,\lambda_2,\ldots,\lambda_m$ corresponding to left multipliers $l_1,l_2,\ldots,l_m$. Define $R:=\bigoplus\limits_{i=1}^m\,R_i$, $\lambda:=\bigotimes\limits_{i=1}^m\,\lambda_i$, and $l:=\prod\limits_{i=1}^m\,\left(l_i\circ \pi_i\right)$, where $\pi_i:R\to R_i$ is the canonical projection. Then, under the product topology, $R$ is a topological ring with Haar measure $\lambda$ with respect to the left multiplier $l$. Hence, if $k_i:=\text{char}\left(R_i\right)$, then the characteristic of $R$ is $\text{lcm}\left(k_1,k_2,\ldots,k_m\right)$. In particular, we know that there exists a ring of a given prime characteristic which possesses a Haar measure, so that there is a ring of a characteristic $k$ which has a Haar measure, where $k$ is a square-free positive integer. It is therefore left to find a ring of characteristic $p^j$ with $p$ being prime and $j>1$ that possesses a Haar measure.
Now, let $R:=\Big(\mathbb{Z}\big/\left(p^j\,\mathbb{Z}\right)\Big)[\![T]\!]$. Each $f(T)\in R$ can be written as $p\,f_0(T)+f_1(T)$, where $f_0(T),f_1(T)\in R$ and the nonzero coefficients of $T^i$ in $f_1(T)$ are not divisible by $p$. Let $\lambda$ be the Haar measure of the group $(R,+)$, which is compact and so we can assume $\lambda(R)=1$. Let $l(f):=l\left(f_1\right)$ for all $f(T)\in R$, $l(f):=1$ if $f(T)$ is an invertible element of $R$, $l(0):=0$, and $l(T):=\frac{1}{p^j}$. Extend the definition of $l$ onto $R$ using the multiplicativity condition. Then, $R$ as a ring has a Haar measure $\lambda$ with respect to the left multiplier $l$ (which is also a right multiplier). Ergo, there exists a topological ring of characteristic $p^j$ with a Haar measure.
Now, how about simple rings with Haar measures? We know that the characteristic of a simple ring is either $0$ or a prime integer. Then, for a fixed prime $p\in\mathbb{N}$, does there exist a simple topological ring of characteristic $p$ with a left, right, or two-sided Haar measure? (For the case of rings of characteristic $0$, $\text{Mat}_{n\times n}(\mathbb{R})$ and $\text{Mat}_{n\times n}(\mathbb{C})$ are simple topological rings with Haar measures.)
Proposition: Any ring with a left, right, or two-sided Haar measure is uncountable.
If $R$ is a countable nonzero ring, then $\left\{0_R\right\}$ must be of zero measure, so that $R=\bigcup_{r\in R}\,\big(r+\left\{0_R\right\}\big)$ has zero measure, but this contradicts positivity of Haar measures of the group $(R,+)$. Hence, the only countable ring with a Haar measure is the zero ring $\{0\}$, which is the unique ring of characteristic $1$, with the counting measure and the two-sided multiplier (or simply the multiplier) $m$ with $m(0):=1$.
Here is another trivial example. A trivial ring (i.e., a ring with trivial multiplication) is a ring with Haar measure. For an uncountable trivial ring $R$ and a Haar measure $\lambda$ of the group $(R,+)$, we can define the multiplier $m$ as $m(x):=0$ for all $x\in R$.
Proposition: For any finite group $G$, the group algebra $\mathbb{C}[G]$ equipped with the standard topology of $\mathbb{C}^{|G|}\cong \mathbb{C}[G]$ (again, $\cong$ means "is homeomorphic to") is a unimodular ring with a Haar measure.
This result is due to the Artin-Wedderburn Theorem. That is, $\mathbb{C}[G]$ decomposes as the direct sum $\bigoplus\limits_{V\,\text{ irrep}}\,\left(V\underset{\mathbb{C}}{\otimes} V^*\right)$ of two-sided ideals $V\underset{\mathbb{C}}{\otimes} V^*$ where $V$ runs over all the (non-isomorphic) simple $\mathbb{C}[G]$-modules and $V^*$ is its algebraic dual. Since $V\underset{\mathbb{C}}{\otimes} V^*$ is isomorphic (and homeomorphic) to the ring $\text{Mat}_{\dim_\mathbb{C}(V)\times \dim_\mathbb{C}(V)}(\mathbb{C})$ for any finite-dimensional $\mathbb{C}$-vector space $V$, the claim follows.
