I am trying to use a modification of group cohomology to **prove the existence and uniqueness** of Haar measure on a compact Hausdorff group.




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I think the best way of introducing the idea I am pursuing is via analogy.
Let $G$ be a finite group and let $A = \mathbb{C}[G]$ be the group algebra. For each $G$-set $X$, there is an $A$-module $[X, \mathbb{C}]$ (functions from $X$ to $\mathbb{C}$). For $g \in G$, $f^g$ sends $x$ to $f(xg)$.


We may view $\mathbb{C}$ as a $\mathbb{C}[G]$-module where $g z = z$ for each $z \in \mathbb{C}$. $\mathbb{C}$ embeds into $[G, \mathbb{C}]$ as constant functions. We may then consider the exact sequence  

$$0 \rightarrow \mathbb{C} \rightarrow [G, \mathbb{C}] \rightarrow [G, \mathbb{C}]/ \mathbb{C} \rightarrow 0$$
We know that this sequence splits because of the symmetrization map $[G, \mathbb{C}] \rightarrow \mathbb{C}$ sending $f \in [G, \mathbb{C}]$ to $\frac{1}{|G|} \sum\_{g \in G} f(g)$.




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For the question at hand, let $G$ be a compact hausdorff group. Let $A = \mathbb{C}[G]$. This is a $\mathbb{C}$-algebra. Let $C$ be the category of topological $\mathbb{C}[G]$-modules (topological abelian groups $A$ with a map of abelian groups $\mathbb{C}[G] \rightarrow \text{End}\_{\text{TopAb}}(A, A)$. We might wish to tweak this later on to get something abelian). For each topological $G$-set $X$, $[X, \mathbb{C}]\_{\text{Top}}$ (continuous functions from $X$ to $\mathbb{C}$) is such a $\mathbb{C}[G]$-module (it is also a $C^\*$-algebra).


We may view $\mathbb{C}$ as a $\mathbb{C}[G]$-module 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
$$0 \rightarrow \mathbb{C} \rightarrow [G, \mathbb{C}]\_{\text{Top}} \rightarrow [G, \mathbb{C}]\_{\text{Top}}/ \mathbb{C} \rightarrow 0$$
We know that this sequence splits because of Haar measure $[G, \mathbb{C}]\_{\text{Top}} \rightarrow \mathbb{C}$ sending $f \in [G, \mathbb{C}]\_{\text{Top}}$ to $\int\_{G} f d \mu$.


On the other hand, we may wish to show that the sequence splits another way, from which it would follow that a unique Haar measure exists. For instance, what is $\text{Ext}^1\_{\mathbb{C}[G]}([G, \mathbb{C}]/\mathbb{C}, \mathbb{C})$? We may need to tweak the category $C$ to make sense of this. But if we succeed, and this cohomology group vanishes, then the sequence will split by the usual characterization of $\text{Ext}^1$ as classifying extensions.


Does anyone see a say to show that $\text{Ext}^1$ vanishes? No doubt we must use somewhere that $G$ is compact hausdorff (or at least locally compact hausdorff), since the theorem does not hold otherwise. Recall that there is an equivalence of categories between $C^\*$-algebras and compact hausdorff topological spaces, [Gelfand duality](https://ncatlab.org/nlab/show/Gelfand+duality). The $C^\*$-algebra structure of $[G, \mathbb{C}]\_{\text{Top}}$ is one way that the compact hausdorff property of $G$ could show up in the proof (there is an equivalence of categories between compact hausdorff topological spaces with a continuous left $G$-action on the one hand, and $C^\*$-algebras with a right $G$-action on the other; the second is equivalent to $C^\*$-algebras which are $\mathbb{C}[G]$-modules in a compatible way).



Existence and uniqueness of Haar measure on compacta; a cohomological approach
Fix a compact group $G$ and consider its category of Banach representations:
the objects are (complex) Banach spaces $X$ endowed with a $G$-action by automorphims (not necessarily isometries) such that the action maps $G\times X\to X$ are jointly continuous and the morphisms are bounded maps $X\to Y$ commuting with the $G$-actions.


Denote the trivial representation by $\bf 1$. It is an injective object in this category, that it for every $0 \to X \to Y$ every $f: X\to \bf 1$ can be extended to $Y\to \bf 1$. This fact could be seen as an equivariant Hahn-Banach theorem.


Let me explain a way to see this. By the classical Hahn-Banach theorem, $f$ extends to a functional $\bar{f}\in Y^\*$, but it might not be invariant. Consider the subset of all such extensions in $Y^\*$ and consider further the collection of all $G$-invariant compact convex subsets of it. This collection is not empty, as it contains the convex hull of the $G$-orbit of $\bar{f}$. Use compactness and Zorn's lemma to find a minimal element $C$ in this collection, with respect to inclusion. $C$ must be a singleton, otherwise it would contain a point $h$ which is not extremal and the convex hull of the $G$-orbit of $h$ would be a proper $G$-invariant comapct convex subset of $C$, contradicting its minimality, as it contains no extreme point of $C$. It follows that $C=\{h\}$ for some $G$-invariant functional, that is a morhism $Y\to \bf 1$ extending $f$.


In particular, the short exact sequence $0\to {\bf 1}\to C(G) \to C(G)/{\bf 1}\to 0$ splits, by taking $X=\bf 1$ and $Y=C(G)$. The morphism $C(G)\to \bf 1$ is the Haar integral.




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**Remark 1:** I wouldn't regard the above as a *categorical* or *cohomological* approach for proving the existence of the Haar measure. Whatever way you prove it, you prove a splitting theorem.


**Remark 2:** The fact that $\bf 1$ is injective in the category of Banach representations characterizes compact groups among all locally compact second countable ones.


**Remark 3:** Take care with standard categorical interpretations of injectivity and alike, as categories of Banach representations are not abelian.




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Let me justify Remark 2. See [here](https://arxiv.org/abs/math/0606794) that every locally compact second countable group has a proper isometric affine action on some reflexive Banach space $V$,
given by $v \mapsto \rho(g)(v)+c(g)$ where $\rho$ is an isometric representation of $G$ on $V$ and $c:G\to V$ is a 1-cocycle. Assuming $G$ is non-compact, this action has no fixed point. Consider the space $V\oplus \mathbb{C}$ and endow it with the linear representation $\pi$ given by $\pi(g)(v,t)=(\rho(g)(v)+tc(g),t)$. Note that the projection $(v,t)\mapsto t$ is $G$-invariant and does not split. Let $Y=(V\oplus \mathbb{C})^\*$ endowed with the contragredient representation. Then we have a morphism ${\bf 1} \to Y$ which does not split.


Alternatively, for a non amenable group we can take $Y$ to be the space of continuous functions over a compact $G$-space having no invariant measure and for a non property (T) group we can take the dual space of a linearization, as above, of an isometric affine action on a Hilbert space having no fixed point, and conclude by the fact that every non-compact group is either amenable or has (T).


