# Integrating covariance with left Haar measure ## OA-FLOW.INT.SETTING — Objects, topology, and integrals Let $G$ be a locally compact Hausdorff group with left Haar measure $ds$. Fix the modular function by $$\int_G k(rs)\,dr=\Delta_G(s)^{-1}\int_G k(r)\,dr.$$ In particular, inversion satisfies $\int k(s^{-1})\,ds=\int k(s)\Delta_G(s)^{-1}\,ds$. No second countability, sigma compactness or unimodularity is assumed. Let $M$ be a von Neumann algebra and let $\alpha:G\to\operatorname{Aut}(M)$ be point-ultraweakly continuous. We use its equivalent predual-continuous action topology from `IMP.CP.REGULAR`. On norm-bounded sets, evaluation $(s,x)\mapsto\alpha_s(x)$ is jointly strong-star continuous in any faithful normal representation. To state precisely the continuity fact needed here: if $s_i\to s$ and a uniformly bounded net $x_i\to x$ strong-star, then $\alpha_{s_i}(x_i)\to\alpha_s(x)$ strong-star. It follows by expanding the defining normal positive-functional seminorms and using norm continuity of $\omega\circ\alpha_s$; it is part of the action-topology contract until that equivalence is proved in the course. Let $\mathcal K_\alpha$ be the vector space of compactly supported strong-star continuous functions $f:G\to M$. Such a function is norm bounded: for each Hilbert vector its continuous image on the compact support is bounded, and the uniform boundedness principle applies. Its norm is lower semicontinuous and therefore Borel, so $$\|f\|_1=\int_G\|f(s)\|\,ds<\infty.$$ For a compactly supported ultraweakly continuous, norm-bounded $M$-valued function, its ultraweak integral is defined by pairing with $M_*$. The bound on those pairings produces an element of $(M_*)^*=M$. If the field is additionally strongly continuous, as are all the strong-star fields integrated in this lesson, the same integral acts on vectors by a Bochner integral: each vector image of its compact support is norm compact in a Hilbert space, hence separable. Ultraweak continuity alone does not supply this norm-compact image argument. The assertion concerns each of these compactly supported strong integrals, not global separability of the representation. We use Haar change of variables and Fubini on compactly supported scalar or Hilbert-vector integrands. Their supports have finite measure. In a locally compact group each compact set is contained in a sigma-compact open subgroup after adjoining an identity neighborhood and taking countably many products; hence these particular applications can also be reduced to sigma-finite Haar spaces. No unrestricted Fubini assertion on nonintegrable functions over a nonsigma-finite product is needed. ### OA-FLOW.INT.IMPORT.RADONFUBINI — The product-integration contract For nonmetrizable locally compact spaces, the integration prerequisite is the Radon/Haar product-measure version of Fubini and Tonelli. It covers compactly supported continuous scalar integrands, the corresponding continuous Hilbert-vector integrands obtained by uniform finite-dimensional approximation on compact support, and nonnegative lower-semicontinuous norm integrands such as those in (I3). In the last case the iterated integrals agree with integration for the Radon product measure, with the appropriate completed/localizable interpretation. No equality between the Borel sigma-algebra of the topological product and the uncompleted tensor-product Borel sigma-algebra is assumed. Restricting support to a sigma-compact open subgroup supplies finite or sigma-finite measures where used; it does not establish that sigma-algebra equality. This is an exact part of the unresolved Haar-integration foundation in IMP.HARMONIC, not an additional proved measure theorem. ## OA-FLOW.INT.OPERATIONS — The right-coefficient convolution For $f,g\in\mathcal K_\alpha$, define $$ (f*g)(r)=\int_G\alpha_s(f(rs))g(s^{-1})\,ds, \qquad f^\sharp(r)=\Delta_G(r)^{-1}\alpha_{r^{-1}}(f(r^{-1})^*). \tag{I1} $$ The left and right $M$-module operations are $$ (a\cdot f)(s)=\alpha_{s^{-1}}(a)f(s),\qquad (f\cdot a)(s)=f(s)a. \tag{I2}$$ **Lemma.