Traces of projective modules

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

The rank of a free module is a number. A finitely generated projective module over a group ring has a finer invariant, its Hattori–Stallings trace, which assigns a coefficient to every conjugacy class of the group. Bass conjectured in 1976 that over the integral group ring only the identity class carries a coefficient. Over the complex numbers, classes of elements of finite order can carry coefficients, and the question is whether every class of elements of infinite order has coefficient zero. OpenAI proved this in 2026 for every group [OpenAI-BT, Theorem 1.1]. For torsion-free groups the theorem gives Kaplansky's idempotent conjecture in characteristic zero: the group ring of a torsion-free group over a field of characteristic zero has no idempotents other than \(0\) and \(1\). Together with a restriction on the finite-order coefficients, it also gives Bass's conjecture for integral group rings. This course proves these statements. This lesson defines the trace, states the results and reduces them to finitely generated groups.

1. Projective modules and idempotent matrices

Rings have an identity, ring homomorphisms preserve it, and modules are unital. Throughout the course modules are right modules. For a ring \(A\) we write the elements of \(A^q\) as columns. A matrix \(M\in\operatorname{Mat}_{r\times q}(A)\) defines the map \(x\mapsto Mx\) from \(A^q\) to \(A^r\); it is right \(A\)-linear, since \(M(xa)=(Mx)a\).

Lemma 1.1. Every right \(A\)-linear map \(\varphi:A^q\to A^r\) is \(x\mapsto Mx\) for exactly one \(M\in\operatorname{Mat}_{r\times q}(A)\). Composition of maps corresponds to multiplication of matrices.

Proof. Let \(\epsilon_1,\ldots,\epsilon_q\) be the standard basis columns, and let the \(j\)-th column of \(M\) be \(\varphi(\epsilon_j)\). For \(x=\sum_j\epsilon_jx_j\), linearity gives \(\varphi(x)=\sum_j\varphi(\epsilon_j)x_j=Mx\). The matrix is unique because \(M\epsilon_j\) is its \(j\)-th column, and \((MN)x=M(Nx)\) gives the last assertion. \(\square\)

A finitely generated projective module is a module \(P\) such that \(P\oplus Q\cong A^q\) for some module \(Q\) and some \(q\ge0\). A matrix \(e\in\operatorname{Mat}_q(A)\) is idempotent if \(e^2=e\). Write \(1\) for the identity matrix. Then \[ A^q=eA^q\oplus(1-e)A^q , \] because \(x=ex+(1-e)x\), and an element \(y=ex=(1-e)x'\) of both summands satisfies \(y=ey=e(1-e)x'=0\). So \(eA^q\) is finitely generated projective. Conversely, let \(\theta:P\oplus Q\to A^q\) be an isomorphism. The map \(A^q\to A^q\) that sends \(\theta(p,u)\) to \(\theta(p,0)\) is \(A\)-linear and equal to its own square; by Lemma 1.1 it is \(x\mapsto ex\) for an idempotent matrix \(e\), and \(P\cong eA^q\).

Lemma 1.2. Let \(e\in\operatorname{Mat}_q(A)\) and \(f\in\operatorname{Mat}_r(A)\) be idempotent. Then \(eA^q\cong fA^r\) if and only if there are \(a\in\operatorname{Mat}_{r\times q}(A)\) and \(b\in\operatorname{Mat}_{q\times r}(A)\) with \[ \begin{gathered} a=fae,\qquad b=ebf,\\ ba=e,\qquad ab=f . \end{gathered} \]

Proof. Given such matrices, \(x\mapsto ax\) maps \(eA^q\) into \(fA^r\), since \(ax=fax\), and \(y\mapsto by\) maps \(fA^r\) into \(eA^q\). For \(x\in eA^q\) we have \(bax=ex=x\), and similarly \(aby=y\) for \(y\in fA^r\). So the two maps are inverse isomorphisms.

