Compact averaging identifies the subgroup fixed algebra

Haar averaging turns a rank-one kernel into an integrated crossed-product coefficient. This gives a direct proof of the subgroup commutation formula: every operator fixed by the combined coefficient action and right translation belongs to the regular crossed product.

Self-checked by the writing AI. Original exposition by GPT-6 Astra (OpenAI), Ultra. Original expression and figures are dedicated to the public domain under CC0-1.0, to the extent of rights held.

The fixed algebra and its regular generators

Let $H$ be an arbitrary compact Hausdorff group with normalized Haar measure, let $N\subseteq B(K)$ be a von Neumann algebra on an arbitrary Hilbert space, and let $V:H\to\mathcal U(K)$ be strongly continuous with $\beta_h=\operatorname{Ad}V_h|_N$ preserving $N$. No countability or separability is assumed. An abstract point-ultraweakly continuous normal action has such a faithful normal implementation by NR1.

The modular function of a compact group is one: its image is a compact subgroup of the positive reals, and taking logarithms leaves only the zero compact additive subgroup. On $K\otimes L^2(H)$, write

$(R_r\xi)(s)=\xi(sr),\qquad (L_h\xi)(s)=\xi(h^{-1}s),\qquad U_r=V_r\otimes R_r. \tag{C1}$

Left and right translation are strongly continuous unitaries by L24. The fixed algebra is

$A=\bigl(N\bar\otimes B(L^2(H))\bigr)\cap U(H)' =\{T:(\beta_r\bar\otimes\operatorname{Ad}R_r)(T)=T \text{ for every }r\in H\}. \tag{C2}$

The regular crossed product $C=N\rtimes_\beta H$ is the von Neumann algebra generated on the same Hilbert space by

$(\pi(n)\xi)(s)=\beta_{s^{-1}}(n)\xi(s),\qquad (\lambda_h\xi)(s)=\xi(h^{-1}s). \tag{C3}$

NR3–4 prove normality and faithfulness of this coefficient representation and identify the regular crossed product independently of the faithful coefficient representation. The normal tensor and multiplier membership required for $\pi(n)\in N\bar\otimes B(L^2(H))$ is also supplied there, with arbitrary $K$.

Both kinds of generator in (C3) lie in $A$. Right and left translations commute. For coefficients, direct calculation gives

$\begin{aligned} (U_r\pi(n)U_r^*\xi)(s) &=V_r\,\beta_{(sr)^{-1}}(n)V_r^*\xi(s)\\ &=\beta_r\beta_{r^{-1}s^{-1}}(n)\xi(s) =\beta_{s^{-1}}(n)\xi(s). \end{aligned} \tag{C4}$

Thus $C\subseteq A$. We prove the opposite inclusion by averaging finite-rank operators.

Why compact averaging is normal

We use a general fact. If a compact group $J$ acts on a concrete von Neumann algebra $B$ by $\gamma_j=\operatorname{Ad}W_j|_B$, with $W$ strongly continuous, then

$E_\gamma(x)=\int_J\gamma_j(x)\,dj \tag{CE1}$

is a faithful normal conditional expectation from $B$ onto $B^\gamma$. Here and below an operator integral without a vector argument is ultraweak.

To prove normality explicitly, for $\omega\in B_*$ set

$E_{\gamma,*}\omega=\int_J\omega\circ\gamma_j\,dj. \tag{CE2}$

The integrand is norm continuous. For a vector functional this follows by moving its two vectors by $W_j^*$ and using $\|\omega_{\xi,\eta}\|\le\|\xi\|\|\eta\|$. Every normal functional is a norm-convergent vector series by CP4 and the concrete predual identification; its uniformly small tail gives the assertion for all $\omega$. The compact image of this continuous map into the Banach space $B_*$ is separable. The finite-measure Bochner construction in QF6 therefore gives (CE2), with $\|E_{\gamma,*}\|\le1$. Its Banach adjoint is precisely (CE1), so $E_\gamma$ is normal.

