If $G$ is a locally compact Hausdorff topological group, with $C_c(G)$ the algebra of compactly supported continuous functions from $C_c(G)$ to $\mathbb{R}$, a Haar integral $\int_G$ is a continuous linear map $C_c(G) \rightarrow \mathbb{R}$ which is invariant under the aparent action of $G$ on $C_c(G)$. It turns out that there exists a unique such integral, up to a scalar multiple. The Riesz representation theorem then allows one to conclude the existence of a unique Haar measure, which is a $G$-invariant Borel measure on $G$.
The archetypal example of Haar measure is the Lebesgue measure on the (additive group underlying) cartesian space $\mathbb{R}^n$.
Let $G$ be a locally compact Hausdorff group. Let $C_c(G)$ denote the vector space of continuous real-valued functionals with compact support on $G$. This is a locally convex topological vector space where the locally convex structure is specified by the family of seminorms
$K$ ranging over compact subsets of $G$. Recall that a Radon measure on $G$ may be described as a continuous linear functional
Such a Radon measure defines a measure $\mu$ on the $\sigma$-algebra of Borel sets in the usual sense of measure theory, where
In this context, a (left) Haar integral on $G$ is a nonzero such linear functional $\int_G$ such that
for each $f \in C_c(G)$ and each $g \in G$, where $f^g : G \rightarrow \mathbb{R}$ sends $x$ to $f(gx)$.
Correspondingly, a (left) Haar measure on $G$ is a nonzero Radon measure $\mu$ such that
for all $g \in G$ and all Borel sets $B$.
There is an analogy between Haar measure and scaled-cardinality on a finite group. In fact, the latter is a special case of the former, as we may view a finite group as a discrete topological group. While measure on a (discrete) finite group is subsumed by the notion of Haar measure, it may be of interest to build intuition.
Let $G_{fin}$ be a finite group. Let $G_{fin} \text{-Rep}$ be the category of $G$-representations of $R$-modules, where $R$ is a ring in which $|G_{fin}|$ is invertible. This category is equivalent to $R[G]$-mod. We can view $R$ as a trivial $G_{fin}$-representation, where $ga = a$ for each $a \in R$ and each $g \in G$.
Let $G$ be a compactum. Let $G \text{-Ban}$ be the category of Banach representations of $G$. Objects in $G \text{-Ban}$ are banach spaces $X$ over $\mathbb{R}$ with a continuous norm preserving action $G \times X \rightarrow X$, i.e. $||gx|| = ||x|| \forall g \in G \forall x \in X$. Maps in $G \text{-Ban}$ are short maps which are $G$-equivariant. (Alternatively, $G \text{-Ban}$ can be viewed as a category of certain $\mathbb{R}[G]$-modules.) We can view $\mathbb{R}$ as a trivial $G$-representation, where $ga = a$ for each $a \in \mathbb{R}$ and each $g \in G$.
Let $C(G_{fin})$ be the ring of set-maps from $G_{fin}$ to $R$. $G$ acts on $C(G_{fin})$ where $f^g : G_{fin} \rightarrow R$ sends $x$ to $f(gx)$. There is a map of abelian groups $\int_{G_{fin}} : C(G_{fin}) \rightarrow R$ sending $f$ to $\frac{1}{|G|} \sum_{g \in G} f(g)$, analogous to the Haar integral. Indeed,
What corresponds to the Haar measure on $G$ is simply cardinality (though we must appropriately divide by the cardinality of $G$, to get a function $\mu_{fin} : P(G) \rightarrow R$ from the power set of $G$ to $R$ sending $S$ to $\frac{|S|}{|G|}$).
Using the Haar integral, we may define convolution product: $* : C(G \times G) \rightarrow C(G)$ sending $f : G \times G \rightarrow \mathbb{R}$ to the map $G \rightarrow \mathbb{R}$ sending $\int_{h \in G} f(gh^{-1}, h) = \int_{hk = g} f(h, k)$. This is analogous to the map $*_{fin} : C(G_{fin} \times G_{fin}) \rightarrow C(G_{fin})$ sending $f$ to the map $C(G_{fin} )$ sending $g$ to $\frac{1}{|G_{fin}|} \sum_{h k = g} f(h, k)$.
In both cases, we get a “bar construction”. For the compact Hausdorff case, we get:
and for the finite case, we get:
Calling this a bar resolution is a slight abuse of terminology; the “monad” involved is actually has no unit, as $C(G)$ has no unit for the convolution product. However, convolution makes $C(G)$ an assocciative nonunitial algebra, so that the resolution is still a unitless monad. Hence there are evident face maps without degeneracies.