** These functions belong to $\mathcal K_\alpha$, and $$\operatorname{supp}(f*g)\subset\operatorname{supp}(f)\operatorname{supp}(g),\quad \|f*g\|_1\leq\|f\|_1\|g\|_1,\quad \|f^\sharp\|_1=\|f\|_1. \tag{I3}$$ **Proof.** For fixed $r$, the convolution integrand is supported in the compact set $\operatorname{supp}(g)^{-1}$. All of its operations are jointly strong-star continuous and uniformly norm bounded on compact parameter sets. It has the integral described in the setting. If a value can be nonzero, there is some $s$ with $rs\in\operatorname{supp}(f)$ and $s^{-1}\in\operatorname{supp}(g)$, giving $r=(rs)s^{-1}$ in the displayed compact product. This proves the support containment. To check strong continuity at $r_0$, fix a Hilbert vector $\eta$ and a compact neighborhood of $r_0$. The vector-valued map $$ (r,s)\longmapsto\alpha_s(f(rs))g(s^{-1})\eta$$ is continuous on that neighborhood times the compact integration support. As $r\to r_0$, its difference from the value at $r_0$ tends uniformly to zero in $s$. This follows from continuity and a finite subcover of the compact integration set; it is valid for nets. Integrating the uniform estimate proves strong continuity. Apply the same argument to the adjoint product $g(s^{-1})^*\alpha_s(f(rs)^*)$ to obtain strong-star continuity. This avoids an invalid appeal to dominated convergence for arbitrary nets. The formulas for $f^\sharp$ and the module operations give compact support and strong-star continuity directly by the action-continuity contract, continuity of inversion and of $\Delta_G$, and bounded-product continuity. For the norm estimate, positive scalar integration and Haar change of variables give $$ \begin{aligned} \|f*g\|_1 &\leq\int_G\int_G\|f(rs)\|\,\|g(s^{-1})\|\,ds\,dr\\ &=\|f\|_1\int_G\Delta_G(s)^{-1}\|g(s^{-1})\|\,ds =\|f\|_1\|g\|_1. \end{aligned} $$ Finally, inversion gives $$\|f^\sharp\|_1=\int_G\Delta_G(r)^{-1}\|f(r^{-1})\|\,dr=\|f\|_1.$$ All norm integrands are Borel and bounded with compact support, so the positive integrals are legitimate. $\square$ ## OA-FLOW.INT.COVARIANT — Integrated operators and their identities Let $\pi:M\to B(K)$ be a normal unital representation and let $U:G\to\mathcal U(K)$ be strongly continuous, with $$U_s\pi(a)U_s^*=\pi(\alpha_s(a)).$$ Define the integrated operator by $$F_{\pi,U}(f)=\int_G U_s\pi(f(s))\,ds. \tag{I4}$$ The vector integrands are continuous and compactly supported. The integral defines a bounded operator with norm at most $\|f\|_1$; its weak integral and vector integral agree. **Proposition.** For $f,g\in\mathcal K_\alpha$ and $a\in M$, $$ F(f*g)=F(f)F(g),\qquad F(f^\sharp)=F(f)^*, \tag{I5} $$ $$F(a\cdot f)=\pi(a)F(f),\qquad F(f\cdot a)=F(f)\pi(a). \tag{I6}$$ **Proof.** Covariance gives $\pi(f(s))U_t=U_t\pi(\alpha_{t^{-1}}(f(s)))$. Therefore $$F(f)F(g)=\int_G\int_GU_{st}\pi(\alpha_{t^{-1}}(f(s))g(t))\,ds\,dt.$$ Put $r=st$. Right translation of left Haar measure gives $ds=\Delta_G(t)^{-1}\,dr$ in this iterated integral. The coefficient of $U_r$ is $$\int_G\Delta_G(t)^{-1}\alpha_{t^{-1}}(f(rt^{-1}))g(t)\,dt.$$ Now put $v=t^{-1}$. Inversion contributes $dt=\Delta_G(v)^{-1}\,dv$ and $\Delta_G(t)^{-1}=\Delta_G(v)$, so the factors cancel. The coefficient becomes $$\int_G\alpha_v(f(rv))g(v^{-1})\,dv=(f*g)(r).