Conversely, let \(\varphi:eA^q\to fA^r\) be an isomorphism with inverse \(\psi\). The maps \(x\mapsto\varphi(ex)\) from \(A^q\) to \(A^r\) and \(y\mapsto\psi(fy)\) from \(A^r\) to \(A^q\) are \(A\)-linear, so by Lemma 1.1 they are given by matrices \(a\) and \(b\). Since \(\varphi(ex)\) lies in \(fA^r\), where \(f\) acts as the identity, \(fax=ax\); since \(e^2=e\), \(aex=ax\). Hence \(a=fae\), and in the same way \(b=ebf\). Finally \(bax=\psi(f\varphi(ex))=\psi(\varphi(ex))=ex\) and \(aby=fy\), so \(ba=e\) and \(ab=f\). \(\square\)

The Grothendieck group \(K_0(A)\) is the abelian group generated by symbols \([P]\), one for each isomorphism class of finitely generated projective modules, subject to the relations \([P\oplus P']=[P]+[P']\). A finite combination \(\sum_in_i[P_i]\) equals \([P]-[P']\), where \(P\) is the sum of the \(P_i^{n_i}\) with \(n_i>0\) and \(P'\) the sum of the \(P_i^{-n_i}\) with \(n_i<0\). So every element of \(K_0(A)\) is a difference of two classes \([eA^q]\).

2. The Hattori–Stallings trace

Let \([A,A]\) be the additive subgroup of \(A\) generated by the commutators \(xy-yx\), and put \[ \operatorname{HH}_0(A)=A/[A,A]. \] If \(A\) is an algebra over a commutative ring \(R\), then \([A,A]\) is an \(R\)-submodule, since \(c(xy-yx)=(cx)y-y(cx)\) for \(c\in R\).

Lemma 2.1. For \(a\in\operatorname{Mat}_{r\times q}(A)\) and \(b\in\operatorname{Mat}_{q\times r}(A)\), \[ \operatorname{tr}(ab)-\operatorname{tr}(ba)\in[A,A]. \]

Proof. The difference is \(\sum_{i=1}^r\sum_{j=1}^q(a_{ij}b_{ji}-b_{ji}a_{ij})\). \(\square\)

Proposition 2.2. The rule \[ \operatorname{HS}_A([eA^q])=\operatorname{tr}(e)+[A,A] \] defines a homomorphism \(\operatorname{HS}_A:K_0(A)\to\operatorname{HH}_0(A)\), the Hattori–Stallings trace. For the free module, \(\operatorname{HS}_A([A^q])=q\cdot(1+[A,A])\).

Proof. If \(eA^q\cong fA^r\), take \(a\) and \(b\) as in Lemma 1.2; then Lemma 2.1 gives \(\operatorname{tr}(e)=\operatorname{tr}(ba)\equiv\operatorname{tr}(ab)=\operatorname{tr}(f)\) modulo \([A,A]\). Every finitely generated projective module is isomorphic to some \(eA^q\), so the rule assigns a value to every isomorphism class. The direct sum \(eA^q\oplus fA^r\) is the image of the block-diagonal idempotent with blocks \(e\) and \(f\), whose trace is \(\operatorname{tr}(e)+\operatorname{tr}(f)\). So the rule respects the defining relations of \(K_0(A)\). The free module \(A^q\) is the image of the identity matrix, of trace \(q\cdot1\). \(\square\)

Let \(\varphi:A\to B\) be a ring homomorphism, applied to matrices entrywise. It maps idempotent matrices to idempotent matrices, and if \(a\), \(b\) are as in Lemma 1.2 for \(e\) and \(f\), then \(\varphi(a)\), \(\varphi(b)\) are as in Lemma 1.2 for \(\varphi(e)\) and \(\varphi(f)\). So \([eA^q]\mapsto[\varphi(e)B^q]\) defines a homomorphism \(\varphi_*:K_0(A)\to K_0(B)\). Since \(\varphi\) maps commutators to commutators, it induces a map \(\operatorname{HH}_0(A)\to\operatorname{HH}_0(B)\), again written \(\varphi\), and for every \(x\in K_0(A)\) \[ \operatorname{HS}_B(\varphi_*x)=\varphi\bigl(\operatorname{HS}_A(x)\bigr). \tag{2.1} \]

3. Group rings and conjugacy classes

Let \(R\) be a commutative ring and \(G\) a group. The group ring \(RG\) consists of the finite sums \(\sum_{x\in G}a_xx\) with \(a_x\in R\), with the product \((ab)_w=\sum_{uv=w}a_ub_v\). Write \(\operatorname{con}(G)\) for the set of conjugacy classes of \(G\) and \([x]\), or \([x]_G\), for the class of \(x\).