Similarly, $\mathbb{R}[G]$ is also a unimodular ring for any finite group $G$. The Artin-Wedderburn Theorem can also be used here to show that $\mathbb{R}[G]$ decomposes as a direct sum of two-sided ideals of the form $\text{Mat}_{n\times n}(\mathbb{R})$ or $\text{Mat}_{n\times n}(\mathbb{C})$.
On the other hand, if $\mathbb{F}$ is a finite field and $G$ is a Lie group, then what can we say about $\mathbb{F}[G]$? Is there a way to equip $\mathbb{F}[G]$ with a "nice" topology, "compatible" with that of $G$? If so, under this topology, is $\mathbb{F}[G]$ a ring with a left, right, or two-sided Haar measure?
Under the usual topology of the quaternion skew-field $\mathbb{H}\cong\mathbb{R}^4$, $\mathbb{H}$ equipped with the Lebesgue measure is a unimodular ring with a Haar measure. Its multiplier is given by $x\mapsto |x|^4$ for all $x\in\mathbb{H}$. One can also verify that $\mathbb{H}[T]/\left(T^n\,\mathbb{H}[T]\right)$ is a unimodular ring. Now, I speculate that $\text{Mat}_{n\times n}(\mathbb{H})$ too is a unimodular ring, whilst $\text{Mat}_{n\times n}(\mathbb{H})\times\mathbb{H}^n$ (better denoted by $\text{Mat}_{n\times n}(\mathbb{H})\,{\supset\!\!\!\!\!\!+}\,\mathbb{H}^n$) is non-unimodular. (Maybe somebody with more experiences in handling $\mathbb{H}$ can help me with my claims.) What can we say about $\mathbb{H}[G]$ for a finite group $G$?
Definition: A symmetric subset $S$ of a ring $R$ is a subset $S$ of $R$ such that $x\cdot S=S\cdot x$ for all $x\in R$.
If there exists a Borel measurable symmetric subset $S$ of a ring $R$ with a Haar measure $\lambda$ such that $0<\lambda(S)<\infty$, then $R$ is unimodular. For $l(x)\,\lambda(S)=\lambda(x\cdot S)=\lambda(S\cdot x)=\lambda(S)\,r(x)$ for all $x\in R$ (where $l$ is the left multiplier and $r$ is the right multiplier). Note that $\mathbb{H}$ is an example of a noncommutative ring $R$ with a Haar measure with such a subset $S$ (by taking $S$ to be the open unit $4$-disc $\big\{x\in\mathbb{H}\,\big|\,|x|<1\big\}$). Ergo, we have a very weak sufficient condition for a ring to be unimodular. Is there a noncommutative topological ring with a Haar measure which is not a division ring and which possesses a Borel measurable symmetric subset?
Some benefits of having the notion of Haar measures for rings are as follows. Firstly, we have, depending on whether the ring $R$ has a left or a right Haar measure, the equalities $$l(t)\,\left(\int_S\,f(t\cdot x)\,\text{d}\lambda(x)\right)=\int_{t\cdot S}\,f(x)\,\text{d}\lambda(x)$$ and $$\left(\int_S\,f(x\cdot t)\,\text{d}\lambda(x)\right)\,r(t)=\int_{ S\cdot t}\,f(x)\,\text{d}\lambda(x)\,,$$ for $t\in R$, for a Borel measurable $S\subseteq R$, and for a measurable function $f:R\to\mathbb{C}$.
Secondly, we may have a way to define the derivative of some function $f:R\to\mathbb{C}$, provided that $R$ is a unital ring, using the formula $$\left(\text{D}_l\,f\right)(x):=\lim_{t\to 1_R}\,\frac{f(t\cdot x)-f(x)}{l(t)-1}$$ or $$\left(\text{D}_r\,f\right)(x):=\lim_{t\to 1_R}\,\frac{f(x\cdot t)-f(x)}{r(t)-1}\,,$$ depending on whether the ring has a left multiplier or a right multiplier. (Note that, in cases where the ring is non-unimodular, we may have two types of derivatives: left derivatives and right derivatives.) These derivatives are not quite the derivatives in the usual sense. For example, for the ring $\mathbb{R}$, $\left(\text{D}_l\,f\right)(x)=x\,f'(x)=\left(\text{D}_r\,f\right)(x)$. (This derivative concept may be totally useless as, in most cases, I would expect that the limit does not exist, or otherwise, we would need some kind of manifold structures on $R$, but then there is already a concept of derivatives on manifolds.)
