The Drinfeld–Kohno theorem for the superalgebra ${\mathfrak {gl}}(1|1)$

Abstract

We revisit the derivation of Knizhnik–Zamolodchikov equations in the case of non-semisimple categories of modules of a superalgebra in the case of the generic affine level and representations parameters. A proof of existence of asymptotic solutions and their properties for the superalgebra ${\mathfrak {gl}}(1|1)$ gives a basis for the proof of existence of associator which satisfy braided tensor categories requirements. Braided tensor category structure of $U_h({\mathfrak {gl}}(1|1))$ quantum algebra is calculated, and the tensor product ring is shown to be isomorphic to ${\mathfrak {gl}}(1|1)$ ring, for the same generic relations between the level and parameters of modules. We review the proof of Drinfeld–Kohno theorem for non-semisimple category of modules suggested by Geer (Adv Math 207:1–38, 2006) and show that it remains valid for the superalgebra ${\mathfrak {gl}}(1|1)$. Examples of logarithmic solutions of KZ equations are also presented.

Appendices

Appendix A

In this Appendix we collect some data about ${\mathfrak {gl}}(1|1)$ and details of solutions of its KZ equations.

1.1 Asymptotic solutions of KZ equation

Lemma 1

If there is an eigenvector (not generalized) v of ${\varOmega }_{12}$ with eigenvalue $\lambda$, and there are no eigenvalues of ${\varOmega }_{12}$ such that $\lambda +n\kappa ,n\in {\mathbb {N}}$, then there exists unique asymptotic solution around $x=0$

$$\begin{aligned} f(x)=x^{\lambda /\kappa }(v+o(x)),\;\;\lim _{x\rightarrow 0}o(x)=0 \end{aligned}$$

Proof

By not generalized eigenvector we mean that v is not a member of a Jordan block. We check existence and uniqueness of asymptotic solution of the form

$$\begin{aligned} f(x)=x^{\lambda /\kappa }(v+xv_{1}+x^{2}v_{2}+\cdots ),\;\;o(v)=\sum _{n=1}x^{n}v_{n} \end{aligned}$$

(7.1)

with some perhaps infinite set of vectors $v_{n}$. After the substitution of it into the left hand side of Eq. (3.5) we get

$$\begin{aligned} lhs=x^{\lambda /\kappa }[\lambda x^{-1}v+(\lambda +\kappa )v_{1}+x(\lambda +2\kappa )v_{2}+x^{2}(\lambda +3\kappa )v_{3}+\cdots ] \end{aligned}$$

(7.2)

On the right hand side we rewrite in the vicinity of $x=0$ as

$$\begin{aligned} \frac{{\varOmega }_{12}}{x}-{\varOmega }_{23}(1+x+x^{2}+\cdots ) \end{aligned}$$

and now we act by it onto (7.1):

$$\begin{aligned} \hbox {rhs}= & {} x^{\lambda /\kappa }\left( \frac{{\varOmega }_{12}}{x}- {\varOmega }_{23}(1+x+x^{2}+\cdots )\right) (v+xv_{1}+x^{2}v_{2}+\cdots ) \nonumber \\= & {} x^{\lambda /\kappa }[\lambda vx^{-1}+({\varOmega }_{12}v_{1}- {\varOmega }_{23}v)+({\varOmega }_{12}v_{2}-{\varOmega }_{23}v_{1}-{\varOmega }_{23}v)x \nonumber \\&+({\varOmega }_{12}v_{3}-{\varOmega }_{23}v_{2}-{\varOmega }_{23}v_{1})x^{2}+\cdots ] \end{aligned}$$

(7.3)

Now we compare the multipliers of the same powers of x in (7.2) and ( 7.3) and get the infinite set of equations

$$\begin{aligned} x^{0}&:&({\varOmega }_{12}-(\lambda +\kappa )Id)v_{1}={\varOmega }_{23}v, \nonumber \\ x^{1}&:&({\varOmega }_{12}-(\lambda +2\kappa )Id)v_{2}={\varOmega }_{23}v_{1}+ {\varOmega }_{23}v, \nonumber \\ x^{2}&:&({\varOmega }_{12}-(\lambda +3\kappa )Id)v_{3}={\varOmega }_{23}v_{2}+ {\varOmega }_{23}v_{1}, \nonumber \\&\ldots \end{aligned}$$

(7.4)

They can be solved one after another. Indeed, the right hand side of the first equation is a known vector. $\det [{\varOmega }_{12}-(\lambda +\kappa )Id]\ne 0$ because $\lambda +\kappa$ is not an eigenvalue of ${\varOmega }_{12}$. Therefore the first equation has a unique solution $v_{1}$. The same arguments can now be applied to the second equation: ${\varOmega }_{23}v_{1}$ is now a known vector. We can solve the second equation for $v_{2}$, which is possible because $\det [{\varOmega }_{12}-(\lambda +2\kappa )Id]\ne 0$, for $\lambda +2\kappa$ is not an eigenvalue of ${\varOmega }_{12}$ and so on. Thus we find uniquely each vector $v_{i}$ by this recurrent procedure, which proves the statement. We do not discuss the convergency question of the infinite sum of vectors in o(v) because we prove only the existence of asymptotic expansion. $\square$

The case of a Jordan block requires more general ansatz. The operator $x^{{\varOmega }_{12}/\kappa }$ is a well-defined operator on any finite-dimensional representation space on which ${\varOmega }_{12}$ acts nilpotently. In this case the operator

$$\begin{aligned} x^{{\varOmega }_{12}/\kappa }=\sum _{i=0}^{n}\frac{(\ln x)^{i}}{i!} \frac{{\varOmega }_{12}^{i}}{\kappa ^{i}} \end{aligned}$$

(7.5)

where n is the degree of nilpotency of ${\varOmega }_{12}$. Then we can reformulate the lemma in the following way.

