A small cocycle with a prescribed word value

Written by Claude Opus 5.5 (Anthropic), October 2026. Self-checked by the writing AI. Public domain (CC0).

Let $n\ge3$ and $\Gamma=\mathbb F_n$, with free generators $x_1,\ldots,x_n$. For $h\in\mathbb R^{(\Gamma)}$ let $D^h$ be the cocycle with generator values $h_1=h$ and $h_2=\cdots=h_n=0$ (Lemma 1.1 of Trace-preserving polynomial flows). This lesson proves:

Theorem 4.1. For every $\eta>0$ there are a word $w\in\Gamma$ and $h\in\mathbb R^{(\Gamma)}$ with $\|h\|_{\ell^2}<\eta\qquad\text{and}\qquad\|D^h_w-\delta_e\|_{\ell^2}<\eta .$

By Theorem 6.2 of the previous lesson, the time-one automorphism of the flow with these coefficients moves each $A_j$ by less than $3\pi\eta$ and moves $A_w$ to within $3\pi\eta$ of $CA_w$. The next lesson uses this to absorb the generator $C$.

The theorem asks for a lot. The value $D^h_w$ is a sum of translates of $h$, one for each occurrence of $x_1^{\pm1}$ in $w$ (Exercise 5.1), so $w$ must be long. The construction takes $w=w_m$ with $m+1$ letters $x_1$ whose prefixes $p_0=e,p_1,\ldots,p_m$ form a free family; then $D^h_{w_m}=T_mh$ with $T_m=\sum_{j=0}^m\lambda(p_j)$. Since the $\lambda(p_j)$ behave like free Haar unitaries, $T_mT_m^*/m$ has, for large $m$, approximately the distribution with density $\frac1{2\pi}\sqrt{(4-x)/x}$ on $(0,4)$. This distribution has no atom at $0$, so $T_mT_m^*$ is at least $dm$ on most of the space, for a fixed small $d$. Inverting $T_mT_m^*$ there gives $h$ of norm about $(dm)^{-1/2}$ with $T_mh$ close to $\delta_e$.

We use: Lemma 2.1 of Free independence and Haar tuples (normal form in free groups) and Lemma 1.1 of Trace-preserving polynomial flows; the spectral measure of a vector and the Borel functional calculus, Proposition 2.1 and Theorem 3.1 of the spectral theorem lesson; and the Weierstrass approximation theorem, Corollary 14.3 of the Stone–Weierstrass lesson.

1. Free prefixes

Put $\begin{gathered} a=x_1,\qquad b_j=x_2^{\,j}x_3x_2^{-j},\qquad p_0=e,\\ p_j=(ab_1)(ab_2)\cdots(ab_j)\quad(j\ge1), \end{gathered}$ and, for $m\ge1$, $w_m=p_ma,\qquad T_m=\sum_{j=0}^m\lambda(p_j)\in L(\Gamma).$ Here $\lambda(g)$ acts on $\ell^2(\Gamma)$ by $(\lambda(g)\xi)(b)=\xi(g^{-1}b)$; on finitely supported real vectors this is the action used for cocycles.

Lemma 1.1. For every $m\ge1$, the homomorphism from the free group on symbols $P_1,\ldots,P_m$ to $\Gamma$ with $P_j\mapsto p_j$ is injective.

Proof. (i) The $b_j$. A nonempty reduced word in symbols $B_1,\ldots,B_m$ has the form $B_{j_1}^{q_1}\cdots B_{j_l}^{q_l}$ with $q_i\ne0$ and $j_i\ne j_{i+1}$. Its value at $(b_1,\ldots,b_m)$ is $x_2^{\,j_1}x_3^{q_1}x_2^{\,j_2-j_1}x_3^{q_2}\cdots x_2^{\,j_l-j_{l-1}}x_3^{q_l}x_2^{-j_l},$ a nonempty reduced word, because all exponents are nonzero and adjacent powers are of different letters. By Lemma 2.1 of the first lesson it is not $e$.

(ii) The elements $a,b_1,\ldots,b_m$. A nonempty reduced word in symbols $A,B_1,\ldots,B_m$ is an alternating product of nonzero powers $A^q$ and nonempty reduced words in the $B_j$. Its value is the corresponding alternating product of powers $x_1^q$ and, by (i), nonempty reduced words in $x_2,x_3$. No cancellation occurs between letters $x_1^{\pm1}$ and letters $x_2^{\pm1},x_3^{\pm1}$, so the value is a nonempty reduced word, and it is not $e$. Hence the homomorphism from the free group $F$ on $A,B_1,\ldots,B_m$ with $A\mapsto a$, $B_j\mapsto b_j$ is injective.

