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If the compact connected group $K$ is second-countable (hence metrizable, hence Polish), one can prove it with pretty much straight measure theory. (I am not sure if this argument remains valid for non-metrizable groups.)

It's based on the following elementary lemma which is due (in some form) to Steinhaus:

Lemma. Let $G$ be a locally compact Polish group with a left-invariant Haar measure $\mu$, and $A \subset G$ be a Borel set with $\mu(A) > 0$. Then the set $A^{-1} A$ contains an open neighborhood of the identity. In particular, if $A$ is a subgroup of $G$ then $A$ is open.

You can find it as Exercise 17.13 (ii) in Kechris, Classical Descriptive Set Theory. You begin by showing that the function $x \mapsto \mu(xA \triangle A)$ is continuous, which uses the regularity of the measure or else Lusin's theorem. (The proof I know uses dominated convergence and hence only shows that this function is sequentially continuous, which is why I am unsure about the non-metrizable case.)

So let $E = \{(x,y) : xy = yx\} \subset K \times K$. The probability of two elements commuting is $p = (\mu \times \mu)(E)$ where $\mu$ is the left Haar probability measure on $K$. Let $C_x$ denote the centralizer of $x$, which is closed. We have $(x,y) \in E$ iff $y \in C_x$, and thus by Fubini's theorem $$p = \iint 1_E(x,y)\,\mu(dy)\,\mu(dx) = \iint 1_{C_x}(y)\,\mu(dy)\,\mu(dx) = \int \mu(C_x) \,\mu(dx).$$

If we suppose $p >0$, then the set $B = \{x : \mu(C_x) > 0\}$ must have positive measure. But if $\mu(C_x) > 0$, then by our lemma, $C_x$ is an open subgroup of $K$. It was already closed, and $K$ is connected, so $C_x = K$, i.e. $x$ is in the center $Z$ of $K$. Thus $B \subset Z$, so $Z$ is a closed subgroup with positive measure. Applying our lemma again, $Z$ is open and so $Z=K$, i.e. $K$ is abelian.

I think there should also be a "Baire category" analogue of this statement: if $\{ (x,y) : xy = yx\}$ is nonmeager in $K \times K$ then $K$ is abelian. This would use the Pettis lemma in place of the unnamed lemma above, and there is a "Fubini theorem" for Baire category which I've seen but can't recall the name or the reference.

If the compact connected group $K$ is second-countable (hence metrizable, hence Polish), one can prove it with pretty much straight measure theory. (I am not sure if this argument remains valid for non-metrizable groups.)

It's based on the following elementary lemma which is due (in some form) to Steinhaus:

Lemma. Let $G$ be a locally compact Polish group with a left-invariant Haar measure $\mu$, and $A \subset G$ be a Borel set with $\mu(A) > 0$. Then the set $A^{-1} A$ contains an open neighborhood of the identity. In particular, if $A$ is a subgroup of $G$ then $A$ is open.

You can find it as Exercise 17.13 (ii) in Kechris, Classical Descriptive Set Theory. You begin by showing that the function $x \mapsto \mu(xA \triangle A)$ is continuous, which uses the regularity of the measure or else Lusin's theorem. (The proof I know uses dominated convergence and hence only shows that this function is sequentially continuous, which is why I am unsure about the non-metrizable case.)

So let $E = \{(x,y) : xy = yx\} \subset K \times K$. This set is closed, and in particular, measurable. The probability of two elements commuting is $p = (\mu \times \mu)(E)$ where $\mu$ is the left Haar probability measure on $K$. Let $C_x$ denote the centralizer of $x$, which is closed. We have $(x,y) \in E$ iff $y \in C_x$, and thus by Fubini's theorem $$p = \iint 1_E(x,y)\,\mu(dy)\,\mu(dx) = \iint 1_{C_x}(y)\,\mu(dy)\,\mu(dx) = \int \mu(C_x) \,\mu(dx).$$

If we suppose $p >0$, then the set $B = \{x : \mu(C_x) > 0\}$ must have positive measure. But if $\mu(C_x) > 0$, then by our lemma, $C_x$ is an open subgroup of $K$. It was already closed, and $K$ is connected, so $C_x = K$, i.e. $x$ is in the center $Z$ of $K$. Thus $B \subset Z$, so $Z$ is a closed subgroup with positive measure. Applying our lemma again, $Z$ is open and so $Z=K$, i.e. $K$ is abelian.

