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).

  • 6
    $\begingroup$ Seems like a novel way to think about it! Can one prove your first exact sequence (involving a finite group $G$) splits without invoking the symmetrisation map? $\endgroup$ Jan 25, 2020 at 0:16
  • 5
    $\begingroup$ Given that topological groups which are not locally compact need not admit any Haar measure, I can't help feeling that analysis of the kind used in e.g. von Neumann's construction of Haar measure will have to enter somewhere $\endgroup$
    – Yemon Choi
    Jan 25, 2020 at 3:07
  • 5
    $\begingroup$ When you say that ${\mathbf C} [G]$ is a topological algebra for $G$ a compact group, what is the topology? How is it detecting the topology of $G$? The space you've denoted by $[G, {\bf C}]$ -- which I guess should be interpreted as the algebra of continuous ${\bf C}$-valued functions on $G$ -- seems to be more natural, but now you have to define convolution... $\endgroup$
    – Yemon Choi
    Jan 25, 2020 at 3:15
  • 3
    $\begingroup$ Also: the claim that $[X, {\bf C}]_{\rm Top}$ is a Hilbert space looks very suspect to me: how are you defining your inner product, which I would -- with my analyst's hat on -- expect to be done using some choice of measure on $X$? $\endgroup$
    – Yemon Choi
    Jan 25, 2020 at 3:18
  • 5
    $\begingroup$ On a constructive note (sorry to swamp the comment thread, but since the question seems to be a moving target, I don't really feel I should write something as an "answer") it seems that your formulation via Ext boils down to wanting the inclusion of the trivial G-module ${\bf C}$ into ${\rm Top}(G, {\bf C})$ to split, with the retraction map being "the" Haar integral. Maybe you can try to show that the trivial 1-dim $G$-module is injective in a suitable category of $G$-modules? In the Banach world, this line of thought leads to a Banach-cohomological characterization of amenability. $\endgroup$
    – Yemon Choi
    Jan 25, 2020 at 18:22

1 Answer 1


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.

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.

Let me justify Remark 2. See here 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).

  • 4
    $\begingroup$ Nice, succinct answer: I strongly support Remarks 1 and 3 in particular. (I hadn't thought of the minimality argument you gave, which certainly avoids measure theory, or at least repackages it inside the statement/proof of the Riesz representation theorem $\endgroup$
    – Yemon Choi
    Jan 29, 2020 at 16:47
  • $\begingroup$ Regarding Remark 2: I am still confused over something, probably because I have the "wrong thing" stuck in my head. If we define the category of Banach representations of ${\bf Z}$ in the analogous way, what goes wrong? Is it that the action map for ${\bf Z} \times \ell^\infty({\bf Z}) \to \ell^\infty({\bf Z})$ has the wrong continuity properties? (separate continuity rather than joint continuity, perhaps) $\endgroup$
    – Yemon Choi
    Jan 29, 2020 at 16:53
  • 3
    $\begingroup$ @YemonChoi, To see that $\bf 1$ is not injective in the category of representations of $\mathbb{Z}$, consider for example the representation of $\mathbb{Z}$ on $\mathbb{C}^2$ where the generator is given by the unipotent transformation $(x,y) \mapsto (x+y,y)$. Then the inclusion $\mathbb{C}\to \mathbb{C}^2$, $x\mapsto (x,0)$ does not split. I promise to expand on Remark 2 in the future. $\endgroup$
    – Uri Bader
    Jan 29, 2020 at 17:22
  • $\begingroup$ Of course: the wrong thing stuck in my head was an implicit assumption that representations are bounded -- and your "shearing" example is the classic one that I should have been thinking of. Thanks! $\endgroup$
    – Yemon Choi
    Jan 29, 2020 at 18:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.