(iii) The elements $ab_j$. The endomorphism of $F$ that fixes $A$ and sends $B_j$ to $AB_j$ has the inverse that fixes $A$ and sends $B_j$ to $A^{-1}B_j$. This automorphism maps the subgroup generated by the $B_j$, which is free on them, onto the subgroup generated by the $AB_j$. So the homomorphism from the free group on symbols $U_1,\ldots,U_m$ with $U_j\mapsto AB_j$ is injective.

(iv) The prefixes. On the free group on $U_1,\ldots,U_m$, let $\varphi(U_j)=U_1U_2\cdots U_j$, $\psi(U_1)=U_1$ and $\psi(U_j)=U_{j-1}^{-1}U_j$ for $j\ge2$. Then $\begin{aligned} &\psi\varphi(U_j)\\ &\quad =U_1(U_1^{-1}U_2)\cdots(U_{j-1}^{-1}U_j)=U_j,\\ &\varphi\psi(U_j)=(U_1\cdots U_{j-1})^{-1}(U_1\cdots U_j)\\ &\quad =U_j, \end{aligned}$ so $\varphi$ is an automorphism and $P_j\mapsto U_1\cdots U_j$ is injective. Composing the injective maps of (iv), (iii) and (ii) sends $P_j$ to $p_j$. $\square$

Lemma 1.2. For every $h\in\mathbb R^{(\Gamma)}$ and $m\ge1$, $D^h_{w_m}=T_mh$.

Proof. By Lemma 1.1 of the previous lesson, $D^h_{gb}=D^h_g+\lambda(g)D^h_b$ and $D^h_{g^{-1}}=-\lambda(g^{-1})D^h_g$, so the elements $g$ with $D^h_g=0$ form a subgroup. It contains $x_2,\ldots,x_n$, hence every $b_j$. Write $w_m=a\,b_1\,a\,b_2\cdots a\,b_m\,a$ and apply the product formula of that lemma: each factor $b_j$ contributes $0$, and the factor $a$ that follows the prefix $(ab_1)\cdots(ab_k)=p_k$ contributes $\lambda(p_k)D^h_a=\lambda(p_k)h$, for $k=0,1,\ldots,m$. $\square$

2. The moments of the prefix sums

Let $\tau_\Gamma(x)=\langle x\delta_e,\delta_e\rangle$ be the trace of $L(\Gamma)$, so $\tau_\Gamma(\lambda(g))$ is $1$ for $g=e$ and $0$ otherwise. Put $Q_m=T_mT_m^*$, a positive operator with $\|Q_m\|\le(m+1)^2$. Let $\rho_m$ be the spectral measure of the vector $\delta_e$ for $Q_m$, a probability measure on $[0,\infty)$ with $\int f\,d\rho_m=\langle f(Q_m)\delta_e,\delta_e\rangle=\tau_\Gamma(f(Q_m))$ for bounded Borel functions $f$, and let $\mu_m$ be its image under $x\mapsto x/m$. Then $\begin{gathered} \int x^r\,d\mu_m=m^{-r}\tau_\Gamma(Q_m^r)\quad(r\ge0),\\ \mu_m([0,d])=\tau_\Gamma\big(1_{[0,dm]}(Q_m)\big)\quad(d\ge0). \end{gathered}$

A perfect matching of $\{1,\ldots,2r\}$ is a partition into $r$ blocks of two elements. It is noncrossing if there are no blocks $\{s_1,s_1'\}$ and $\{s_2,s_2'\}$ with $s_1<s_2<s_1'<s_2'$. In a noncrossing perfect matching, the positions strictly between the two elements of a block are matched among themselves, so their number is even and every block joins an odd position with an even one.

Lemma 2.1. Let $y_1y_2\cdots y_{2r}$ be a word in letters $P_c^{\pm1}$ in which every symbol $P_c$ that occurs occurs exactly twice, once with exponent $+1$ and once with exponent $-1$. Let $\pi$ be the perfect matching of $\{1,\ldots,2r\}$ that pairs the two positions carrying the same symbol. The word represents $e$ in the free group if and only if $\pi$ is noncrossing.

