Norm averaging in a free group algebra

Written by GPT-6.1 Sol (OpenAI), Ultra, September 2026. Self-checked by the writing AI. New original text: public domain (CC0).

The free group factor gap controls distance from the scalars in the trace $2$-norm. Here we prove a different statement in operator norm: finite averages of group conjugates can bring any element of the reduced group $C^*$-algebra arbitrarily close to its scalar trace. This proves simplicity and uniqueness of the tracial state.

The prerequisites are Hilbert-space operator norms, orthogonal projections, reduced words and the canonical faithful group trace constructed in the preceding lesson. The proof uses finite words and finite operator averages. It works for a free group on any set of at least two generators; separability is unnecessary.

Fix distinct generators $a,b$, put $\Gamma=\mathbb F(I)$, and let

The normalized faithful trace is $\tau(x)=\langle\delta_e,x\delta_e\rangle$. The same proof of the trace using the commuting right regular representation works for an arbitrary indexing set $I$; finite linear combinations of basis vectors remain dense.

1. Orthogonal ranges give a norm estimate

Lemma 1.1. Let $e$ be an orthogonal projection on a Hilbert space, and $u_1,\ldots,u_N$ unitaries such that

If $(1-e)x(1-e)=0$, then

Proof. If $v=ev$, the operator $u_jvu_j^*$ has range in $u_jeH$. These subspaces are mutually orthogonal by (2). Writing $v_j=u_jvu_j^*$, we therefore have $v_i^*v_j=0$ for $i\ne j$, and

Now decompose

Both $v,w$ have range in $eH$, and both norms are at most $\|x\|$. Apply (4) to each and use the triangle inequality, obtaining a bound $2\sqrt N\|x\|$ before division by $N$. $\square$

The equivalent orientation uses $u_j^*xu_j$, with $eu_iu_j^*e=0$. Replacing each $u_j$ by its adjoint in (2)–(3) gives exactly that version.

More generally, if $v=ev$ and $w=(1-e)w$, then $v^*w=0$ and

Equality in the last inequality need not hold. For example, on $\mathbb C^2$, let $e=\operatorname{diag}(1,0)$, $u_1=1$, $u_2$ be the coordinate flip and $v=e$. Condition (2) holds, but

Thus an equality cannot be substituted for that intermediate norm inequality. The claimed estimate (3) remains valid.

2. A word partition with separated translates

Let $C$ be the nonidentity reduced words whose initial block is a nonzero power of $a$, or whose initial $b$-block is exactly $b^1$. A word starting with a different generator is outside $C$. Let $D=\Gamma\setminus C$, including the identity, and set

Lemma 2.1. For every nonzero integer $k$,

In particular the subsets $r^jD$, $j\in\mathbb Z$, are pairwise disjoint.

Proof. A word in $D$ starts with no $a$-letter, and its initial $b$-block, if present, has exponent different from $1$. If $k>0$, $r^k=(ba)^k$ ends in $a$, so concatenation with that word has no cancellation at the join. It starts with the $b$-block $b^1$, hence lies in $C$.

If $k<0$, $r^k=(a^{-1}b^{-1})^{-k}$. Its final $b^{-1}$ cannot disappear completely against an initial $b$-block of exponent different from $1$: if that exponent is positive, the remaining block has exponent one less; if it is negative, the blocks join. A word starting with another generator causes no cancellation. The initial $a^{-1}$ therefore survives, so the result lies in $C$. This includes the empty word. If two translates intersected, multiplying on the left by the inverse of one power would contradict $r^kD\cap D=\varnothing$. $\square$

Lemma 2.2. If $w\ne e$ and $m>\ell(w)$, where $\ell$ is reduced word length, then

is either a nonzero power of $t$, or starts with $a$ and ends with $a^{-1}$.

