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• CommentRowNumber1.
• CommentAuthorDean
• CommentTimeFeb 23rd 2020
• (edited Feb 23rd 2020)
Unfortunately, my edits were mistaken by the spam detector and I cannot edit the Haar measure page. But I think I have something valuable to add. If it's not too much trouble, could someone add these edits?

# Haar measure
* table of contents
{: toc}

## Idea

If $G$ is a topological group, a _Haar measure_ is a translation-invariant measure on the Borel sets of $G$. The archetypal example of Haar measure is the Lebesgue measure on the (additive group underlying) cartesian space $\mathbb{R}^n$.

## Definition

The proper generality in which to discuss Haar measure is where the topological group $G$ is assumed to be locally compact Hausdorff, and from here on we assume this. (For topological groups, the Hausdorff assumption is rather mild; it is equivalent to the $T_0$ separation condition. See the discussion at uniform space.)

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

$$\rho_K(f) = \sup_{x \in K} |f(x)|,$$

$K$ ranging over compact subsets of $G$. Recall that a Radon measure on $G$ may be described as a continuous linear functional

$$\mu: C_c(G) \to \mathbb{R}$$

which is _positive_ in the sense that $\mu(f) \geq 0$ whenever $f \geq 0$. This defines a measure $\hat{\mu}$ on the $\sigma$-algebra of Borel sets in the usual sense of measure theory, where

$$\hat{\mu}(B) = sup \{\mu(f): supp(f) = K \subseteq B, \rho_K(f) = 1\}$$

By abuse of notation, we generally conflate $\mu$ and $\hat{\mu}$.

A **left Haar measure** on $G$ is a nonzero Radon measure $\mu$ such that

$$\mu(g B) = \mu(B)$$

for all $g \in G$ and all Borel sets $B$.

### The Haar Integral

Let $G$ be a topological group, and let $\mathbb{C}[G]$, the group ring over $G$. Let $G \text{-Ban}$ be the category of Banach representations of $G$. Objects in $G \text{-Ban}$ are banach spaces $X$ over $\mathbb{C}$ with a continuous action $G \times X \rightarrow X$. Maps in $C$ are bounded, $G$-equivariant maps. (Alternatively, $G \text{-Ban}$ can be viewed as a category of certain $\mathbb{C}[G]$-modules.)

Let $\text{Top}$ be the category of topological spaces, and consider $[G, \mathbb{C}]_{ \text{Top}}$, a Banach representation of $G$ with action $G \times [G, \mathbb{C}]_{ \text{Top}} \rightarrow [G, \mathbb{C}]_{ \text{Top}}$.

We may view $\mathbb{C}$ as a Banach representation of $G$ where $gz = z$ for each $z \in \mathbb{C}$ and each $g \in G$. $\mathbb{C}$ embeds into $[G, \mathbb{C}]_{\text{Top}}$ as constant functions. We may then consider the exact sequence
$$0 \rightarrow \mathbb{C} \rightarrow [G, \mathbb{C}]_{\text{Top}} \rightarrow [G, \mathbb{C}]_{\text{Top}}/ \mathbb{C} \rightarrow 0$$

A Haar integral on the $G$-representation $[G, \mathbb{C}]_{\text{Top}}$ is a retract $\int_G : [G, \mathbb{C}]_{\text{Top}} \rightarrow \mathbb{C}$ for the injection $\mathbb{C} \rightarrow [G, \mathbb{C}]_{\text{Top}}$. In other words, it is a function $\int_G : [G, \mathbb{C}]_{\text{Top}} \rightarrow \mathbb{C}$ such that

$$\int_G (f_1 + f_2) = \int_G f_1 + \int_G f_2 \ \ \ \forall f_1, f_2 \in [G, \mathbb{C}]_{\text{Top}}$$
$$\int_G a f = a \int_G f \ \ \ \forall f \in [G, \mathbb{C}]_{\text{Top}}, a \in \mathbb{C}$$
$$\int_G f^g = \int_G f \ \ \ \forall f \in [G, \mathbb{C}]_{\text{Top}}, g \in G$$
$$\exists C \in \mathbb{R}_{\geq 0 } : \left| \left| \int_G f \right| \right| \leq C \int_G ||f|| \ \ \ \forall [G, \mathbb{C}]_{\text{Top}}$$
The last of these requirements, given the others, is equivalent to continuity of $\int_G$.

It is a fundamental theorem, which we will now show, that there is precisely one Haar Measure.