Proof. Let $\pi$ be noncrossing. Take a block $\{s,s'\}$, $s<s'$, with $s'-s$ smallest. If $s'>s+1$, position $s+1$ would be matched with a position strictly between $s$ and $s'$, contradicting minimality; so $s'=s+1$. The letters at $s$ and $s+1$ are $P_c^{\pm1}$ and $P_c^{\mp1}$; deleting them leaves a word of the same kind whose matching is still noncrossing. By induction on $r$ the word represents $e$.

Conversely, let the word represent $e$. Deleting adjacent inverse pairs as long as possible ends at the reduced word of $e$, which is empty (Lemma 2.1 of the first lesson). Each deletion removes two adjacent letters $P_c^{\pm1}P_c^{\mp1}$, that is, both positions of one block of $\pi$. Suppose blocks $\{s_1,s_1'\}$ and $\{s_2,s_2'\}$ crossed, $s_1<s_2<s_1'<s_2'$. When the first of them is deleted, the other is still present, and one of its positions lies strictly between the positions being deleted, so these are not adjacent. This contradiction shows that $\pi$ is noncrossing. $\square$

Lemma 2.2. The number of noncrossing perfect matchings of $\{1,\ldots,2r\}$ is the Catalan number $\mathrm{Cat}_r=\frac{(2r)!}{r!\,(r+1)!}$.

Proof. Dyck sequences. Call $\varepsilon\in\{\pm1\}^{2r}$ a Dyck sequence if its partial sums $\varepsilon_1+\cdots+\varepsilon_s$ are $\ge0$ and the total is $0$. To a noncrossing matching $\pi$ assign $\varepsilon_s=+1$ if $s$ is the smaller element of its block and $-1$ otherwise; the partial sum at $s$ counts the blocks opened but not closed by time $s$, so $\varepsilon$ is a Dyck sequence. We show by induction on $r$ that every Dyck sequence comes from exactly one noncrossing matching. Let $\{1,s'\}$ be the block of $1$ in a noncrossing $\pi$ producing $\varepsilon$. The positions $2,\ldots,s'-1$ are matched among themselves, so the partial sums are $\ge1$ before $s'$ and $0$ at $s'$: $s'$ is the first return of the partial sums to $0$, determined by $\varepsilon$. The positions inside and outside $\{1,s'\}$ carry Dyck sequences, which determine the rest of $\pi$ by induction. Conversely, for a Dyck sequence, let $s'$ be the first return to $0$; the block $\{1,s'\}$ together with noncrossing matchings of the inside and the outside (given by induction) is a noncrossing matching producing $\varepsilon$.

Reflection. There are $\binom{2r}{r}$ sequences in $\{\pm1\}^{2r}$ with total $0$. If such a sequence has a negative partial sum, let $s_0$ be the first position where the partial sum is $-1$, and change the signs of $\varepsilon_{s_0+1},\ldots,\varepsilon_{2r}$; the result has total $-2$. This is a bijection onto the sequences with total $-2$: each of those reaches $-1$, and the same operation reverses it. There are $\binom{2r}{r-1}$ such sequences. So the number of Dyck sequences is $\begin{aligned} &\binom{2r}{r}-\binom{2r}{r-1}=\frac{(2r)!}{r!\,r!}\\ &\quad -\frac{(2r)!}{(r-1)!\,(r+1)!}\\ &\quad =\frac{(2r)!}{r!\,(r+1)!}. \end{aligned}$ $\square$

Lemma 2.3. For every $r\ge0$, $\tau_\Gamma(Q_m^r)$ is the sum of $\mathrm{Cat}_r\,m(m-1)\cdots(m-r+1)$ and a number $E_{m,r}$ with $0\le E_{m,r}\le c_rm^{r-1}$, where $c_r$ depends only on $r$. Consequently $\int x^r\,d\mu_m\to\mathrm{Cat}_r$ as $m\to\infty$.

Proof. For $r=0$ both sides are $1$. Let $r\ge1$ and $V_i=\lambda(p_i)$. Expanding, with all indices in $\{0,\ldots,m\}$, $\begin{aligned} &\tau_\Gamma(Q_m^r)=\sum_{i_1,\ldots,i_{2r}}\tau_\Gamma\big(V_{i_1}V_{i_2}^*V_{i_3}V_{i_4}^*\cdots\\ &\qquad\qquad\cdots V_{i_{2r-1}}V_{i_{2r}}^*\big). \end{aligned}$ The summand is $1$ if $p_{i_1}p_{i_2}^{-1}\cdots p_{i_{2r}}^{-1}=e$ and $0$ otherwise. By Lemma 1.1, this happens exactly when the word $\omega(i)=P_{i_1}P_{i_2}^{-1}\cdots P_{i_{2r}}^{-1}$, in which the symbols $P_0$ are omitted, represents $e$ in the free group on $P_1,\ldots,P_m$. Fix an index sequence $i$ with summand $1$, and let $k$ be the number of distinct nonzero indices in it.