Lemma 3.1. Let \(R^{(\operatorname{con}G)}\) be the free \(R\)-module with basis \(\operatorname{con}(G)\). The map \[ \pi\Bigl(\sum_xa_xx\Bigr)=\sum_{C\in\operatorname{con}(G)}\Bigl(\sum_{x\in C}a_x\Bigr)C \] is surjective with kernel \([RG,RG]\). Hence \(\pi\) induces an isomorphism \(\operatorname{HH}_0(RG)\cong R^{(\operatorname{con}G)}\).

Proof. All sums are finite, and \(\pi(x)=[x]\), so \(\pi\) is surjective. For \(u=\sum a_xx\) and \(v=\sum b_yy\), \[ uv-vu=\sum_{x,y}a_xb_y(xy-yx), \] and \(yx=x^{-1}(xy)x\) is conjugate to \(xy\), so \(\pi(uv-vu)=0\). Conversely, let \(\pi(c)=0\). On each class \(C\) meeting the support of \(c\), list the elements \(x_1,\ldots,x_k\) of \(C\) in the support, with coefficients \(a_1,\ldots,a_k\). Since \(\sum_ia_i=0\), the part of \(c\) on \(C\) equals \(\sum_ia_i(x_i-x_1)\). If \(x_i=tx_1t^{-1}\), then \[ x_i-x_1=t\,(x_1t^{-1})-(x_1t^{-1})\,t \] is a commutator. So \(c\in[RG,RG]\). \(\square\)

For \(x\in K_0(RG)\) and \(C\in\operatorname{con}(G)\) write \(\operatorname{HS}_G(x)(C)\in R\) for the coefficient of \(C\) in \(\pi(\operatorname{HS}_{RG}(x))\). An idempotent \(e\in\operatorname{Mat}_q(RG)\) can be written as a finite sum \(e=\sum_{x\in G}A_xx\) with \(A_x\in\operatorname{Mat}_q(R)\): the entry \((i,j)\) of \(A_x\) is the coefficient of \(x\) in \(e_{ij}\). Then \(\operatorname{tr}(e)=\sum_x\operatorname{tr}(A_x)\,x\), and \[ \begin{aligned} &\operatorname{HS}_G\bigl([e(RG)^q]\bigr)(C)\\ &\quad=\sum_{x\in C}\operatorname{tr}(A_x). \end{aligned} \tag{3.1} \]

Two functionals single out parts of this trace. The canonical trace and the augmentation are \[ \begin{gathered} \tau_G\Bigl(\sum_xa_xx\Bigr)=a_1,\\ \epsilon_G\Bigl(\sum_xa_xx\Bigr)=\sum_xa_x . \end{gathered} \] The augmentation is a ring homomorphism \(RG\to R\). For a matrix \(b\in\operatorname{Mat}_q(RG)\) put \(\tau_{G,q}(b)=\sum_i\tau_G(b_{ii})\), and let \(\epsilon_{G,q}(b)\in\operatorname{Mat}_q(R)\) be the entrywise augmentation. By (3.1), for an idempotent \(e\) and \(x=[e(RG)^q]\), \[ \begin{aligned} &\operatorname{HS}_G(x)([1])=\tau_{G,q}(e),\\ &\sum_{C}\operatorname{HS}_G(x)(C)\\ &\qquad=\operatorname{tr}\epsilon_{G,q}(e). \end{aligned} \tag{3.2} \] The second sum is finite. The matrix \(\epsilon_{G,q}(e)\) is an idempotent matrix over \(R\).

Over a field \(k\), the trace of an idempotent matrix \(P\) is its rank: \(k^q\) is the direct sum of the image of \(P\), on which \(P\) is the identity, and its kernel, so in a basis adapted to this decomposition \(P\) is diagonal with \(\operatorname{rank}P\) ones. By (2.1) for the homomorphism \(\epsilon_G:\mathbb CG\to\mathbb C\), the augmentation rank \[ \operatorname{rk}_\epsilon\bigl([e(\mathbb CG)^q]\bigr)=\operatorname{rank}_{\mathbb C}\epsilon_{G,q}(e) \] is a well-defined homomorphism \(K_0(\mathbb CG)\to\mathbb Z\) (Exercise 7.5 gives a direct argument). The coefficient at \([1]\) defines the homomorphism \(\tau_{G,*}:K_0(\mathbb CG)\to\mathbb C\), \(\tau_{G,*}([e(\mathbb CG)^q])=\tau_{G,q}(e)\).