Positivity and unitality follow by testing the integral against normal positive functionals. Left invariance of Haar gives $\gamma_kE_\gamma=E_\gamma$, while the integral fixes every element of $B^\gamma$. Hence it is a projection with range $B^\gamma$. For $a,b\in B^\gamma$, normality of multiplication and the integral give $E_\gamma(axb)=aE_\gamma(x)b$, the conditional-expectation identity. Finally, if $x\ge0$ and $E_\gamma(x)=0$, then, for every vector $\xi$, the continuous nonnegative function $j\mapsto\langle\gamma_j(x)\xi,\xi\rangle$ has zero Haar integral. Haar measure is positive on every nonempty open set, so this function vanishes everywhere. Its value at the identity is zero for every $\xi$, whence $x=0$. This proves faithfulness.

Apply this fact to $B=N\bar\otimes B(L^2(H))$ and $W_r=U_r$. Its action preserves the tensor algebra, first on elementary tensors and then by the normal tensor construction. We obtain

$\mathcal E(T)=\int_H U_rTU_r^*\,dr,\qquad \mathcal E\bigl(N\bar\otimes B(L^2(H))\bigr)=A. \tag{C5}$

Averaging a rank-one kernel

For $f,g\in C(H)$, write $|f\rangle\langle g|$ for the operator $\zeta(s)\longmapsto f(s)\int_H\zeta(t)\overline{g(t)}\,dt. \tag{CE3}$ The inner product is linear in its first variable. For $n\in N$, the averaged operator $\mathcal E(n\otimes|f\rangle\langle g|)$ has kernel

$K_{n,f,g}(s,t) =\int_H\beta_r(n)f(sr)\overline{g(tr)}\,dr. \tag{C6}$

This formula is a weak operator integral. It can be checked against vectors $\eta u(s)$ and $\zeta v(t)$, with $u,v\in C(H)$: the coefficient of $\beta_r(n)$ is continuous, and every scalar integral is over a finite compact Radon product. HR5 justifies the interchanges. Finite sums of these vectors are dense in $K\otimes L^2(H)$, so the resulting bounded-operator equality holds on the whole Hilbert space. No operator-norm measurability of $r\mapsto\beta_r(n)$ has been presumed.

For $h\in H$, define

$a_h=\int_H\beta_u(n)f(u)\overline{g(h^{-1}u)}\,du\ \in N. \tag{C7}$

The ultraweak integral exists by predual duality, or by the bounded vector coefficients of the strongly implemented action. It belongs to $N$, since every element of $N'$ commutes with the integral. Its dependence on $h$ is norm continuous:

$\|a_h-a_k\| \le\|n\|\,\|f\|_2 \|g(h^{-1}\,\cdot)-g(k^{-1}\,\cdot)\|_2 \longrightarrow0\qquad(k\to h). \tag{C8}$

This is Cauchy–Schwarz and strong continuity of left translation on $L^2(H)$. The same estimate gives $\|a_h\|\le\|n\|\|f\|_2\|g\|_2$.

The operator

$X_{n,f,g}=\int_H\pi(a_h)\lambda_h\,dh \tag{C9}$

is well defined by integrating its action on each vector. The integrand is strongly continuous because $a_h$ is norm continuous and $\lambda_h$ is strongly continuous, and it has a uniform operator bound. Its continuous vector orbits have compact, hence separable, image, so finite-measure vector integration applies even on a nonseparable Hilbert space. It belongs to $C$: every operator in $C'$ commutes with each integrand and consequently with its integral.

In the kernel of (C9), the input coordinate is $t=h^{-1}s$, so $h=st^{-1}$. Inversion and translations preserve normalized compact Haar measure. Therefore its kernel is

$\begin{aligned} \beta_{s^{-1}}(a_{st^{-1}}) &=\int_H\beta_{s^{-1}u}(n) f(u)\overline{g(ts^{-1}u)}\,du\\ &=\int_H\beta_r(n)f(sr)\overline{g(tr)}\,dr =K_{n,f,g}(s,t), \end{aligned} \tag{C10}$

where $u=sr$. Testing on the same dense compact scalar tensors and using finite Radon Fubini proves the equality of bounded operators, not just a formal kernel identity. We conclude that $\mathcal E(n\otimes|f\rangle\langle g|)=X_{n,f,g}\in C$.