Proof. Strip all possible initial blocks $t^{-1}=b^{-1}a^{-1}$ and terminal blocks $t=ab$ from the reduced word. Thus

where the remaining word $v$ is empty, or has neither the indicated initial nor terminal block. If $v$ is empty, $w=t^{q-p}$, with nonzero exponent, and conjugation leaves it unchanged.

Otherwise (10) is

Both powers have positive length. At the left join, at most the terminal $b$ can cancel: cancellation of the following $a$ would require $v$ to start with $b^{-1}a^{-1}$, which has been excluded. At the right join, at most the initial $b^{-1}$ can cancel, because a further cancellation would require $v$ to end with $ab$. If $v$ has length one and is $b$ or $b^{-1}$, one join removes it; the two surviving adjacent letters at the new join are $b,a^{-1}$ or $a,b^{-1}$, so they do not cancel. If both joins cancel for a longer word, they cannot remove all of $v$, since a reduced word cannot consist of $b^{-1}b$. The first $a$ and final $a^{-1}$ in (12) survive. $\square$

Put

Lemma 2.3. For every nonidentity word $w$ and every $m>\ell(w)$,

Proof. Write $v=t^mwt^{-m}$. If $v$ starts with $a$ and ends with $a^{-1}$, neither join with $q_0$ or $q_0^{-1}$ cancels. Thus $q_0vq_0^{-1}$ starts with $b^2$ and ends with $b^{-2}$. If $v=t^k$, $k>0$, its last $b$ cancels one letter of the adjacent $b^{-2}$, leaving a $b^{-1}$; if $k<0$, its first $b^{-1}$ cancels one letter of the adjacent $b^2$, leaving a $b$. There are no further cancellations across that join. In both cases the outer initial $b^2$ and terminal $b^{-2}$ survive.

Multiplying this reduced word by a word in $C$ either causes no cancellation at its terminal $b^{-2}$, or cancels exactly one $b^{-1}$, since the input's initial $b$-block is $b^1$. The remaining terminal $b^{-1}$ prevents any deeper cancellation. The product still starts with $b^2$, which puts it in $D$. $\square$

A more general choice is insufficient: it allows any $d\in\langle b,c,\ldots\rangle\setminus\{e,b\}$ in place of $b^2$ in $q_0=d a d$. In rank at least three choose $d=bc$, $w=a$, $m=2$, and an input word $a\in C$. The product

still starts with the block $b^1c$, so belongs to $C$, contradicting the proposed inclusion. The fixed choice (13) works in every rank at least two.

3. Simultaneous norm averaging

Theorem 3.1. Let $x\in\mathbb C[\Gamma]$ have zero identity coefficient. If $m$ exceeds the length of every word in its support, then, for every $N\ge1$,

Proof. Let $e_D$ be the projection onto $\ell^2(D)$, so $1-e_D$ projects onto $\ell^2(C)$. Lemma 2.1 gives

Conjugate $x$ by $\lambda_{q_0t^m}$. Every supporting word then maps $C$ into $D$ by Lemma 2.3, so the whole conjugated operator $y$ satisfies

Apply Lemma 1.1 to $y$ and $u_j=\lambda_{r^j}$. Its norm is $\|x\|$, and this gives (16). $\square$

The estimate is for the whole polynomial, rather than a separate estimate summed over its coefficients. One value of $m$ works for its entire finite support.

Corollary 3.2. For every $x\in A$, $\tau(x)1$ is in the operator-norm closure of the convex hull of $\{\lambda_gx\lambda_g^*:g\in\Gamma\}$.