What can we say about the quotient $R/I$, where $R$ is a topological ring with a left, right, or two-sided Haar measure and $I$ is a two-sided ideal of $R$? Do we have anything similar to https://mathoverflow.net/questions/21704/haar-measure-on-a-quotient? What are good conditions on $I$ such that the Haar measure on $R$ will "carry over" onto $R/I$? What are good conditions on $R$ such that a good $I$ exists?
What can we say about topological unital rings $R$ with Haar measures if we add this strong condition to the topology: $x\mapsto x^{-1}$ is a continuous map from $\text{Units}(R)$ into itself? Here, $\text{Units}(R)$ is the multipliciative group of units in $R$, equipped with the subspace topology. Topological division rings are examples of such rings with this stronger condition. My examples as well as others' examples so far obey this condition. Do we have a situation where a topological unital ring with this strong condition doesn't have a Haar measure, while it will have a Haar measure if this condition is dropped, and vice versa? What are examples of non-unital rings with Haar measures which are non-trivial or non-commutative?
Now, regarding representations of a topological ring $R$ with a left, right, or two-sided Haar measure, is it possible to introduce a representation theory of $R$ in parallel to the representation theory of topological groups? I can think of the case where $R$ is a unital $\mathbb{R}$-algebra, where we may use a unitary left $R$-module $V$ of finite $\mathbb{R}$-dimension and define an $R$-invariant inner product $\langle \bullet,\bullet\rangle$ on $V$ via $$\langle u,v\rangle:=\int\limits_R\,\langle\!\langle x\cdot u,x\cdot v\rangle\!\rangle\,\text{d}\lambda(x)$$ for $u,v\in V$ and for a fixed inner product $\langle\!\langle\bullet,\bullet\rangle\!\rangle$ on $V$. I haven't tried to make anything out of this $R$-invariant inner product yet. It might not even make sense at all because $\mathbb{R}$-algebras with Haar measures are unlikely to be compact, whence the integral may not be defined or have a finite value in most cases. We can also look at the Banach space $\mathcal{L}^\kappa(R,\lambda)$ for $\kappa\in[1,\infty]$.
I want to give a little discussion: Let $R$ be a topological ring and i assume the underlying group to be locally compact with Haar-measure $\lambda$. For $x \in R$ we set $\mu(A):=\lambda(xA)$ and observe $$\mu(y + A) = \lambda(x(y + A)) = \lambda(xy + xA) = \lambda(xA) = \mu(A),$$ where we assumed that every occuring set is measurable (maybe someone knows a short proof or some sufficient conditions? I don't know if it's possible). So the map $\mu$ is translation invariant.
This is not enough for being a measure. Let $\bigcup_n A_n = A$ be a disjoint union. We obtain $$\mu(A) = \mu(\bigcup_n A_n) = \lambda(x \bigcup_n A_n) = \lambda(\bigcup_n xA_n) \overset{?}{=} \sum_n \lambda(xA_n) = \sum_n \mu(A_n).$$ The problem: The last occuring union doesn't have to be disjoint (take i.e. $A_0 = {0},A_1 = {a},x$ with $xa = 0$). This is a problem which has to be fixed somehow, maybe with excluding elements which don't induce an injection via multiplication. But this is a stupid fix. There might be elements which don't induce an injection, for which the equality is still true, i.e. if the occuring intersections have measure zero.
Let's assume that this equality holds for all measureable sets. If one can show that $\mu$ is a regular borelmeasure, it would also be a Haar-measure and by uniqueness there exists $l(x) > 0$ such that $\mu(A) = l(x) \lambda(A)$. We see that $l(xy) = l(x)l(y)$.
Now we only have to discuss when $\mu$ is a regular borelmeasure.