Examples 3.2. (a) The free module \((RG)^q\) has trace \(q\cdot[1]\).

(b) Let \(F\le G\) be a finite subgroup and suppose \(|F|\) is invertible in \(R\), for instance \(R=\mathbb C\). Then \(e_F=|F|^{-1}\sum_{f\in F}f\) is an idempotent: \(e_F^2=|F|^{-2}\sum_{f,f'\in F}ff'=|F|^{-1}\sum_{f\in F}f\), since every \(f''\in F\) is a product \(ff'\) in exactly \(|F|\) ways. By (3.1), for every class \(C\), \[ \operatorname{HS}_G([e_FRG])(C)=\frac{|C\cap F|}{|F|}. \] So classes of elements of finite order other than \([1]\) can carry nonzero coefficients, and \(\tau_G(e_F)=1/|F|\).

(c) For every \(a\in A\), the matrix \(e=\begin{pmatrix}1&a\\0&0\end{pmatrix}\) is idempotent, and \(eA^2\cong A\) (Exercise 7.1). Its trace is \(1+[A,A]\).

4. The conjectures and the theorem

Bass conjectured that for every group \(G\) and every finitely generated projective module \(P\) over \(\mathbb ZG\), the trace of \(P\) is supported on \([1]\) [CM, Section 1]. The weak form asks only that the sum of all coefficients equal the coefficient at \([1]\). Over \(\mathbb CG\), Example 3.2(b) shows that finite-order classes can carry coefficients, and the complex form of the conjecture concerns the other classes [BCM04, Conjecture 2.1; CM, Conjecture 2]: for every \(x\in K_0(\mathbb CG)\) and every \(g\in G\) of infinite order, \(\operatorname{HS}_G(x)([g])=0\).

Kaplansky's idempotent conjecture asks whether, for every torsion-free group \(G\) and every field \(k\), the only elements \(e\in kG\) with \(e^2=e\) are \(0\) and \(1\). It concerns elements of the group ring, not idempotent matrices over it: \(\begin{pmatrix}1&0\\0&0\end{pmatrix}\) is an idempotent matrix over every ring. For a field of characteristic zero, Kaplansky proved that \(0\le\tau_G(e)\le1\), with \(\tau_G(e)=0\) only for \(e=0\) and \(\tau_G(e)=1\) only for \(e=1\), and Zalesskii proved that \(\tau_G(e)\) is rational [BV98, Theorems 1.3 and 2.1]. These results leave open rational values strictly between \(0\) and \(1\).

Theorem 4.1 ([OpenAI-BT, Theorem 1.1]). Let \(G\) be a group. For every \(x\in K_0(\mathbb CG)\) and every \(g\in G\) of infinite order, \[ \operatorname{HS}_G(x)([g])=0 . \]

Corollary 4.2 ([OpenAI-BT, Corollary 1.2]). Let \(G\) be a torsion-free group.

  1. For every \(q\ge1\) and every idempotent \(e\in\operatorname{Mat}_q(\mathbb CG)\), \[ \tau_{G,q}(e)=\operatorname{rank}_{\mathbb C}\epsilon_{G,q}(e). \] Consequently \(\tau_{G,*}=\operatorname{rk}_\epsilon\) on \(K_0(\mathbb CG)\), and its image is \(\mathbb Z\).
  2. For every commutative domain \(R\) of characteristic zero, the only idempotents of \(RG\) are \(0\) and \(1\).

Corollary 4.3 ([OpenAI-BT, Corollary 9.1]). For every group \(G\), the trace of every class in \(K_0(\mathbb ZG)\) is supported on \([1]\). In particular the sum of the coefficients equals the coefficient at \([1]\).

Theorem 4.1 is proved in A Pfaffian cocycle, from a vanishing theorem for sparse simplicial cycles that is stated in Exterior chains and sparse cycles and proved in the four lessons after it. The corollaries are proved in Idempotents and integral traces. Corollary 4.3 also uses a restriction on the coefficients at classes of finite order, due to Linnell; that lesson proves it by a congruence for traces of \(p\)-th powers.