Every fixed operator is obtained

Continuous scalar functions are dense in $L^2(H)$, by finite Radon regularity and compact cutoffs. Finite-dimensional subspaces spanned by continuous functions therefore increase, as a directed family, to a dense subspace of $L^2(H)$. Their orthogonal projections converge strongly to the identity. Each such projection and every operator between two such subspaces is a finite sum of rank-one operators with continuous vectors, by Gram–Schmidt within the finite span.

Consequently the star algebra spanned by $n\otimes|f\rangle\langle g|$, $n\in N$, $f,g\in C(H)$, is strongly and ultraweakly dense in $N\bar\otimes B(L^2(H))$. One can see this first for elementary tensors by finite compressions, then for the spatial tensor algebra by BD1 and BD4–5. These are nets over finite subspaces; no countable basis of $L^2(H)$ is required.

For $T\in A$, take a bounded ultraweakly convergent net $T_i\to T$ from that star algebra, using BD4–5. Formula (C10) gives $\mathcal E(T_i)\in C$. Normality of $\mathcal E$ gives $\mathcal E(T_i)\to\mathcal E(T)=T$ ultraweakly. Since $C$ is ultraweakly closed, $T\in C$. Together with (C4), this proves

$\boxed{\bigl(N\bar\otimes B(L^2(H))\bigr) \cap\{V_h\otimes R_h:h\in H\}' =N\rtimes_\beta H.} \tag{C11}$

This is equality in the faithful regular representation (C3), and hence a normal identification with normal inverse. It holds for every compact Hausdorff $H$ and arbitrary $N,K$. For $H=\{e\}$ it is $N=N$; for $N=\mathbb C$ and $V=1$ it gives $R(H)'=L(H)''$. The general locally compact version is proved by a different method in NCF2–3.

A finite kernel shows the displacement

Take $H=S_3=\langle a,b:a^3=b^2=e,\ bab=a^{-1}\rangle$, $N=\mathbb C$, and normalized Haar measure $1/6$ per element. The vectors $e_x=\sqrt6\,1_{\{x\}}$ are orthonormal. For $Q=|e_e\rangle\langle e_a|$,

$R_r Q R_r^*=|e_{r^{-1}}\rangle\langle e_{ar^{-1}}|, \qquad \mathcal E(Q)=\frac16\sum_{r\in S_3} |e_{r^{-1}}\rangle\langle e_{ar^{-1}}| =\frac16 L_{a^{-1}}. \tag{C12}$

Each surviving matrix position has row $s=r^{-1}$, column $t=ar^{-1}$, and hence $st^{-1}=a^{-1}$. In contrast, $t^{-1}s=ra^{-1}r^{-1}$ changes with $r$; the noncommutativity is visible. Formula (C7) gives $a_h=1_{\{a^{-1}\}}(h)$, whose normalized integral (C9) is exactly $\frac16L_{a^{-1}}$.

Right-translation averaging of a rank-one operator on the six elements of S3.

Figure 52.1. Exact matrices in the orthonormal basis $e,a,a^2,b,ab,a^2b$. The rank-one matrix has one entry equal to $1$. Averaging its six right-translation conjugates places $1/6$ at each of the six positions with $st^{-1}=a^{-1}$, giving $\frac16L_{a^{-1}}$. This is the scalar finite instance of the kernel identity (C6)–(C10), with the Haar normalization included.

Problem. Why does the integrated coefficient use $h=st^{-1}$ rather than $t^{-1}s$?

Solution. The regular group operator acts by $\lambda_h\xi(s)=\xi(h^{-1}s)$. Its input coordinate is therefore $t=h^{-1}s$, which gives $h=st^{-1}$. That quantity is unchanged by simultaneous right translation of $s,t$; the finite example shows that $t^{-1}s$ need not be unchanged.

Problem. Does the average in (C5) establish the same proof for a noncompact subgroup?

Solution. No normalized finite Haar measure is available there. The argument proving (CE2) as a bounded preadjoint uses total mass one. The noncompact equality has the separate proof linked after (C11).

The classical fixed-algebra identification occurs in Masamichi Takesaki, Theory of Operator Algebras II, proof of Theorem X.4.12. The compact kernel-averaging proof above gives its own normal expectation, coefficient formula and density argument. Cited works retain their own rights.