Open MathsAll courses

Idempotents and integral traces

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

The trace theorem (Theorem 4.1 of Traces of projective modules, proved in A Pfaffian cocycle and From filtrations to chains) says that the Hattori–Stallings trace of every projective module over \(\mathbb CG\) vanishes at the classes of elements of infinite order. This lesson draws its consequences. For a torsion-free group only the identity class remains, its coefficient equals the rank of the augmentation, and a faithful trace then shows that the group ring has no idempotents other than \(0\) and \(1\), over every commutative domain of characteristic zero. For integral group rings, a congruence for traces of \(p\)-th powers shows that the trace of a projective module is unchanged by the \(p\)-th power map on conjugacy classes; this excludes the classes of nontrivial elements of finite order, and with the trace theorem it proves Bass's conjecture. Consequences for fixed points of homotopy idempotents follow from results of Berrick, Chatterji and Mislin. The lesson follows [OpenAI-BT, Sections 9 and 10].

1. Trace and augmentation rank

Proof of Corollary 4.2(1) of Traces of projective modules. Let \(G\) be torsion-free and \(e\in\operatorname{Mat}_q(\mathbb CG)\) idempotent, \(x=[e(\mathbb CG)^q]\). The only element of finite order is \(1\), so by Theorem 4.1 the trace of \(x\) is supported on \([1]\). By (3.2) of that lesson, \[ \begin{aligned} \tau_{G,q}(e)&=\operatorname{HS}_G(x)([1])\\ &=\sum_C\operatorname{HS}_G(x)(C)\\ &=\operatorname{tr}\epsilon_{G,q}(e), \end{aligned} \] and \(\epsilon_{G,q}(e)\) is an idempotent complex matrix, whose trace is its rank. Taking differences, \(\tau_{G,*}=\operatorname{rk}_\epsilon\) on \(K_0(\mathbb CG)\). The values are integers, and the free module of rank one has value \(1\), so the image is \(\mathbb Z\). \(\square\)

For matrices this is a numerical statement only: \(\begin{pmatrix}1&0\\0&0\end{pmatrix}\) is an idempotent matrix other than \(0\) and \(1\), of augmentation rank \(1\). For a single element of \(\mathbb CG\) the augmentation rank is \(0\) or \(1\), and that is what forces the element itself.

2. Kaplansky's idempotent conjecture in characteristic zero

The proof uses the trace of the group von Neumann algebra from The regular trace and spectral phase, Section 1, as in Directly finite group algebras, Section 3. There, \(\mathcal M\) is the algebra of bounded operators on \(\ell^2(G)\) commuting with the right translations, \(\lambda:\mathbb CG\to\mathcal M\) is the injective homomorphism given by left convolution, and \(\tau(T)=\langle T\delta_1,\delta_1\rangle\) satisfies \(\tau(ST)=\tau(TS)\), \(\tau(T)\ge0\) for \(T\ge0\) with equality only for \(T=0\), and \(\tau(1)=1\) (Lemma 1.3 of The regular trace and spectral phase). On \(\mathbb CG\), \(\tau\circ\lambda\) is the canonical trace \(\tau_G\).

Theorem 2.1 (Corollary 4.2(2) of Traces of projective modules; [OpenAI-BT, Corollary 1.2(ii)]). Let \(G\) be torsion-free and \(R\) a commutative domain of characteristic zero. Every idempotent of \(RG\) is \(0\) or \(1\).

Proof. First let \(R=\mathbb C\) and \(e\in\mathbb CG\) idempotent. By Section 1 with \(q=1\), \(\tau_G(e)=\epsilon_G(e)\), and \(\epsilon_G(e)\in\{0,1\}\) because it is an idempotent complex number. Let \(P\) be the orthogonal projection onto the image of \(\lambda(e)\). By Lemma 3.2 of Directly finite group algebras, \(P\in\mathcal M\) and \(\tau(P)=\tau(\lambda(e))=\tau_G(e)\). If this value is \(0\), faithfulness applied to \(P=P^*P\ge0\) gives \(P=0\); then the image of \(\lambda(e)\) is \(0\), so \(\lambda(e)=0\) and \(e=0\). If it is \(1\), then \(\tau(1-P)=0\) and \(1-P\ge0\), so \(P=1\); the image of \(\lambda(e)\) is all of \(\ell^2(G)\), and \(\lambda(e)\), being the identity on its image, is the identity, so \(e=1\).

