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

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 $\frac{1}{|G|} \sum_{g \in G} f(g)$.

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. 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).

amenability. $\endgroup$ – Yemon Choi Jan 25 '20 at 18:2223more comments