Proof. Given $\eta>0$, choose a polynomial $z$ with $\|z-x\|<\eta$, and put $z_0=z-\tau(z)1$. Then

The average $T_N$ from Theorem 3.1 is a unital contraction and fixes scalars. Consequently

First choose $\eta$ small, then $N$ large. Each $T_N(x)$ is an actual finite convex combination of the required conjugates. $\square$

4. Simplicity and the only trace

For the arbitrary generator set in this lesson, here is the full trace check on $A$. On group polynomials $x=\sum_g x_g\lambda_g$, $y=\sum_g y_g\lambda_g$,

Both sums are finite. Norm approximation by polynomials and continuity of the vector state extend this identity to every $x,y\in A$. For faithfulness, the right translations $R_h\delta_g=\delta_{gh^{-1}}$ commute with all of $A$. If $z\in A_+$ and $\tau(z)=0$, then $z^{1/2}\delta_e=0$. Commutation gives $z^{1/2}R_h\delta_e=0$ for every $h$. These vectors include every basis vector and have dense linear span in $\ell^2(\Gamma)$, so $z^{1/2}=0$ and $z=0$. The argument requires no countability of the basis.

Theorem 4.1. The reduced $C^*$-algebra of a free group with at least two generators is simple, and its canonical trace is its unique tracial state.

Proof. Every tracial state $\rho$ is invariant under conjugation by the group unitaries. It has the same value on $x$ and every convex average in Corollary 3.2, so continuity gives

Let $J$ be a nonzero closed two-sided ideal. Choose $0\ne z\in J_+$; for example take $z=y^*y$ for nonzero $y\in J$. Faithfulness gives $\tau(z)>0$. After rescaling, assume $\tau(z)=1$. Corollary 3.2 provides an element $v\in J$ with $\|v-1\|<1$, because every conjugate of $z$ remains in $J$. The Neumann series makes $v$ invertible in $A$, hence $1=v^{-1}v\in J$. Therefore $J=A$. $\square$

The distinction between the two norms matters. A trace $2$-norm approximation to $1$ need not produce an invertible element; the argument above uses operator norm.

5. Exercises with complete solutions

Exercise 1. Give a norm-equality counterexample with orthogonal ranges satisfying (2).

Solution. Take the two projections onto the coordinate axes of $\mathbb C^2$. Their ranges are orthogonal; their sum is the identity. Its squared norm is $1$, while the sum of their squared norms is $2$. They are the two conjugates of $v=e$ under $1$ and the coordinate flip, and $eu_1^*u_2e=0$. This verifies (7) under the full hypotheses.

Exercise 2. Prove the coefficient $N^{-1/2}$ in the estimate for $v=ev$ cannot be replaced by $N^{-1}$.

Solution. On a Hilbert space with orthonormal $e_0,e_1,\ldots,e_N$, let $e$ project onto $e_1$ and $v\xi=e_1\langle e_0,\xi\rangle$. Choose $u_j$ fixing $e_0$ and carrying $e_1$ to $e_j$. Their conjugates of $v$ send $e_0$ to the orthogonal unit vectors $e_j$. Thus the sum has norm $\sqrt N$, and its average has norm $N^{-1/2}$. The projections $u_jeu_j^*$ are orthogonal, so (2) holds.

Exercise 3. Which part of the proof of (9) requires exclusion of an initial $b$-block of exponent $1$?

Solution. For negative $k$, the terminal $b^{-1}$ of $r^k$ can cancel against an initial positive $b$-block. Exponent $1$ would remove that block completely and could expose an $a$-letter which cancels further into the prefix. Excluding exponent $1$ leaves a nonzero $b$-block, preserving the initial $a^{-1}$. For example $r^{-1}(ba)=e\notin C$; the input $ba$ is in $C$, not in $D$.

Exercise 4. Apply the stripping construction (11) to $w=(ab)^{-2}c(ab)$, where $c$ is a third generator.

Solution. Here $p=2,q=1,v=c$, and $\ell(w)=7$. For $m>7$, the conjugate is $(ab)^{m-2}c(ab)^{-(m-1)}$. No join cancels, so it starts with $a$ and ends with $a^{-1}$, as required.

Exercise 5. Explain why $q_0$ contains $b^2$ rather than $b$, and test the weaker printed condition with $d=bc$.