Next let \(R\) be a commutative domain of characteristic zero and \(f=\sum_{g\in S}r_gg\in RG\) idempotent, \(S\) finite. Let \(R_0\subseteq R\) be the subring generated by the \(r_g\), and \(F\) its field of fractions, a finitely generated field of characteristic zero. By Lemma 3.3 of Directly finite group algebras, \(F\) has an injective homomorphism into \(\mathbb C\). The coefficientwise maps \(R_0G\to RG\) and \(R_0G\to\mathbb CG\) are injective ring homomorphisms. The equation \(f^2=f\) holds in \(R_0G\), so the image of \(f\) in \(\mathbb CG\) is an idempotent, hence \(0\) or \(1\), and injectivity gives \(f=0\) or \(f=1\). \(\square\)

Only the finitely generated coefficient field is embedded into \(\mathbb C\); no embedding of \(R\) is needed. Because \(e(1-e)=0\), an idempotent other than \(0\), \(1\) gives a pair of nonzero zero divisors, so the zero-divisor conjecture over a field implies the idempotent conjecture over that field. Theorem 2.1 proves the idempotent statement in characteristic zero directly and says nothing about zero divisors; over the field of two elements, torsion-free group algebras can have zero divisors (A torsion-free group algebra with zero divisors). The argument uses complex coefficients in two places, the faithful trace and the embedding of the coefficient field, and says nothing about fields of positive characteristic.

The same range projection, applied to an idempotent matrix over \(\mathbb CG\) in \(\operatorname{Mat}_q(\mathcal M)\), has trace \(\tau_{G,q}(e)\), an integer by Section 1. In the reduced group \(C^*\)-algebra, OpenAI shows that the canonical trace takes integer values on the image of \(K_0(\mathbb CG)\) [OpenAI-BT, Remark 10.1]; this is a statement about that image, not about all projections of the completion (Section 6).

3. A congruence for traces of powers

For a matrix \(B\in\operatorname{Mat}_q(\mathbb ZG)\) write \(r(B)\in\mathbb Z^{(\operatorname{con}G)}\) for the class sums of its trace, \(r(B)(C)=\sum_{x\in C}\operatorname{tr}B_x\), where \(B=\sum_xB_xx\) with integer matrices \(B_x\). For \(s\ge1\) let \(\psi^s\) be the linear map of \(\mathbb Z^{(\operatorname{con}G)}\) with \(\psi^s[x]=[x^s]\); it is well defined because \((yxy^{-1})^s=yx^sy^{-1}\), and \(\psi^s\psi^{s'}=\psi^{ss'}\).

Lemma 3.1. Let \(p\) be a prime, \(a\) an integer and \(t\ge1\). Then \(a^{p^t}\equiv a^{p^{t-1}}\pmod{p^t}\).

Proof. For \(t=1\) this is Fermat's congruence \(a^p\equiv a\pmod p\): for \(a\ge0\) by induction on \(a\), since \(p\) divides the binomial coefficients \(\binom pi\), \(0<i<p\), so \((a+1)^p\equiv a^p+1\); for \(a<0\) use \((-a)^p\equiv-a^p\) modulo \(p\) (an equality for odd \(p\), and \(a^2\equiv a\) modulo \(2\)). If \(a^{p^t}=a^{p^{t-1}}+p^tm\), raise to the power \(p\): \[ \begin{aligned} &(a^{p^{t-1}}+p^tm)^p\\ &\quad=a^{p^t}+\sum_{i\ge1}\binom pia^{p^{t-1}(p-i)}(p^tm)^i , \end{aligned} \] where the term \(i=1\) is divisible by \(p\cdot p^t\) and the terms \(i\ge2\) by \(p^{2t}\). So \(a^{p^{t+1}}\equiv a^{p^t}\pmod{p^{t+1}}\). \(\square\)

Proposition 3.2. For \(A\in\operatorname{Mat}_q(\mathbb ZG)\), a prime \(p\) and \(k\ge1\), \[ r\bigl(A^{p^k}\bigr)\equiv\psi^p\,r\bigl(A^{p^{k-1}}\bigr)\pmod{p^k}, \] coefficient by coefficient.

