Projections in the reduced 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 reduced $C^*$-algebra $A=C_r^*(\mathbb F(a,b))$ has only two projections: $0$ and $1$. Its von Neumann closure is a factor of type $\mathrm{II}_1$ with many projections, so the assertion concerns the norm-closed algebra. We prove it by comparing two representations that differ by trace-class operators on a dense algebra.

We use the faithful canonical trace from the group factor lesson and the norm-averaging result. Foundational prerequisites are the compact self-adjoint spectral theorem, trace-class completeness and its two-sided ideal bound, absolute diagonal summation for the trace, and the holomorphic functional calculus with surrounding cycles. The exact foundational proof scopes remain prerequisites; selected free trace-class methods are identified below. The representation comparison, trace-class closure, projection perturbation and integer trace calculation are proved below.

Write $\Gamma=\mathbb F(a,b)$, $H=\ell^2(\Gamma)$, $P$ for the projection onto $\mathbb C\delta_e$, and $p=1-P$. Inner products are linear in the second variable.

1. Removing the identity from the regular representation

On $H'=pH$, define two permutations of its basis by

The ordinary left translation would send $\delta_{s^{-1}}$ to the missing vector $\delta_e$; formula (1) sends it to $\delta_s$, which had the missing predecessor $e$. This is a bijection of the remaining basis. Its inverse has the same formula with $s$ replaced by $s^{-1}$. The free generators therefore define a unitary representation $\lambda'$ of $\Gamma$ on $H'$.

We need more than a representation of the abstract group: we must show it extends to the reduced norm.

For $c=a,b$, define a bijection $\phi_c:\Gamma\to\Gamma_c$, where $\Gamma_c$ is the set of words ending in $c$ or $c^{-1}$, by

In the second case appending $c$ causes no cancellation. Words ending in $c^{-1}$ are their own preimage; words ending in $c$ have preimage obtained by removing their last $c$. These two rules prove bijectivity.

Lemma 1.1. The map

is unitary and intertwines $\lambda\oplus\lambda$ with $\lambda'$.

Proof. The disjoint sets $\Gamma_a,\Gamma_b$ partition the nonidentity words, so (3) sends an orthonormal basis bijectively onto an orthonormal basis. For each letter $s=a^{\pm1},b^{\pm1}$, direct reduction gives

Here $T_{s^{-1}}=T_s^*$. Away from $w=e$ and $w=s^{-1}$, multiplication on the left preserves the terminal letter, so (4) is ordinary concatenation and reduction. The exceptional modified step can occur only when $\phi_c(w)=s^{-1}$, which means $w=e,s=c^{-1}$, or $w=c^{-1},s=c$; these are among the two boundary cases. They give respectively $\phi_c(c^{-1})=c^{-1}$ and $\phi_c(e)=c$, exactly as (1) requires. In the other boundary cases, cancelling $s^{-1}$ against $s$ leaves the appended $c$, again $\phi_c(e)$. Thus (4) holds for every letter, and hence every word. $\square$

Consequently $\lambda'$ extends isometrically to a representation of $A$. On $H$ let $\rho(x)$ denote this representation extended by zero on $\mathbb C\delta_e$. It is a star homomorphism supported on $p$, with

The original representation is $\pi(x)=x$. Define

In particular $D(1)=P$, rather than zero.

2. Finite differences and a dense algebra

For vectors $\xi,\eta$, put $\theta_{\xi,\eta}\zeta=\xi\langle\eta,\zeta\rangle$. Formula (1), also for inverse letters, gives

It vanishes on every other basis vector and has rank $2$. For arbitrary $x,y\in A$, multiplicativity gives the exact identity

Thus $D(\lambda_g)$ has finite rank for every group element, by induction on word length; for $g\ne e$ its rank is at most $2\ell(g)$.

Let $\mathcal S_1(H)$ be the trace-class operators and define

Recall the prerequisite inequalities and trace formula:

where the last sum is absolutely convergent.