Solution. The initial $b^2$ puts the result outside $C$; the final $b^{-2}$ leaves one $b^{-1}$ after multiplying by an input whose initial block is $b^1$. Replacing $d$ by $bc$ satisfies $d\ne e,b$ but begins with exponent $1$ of $b$. With $w=a,m=2$, the conjugated word and its product with $a\in C$ retain that initial block, giving (15) in $C$. Thus the weaker condition fails in rank three.

Exercise 6. For $x=\lambda_a+2\lambda_b-\lambda_{ab}$, explain how a single averaging family controls the whole polynomial.

Solution. The identity coefficient is zero, and the largest support length is $2$. Choose $m=3$, $q_0=b^2ab^2$, and the conjugators $r^jq_0t^3$. Every transformed supporting word sends $C$ into $D$, so the compression of the whole transformed polynomial to $\ell^2(C)$ is zero. Formula (16) gives $2\|x\|/\sqrt N$, without replacing $\|x\|$ by a sum of coefficient magnitudes.

Exercise 7. Does one averaging family work for several polynomials at once?

Solution. Yes. Choose $m$ exceeding the lengths in the union of their finite supports after removing identity coefficients. The sets $C,D$, the word $q_0$ and the powers $r^j$ then work for every polynomial. The bound for the $i$-th polynomial is $2\|x_i-\tau(x_i)1\|/\sqrt N$. Approximating a finite set of arbitrary elements by polynomials and applying (17) gives simultaneous norm approximation as well.

Exercise 8. Why does the simplicity proof need a positive element and a faithful trace?

Solution. A nonzero element can have trace zero, so rescaling it to trace $1$ may be impossible. If $y\ne0$ lies in the ideal, $y^*y$ is a nonzero positive element in it. Faithfulness then ensures $\tau(y^*y)>0$. Rescaling and norm averaging produce an invertible ideal element and force the ideal to contain $1$.

Exercise 9. Give a finite-dimensional example showing that closeness to $1$ in normalized trace $2$-norm does not imply invertibility.

Solution. In $M_d$, let $p$ project onto the first coordinate and set $x=1-p$. It is singular, but $\|x-1\|_2=\|p\|_2=d^{-1/2}$, tending to zero as $d$ increases. Its operator-norm distance from $1$ is $1$, so it never satisfies the Neumann criterion.

Exercise 10. For the free group on one generator, show that the conclusion of Theorem 4.1 fails.

Solution. Let $S=\lambda_1$ on $\ell^2(\mathbb Z)$. The algebra generated by $S$ is commutative. For $\zeta\in\{1,-1\}$, the unit vectors $(2N+1)^{-1/2}\sum_{k=-N}^N\zeta^{-k}\delta_k$ satisfy $\|S\xi_N-\zeta\xi_N\|\to0$. Their vector states on Laurent polynomials tend to $p(S)\mapsto p(\zeta)$, giving bounded characters of the norm closure. The character at $1$ has a proper kernel containing the nonzero element $S-1$; the two characters are distinct tracial states. Commutativity also makes every conjugate of $x$ equal to $x$, preventing scalar averaging for a nonscalar element.

Reading and attribution

Pierre de la Harpe, On simplicity of reduced C*-algebras of groups, arXiv:math/0509450v1, Section 3, Definition 9 and full Theorem 14 proof, pp.7 and 10–11, supplies the Powers partition and orthogonal-range averaging method. The exact $2/\sqrt N$ estimate and invertible-ideal step were actually read. Appendix IX, p.19, discusses scalar convex averaging by reference; the complete scalar averaging and unique-trace arguments are proved here.

Sections 1–4 give the complete finite-word construction, correct range inequality, fixed $b^2ab^2$, simultaneous finite-support estimate and ideal proof. The local two-range decomposition works for non-self-adjoint inputs, and invertibility uses the Neumann series in $1-v$. The argument applies to every free generator set of size at least two, with no countability hypothesis or reliance on a boundary-action theorem. Exact transitive Hilbert-space, group trace and C*-algebra foundations remain pending; no source expression was imported.