Proof. For \(m\ge1\), expanding the matrix product, \[ r(A^m)=\sum_ww(w)\,[x_1x_2\cdots x_m], \] the sum over words \(w=((i_1,x_1),\ldots,(i_m,x_m))\) with \(i_l\in\{1,\ldots,q\}\), \(x_l\in G\), with weight \(w(w)=\prod_{l=1}^m(A_{x_l})_{i_li_{l+1}}\), where \(i_{m+1}=i_1\); finitely many words have nonzero weight. The cyclic rotation \(((i_2,x_2),\ldots,(i_m,x_m),(i_1,x_1))\) has the same weight, a product of the same integers, and its product of group elements \(x_2\cdots x_mx_1=x_1^{-1}(x_1\cdots x_m)x_1\) lies in the same class. A word of length \(m\) has a least period \(d\), a divisor of \(m\), and is \(u^{m/d}\) for a word \(u\) of length \(d\) that is primitive, of least period \(d\); its rotation orbit has \(d\) elements, which correspond to the rotations of \(u\). Its weight is \(w(u)^{m/d}\): the factor between consecutive copies of \(u\) is the closing factor of \(u\). Its class is \([\pi(u)^{m/d}]=\psi^{m/d}[\pi(u)]\), with \(\pi(u)\) the product of the group elements of \(u\). Let \(\mathcal N_d\) be the finite set of rotation orbits of primitive words of length \(d\) with nonzero weight; weight and class are constant on orbits. Then \[ r(A^m)=\sum_{d\mid m}d\sum_{\nu\in\mathcal N_d}w(\nu)^{m/d}\,\psi^{m/d}[\nu]. \] For \(m=p^k\) the divisors are \(d=p^j\), \(0\le j\le k\), and applying \(\psi^p\) to the same formula for \(m=p^{k-1}\), \[ \begin{aligned} &r(A^{p^k})-\psi^pr(A^{p^{k-1}})\\ &\ =p^k\sum_{\nu\in\mathcal N_{p^k}}w(\nu)[\nu]\\ &\ \ +\sum_{j=0}^{k-1}p^j\sum_{\nu\in\mathcal N_{p^j}}\delta_j(\nu)\,\psi^{p^{k-j}}[\nu], \end{aligned} \] where \(\delta_j(\nu)=w(\nu)^{p^{k-j}}-w(\nu)^{p^{k-j-1}}\). By Lemma 3.1 with \(t=k-j\), each \(\delta_j(\nu)\) is divisible by \(p^{k-j}\), so every term is divisible by \(p^k\). \(\square\)

Corollary 3.3. If \(e\in\operatorname{Mat}_q(\mathbb ZG)\) is idempotent, then \(r(e)=\psi^s\,r(e)\) for every \(s\ge1\).

Proof. For a prime \(p\), \(e^{p^k}=e^{p^{k-1}}=e\), so Proposition 3.2 gives \(r(e)\equiv\psi^pr(e)\) modulo \(p^k\) for every \(k\). Both sides have finitely many nonzero integer coefficients, so they are equal. For general \(s\), write \(s\) as a product of primes and use \(\psi^{ss'}=\psi^s\psi^{s'}\); \(\psi^1\) is the identity. \(\square\)

Theorem 3.4 (Linnell's restriction; [BH, Theorem A]). Let \(P\) be a finitely generated projective module over \(\mathbb ZG\), and let \(g\in G\) with \(\operatorname{HS}_G([P])([g])\ne0\). There is an integer \(m\ge1\) such that \(g\) is conjugate to \(g^{s^m}\) for every \(s\ge1\). In particular, if \(g\ne1\), then \(g\) has infinite order.