Lemma 2.1. The set $A_0$ is a norm-dense unital star subalgebra of $A$. It is complete in the graph norm

Proof. The ideal inequality and (8) give closure under products and adjoints; $D(1)=P$ is trace class. Finite group polynomials belong to $A_0$ by (7)–(8), so it is norm dense. Also (8) bounds the graph norm of a product by the product of the graph norms. If $x_n$ is graph-norm Cauchy, then $x_n\to x$ in $A$ and $D(x_n)\to T$ in trace norm. Since $D:A\to B(H)$ is operator-norm bounded and trace norm dominates operator norm, $D(x)=T$. Thus $x\in A_0$, proving completeness. $\square$

Density alone does not show that every element of $A$ has trace-class difference. We will use a spectral-gap projection construction inside $A_0$.

3. Holomorphic closure with the missing unit retained

Theorem 3.1. If $x\in A_0$ and $f$ is holomorphic on a neighborhood of $\operatorname{Sp}_A(x)$, then $f(x)\in A_0$.

Proof. For $z\notin\operatorname{Sp}_A(x)$, put

The latter is the resolvent on $pH$, extended by zero. Its corner identities are

Together with $\pi(x)R(z)=zR(z)-1$, they give

Indeed the second term is $R(z)p-R'(z)$; adding the first gives the claimed difference. The rank-one term is essential because the representations have different units.

The right side is trace class by (10), and is continuous in trace norm along every compact resolvent contour. Let $\mathcal C$ be a surrounding cycle within the domain of $f$. Computing the functional calculus in $B(H)$ and in the unital corner $B(pH)$ gives

This is a trace-norm integral in the Banach space $\mathcal S_1(H)$. It lies in that space, proving the theorem. No division by $z$ is used, so the formula is valid even if a contour passes through $0$ outside the spectrum. $\square$

4. Recovering the canonical trace

Lemma 4.1. For every $x\in A_0$,

Proof. Every basis diagonal of $\pi(x)$ equals $\tau(x)$: this holds for group polynomials and extends by operator-norm continuity. Lemma 1.1 makes $\rho$ the sum of two regular representations on $pH$, with every basis vector corresponding to a regular basis vector. Thus

The diagonal of $D(x)$ is zero except at $e$, where it is $\tau(x)$. Absolute diagonal summation in (10) proves (14). This avoids an unjustified trace-norm limit of arbitrary polynomial approximants. $\square$

Lemma 4.2. If $E,F$ are projections on any Hilbert space and $E-F$ is trace class, then

Proof. Put $B=E-F$, $C=E+F-1$. Multiplication gives

The self-adjoint operator $B$ is compact, and $-1\le B\le1$. Its nonzero eigenvalues have finite multiplicities. If $B\xi=t\xi$, $0<|t|<1$, then

Hence $C/\sqrt{1-t^2}$ identifies the eigenspaces for $t$ and $-t$ isometrically. Their dimensions agree. The trace-class condition permits absolute summation of the eigenvalues, so all these pairs cancel.

The eigenspace for $1$ is $EH\cap\ker F$: equality in

forces $E\xi=\xi,F\xi=0$. The eigenspace for $-1$ is obtained by exchanging $E,F$. Both are finite-dimensional by compactness. Only their dimension difference remains in the trace, giving (15). $\square$

Thus every projection $q\in A_0$ satisfies

Both $\pi(q)$ and $\rho(q)$ are genuine orthogonal projections on $H$, even though the latter representation is supported on $p$.

5. Bringing every projection into the dense algebra

Lemma 5.1. Every projection $e\in A$ is unitarily conjugate within $A$ to a projection $q\in A_0$.

Proof. Choose a self-adjoint group polynomial $h$ with

Such a polynomial is obtained by symmetrizing a polynomial approximant. For real $z$ at distance greater than $\delta$ from $\{0,1\}$, the resolvent identity and Neumann series show $z-h$ invertible. Since $h$ is self-adjoint,

Choose $f$ equal to $0$ on a neighborhood of the first interval and $1$ on a disjoint neighborhood of the second. This function is holomorphic on their disconnected union. Theorem 3.1 gives $q=f(h)\in A_0$, and the functional calculus makes $q$ a self-adjoint projection.