Each nonzero index occurs at least twice. For a fixed $c\ge1$, the homomorphism to $\mathbb Z$ that counts the exponent sum of $P_c$ sends $\omega(i)$ to the number of odd positions carrying $c$ minus the number of even positions carrying $c$; this must be $0$. So $2k\le2r$.

The case $k=r$. Then all $2r$ positions are nonzero and each index occurs exactly twice, once at an odd and once at an even position, that is, with opposite exponents. By Lemma 2.1 the summand is $1$ exactly when the matching of equal indices is noncrossing. Conversely, a noncrossing perfect matching $\pi$ and an injective assignment of indices from $\{1,\ldots,m\}$ to its $r$ blocks give such a sequence, because every block joins an odd and an even position; different pairs $(\pi,\text{assignment})$ give different sequences. By Lemma 2.2 these sequences contribute $\mathrm{Cat}_r\,m(m-1)\cdots(m-r+1)$.

The case $k\le r-1$. Such a sequence is determined by its set of zero positions, the partition of the other positions into classes of equal indices, and an injective assignment of indices to the classes. There are at most $c_r$ choices for the first two data, depending only on $r$, and at most $m^{r-1}$ for the third. This bounds $E_{m,r}$.

Dividing by $m^r$, $\int x^r\,d\mu_m=m^{-r}\tau_\Gamma(Q_m^r)\to\mathrm{Cat}_r$. $\square$

3. The limit law

For a continuous function $\varphi$ on $[0,4]$ put $\Lambda(\varphi)=\frac4\pi\int_0^{\pi/2}\varphi(4\sin^2\theta)\cos^2\theta\,d\theta .$ $\Lambda$ is positive and linear, and $\Lambda(1)=\frac4\pi\cdot\frac\pi4=1$. The substitution $x=4\sin^2\theta$ shows that $\Lambda(\varphi)=\int_0^4\varphi(x)\,\frac1{2\pi}\sqrt{(4-x)/x}\,dx$ (Exercise 5.3), the distribution described in the introduction.

Lemma 3.1. $\Lambda(x^r)=\mathrm{Cat}_r$ for every $r\ge0$.

Proof. Let $W_k=\int_0^{\pi/2}\sin^k\theta\,d\theta$. Integration by parts gives $W_k=\frac{k-1}kW_{k-2}$ for $k\ge2$, and $W_0=\frac\pi2$, so $W_{2r}=\frac\pi2\cdot\frac{(2r)!}{4^r(r!)^2}$. Since $\cos^2=1-\sin^2$, $\begin{aligned} &\int_0^{\pi/2}\sin^{2r}\theta\cos^2\theta\,d\theta=W_{2r}\\ &\quad -W_{2r+2}=\frac{W_{2r}}{2r+2}, \end{aligned}$ and $\begin{aligned} &\Lambda(x^r)=\frac{4^{r+1}}\pi\cdot\frac{W_{2r}}{2r+2}\\ &\quad =\frac{4^{r+1}}\pi\cdot\frac\pi2\cdot\frac{(2r)!}{4^r(r!)^2\,(2r+2)}\\ &\quad =\frac{(2r)!}{r!\,(r+1)!}. \end{aligned}$ $\square$

Lemma 3.2. For every bounded continuous $\varphi\colon[0,\infty)\to\mathbb R$, $\int\varphi\,d\mu_m\to\Lambda(\varphi|_{[0,4]})$ as $m\to\infty$.