**Remark:** In some sense, we might wish to show that $\text{Ext}^1_{\mathbb{C}[G]}([G, \mathbb{C}]_{\text{Top}}, \mathbb{C})$ vanishes in an appripriate category; this would show that the sequence
$$0 \rightarrow \mathbb{C} \rightarrow [G, \mathbb{C}]_{\text{Top}} \rightarrow [G, \mathbb{C}]_{\text{Top}}/ \mathbb{C} \rightarrow 0$$
splits by the usual characterization of extensions via $\text{Ext}^1$. On further contemplation, however, it is sufficient only to show that the trivial $G$-representation $\mathbb{C}$ is an injective object in $G \text{-Ban}$. This could be seen as an equivariant Hahn-Banach theorem.

**Proof:** From the remark, it is sufficient to show that $\mathbb{C}$ is an injective object in $G \text{-Ban}$. Take an injection of Banach representations of $G$, $X \rightarrow Y$. Let $f : X \rightarrow \mathbb{C}$ be a map of Banach representations of $G$. By the (usual) Hahn-Banach theorem, there exists a functional $g : Y \rightarrow \mathbb{C}$ 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 $\mathbb{C}$ has been shown to be injective, the map $\text{Id}_{\mathbb{C}} : \mathbb{C} \rightarrow \mathbb{C}$ lifts along the inclusion
$$0 \rightarrow \mathbb{C} \rightarrow [G, \mathbb{C}]_{\text{Top}}$$

**Remark:** this alone does not show uniqueness. However, uniqueness is not hard.

**Remark:** by the Riesz-Markov-Kakutani representation theorem, it follows that there is a unique Haar measure on $G$. This result was first proven by Weil. A proof along different lines can be found in these online [notes](http://simonrs.com/HaarMeasure.pdf) by Rubinstein-Salzedo.

### Left and Right Haar Measures that Differ

The left and the right Haar measure may or may not coincide, groups for which they coincide are called **unimodular**.
Consider the matrix subgroup
$$G := \left\{ \left.\, \begin{pmatrix} y & x \\ 0 & 1 \end{pmatrix}\,\right|\, x, y \in \mathbb{R}, y \gt 0 \right\}$$

The left and right invariant measures are, respectively,
$$\mu_L = y^{-2} \,\mathrm{d}x \,\mathrm{d}y,\quad \mu_R = y^{-1} \,\mathrm{d}x \,\mathrm{d}y$$

and so G is not unimodular.

Abelian groups are obviously unimodular; so are compact groups and discrete groups.

[1]: https://arxiv.org/abs/math/0606794

!redirects Haar measure
!redirects Haar measures
!redirects haar measure
!redirects haar measures
• CommentRowNumber2.
• CommentAuthorDean
• CommentTimeFeb 23rd 2020
Also, it might be worth considering to design this article around the Haar Integral instead of the Haar Measure. It seems like the simpler proof of uniqueness is for Haar integral, and that the uniqueness of Haar measure follows from the Riesz Representation theorem.

I was going to add a section about an analogy with the case for finite groups, and $\frac{1}{|G|} \sum_{g \in G} f^g$. But we'll see what the spam detector thinks. (Sigh)
• CommentRowNumber3.
• CommentAuthorDavid_Corfield
• CommentTimeFeb 23rd 2020

Thanks for the edit (attempt) Dean, and sorry for the trouble. I think it’s those three backslashes:

Invalid LaTeX block: \int_G (f_1 + f_2) = \int_G f_1 + \int_G f_2 \ \ \ \forall f_1, f_2 \in [G, \mathbb{C}]_{\text{Top}}

Did you just want a horizontal space there?

• CommentRowNumber4.
• CommentAuthorDavid_Corfield
• CommentTimeFeb 23rd 2020

If so, I think we just use \;, so

$\int_G (f_1 + f_2) = \int_G f_1 + \int_G f_2 \; \; \; \forall f_1, f_2 \in [G, \mathbb{C}]_{\text{Top}}$
• CommentRowNumber5.
• CommentAuthorDean
• CommentTimeFeb 23rd 2020
Oh, I get it now. The spam blocker must be keeping the page from becoming incompilable code.

Yes, I wanted a horizontal space. I'll change it and see if that works.
• CommentRowNumber6.
• CommentAuthorDavid_Corfield
• CommentTimeFeb 23rd 2020
• (edited Feb 23rd 2020)

There are differences between the itex used here and Latex. A list of commands is here.

You can also try things out at Sandbox.

• CommentRowNumber7.
• CommentAuthorDavidRoberts
• CommentTimeFeb 23rd 2020

Or why not \quad or \qquad, which are bigger?

• CommentRowNumber8.
• CommentAuthorDean
• CommentTimeFeb 23rd 2020
Thanks for your suggestions, guys. I will keep them in mind. \quad is good, I'll add it in my next edit.
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