Proof. Let \(P\cong e(\mathbb ZG)^q\), so the trace of \(P\) is \(r(e)=a_0[1]+\sum_{i=1}^Na_i[g_i]\) with distinct classes \([g_i]\ne[1]\) and \(a_i\ne0\). Let \(\Sigma=\{[g_1],\ldots,[g_N]\}\) and \(s\ge1\). By Corollary 3.3, \(r(e)=\psi^sr(e)=a_0[1]+\sum_ia_i[g_i^s]\). Every class in \(\Sigma\) has a nonzero coefficient on the left, so it is \([g_i^s]\) for some \(i\), because \(\psi^s[1]=[1]\notin\Sigma\). So \(\psi^s(\Sigma)\supseteq\Sigma\), and since \(\Sigma\) is finite, \(\psi^s\) restricts to a permutation of \(\Sigma\). With \(m\) the least common multiple of \(1,\ldots,N\), \((\psi^s)^m=\psi^{s^m}\) is the identity on \(\Sigma\). If \(g\ne1\) had finite order \(t\), take \(s=t\): \(g\) would be conjugate to \(g^{t^m}=1\). For \(g=1\) any \(m\) works. \(\square\)

Linnell proved this restriction with a theory of matrices over group rings due to Cliff, and Berrick and Hesselholt recovered it through topological Hochschild homology [BH, Theorem A and the remark after it]. The congruence of Proposition 3.2 is the group-ring form of the congruences \(\operatorname{tr}(A^{p^k})\equiv\operatorname{tr}(A^{p^{k-1}})\) for integer matrices. For projective modules over \(\mathbb CG\), Bass had proved that a nonzero coefficient at \([g]\) forces \(g\) to be conjugate to \(g^{p^N}\), for a fixed \(N\) and almost all primes \(p\) [CM, Section 1].

4. Bass's conjecture for integral group rings

Theorem 4.1 (Corollary 4.3 of Traces of projective modules; [OpenAI-BT, Corollary 9.1]). For every group \(G\) and every \(x\in K_0(\mathbb ZG)\), the trace \(\operatorname{HS}_G(x)\) is supported on \([1]\).

Proof. It suffices to treat \(x=[e(\mathbb ZG)^q]\). The same matrix \(e\) is idempotent over \(\mathbb CG\), and its trace coefficients over \(\mathbb CG\) are the images of the integer coefficients under \(\mathbb Z\subseteq\mathbb C\). By Theorem 4.1 of Traces of projective modules, the coefficients at classes of elements of infinite order vanish. By Theorem 3.4, so do the coefficients at classes of nontrivial elements of finite order. Differences of classes give the statement on \(K_0(\mathbb ZG)\). No comparison between \(K_0(\mathbb ZG)\) and \(K_0(\mathbb CG)\) is needed. \(\square\)

In particular the sum of the coefficients equals the coefficient at \([1]\): the augmentation trace and the Kaplansky trace \(\tau_G\) agree on \(K_0(\mathbb ZG)\). This is the weak Bass conjecture.

The statements do not depend on the side of the modules. The anti-automorphism \(\iota(\sum a_xx)=\sum a_xx^{-1}\) of \(\mathbb ZG\) turns left modules into right modules; a projective left module \((\mathbb ZG)^qe\), with \(e\) acting on rows, corresponds to the right module of the transposed matrix \(\iota(e)^T\), whose trace coefficient at \([x]\) is the original coefficient at \([x^{-1}]\). Support on \([1]\), the augmentation trace and \(\tau_G\) are preserved.

5. Homotopy idempotents

A self-map \(f\) of a space is a homotopy idempotent if \(f\circ f\) is homotopic to \(f\). Berrick, Chatterji and Mislin translated Bass's conjecture into fixed-point statements for such maps on manifolds; the following implications, proved in [BCM07, Theorems 1 and 2] with the theory of Nielsen numbers and \(L^2\)-Lefschetz numbers, are named statements of this lesson and are not proved in the programme.