There is also be a "Baire category" analogue of this statement in the Polish case: if $E$ is nonmeager in $K \times K$, then $K$ is abelian. You could prove it by a nearly identical argument: use the Pettis lemma (Kechris 9.9) in place of Steinhaus above, and the Kuratowski-Ulam Theorem (Kechris 8.41) in place of Fubini. (Though maybe there's a simpler argument, since a nonmeager closed set has to have nonempty interior.)

If the compact connected group $K$ is second-countable (hence metrizable, hence Polish), one can prove it with pretty much straight measure theory. (I am not sure if this argument remains valid for non-metrizable groups.)

It's based on the following elementary lemma which is due (in some form) to Steinhaus:

Lemma. Let $G$ be a locally compact Polish group with a left-invariant Haar measure $\mu$, and $A \subset G$ be a Borel set with $\mu(A) > 0$. Then the set $A^{-1} A$ contains an open neighborhood of the identity. In particular, if $A$ is a subgroup of $G$ then $A$ is open.

You can find it as Exercise 17.13 (ii) in Kechris, Classical Descriptive Set Theory. You begin by showing that the function $x \mapsto \mu(xA \triangle A)$ is continuous, which uses the regularity of the measure or else Lusin's theorem. (The proof I know uses dominated convergence and hence only shows that this function is sequentially continuous, which is why I am unsure about the non-metrizable case.)

So suppose that there is a positive probability $p > 0$ of two elements commuting. Let $C_x$ denote the centralizer of $x \in K$, which is a closed subgroup of $K$. By Fubini's theorem, we have $p = \int_K \mu(C_x) \mu(dx)$. So if $p>0$, the set $A = \{x : \mu(C_x) > 0\}$ must have positive measure: $\mu(A) > 0$.

  But if $x \in A$, so that $\mu(C_x) > 0$, then by our lemma, $C_x$ is an open subgroup of $K$. It was already closed, and $K$ is connected, so $C_x = K$, i.e. $x$ is in the center $Z$ of $K$.

We have thus shown $\mu(Z) > 0$. Noting that $Z$ is a closed subgroup and applying our lemma again, $Z$ is open and so $Z=K$, i.e. $K$ is abelian.

I think there should also be a "Baire category" analogue of this statement: if $\{ (x,y) : xy = yx\}$ is nonmeager in $K \times K$ then $K$ is abelian. This would use the Pettis lemma in place of the unnamed lemma above, and there is "Fubini theorem" for Baire category which I've seen but can't recall the name or the reference.

If the compact connected group $K$ is second-countable (hence metrizable, hence Polish), one can prove it with pretty much straight measure theory. (I am not sure if this argument remains valid for non-metrizable groups.)

It's based on the following elementary lemma which is due (in some form) to Steinhaus:

Lemma. Let $G$ be a locally compact Polish group with a left-invariant Haar measure $\mu$, and $A \subset G$ be a Borel set with $\mu(A) > 0$. Then the set $A^{-1} A$ contains an open neighborhood of the identity. In particular, if $A$ is a subgroup of $G$ then $A$ is open.

You can find it as Exercise 17.13 (ii) in Kechris, Classical Descriptive Set Theory. You begin by showing that the function $x \mapsto \mu(xA \triangle A)$ is continuous, which uses the regularity of the measure or else Lusin's theorem. (The proof I know uses dominated convergence and hence only shows that this function is sequentially continuous, which is why I am unsure about the non-metrizable case.)

So let $E = \{(x,y) : xy = yx\} \subset K \times K$. The probability of two elements commuting is $p = (\mu \times \mu)(E)$ where $\mu$ is the left Haar probability measure on $K$. Let $C_x$ denote the centralizer of $x$, which is closed. We have $(x,y) \in E$ iff $y \in C_x$, and thus by Fubini's theorem $$p = \iint 1_E(x,y)\,\mu(dy)\,\mu(dx) = \iint 1_{C_x}(y)\,\mu(dy)\,\mu(dx) = \int \mu(C_x) \,\mu(dx).$$

If we suppose $p >0$, then the set $B = \{x : \mu(C_x) > 0\}$ must have positive measure. But if $\mu(C_x) > 0$, then by our lemma, $C_x$ is an open subgroup of $K$. It was already closed, and $K$ is connected, so $C_x = K$, i.e. $x$ is in the center $Z$ of $K$. Thus $B \subset Z$, so $Z$ is a closed subgroup with positive measure. Applying our lemma again, $Z$ is open and so $Z=K$, i.e. $K$ is abelian.