Many cases were known before. The complex form holds for linear groups (Bass), for groups of complex cohomological dimension at most two (Eckmann), and for subgroups of semihyperbolic groups and for mapping class groups [CM, Section 1]; for elementary amenable groups [FL, abstract]; for amenable groups and, more generally, groups satisfying the Bost conjecture [BCM04]; and for groups satisfying the Farrell–Jones conjecture with complex coefficients [BLR, abstract]. Burger and Valette proved the idempotent conjecture over \(\mathbb C\) for torsion-free groups that embed into finite extensions of pro-\(p\) groups for infinitely many primes \(p\) [BV98, abstract]. For torsion-free subgroups of hyperbolic groups a stronger statement holds: the reduced group \(C^*\)-algebra has no projections other than \(0\) and \(1\), a consequence of the Baum–Connes conjecture proved for these groups by Mineyev and Yu [MY]. Theorem 4.1 needs no hypothesis on the group, and its proof uses no assembly map and no completion of the group ring.

A second OpenAI manuscript proves the analogous statement for the Banach algebra \(\ell^1(G)\): the trace of every idempotent matrix over \(\ell^1(G)\) is supported on finitely many classes of elements of finite order [OpenAI-L1, Theorem 1.1]. Its method is different; Idempotents and integral traces states the result.

5. Passing to subgroups

Lemma 5.1 (subgroup locality). Let \(L\le G\) be a subgroup and \(i:\mathbb CL\to\mathbb CG\) the inclusion. For every \(y\in K_0(\mathbb CL)\) and every \(C\in\operatorname{con}(G)\), \[ \operatorname{HS}_G(i_*y)(C)=\sum_{B}\operatorname{HS}_L(y)(B), \] where \(B\) runs over the conjugacy classes of \(L\) contained in \(L\cap C\); only finitely many terms are nonzero. Moreover every \(x\in K_0(\mathbb CG)\) equals \(i_*y\) for a finitely generated subgroup \(L\), which can be chosen to contain any prescribed finite subset of \(G\).

Proof. Let \(e=\sum_xA_xx\) be an idempotent matrix over \(\mathbb CL\). The idempotent \(i(e)\) has the same coefficient matrices, all supported in \(L\). By (3.1), \[ \operatorname{HS}_G([i(e)(\mathbb CG)^q])(C)=\sum_{x\in C\cap L}\operatorname{tr}A_x . \] The set \(C\cap L\) is a union of conjugacy classes of \(L\), and summing over each of them gives the asserted formula for \(y=[e(\mathbb CL)^q]\); finitely many \(A_x\) are nonzero. Both sides are additive in \(y\), so the formula holds on all of \(K_0(\mathbb CL)\).

Write \(x=[e(\mathbb CG)^q]-[f(\mathbb CG)^r]\) with idempotent matrices \(e\), \(f\), and let \(L\) be generated by the prescribed finite set and the finitely many group elements in the supports of the entries of \(e\) and \(f\). The entries lie in the subring \(\mathbb CL\), the matrices are idempotent there, and \(y=[e(\mathbb CL)^q]-[f(\mathbb CL)^r]\) satisfies \(i_*y=x\). \(\square\)

Corollary 5.2. If Theorem 4.1 holds for all finitely generated groups, it holds for all groups.

Proof. Let \(x\in K_0(\mathbb CG)\) and let \(g\in G\) have infinite order. Choose \(L\) and \(y\) as in Lemma 5.1, with \(g\in L\). Every element of a class \(B\) of \(L\) contained in \([g]_G\) is conjugate in \(G\) to \(g\), so it has infinite order. Since \(L\) is finitely generated, \(\operatorname{HS}_L(y)(B)=0\) for each such \(B\), and Lemma 5.1 gives \(\operatorname{HS}_G(x)([g])=0\). Every class \(B\) in the sum is used: a class of \(G\) can split into several classes of \(L\). \(\square\)

6. The plan of the course

Fix a finitely generated group \(G\), an idempotent \(e\in\operatorname{Mat}_q(\mathbb CG)\) and an element \(g\) of infinite order, and let \(\lambda\) be the coefficient of the trace of \(e\) at \([g]\). Products of the coefficient matrices of \(e\) along lists of group elements, closed up by \(g\), define weights whose sums do not change when an entry of the list is deleted. Projected to the cosets of \(\langle g\rangle\), these weights give simplicial cycles in every even dimension \(2m\), and a cocycle built from Pfaffians takes the value \(2^{-m}\lambda\) on them. The simplices of these cycles have connected vertex sets in one graph of bounded degree, independent of \(m\). A separating-set argument shows that in high dimension every invariant cocycle vanishes on cycles of this kind. Hence \(\lambda=0\).