$$ This proves the first identity. To take adjoints, covariance gives $$\begin{aligned} F(f)^*&=\int_G\pi(f(s)^*)U_{s^{-1}}\,ds\\ &=\int_G U_{s^{-1}}\pi(\alpha_s(f(s)^*))\,ds\\ &=\int_G U_r\pi(\Delta_G(r)^{-1}\alpha_{r^{-1}}(f(r^{-1})^*))\,dr. \end{aligned}$$ The last line is exactly $F(f^\sharp)$. Products and adjoints commute with these bounded weak integrals in the indicated variables; every double integral can be checked against two vectors on its compact support. Finally $\pi(a)U_s=U_s\pi(\alpha_{s^{-1}}(a))$ proves the left module identity, and the right module identity follows immediately from (I4). $\square$ The two modular factors cancel in the convolution change of variables. There remains a modular factor in the involution. Its absence from one formula is not a reason to omit it from the other. ## OA-FLOW.INT.KERNEL — The regular action on vector sections Represent $M\subset B(H)$ faithfully and normally. On $L^2(G,H)$ use the course's regular pair $$ (\pi_\alpha(a)\xi)(r)=\alpha_{r^{-1}}(a)\xi(r),\qquad (\lambda_s\xi)(r)=\xi(s^{-1}r).$$ Let $F_\alpha=F_{\pi_\alpha,\lambda}$. **Proposition.** If $f\in\mathcal K_\alpha$ and $\xi\in C_c(G,H)$, then $F_\alpha(f)\xi$ has the continuous compactly supported representative $$ F_\alpha(f)\xi =\int_G\alpha_{r^{-1}s}(f(s))\xi(s^{-1}r)\,ds. \tag{I7} $$ Its support is contained in $\operatorname{supp}(f)\operatorname{supp}(\xi)$. More generally, for any normal unital representation $\rho:M\to B(H_\rho)$, the regular model built from $\rho$ satisfies $$F_{\alpha,\rho}(f)\xi=\int_G\rho\bigl(\alpha_{r^{-1}s}(f(s))\bigr)\xi(s^{-1}r)\,ds,$$ with the same continuity and support conclusion. Faithfulness of $\rho$ is not required for this formula. When $\rho$ is faithful, the map $f\mapsto F_{\alpha,\rho}(f)$ is injective on $\mathcal K_\alpha$. **Proof.** Applying $\lambda_s\pi_\alpha(f(s))$ to $\xi$ gives $$\lambda_s\pi_\alpha(f(s))\xi=\alpha_{(s^{-1}r)^{-1}}(f(s))\xi(s^{-1}r),$$ which is the integrand of (I7). On a compact neighborhood in the $r$ variable, the vector integrand is jointly continuous on a fixed compact $s$ support. The same uniform-on-compact argument used for convolution proves continuity of the integral. A nonzero integrand requires $r=s(s^{-1}r)$ in the displayed support product. Equality with the integrated bounded operator follows by testing against $C_c(G,H)$ vector sections, using compact Fubini and then density in $L^2$. For injectivity suppose $F_\alpha(f)=0$. Fix $\eta\in H$. For each sufficiently small identity neighborhood $V$, choose $h_V\in C_c(G)$, $h_V\geq0$, supported in $V$, with $$\int_G\Delta_G(t)^{-1}h_V(t)\,dt=1.$$ These functions exist by continuous compactly supported cutoffs and positivity of Haar measure on nonempty open sets; rescale each nonzero cutoff by the displayed integral. Put $\xi_V(t)=h_V(t)\eta$. The change of variables $s=rt^{-1}$ in (I7) gives $$F_\alpha(f)\xi_V =\int_G\Delta_G(t)^{-1}\alpha_{t^{-1}}(f(rt^{-1}))h_V(t)\eta\,dt. \tag{I8}$$ For each fixed $r$, the vector $\alpha_{t^{-1}}(f(rt^{-1}))\eta$ tends to $f(r)\eta$ as $t\to e_G$. Positivity, normalization and shrinking support of $h_V$ show that (I8) tends to $f(r)\eta$. On the other hand $F_\alpha(f)=0$ says that every function in (I8) is zero in $L^2$. Its representative is continuous, so it is zero at every point: a continuous Hilbert-valued function nonzero at a point is bounded away from zero on a nonempty open set of positive Haar measure. Thus $f(r)\eta=0$. Since $r$ and $\eta$ were arbitrary, $f=0$. $\square$ The proof of the kernel formula for a general $\rho$ is identical with $\rho$ applied to each coefficient: normal representations preserve bounded strong-star convergence, so the compact continuity and vector-integration arguments still apply. The injectivity argument then recovers $\rho(f(r))=0$; precisely at this last step, faithfulness is needed to conclude $f(r)=0$. ## OA-FLOW.INT.ALGEBRA — Algebra laws without formal rearrangements **Corollary.