Modified trace is a symmetrised integral

Abstract

A modified trace for a finite ${\mathbb {k}}$-linear pivotal category is a family of linear forms on endomorphism spaces of projective objects which has cyclicity and so-called partial trace properties. The modified trace provides a meaningful generalisation of the categorical trace to non-semisimple categories and allows to construct interesting topological invariants. We show that a non-degenerate modified trace defines a compatible with duality Calabi–Yau structure on the subcategory of projective objects. We prove, that for any finite-dimensional unimodular pivotal Hopf algebra over a field ${\mathbb {k}}$, a modified trace is determined by a symmetric linear form on the Hopf algebra constructed from an integral. More precisely, we prove that shifting with the pivotal element defines an isomorphism between the space of right integrals, which is known to be 1-dimensional, and the space of modified traces. This result allows us to compute modified traces for all simply laced restricted quantum groups at roots of unity.

Appendices

Appendix A: Proof of Proposition 2.3

From the definitions of ${\text {HH}}_0(A)$ and ${\text {HH}}_0(A\text {{-pmod}})$ the map $x\mapsto r_x$ induces a linear map $\Phi :{\text {HH}}_0(A)\rightarrow {\text {HH}}_0(A\text {{-pmod}})$ on the corresponding classes. We need to construct its inverse. By Lemma 2.2, for $P\in A\text {{-pmod}}$ we have a decomposition:

$$\begin{aligned} \mathrm {id}_P=\sum _{i=1}^k a_i\circ \mathrm {id}_A\circ b_i\ , \quad {\text {with}}\quad b_i:P\rightarrow A, \;a_i:A\rightarrow P\ .\end{aligned}$$

(A.1)

Let us define a map $\psi _P:{{\,\mathrm{End}\,}}_A(P)\rightarrow {\text {HH}}_0(A)$ by

$$\begin{aligned} \psi _P(f):= \sum _i \left[ (b_i\circ f\circ a_i)(\varvec{1})\right] \ .\end{aligned}$$

(A.2)

We will check that the map

$$\begin{aligned} \Psi :\; {\text {HH}}_0(A\text {{-pmod}})&\xrightarrow {\sim }{\text {HH}}_0(A)\nonumber \\ [P,f]&\mapsto \psi _P(f) \end{aligned}$$

(A.3)

is well-defined, i.e. it does not depend on the choice of the decomposition (A.1) and descends on the class of f in ${\text {HH}}_0(A\text {{-pmod}})$.

Assume we have another decomposition $\mathrm {id}_P=\sum _{i'}a'_{i'}\circ \mathrm {id}_A \circ b'_{i'}$, with the associated map

$$\begin{aligned} \psi '_P(f)=\sum _{i'}[(b'_{i'}\circ f\circ a'_{i'})(\varvec{1})]\ .\end{aligned}$$

Inserting the identity (A.1), we have

$$\begin{aligned} \psi '_P(f)= & {} \sum _{i,i'} \left[ (b'_{i'}\circ f\circ a_i\circ b_i\circ a'_{i'})(\varvec{1})\right] \nonumber \\= & {} \sum _{i,i'} \left[ ( b_i\circ a'_{i'})(\varvec{1})\,(b'_{i'}\circ f\circ a_i)(\varvec{1})\right] \end{aligned}$$

(A.4)

where we applied the algebra isomorphism from Lemma 2.1 to the composition of A-endomorphisms $(b'_{i'}\circ f\circ a_i)$ and $(b_i\circ a'_{i'})$. Similarly,

$$\begin{aligned} \psi _P(f)= & {} \sum _{i,i'} \left[ (b_{i}\circ a'_{i'}\circ b'_{i'}\circ f\circ a_{i})(\varvec{1})\right] \nonumber \\= & {} \sum _{i,i'} \left[ (b'_{i'}\circ f\circ a_i)(\varvec{1})\, ( b_i\circ a'_{i'})(\varvec{1})\right] \end{aligned}$$

(A.5)

which is equal to the second line in (A.4) because the summands are classes in ${\text {HH}}_0(A)$. We thus get the equality $\psi '_P(f)=\psi _P(f)\in {\text {HH}}_0(A)$.