The circle $|z-1|=1/2$ isolates the second spectral cluster. The Riesz formulas for $q,e$ and their resolvent difference give

the circle has length $\pi$, the resolvent of $e$ has norm at most $2$ there, and that of $h$ has norm at most $(1/2-\delta)^{-1}$.

Set $v=qe+(1-q)(1-e)$. Direct multiplication gives

The last operator is invertible by (19) and commutes with $e,q,v$. Therefore

is unitary and $ueu^*=q$. $\square$

Theorem 5.2. Every projection $e\in C_r^*(\mathbb F(a,b))$ has integer canonical trace. Consequently the only projections are $0$ and $1$.

Proof. Lemma 5.1 and traciality give $\tau(e)=\tau(q)$, which is integer by (17). Since $0\le\tau(e)\le1$, it is $0$ or $1$. Faithfulness gives $e=0$ in the first case and $1-e=0$ in the second. $\square$

This says nothing comparable about projections in matrix algebras over $A$: $\operatorname{diag}(1,0)\in M_2(A)$ is already a nontrivial projection. Nor does it exclude projections in the von Neumann closure. The norm density and spectral-gap step above explain why the assertion concerns $A$ itself.

6. Exercises with complete solutions

Exercise 1. Check that the modified generator $T_a$ is a permutation, including its exceptional inverse step.

Solution. Ordinary left multiplication sends $a^{-1}$ to the omitted $e$, and $e$ to $a$. After removing $e$, replace these two steps by $a^{-1}\mapsto a$; every other input and output retains its ordinary predecessor. Thus the map is bijective on nonidentity words. Its inverse sends $a\mapsto a^{-1}$ and every other word $g$ to $a^{-1}g$. Both preserve the orthonormal basis, giving inverse unitaries.

Exercise 2. Compute $\phi_a(e),\phi_a(a^{-1}),\phi_a(a),\phi_a(b)$, and explain how the two maps in (3) fill $H'$.

Solution. These values are $a,a^{-1},a^2,ba$. The image of $\phi_a$ consists exactly of words ending in $a^{\pm1}$; the analogous image of $\phi_b$ consists of those ending in $b^{\pm1}$. Every nonidentity reduced word ends in exactly one of these two kinds, and (2) has a unique inverse on each. Thus (3) is a basis bijection and not merely an embedding.

Exercise 3. Compute the rank, trace and trace norm of $D(\lambda_a)$ on its finite-dimensional support.

Solution. Its only nonzero columns are the inputs $\delta_e,\delta_{a^{-1}}$, with outputs $\delta_a,\delta_e-\delta_a$. These are linearly independent, giving rank $2$. Every diagonal entry is zero, so its trace is zero. The Gram matrix of these columns is

Its eigenvalues are $(3\pm\sqrt5)/2$, with sum $3$ and product $1$. The two singular values therefore have squared sum $3$ and product $1$, so their sum, the trace norm, is $\sqrt5$.

Exercise 4. Why does finite rank for the generators imply finite rank for every group word, without summing an infinite Fourier series?

Solution. Formula (8) expresses the difference for a product of two words as bounded operators multiplying their two differences. Finite-rank operators form a two-sided ideal, so induction on the finite word length proves the claim. The same formula bounds the rank by the sum of the generator difference ranks, hence $2\ell(g)$ for a nonempty reduced word. The identity has difference $P$, of rank $1$.

Exercise 5. Test (12) with $x=1$. What error would omitting the term $R(z)P$ cause?

Solution. For $z\ne1$, $R(z)=(z-1)^{-1}1$, $R'(z)=(z-1)^{-1}p$, and their difference is $(z-1)^{-1}P$. Since $D(1)=P$ and $Pp=0$, the product $R(z)D(1)R'(z)$ is zero. The first term in (12) gives the entire difference. Omitting it would incorrectly identify the different units of the two representations.