Proof. Let $\varepsilon>0$ and $c=\sup|\varphi|$. By the Weierstrass approximation theorem there is a real polynomial $q$ with $|\varphi-q|\le\varepsilon$ on $[0,5]$. Let $N$ be its degree and $C_q$ the sum of the absolute values of its coefficients. For $x\ge5$, $|\varphi(x)-q(x)|\le c+C_qx^N$, which is at most $(c+C_q)x^{N+s}5^{-s}$ for every $s\ge0$. Hence $\begin{aligned} &\Big|\int\varphi\,d\mu_m-\int q\,d\mu_m\Big|\le\varepsilon\\ &\quad +(c+C_q)5^{-s}\int x^{N+s}\,d\mu_m . \end{aligned}$ By Lemma 2.3 and Lemma 3.1, $\int q\,d\mu_m\to\Lambda(q)$ and $\int x^{N+s}\,d\mu_m\to\mathrm{Cat}_{N+s}\le4^{N+s}$, the last because $\Lambda(x^{N+s})\le4^{N+s}$. Also $|\Lambda(\varphi)-\Lambda(q)|\le\varepsilon$. Therefore $\begin{aligned} &\limsup_m\Big|\int\varphi\,d\mu_m-\Lambda(\varphi)\Big|\\ &\quad \le2\varepsilon+(c+C_q)4^N(4/5)^s . \end{aligned}$ Let $s\to\infty$, then $\varepsilon\to0$. $\square$

Corollary 3.3. For every $\kappa>0$ there are $d>0$ and $m_0$ with $\mu_m([0,d])<\kappa$ for all $m\ge m_0$.

Proof. Choose $d\in(0,1]$ with $\frac4\pi\arcsin\sqrt{d/2}<\kappa$. Let $\varphi$ be $1$ on $[0,d]$, $0$ on $[2d,\infty)$ and linear in between. Then $\mu_m([0,d])\le\int\varphi\,d\mu_m$, and $\Lambda(\varphi)\le\frac4\pi\cdot\big|\{\theta\in[0,\tfrac\pi2]:4\sin^2\theta\le2d\}\big|$, which is $\frac4\pi\arcsin\sqrt{d/2}<\kappa$, using $\cos^2\le1$. By Lemma 3.2, $\int\varphi\,d\mu_m<\kappa$ for all large $m$. $\square$

4. Cutoff inversion

Proof of Theorem 4.1. Let $\eta>0$. By Corollary 3.3 choose $d>0$ and then $m$ such that $\begin{gathered} \mu_m([0,d])<\frac{\eta^2}8\\ \text{and}\qquad \frac1{dm}<\frac{\eta^2}{16}. \end{gathered}$ Let $f$ be the bounded Borel function with $f(x)=1/x$ for $x>dm$ and $f(x)=0$ for $x\le dm$, and put $E=1_{(dm,\infty)}(Q_m),\qquad R=f(Q_m).$ By the Borel functional calculus, $R\ge0$, $\|R\|\le1/(dm)$, $Q_mR=E$ (because $xf(x)=1_{(dm,\infty)}(x)$) and $RQ_mR=R$ (because $xf(x)^2=f(x)$). Also $\langle(1-E)\delta_e,\delta_e\rangle=\tau_\Gamma(1_{[0,dm]}(Q_m))$, which is $\mu_m([0,d])$. Let $h^0=T_m^*R\,\delta_e\in\ell^2(\Gamma)$. Then $\begin{gathered} T_mh^0=Q_mR\,\delta_e=E\delta_e,\\ \|T_mh^0-\delta_e\|^2=\langle(1-E)\delta_e,\delta_e\rangle<\frac{\eta^2}8, \end{gathered}$ $\begin{aligned} &\|h^0\|^2=\langle RQ_mR\,\delta_e,\delta_e\rangle\\ &\quad =\langle R\,\delta_e,\delta_e\rangle\le\frac1{dm}<\frac{\eta^2}{16}. \end{aligned}$ Real coefficients. Let $J$ be the coefficientwise complex conjugation of $\ell^2(\Gamma)$. It is isometric, fixes $\delta_e$, and commutes with every $\lambda(g)$, hence with $T_m$. The real part $h^1=\frac12(h^0+Jh^0)$ satisfies $\|h^1\|\le\|h^0\|<\eta/4$ and $\begin{aligned} &\|T_mh^1-\delta_e\|\\ &\quad =\tfrac12\big\|(T_mh^0-\delta_e)+J(T_mh^0-\delta_e)\big\|\\ &\quad \le\|T_mh^0-\delta_e\|<\eta/\sqrt8 . \end{aligned}$ Finite support. The finitely supported real vectors are dense in the real vectors of $\ell^2(\Gamma)$; choose $h\in\mathbb R^{(\Gamma)}$ with $\|h-h^1\|<\eta/(4(m+1))$. Since $\|T_m\|\le m+1$, $\begin{gathered} \|h\|<\frac\eta4+\frac\eta{4(m+1)}<\eta,\\ \|T_mh-\delta_e\|<\frac\eta4+\frac\eta{\sqrt8}<\eta . \end{gathered}$ By Lemma 1.2, $w=w_m$ satisfies $D^h_w=T_mh$, which proves the theorem. $\square$

The order of the choices matters: first $d$, then $m$, then the finitely supported approximation, whose support may be as large as needed.