I think there should also be a "Baire category" analogue of this statement: if $\{ (x,y) : xy = yx\}$ is nonmeager in $K \times K$ then $K$ is abelian. This would use the Pettis lemma in place of the unnamed lemma above, and there is a "Fubini theorem" for Baire category which I've seen but can't recall the name or the reference.

If the compact connected group $K$ is second-countable (hence metrizable, hence Polish), one can prove it with pretty much straight measure theory. (I am not sure if this argument remains valid for non-metrizable groups.)

It's based on the following elementary lemma, whose attribution I'm not sure about:

Lemma. Let $G$ be a locally compact Polish group with a left-invariant Haar measure $\mu$, and $A \subset G$ be a Borel set with $\mu(A) > 0$. Then the set $A^{-1} A$ contains an open neighborhood of the identity. In particular, if $A$ is a subgroup of $G$ then $A$ is open.

You can find it as Exercise 17.13 (ii) in Kechris, Classical Descriptive Set Theory. You begin by showing that the function $x \mapsto \mu(xA \triangle A)$ is continuous, which uses the regularity of the measure or else Lusin's theorem. (The proof I know uses dominated convergence and hence only shows that this function is sequentially continuous, which is why I am unsure about the non-metrizable case.)

So suppose that there is a positive probability $p > 0$ of two elements commuting. Let $C_x$ denote the centralizer of $x \in K$, which is a closed subgroup of $K$. By Fubini's theorem, we have $p = \int_K \mu(C_x) \mu(dx)$. So if $p>0$, the set $A = \{x : \mu(C_x) > 0\}$ must have positive measure: $\mu(A) > 0$.

But if $x \in A$, so that $\mu(C_x) > 0$, then by our lemma, $C_x$ is an open subgroup of $K$. It was already closed, and $K$ is connected, so $C_x = K$, i.e. $x$ is in the center $Z$ of $K$.

We have thus shown $\mu(Z) > 0$. Noting that $Z$ is a closed subgroup and applying our lemma again, $Z$ is open and so $Z=K$, i.e. $K$ is abelian.

I think there should also be a "Baire category" analogue of this statement: if $\{ (x,y) : xy = yx\}$ is nonmeager in $K \times K$ then $K$ is abelian. This would use the Pettis lemma in place of the unnamed lemma above, and there is "Fubini theorem" for Baire category which I've seen but can't recall the name or the reference.

If the compact connected group $K$ is second-countable (hence metrizable, hence Polish), one can prove it with pretty much straight measure theory. (I am not sure if this argument remains valid for non-metrizable groups.)

It's based on the following elementary lemma which is due (in some form) to Steinhaus:

Lemma. Let $G$ be a locally compact Polish group with a left-invariant Haar measure $\mu$, and $A \subset G$ be a Borel set with $\mu(A) > 0$. Then the set $A^{-1} A$ contains an open neighborhood of the identity. In particular, if $A$ is a subgroup of $G$ then $A$ is open.

You can find it as Exercise 17.13 (ii) in Kechris, Classical Descriptive Set Theory. You begin by showing that the function $x \mapsto \mu(xA \triangle A)$ is continuous, which uses the regularity of the measure or else Lusin's theorem. (The proof I know uses dominated convergence and hence only shows that this function is sequentially continuous, which is why I am unsure about the non-metrizable case.)

So suppose that there is a positive probability $p > 0$ of two elements commuting. Let $C_x$ denote the centralizer of $x \in K$, which is a closed subgroup of $K$. By Fubini's theorem, we have $p = \int_K \mu(C_x) \mu(dx)$. So if $p>0$, the set $A = \{x : \mu(C_x) > 0\}$ must have positive measure: $\mu(A) > 0$.

But if $x \in A$, so that $\mu(C_x) > 0$, then by our lemma, $C_x$ is an open subgroup of $K$. It was already closed, and $K$ is connected, so $C_x = K$, i.e. $x$ is in the center $Z$ of $K$.

We have thus shown $\mu(Z) > 0$. Noting that $Z$ is a closed subgroup and applying our lemma again, $Z$ is open and so $Z=K$, i.e. $K$ is abelian.

I think there should also be a "Baire category" analogue of this statement: if $\{ (x,y) : xy = yx\}$ is nonmeager in $K \times K$ then $K$ is abelian. This would use the Pettis lemma in place of the unnamed lemma above, and there is "Fubini theorem" for Baire category which I've seen but can't recall the name or the reference.