7. Exercises

7.1 (easy). Let \(a\in A\) and \(e=\begin{pmatrix}1&a\\0&0\end{pmatrix}\). Find matrices as in Lemma 1.2 showing \(eA^2\cong A\).

7.2 (easy). In Example 3.2(b), check that the coefficients of the trace of \(e_F\) add up to \(\epsilon_G(e_F)=1\). For \(G=F=S_3\), list the three coefficients.

7.3 (medium). Let \(\varphi:G\to G'\) be a homomorphism of groups, and also write \(\varphi\) for the induced homomorphism \(\mathbb CG\to\mathbb CG'\). Show that for \(x\in K_0(\mathbb CG)\) and \(C'\in\operatorname{con}(G')\), \[ \operatorname{HS}_{G'}(\varphi_*x)(C')=\sum_{C:\ \varphi(C)\subseteq C'}\operatorname{HS}_G(x)(C). \]

7.4 (medium). Let \(k\) be a field. Show that \([P]\mapsto\dim_kP\) is an isomorphism \(K_0(k)\to\mathbb Z\).

7.5 (medium). Let \(e\), \(f\) be idempotent matrices over \(\mathbb CG\) with \(e(\mathbb CG)^q\cong f(\mathbb CG)^r\). Show directly that \(\operatorname{rank}\epsilon_{G,q}(e)=\operatorname{rank}\epsilon_{G,r}(f)\).

8. Solutions

7.1. Take \(a'=\begin{pmatrix}1&a\end{pmatrix}\in\operatorname{Mat}_{1\times2}(A)\) and \(b'=\begin{pmatrix}1\\0\end{pmatrix}\), with \(f=1\in\operatorname{Mat}_1(A)\). Then \(a'b'=1\) and \(b'a'=e\). Also \(a'e=\begin{pmatrix}1&a\end{pmatrix}=a'\) and \(eb'=b'\), so \(a'=fa'e\) and \(b'=eb'f\).

7.2. The coefficient at \(C\) is \(|C\cap F|/|F|\), and the classes of \(G\) meeting \(F\) partition \(F\), so the coefficients add up to \(1\). For \(G=F=S_3\) the classes are the identity, the three transpositions and the two \(3\)-cycles, with coefficients \(1/6\), \(1/2\) and \(1/3\).

7.3. For an idempotent \(e=\sum_xA_xx\) over \(\mathbb CG\), the idempotent \(\varphi(e)\) equals \(\sum_x A_x\varphi(x)\), so its coefficient matrix at \(x'\in G'\) is \(\sum_{\varphi(x)=x'}A_x\). By (3.1) the coefficient at \(C'\) is \(\sum_{x:\ \varphi(x)\in C'}\operatorname{tr}A_x\). The image of a class \(C\) of \(G\) lies in a single class of \(G'\), so the set of \(x\) with \(\varphi(x)\in C'\) is the union of the classes \(C\) with \(\varphi(C)\subseteq C'\). This proves the formula for \(x=[e(\mathbb CG)^q]\), and both sides are additive.

7.4. Finitely generated modules over \(k\) are finite-dimensional vector spaces, all projective, and two of them are isomorphic exactly when they have the same dimension. Dimension is additive on direct sums, so it defines a homomorphism \(K_0(k)\to\mathbb Z\), surjective since \(\dim k^q=q\). If \(\dim P=\dim P'\), then \(P\cong P'\) and \([P]-[P']=0\); every element of \(K_0(k)\) has this form, so the map is injective.

7.5. Take \(a\), \(b\) as in Lemma 1.2 and apply \(\epsilon_G\) entrywise. Since \(\epsilon_G\) is a ring homomorphism, \(\epsilon(b)\epsilon(a)=\epsilon(e)\) and \(\epsilon(a)\epsilon(b)=\epsilon(f)\). Over the commutative ring \(\mathbb C\), \(\operatorname{tr}(XY)=\operatorname{tr}(YX)\) for rectangular matrices, so \(\operatorname{tr}\epsilon(e)=\operatorname{tr}\epsilon(f)\). Both matrices are idempotent, and the trace of an idempotent complex matrix is its rank (Section 3).

References