Let us now show that the family

$$\begin{aligned} \{\psi _P:{{\,\mathrm{End}\,}}_A(P)\rightarrow {\text {HH}}_0(A)\, | \, P\in A\text {{-pmod}}\} \end{aligned}$$

has cyclicity property. Let $f:P\rightarrow P'$ and $g:P'\rightarrow P$, and $\mathrm {id}_P$ as in (A.1) and let $\mathrm {id}_{P'}=\sum _{i'}a'_{i'}\circ \mathrm {id}_A \circ b'_{i'}$. We then have

$$\begin{aligned} \psi _{P'}(f\circ g)= & {} \sum _{i,i'} \left[ (b'_{i'}\circ f\circ a_i\circ \mathrm {id}_A\circ b_i\circ g\circ a'_{i'})(\varvec{1})\right] \nonumber \\= & {} \sum _{i,i'} \left[ ( b_i\circ g\circ a'_{i'})(\varvec{1})\, (b'_{i'}\circ f\circ a_i)(\varvec{1})\right] \nonumber \\= & {} \sum _{i,i'} \left[ (b'_{i'}\circ f\circ a_i)(\varvec{1})\,( b_i\circ g \circ a'_{i'})(\varvec{1})\right] \nonumber \\= & {} \sum _{i,i'} \left[ ( b_i\circ g\circ a'_{i'}\circ b'_{i'}\circ f\circ a_i)(\varvec{1})\right] \nonumber \\= & {} \psi _P(g\circ f) \end{aligned}$$

(A.6)

where we again used the algebra isomorphism in Lemma 2.1. From this cyclicity property, we see that the map $\psi _P$ does not depend on representatives f in the class $[f]\in {\text {HH}}_0(A\text {{-pmod}})$, for $f\in {{\,\mathrm{End}\,}}_A(P)$. Therefore, the map $\Psi$ in (A.3) is well-defined.

To see that $\Psi \circ \Phi =\mathrm {id}_{{\text {HH}}_0(A)}$ we have to check that the composition $[x]\mapsto [r_x]\mapsto \psi _A(r_x)$ is identity. Note that here we use only $P=A$ component in the quotient (2.4). Using the trivial decomposition of $\mathrm {id}_A$ from (A.1), we indeed get the expected identity, and so $\Psi$ is a left inverse of $\Phi$.

To show that $\Psi$ is also a right inverse of $\Phi$, assume $P\in A\text {{-pmod}}$ and $f\in {{\,\mathrm{End}\,}}_A(P)$. Then $\Psi$ maps [P, f] to the class of $x=\sum _i (b_i\circ f\circ a_i)(\varvec{1})\in A$. We note that the corresponding endomorphism of A by right multiplication with x is $r_x=\sum _i (b_i\circ f\circ a_i)$. And by cyclicity we have $[r_x]=[f]\in {\text {HH}}_0(A\text {{-pmod}})$. We thus get $\Phi \circ \Psi =\mathrm {id}_{{\text {HH}}_0(A\text {{-pmod}})}$, which completes the proof of the proposition.

Appendix B: Proof of Lemma 7.2

The fact that $w(\alpha _i)$ is a positive root if $l(ws_i)=l(w)+1$ follows from [3, VI.1.6, Cor. 2]. We will prove the formula for ${\mathsf {w}}{\mathsf {t}}(T_w(E_i))$ by induction on the length $l(w)=\nu \ge 0$. A proof for ${\mathsf {w}}{\mathsf {t}}(T_w(F_i))$ works similarly. For $\nu =0$, w is the unit element and the statement holds by definition of the L-grading. We suppose that the statement holds for $\nu \ge 0$, i.e. that $T_w(E_i)$ has L-grading ${\mathsf {w}}{\mathsf {t}}\bigl (T_w(E_i)\bigr )=w(\alpha _i)$ if $l(w)\le \nu$ and $l(ws_i)=l(w)+1$.