** The operations in (I1) make $\mathcal K_\alpha$ an involutive algebra. Its $L^1$ norm is submultiplicative and its involution is isometric. Every $F_{\pi,U}$ is a contractive representation of this normed involutive algebra. **Proof.** The preceding results establish closure of the operations and their integrated identities. In the faithful regular model, associativity of bounded-operator multiplication gives $$F_\alpha((f*g)*h)=F_\alpha(f)F_\alpha(g)F_\alpha(h)=F_\alpha(f*(g*h)).$$ Injectivity of $F_\alpha$ implies associativity of convolution. The same argument with adjoints gives $(f^\sharp)^\sharp=f$ and $(f*g)^\sharp=g^\sharp*f^\sharp$. Linearity and conjugate linearity follow from (I1). The norm assertions are (I3), and the representation assertions are (I5) and the norm bound on (I4). $\square$ The coefficient module operations also satisfy, for $a\in M$, $$ (a\cdot f)*g=a\cdot(f*g),\quad (f\cdot a)*g=f*(a\cdot g),\quad (f*g)\cdot a=f*(g\cdot a),$$ $$ (a\cdot f)^\sharp=f^\sharp\cdot a^*,\qquad (f\cdot a)^\sharp=a^*\cdot f^\sharp.$$ For completeness, apply the faithful regular $F_\alpha$ to the first row. By (I5)–(I6), the respective pairs of expressions become $\pi_\alpha(a)F_\alpha(f)F_\alpha(g)$, $F_\alpha(f)\pi_\alpha(a)F_\alpha(g)$, and $F_\alpha(f)F_\alpha(g)\pi_\alpha(a)$. Applying it to the second row gives $F_\alpha(f)^*\pi_\alpha(a)^*$ and $\pi_\alpha(a)^*F_\alpha(f)^*$. Injectivity proves all five identities. The ordinary bimodule laws follow directly from (I2); their compatibility with convolution has thus also been verified. One may take its norm completion and extend each contractive integrated representation continuously. This observation does not identify its full and reduced C* completions. Those completions, their universal properties and the conditions under which they coincide remain separate course obligations. ## OA-FLOW.INT.GENERATION — Recovering both parts of covariance **Theorem.** For a normal unital covariant pair $(\pi,U)$, $$\{F_{\pi,U}(f):f\in\mathcal K_\alpha\}'' =(\pi(M)\cup U(G))''. \tag{I9}$$ **Proof.** Denote the left side by $A$ and the right side by $B$. Every integrated operator belongs to $B$: any operator commuting with $\pi(M)$ and $U(G)$ commutes with the integrand and its bounded weak integral, and the bicommutant theorem applies. Thus $A\subset B$. Choose nonnegative $h_i\in C_c(G)$ with integral one and supports tending to the identity. Since $\pi$ is unital, $$F(h_i1)=\int_G h_i(s)U_s\,ds\longrightarrow1$$ strongly. Indeed, on each vector the error is bounded by the supremum of $\|(U_s-1)\eta\|$ over the shrinking support, which tends to zero by strong continuity. The module identity now gives $$F(h_i1\cdot a)=F(h_i1)\pi(a)\longrightarrow\pi(a)$$ strongly for every $a\in M$. Hence $\pi(M)\subset A$. For $s\in G$ set $h_{i,s}(r)=h_i(s^{-1}r)$. Left invariance gives $$F(h_{i,s}1)=U_sF(h_i1)\longrightarrow U_s$$ strongly. Therefore $U(G)\subset A$, so $B\subset A$. $\square$ **Degenerate representations.** If $\pi$ is normal but not unital, let $p=\pi(1)$. Covariance makes $p$ commute with every $U_s$, so the pair restricts to a unital pair on $pK$. Every integrated operator vanishes on $(1-p)K$, while $U$ may be arbitrary there. Consequently (I9) holds on $pK$, but its assertion on all of $K$ would generally be false. For instance $\pi=0$ makes every integrated operator zero, while a nontrivial $U$ need not generate the scalar algebra. The theorem names unitality because that is what its recovery argument uses.