While $C(G)$ has no unit, there is a canonical way of adding units to it: take products with $\mathbb{R}[G]$. Write $\sum_{g \in G} a_g \delta_g$ for the elements of $\mathbb{R}[G]$; since $\delta_1$ is a formal unit for convolution, we may think of it as a Dirac delta function. Now the bar resolution
has degeneracies as well.
Any locally compact Hausdorff topological group $G$ admits a Haar integral (and therefore Haar measure) that is unique up to scalar multiple. This result was first proven by Weil. A proof can be found in these online notes by Rubinstein-Salzedo. A different, constructive proof, due to E.M. Alfsen, can be found in the article “A simplified and constructive proof of the existence and uniqueness of haar measure”.
We here give a lesser known proof of the existence of the Haar integral, specifically on compact Hausdorff groups $G$, which uses convex sets and the Krein Milman theorem instead of measure theory. Let $G \text{-Ban}$ be the category of Banach representations of $G$ (see “Analogy with the Finite Case”).
$C_c(G) = C(G)$ is such a Banach representation. We may view $\mathbb{R}$ as a Banach representation of $G$ as well, where $gz = z$ for each $z \in \mathbb{R}$ and each $g \in G$. $\mathbb{R}$ embeds into $C(G)$ as constant functions. We may then consider the exact sequence
A Haar integral on the $G$-representation $C(G)$ is equivalently a retract $\int_G : C(G) \rightarrow \mathbb{R}$ for the injection $\mathbb{R} \rightarrow C(G)$. In other words, it is a function $\int_G : C(G) \rightarrow \mathbb{R}$ such that
The last of these requirements, given the others, is equivalent to continuity of $\int_G$.
In some sense, we might wish to show that $\text{Ext}^1_{G \text{-Ban}}(C(G), \mathbb{R})$ vanishes; this would show that the sequence
splits by the usual characterization of extensions via $\text{Ext}^1$. On further contemplation, it is sufficient to show that the trivial $G$-representation $\mathbb{R}$ is an injective object in $G \text{-Ban}$. This could be seen as an equivariant Hahn-Banach theorem.
Proof: We show that $\mathbb{R}$ is an injective object in $G \text{-Ban}$. Take an injection of Banach representations of $G$, $X \rightarrow Y$. Let $f : X \rightarrow \mathbb{R}$ be a map of Banach representations of $G$. By the (usual) Hahn-Banach theorem, there exists a map $g : Y \rightarrow \mathbb{R}$ in the category of Banach spaces and short maps extending $f$, though it may lack $G$-invariance.
Consider the subset of all extensions of $f$ to $Y$. Let $S$ be the collection of $G$-invariant compact convex subsets of this set. $S$ contains the convex hull of $G g$, where $g$ is some chosen extension of $f$ to $Y$, so $S$ is nonempty. Using compactness and Zorn's Lemma, we may find a minimal element of $S$ in this collection, where $S$ is ordered where $A \leq B$ when $A \subset B$. Call this element $H$. $H$ must be a singleton. If $H$ contains a point which is not extremal then it contains the convex hull of the orbit of that point, which would be a proper $G$-invariant compact convex subset of $H$ (see Krein Milman theorem). Therefore $H$ is a singleton, and its unique element is a $G$-invariant functional extending $f$.
In particular, since $\mathbb{R}$ has been shown to be injective, the map $\text{Id}_{\mathbb{R}} : \mathbb{R} \rightarrow \mathbb{R}$ lifts along the inclusion
giving a retract $\int_G : C(G) \rightarrow \mathbb{R}$ for $0 \rightarrow \mathbb{R} \rightarrow C(G)$ in $G \text{-Ban}$.
It follows that $\int_G$ has norm $1$, and from this positivity follows immediately.
In Lawvere’s thinking about extensive and intensive quantities,
$C(G)$ is a space of intensive quantities on $G$.
$[C(G), \mathbb{R}]_{\text{Ban}}$ is the space of extensive quantities on $X$, where $\text{Ban}$ is the category of Banach spaces with bounded maps as maps. The Haar integral is the unique $G$-invariant element of this space of norm $1$.
the integration map is the canonical evaluation pairing
If we suggestively write $\int_G f d \phi$ for $\int_G (f, \phi) = \phi(f)$, then $\int_G - d \phi$ becomes a way of writing $\phi$. In particular, choosing $\phi$ to be the Haar measure, we can write $\phi$ as $\int_G - d \phi$.
The left and the right Haar measure may or may not coincide, groups for which they coincide are called unimodular. Consider the matrix subgroup
The left and right invariant measures are, respectively,
and so G is not unimodular.
Abelian groups are obviously unimodular; so are compact groups and discrete groups.
Last revised on April 8, 2021 at 02:23:49. See the history of this page for a list of all contributions to it.