Let $w\in \mathcal {W}$ be an element with length $l(w)=\nu +1$ and i be such that $l(ws_i)=\nu +2$. Recall that $w(\alpha _i) \in \Delta _+$. We claim that there exists $j\ne i$ such that $w(\alpha _j)$ is a negative root. This follows from [3, Sec. V.4.4, Thm 1], indeed if w permutes the positive roots, then w fixes the positive chamber $C=\{x\in L \;|\;(\alpha _i|x)>0, 1\le i\le n\}$ and hence is identity. Let us choose such j. Recall that $l(ws_j)=l(w)+1$ would imply that $w(\alpha _j)$ is a positive root, hence we have that $l(ws_j)<\nu +2$. From the defining relations, multiplication with $s_j$ changes the length by $\pm 1$, we then clearly have $l(ws_j)\ne l(w)$, therefore $l(ws_j)=\nu$. Denote by $\langle s_i,s_j\rangle \subset \mathcal {W}$ the subgroup generated by $s_i$ and $s_j$. The idea is to use elements from the orbit $w\langle s_i,s_j\rangle$ to construct an appropriate pair $(w',k)$ to which the induction hypothesis applies. For a given choice of j above, we have 3 cases: $a_{ij}=0$ or if $a_{ij}=-1$ then $w s_j s_i$ might have length $\nu \pm 1$. We analyse all of these cases:

Case 1: $a_{ij}=0$. We can choose $(w',k)=(w s_j,i)$. Indeed, $l(w')=\nu$ and since $l(w s_i)=\nu +2$ then $w's_i=ws_is_j$ has length $\nu +1$, and so we can apply the induction hypothesis. We then get $T_w(E_i)=(T_{w'}\circ T_j)(E_i)=T_{w'}(E_i)$ because $T_j(E_i)=E_i$, see (7.5). Using that $s_j(\alpha _i)=\alpha _i$ we get ${\mathsf {w}}{\mathsf {t}}\bigl (T_w(E_i)\bigr )=w'(\alpha _i)=w(\alpha _i)$.

Case 2a: $a_{ij}=-1$ and $l(w s_j s_i)=\nu +1$. We choose $w'=ws_j$ and to both $(w',i)$, $(w',j)$ the induction hypothesis applies. We have $T_j(E_i)=-E_iE_j+q^{-1}E_jE_i$, $s_j(\alpha _i)=\alpha _i+\alpha _j$, hence

$$\begin{aligned} {\mathsf {w}}{\mathsf {t}}(T_w(E_i))&={\mathsf {w}}{\mathsf {t}}(T_{w'}\circ T_j(E_i))= {\mathsf {w}}{\mathsf {t}}(T_{w'}(E_i))+{\mathsf {w}}{\mathsf {t}}(T_{w'}(E_j))\nonumber \\&=w'(\alpha _i)+w'(\alpha _j)=(w'\circ s_j)(\alpha _i)= w(\alpha _i)\ , \end{aligned}$$

(B.1)

where we used that $T_{w'}$ is an automorphism of the algebra and that ${\mathsf {w}}{\mathsf {t}}$ makes the algebra graded.

Case 2b: $a_{ij}=-1$ and $l(w s_j s_i)=\nu -1$. We choose $w'=w s_j s_i$ and check that $l(w's_j)= l(w s_i s_j s_i) =\nu$ because on one side it is at most $\nu$ and on the other side it is at least $l(w s_i)-l(s_j s_i)=\nu$. Therefore, we can apply the induction hypothesis to $(w',j)$. We have $(T_i\circ T_j)(E_i)=E_j$ and $(s_js_i)(\alpha _j)=\alpha _i$, hence

$$\begin{aligned} {\mathsf {w}}{\mathsf {t}}(T_w(E_i))={\mathsf {w}}{\mathsf {t}}(T_{w'}(E_j))=w'(\alpha _j)=w(\alpha _i)\ . \end{aligned}$$

This finishes the proof.

Appendix C: Proof of Lemma 7.4

We will use the following result [41, Thm. 2.3]⁹ stated for $1\le j\le k$, and $1\le a,b\le p-1$:

$$\begin{aligned} E_{\beta _k}^aE_{\beta _j}^b=q^{ab(\beta _j|\beta _k)} E_{\beta _j}^bE_{\beta _k}^a + \sum _{\begin{array}{c} {0\le a_j,a_{j+1},\dots ,a_k\le p-1}\nonumber \\ {a_j<b\ ,\ a_k<a} \end{array}} \rho (a_j,\dots ,a_k) E_{\beta _j}^{a_j}E_{\beta _{j+1}}^{a_{j+1}}\dots E_{\beta _k}^{a_k}\nonumber \\ \end{aligned}$$

(C.1)

where the coefficients $\rho (a_j,\dots ,a_k)\in {\mathbb {k}}$ vanish if the corresponding monomials do not have the expected L-grading:

$$\begin{aligned} \rho (a_j,\dots ,a_k)=0\quad \text { if } \quad a_j\beta _j+a_{j+1}\beta _{j+1}+\dots +a_k\beta _k\ne b\beta _j+a\beta _k \ .\end{aligned}$$

(C.2)

We prove the lemma by induction on $\nu =k-j$.