Exercise 6. Does operator-norm density of group polynomials prove (14) by continuity of $\operatorname{Tr}$?

Solution. No. The trace is continuous for trace norm, and operator-norm convergence does not control that norm. Rank-$N$ projections divided by $N$ have operator norm $1/N$ but trace $1$. The proof instead computes every diagonal of the actual trace-class difference $D(x)$, and uses its absolute diagonal sum.

Exercise 7. Compute (15) for two rank-one projections whose ranges make an angle $0<\theta<\pi/2$.

Solution. Choose unit vectors $e_1$ and $\cos\theta\,e_1+\sin\theta\,e_2$. On their two-dimensional span the projection difference has trace zero and determinant $-\sin^2\theta$, so its eigenvalues are $\pm\sin\theta$. There is no $1$ or $-1$ eigenspace. The generic pair cancels, giving trace zero. Adding an orthogonal one-dimensional summand to the range of the first projection adds one unmatched $1$ eigenvalue and changes the trace to $1$.

Exercise 8. Show that compactness, or even the Hilbert–Schmidt condition, cannot replace trace class in the trace calculation for projection differences.

Solution. Take the direct sum of the rank-one pairs from Exercise 7 with $\sin\theta_n=1/n$, $n\ge2$. The difference has eigenvalues $\pm1/n$, so is compact and Hilbert–Schmidt because $\sum_n2/n^2<\infty$. It is not trace class because $\sum_n2/n=\infty$. The formal paired sum zero is not an absolutely convergent trace. Thus the trace in (15) would be undefined.

Exercise 9. With $\|h-e\|=0.1$, estimate the projection error and explain why (20) stays inside $A$.

Solution. Formula (19) gives $\|q-e\|\le0.1/(0.5-0.1)=0.25$. Thus $1-(e-q)^2$ is positive with spectrum in $[1-0.25^2,1]$, so its inverse square root is obtained by the continuous functional calculus in $A$. Multiplying it by $v\in A$ gives $u\in A$. The algebraic identities in the proof make this operator a unitary that aligns the projections.

Exercise 10. Explain why the projection $\operatorname{diag}(1,0)$ in $M_2(A)$ does not contradict Theorem 5.2.

Solution. The theorem concerns projections in $A$. In $M_2(A)$, the indicated matrix is a nonzero projection different from the matrix identity. Its unnormalized matrix trace composed with $\tau$ is $1$, an integer, while its normalized trace is $1/2$. Matrix amplification changes the unit's unnormalized trace from $1$ to $2$, so integrality no longer forces a projection to be $0$ or the unit.

Reading and attribution

Thomas Schick, The trace on the K-theory of group C*-algebras, arXiv:math/9907121v5, Lemmas 2.6, 2.8 and 2.9, pp.7–11, supplies the actually read diagonal trace, nonunital resolvent and projection-difference methods. The source cites trace ideal facts and holomorphic-density K-theory results by reference. Here the full ordinary-Hilbert-space argument includes the explicit basis intertwiner, graph-norm closure, missing-unit resolvent term, absolute diagonal trace, eigenvalue pairing and unitary alignment of nearby projections. It does not require a K-theory computation.

Mihai Pimsner and Dan Voiculescu, K-groups of reduced crossed products by free groups, Journal of Operator Theory 8(1) (1982), 131–156, introduction pp.131–132, states projection absence. Those pages were actually visually read; its full K-theory proof is not claimed read or used here. The source distinguishes this published paper from an earlier preprint with a false key lemma.

The proof concerns projections in $C_r^*(\mathbb F(a,b))$ itself. Section 5 supplies the full spectral-gap and conjugacy argument, so mere norm density is not used as trace-norm continuity. Matrix projections and von Neumann projections remain distinct. Complete freely accessible compact spectral, trace-class and holomorphic functional calculus foundations are still pending; no source expression was imported.