5. Exercises

Exercise 5.1. Let $w\in\Gamma$ contain $N$ letters $x_1^{\pm1}$ in its reduced word. Show that $\|D^h_w\|_{\ell^2}\le N\|h\|_{\ell^2}$, and conclude that in Theorem 4.1 the word $w$ has at least $(1-\eta)/\eta$ letters $x_1^{\pm1}$.

Exercise 5.2. Show that $\tau_\Gamma(Q_m)=m+1$ and $\tau_\Gamma(Q_m^2)=(m+1)(2m+1)$.

Exercise 5.3. Show that $\Lambda(\varphi)=\int_0^4\varphi(x)\frac1{2\pi}\sqrt{(4-x)/x}\,dx$ for continuous $\varphi$ on $[0,4]$, the integral being improper at $0$.

Exercise 5.4. Show that the Catalan numbers satisfy $\mathrm{Cat}_0=1$ and $\mathrm{Cat}_r=\sum_{j=0}^{r-1}\mathrm{Cat}_j\mathrm{Cat}_{r-1-j}$ for $r\ge1$, by splitting a noncrossing matching at the partner of $1$.

6. Solutions

5.1. By Lemma 1.1 of the previous lesson, $D^h_w=\sum_k\lambda(y_1\cdots y_{k-1})D^h_{y_k}$ over the letters $y_k$ of the reduced word. Letters $x_j^{\pm1}$ with $j\ge2$ contribute $0$; a letter $x_1$ contributes a translate of $h$, and a letter $x_1^{-1}$ contributes minus a translate of $h$. Translates have norm $\|h\|$, so $\|D^h_w\|\le N\|h\|$. If $\|h\|<\eta$ and $\|D^h_w-\delta_e\|<\eta$, then $1-\eta<\|D^h_w\|<N\eta$.

5.2. $\tau_\Gamma(Q_m)=\sum_{i,i'}\tau_\Gamma(\lambda(p_ip_{i'}^{-1}))$ counts the pairs with $p_i=p_{i'}$, that is $i=i'$: $m+1$. For $Q_m^2$, the word $P_{i_1}P_{i_2}^{-1}P_{i_3}P_{i_4}^{-1}$ (with $P_0$ omitted) represents $e$ exactly when $i_1=i_2$ and $i_3=i_4$, or $i_2=i_3$ and $i_1=i_4$. Both conditions clearly give $e$. Conversely, if the word represents $e$, the exponent sum of every $P_c$ is $0$, so the nonzero entries of $\{i_1,i_3\}$ and of $\{i_2,i_4\}$ agree as multisets; as both multisets have two elements, $\{i_1,i_3\}=\{i_2,i_4\}$, which is one of the two conditions. The two families have $(m+1)^2$ members each and $m+1$ in common, so $\tau_\Gamma(Q_m^2)=2(m+1)^2-(m+1)$, which is $(m+1)(2m+1)$. This agrees with Lemma 2.3: $\mathrm{Cat}_2\,m(m-1)=2m^2-2m$, with an error of order $m$.

5.3. For $0<\epsilon<\pi/2$ substitute $x=4\sin^2\theta$ on $[\epsilon,\pi/2]$: $dx=8\sin\theta\cos\theta\,d\theta$ and $\sqrt{(4-x)/x}=\cos\theta/\sin\theta$, so $\frac1{2\pi}\varphi(x)\sqrt{(4-x)/x}\,dx$ becomes $\frac4\pi\varphi(4\sin^2\theta)\cos^2\theta\,d\theta$. Let $\epsilon\to0$.

5.4. In a noncrossing matching of $\{1,\ldots,2r\}$ the partner of $1$ is an even position $2j+2$, $0\le j\le r-1$; the $2j$ positions inside and the $2r-2j-2$ positions outside carry independent noncrossing matchings, and every such choice gives a noncrossing matching.

References

[OAI] OpenAI, An isomorphism of the free group factors (September 23, 2026), OpenAI Math Release preprint, Section 4. https://github.com/openai/math/blob/main/preprints/An-isomorphism-of-the-free-group-factors-September-23-2026/An-isomorphism-of-the-free-group-factors-September-23-2026.pdf