Let us consider the case $\nu =1$. The formula (C.1) gives

$$\begin{aligned} E_{\beta _{j+1}}^{p-1}E_{\beta _j}=q^{(p-1)(\beta _j|\beta _{j+1})} E_{\beta _j}E_{\beta _{j+1}}^{p-1} \ ,\end{aligned}$$

(C.3)

where we used that the second term in (C.1) vanishes because of the condition (C.2), which is in our case

$$\begin{aligned} a_{j+1}\beta _{j+1}\ne \beta _j+(p-1)\beta _{j+1}\ ,\end{aligned}$$

(C.4)

holds for all $a_{j+1}< p-1$. Equality (C.3) shows that (7.12) is true for $k-j=1$.

Assume the induction hypothesis that for $1\le \nu <N$ the formula (7.12) is true if $k-j\le \nu$. We consider the case where $k-j=\nu +1$. From (C.1), we get

$$\begin{aligned} E_{\beta _k}^{p-1}E_{\beta _j}=q^{(p-1)(\beta _j|\beta _k)} E_{\beta _j}E_{\beta _{k}}^{p-1} + \sum _{\begin{array}{c} {0\le a_{j+1},\dots ,a_k\le p-1}\nonumber \\ {a_k<p-1} \end{array}} \rho (0,a_{j+1},\dots ,a_k) E_{\beta _{j+1}}^{a_{j+1}}\dots E_{\beta _k}^{a_k}\ .\nonumber \\ \end{aligned}$$

(C.5)

We then use the condition (C.2) on vanishing coefficients $\rho (0,a_{j+1},\dots ,a_k)$, which is in our case

$$\begin{aligned} a_{j+1}\beta _{j+1}+\dots +a_k\beta _k\ne \beta _j+(p-1)\beta _k\ .\end{aligned}$$

We see that it certainly holds if all the integers $a_{j+1}, \dots , a_{k-1}$ are zero – in this case we get the inequality $a_k\beta _k\ne \beta _j+(p-1)\beta _k$, similar to (C.4). Therefore, for non-vanishing coefficients $\rho$ in the sum (C.5) we have to necessarily assume that at least one of the integers $a_{j+1}, \dots , a_{k-1}$ is non zero. Let l be the smallest index for which $a_l$ is non zero. We have $j+1\le l<k$ hence $|k-l|<\nu$. The induction hypothesis gives us commutation relation for the root vector $E_{\beta _l}$, and we get

$$\begin{aligned} E_{\beta _{l}}^{p-1}E_{\beta _{l+1}}^{p-1}\dots E_{\beta _{k-1}}^{p-1}E_{\beta _l}= q^{(p-1)(\beta _l|\beta _{l+1}+\dots +\beta _{k-1})} E_{\beta _l}^{p-1} E_{\beta _l}E_{\beta _{l+1}}^{p-1}\dots E_{\beta _{k-1}}^{p-1}=0\ . \nonumber \\ \end{aligned}$$

(C.6)

This gives the following vanishing result for terms in the sum (C.5) corresponding to non-zero coefficients $\rho (0,a_{j+1},\dots ,a_k)$:

$$\begin{aligned} E_{\beta _{j+1}}^{p-1}E_{\beta _{j+2}}^{p-1}\dots E_{\beta _{k-1}}^{p-1}E_{\beta _l}=0\ , \end{aligned}$$

and therefore these terms do not contribute while moving $E_{\beta _j}$ to the left in LHS of (7.12). We have thus obtained

$$\begin{aligned} E_{\beta _{j+1}}^{p-1}E_{\beta _{j+2}}^{p-1}\dots E_{\beta _{k}}^{p-1}E_{\beta _j}= q^{(p-1)(\beta _j|\beta _k)} E_{\beta _{j+1}}^{p-1}E_{\beta _{j+2}}^{p-1}\dots E_{\beta _{k-1}}^{p-1} E_{\beta _j}E_{\beta _{k}}^{p-1}\ . \end{aligned}$$

Using again the induction hypothesis, we move $E_{\beta _j}$ to the left using (C.1) and get the expected formula (7.12), which completes the proof.