Corollary 5.1 ([OpenAI-BT, Corollary 9.2]). Given (BCM1) and (BCM2), Theorem 4.1 implies: every homotopy idempotent of a closed, smooth, oriented, connected manifold of dimension greater than two is homotopic to a map with exactly one fixed point; and every pointed homotopy idempotent of such a manifold inducing the identity on the fundamental group has Lefschetz number equal to its \(L^2\)-Lefschetz number.

Proof. Theorem 4.1 and the weak form after it give the hypotheses of (BCM1) and (BCM2), in either convention for the side of modules (Section 4). \(\square\)

The connection runs through finitely dominated spaces. A homotopy idempotent \(f\) of a connected finite complex \(X\) splits up to homotopy through a space \(Y\) dominated by \(X\), and when \(f\) induces the identity on \(G=\pi_1(X)\), the cellular chains of the universal cover of \(Y\) are chain equivalent to a finite complex of finitely generated projective \(\mathbb ZG\)-modules. The trace of its Euler class \(\sum_i(-1)^i[P_i]\) records the fixed-point indices of \(f\) class by class [BCM07, Sections 3, 4, 7, 8].

7. Exercises

7.1 (easy). Show that the idempotent \(e_F=|F|^{-1}\sum_{f\in F}f\) of a finite subgroup \(F\ne1\) satisfies \(\tau_G(e_F)\ne\epsilon_G(e_F)\), and explain why this does not contradict Section 1.

7.2 (easy). Check Proposition 3.2 for \(G=1\), \(q=1\), \(A=(a)\): it says \(a^{p^k}\equiv a^{p^{k-1}}\pmod{p^k}\).

7.3 (medium). Let \(G=\mathbb Z/2=\{1,u\}\) and \(A=u\in\mathbb ZG\). Compute \(r(A^2)\) and \(\psi^2r(A)\) and verify Proposition 3.2 for \(p=2\), \(k=1\).

7.4 (medium). Let \(R\) be a commutative domain of characteristic zero and \(G\) torsion-free. Show that \(RG\) has no idempotent matrices \(e\in\operatorname{Mat}_q(RG)\) with \(\epsilon_{G,q}(e)=0\) other than \(e=0\).

7.5 (medium). In the proof of Theorem 3.4, show that \((\psi^s)^m=\psi^{s^m}\) and that every permutation of a set of \(N\) elements has order dividing the least common multiple of \(1,\ldots,N\).

8. Solutions

7.1. \(\tau_G(e_F)=1/|F|\) and \(\epsilon_G(e_F)=1\). The group contains the nontrivial finite subgroup \(F\), so it is not torsion-free, and Section 1 does not apply; the trace of \(e_F\) has coefficients at classes of nontrivial elements of \(F\) (Example 3.2 of Traces of projective modules).

7.2. Here \(r(A^m)=a^m[1]\) and \(\psi^p\) is the identity, so the statement is Lemma 3.1 with \(t=k\).

7.3. \(A^2=u^2=1\), so \(r(A^2)=[1]\). Also \(r(A)=[u]\) and \(\psi^2[u]=[u^2]=[1]\). The congruence is an equality here. In the proof, the words of length \(2\) with nonzero weight are \((u,u)\) only, of period \(1\): \(\mathcal N_1=\{(u)\}\) with weight \(1\), contributing \(1\cdot\psi^2[u]=[1]\).

7.4. By Section 1 applied over \(\mathbb C\) after embedding the finitely many coefficients as in the proof of Theorem 2.1, \(\tau_{G,q}(e)=\operatorname{rank}\epsilon_{G,q}(e)=0\). The range projection \(P\) of \(e\) in \(\operatorname{Mat}_q(\mathcal M)\) has \(\tau_q(P)=0\), so \(P=0\) by faithfulness, hence the image of \(e\) is \(0\) and \(e=0\); injectivity of the coefficient maps gives \(e=0\) over \(R\).

7.5. \(\psi^s\psi^{s'}[x]=[x^{s's}]\), so by induction \((\psi^s)^m[x]=[x^{s^m}]\). A permutation of \(N\) elements decomposes into disjoint cycles of lengths at most \(N\); a cycle of length \(\ell\) has order \(\ell\), which divides \(\operatorname{lcm}(1,\ldots,N)\).

References