Lemma 2

If there is a Jordan block of ${\varOmega }_{12}$ with eigenvalue $\lambda$ with the set of eigenvectors $v^{(i)},i=0,\ldots ,n-1$, ${\varOmega }_{12}v^{(i)}=\lambda v^{(i)}+v^{(i-1)}$, ($v^{(-1)}=0$) and there are no eigenvalues of ${\varOmega }_{12}$ such that $\lambda +n\kappa ,n\in {\mathbb {N}}$, then there exist n asymptotic solutions around $x=0$ of the form

$$\begin{aligned}&f_{i}(x)=x^{\lambda /\kappa }(v^{(i)}(\ln x)^{i}+\kappa ^{-1}v^{(i-1)}(\ln x)^{i-1}+o^{(i)}(x)), \nonumber \\&\lim _{x\rightarrow 0^{+}}o^{(i)}(x)=0,\quad i=0,\ldots ,n-1 \end{aligned}$$

(7.6)

Proof

To make the presentation more clear we put $\kappa =1$ and prove the statement for the case of rank $n=2$ Jordan block. With a more lengthy formulas the same proof can be repeated for $n>2$. The claim of the lemma for $f_{0}(x)$ becomes identical to the claim of Lemma 1, with the same proof and the same form of the vector $o^{(0)}(x)=xv_{1}+x^{2}v_{2}+\cdots$. Now we prove the lemma for $f_{1}(x).$ We show existence and uniqueness of $v_{j},u_{j},j=1,2,\ldots$ such that

$$\begin{aligned} f_{1}(x)= & {} x^{{\varOmega }_{12}}(v^{(1)}\ln x+v^{(0)}+o^{(1)}(x)), \nonumber \\ o^{(1)}(x)= & {} \sum _{j=1}^{\infty }v_{j}x^{j}\ln x+\sum _{j=1}^{\infty }u_{j}x^{j} \end{aligned}$$

(7.7)

First we prove existence of the vectors $v_{j}$. We substitute this ansatz for $o^{(1)}(x)$ into the KZ equation (3.5). We see that the terms proportional to $\ln x/x$ and 1/x cancel. Using the same expansion in powers of x of the term ${\varOmega }_{23}/(x-1)$ in before and extracting the terms containing $\ln x$ we get the equations

$$\begin{aligned} \ln x&:&({\varOmega }_{12}-(\lambda +1)Id)v_{1}={\varOmega }_{23}v^{(0)}, \nonumber \\ x\ln x&:&({\varOmega }_{12}-(\lambda +2)Id)v_{2}={\varOmega }_{23}(v^{(0)}+v_{1}) \nonumber \\&\ldots \end{aligned}$$

(7.8)

As before we can solve these equations for $v_{1},v_{2},\ldots$ sequentially because $\lambda +n,n\ge 1$ is not an eigenvalue of ${\varOmega }_{12}$ and the right hand side of these equations are known vectors. After we found $v_{i}$ s we do the same extracting on both hand side of KZ equation the terms which are not proportional to $\ln x$. We get

$$\begin{aligned} x&:&({\varOmega }_{12}-(\lambda +1)Id)u_{1}={\varOmega }_{23}v^{(1)}+v_{1}, \nonumber \\ x^{2}&:&({\varOmega }_{12}-(\lambda +2)Id)u_{2}={\varOmega }_{23}v^{(1)}+v_{2}+ {\varOmega }_{23}u_{1}, \nonumber \\&\ldots \end{aligned}$$

(7.9)

By the same reasons as before the equations can be uniquely solved sequentially for $u_{i}$. This completes the proof. $\square$

In the same way we can prove similar statements about existence of unique asymptotic solutions of the 3.5 equation around $x=1,x<1$.

Lemma 3

If there is an eigenvector v of ${\varOmega }_{23}$ with eigenvalue $\lambda$, and there are no eigenvalues of ${\varOmega }_{23}$ such that $\lambda +n\kappa ,n\in {\mathbb {N}}$, then there exists unique asymptotic solution around $x=1$ of the form

$$\begin{aligned} f(x)=(1-x)^{-\lambda /\kappa }(v+o(x)),\;\;\lim _{x\rightarrow 1^{-}}o(x)=0 \end{aligned}$$

(7.10)

in the case this eigenvector is not a member of a Jordan block. For the case of Jordan block of the size n the n asymptotic solutions are of the form

$$\begin{aligned}&f_{i}(x) =(1-x)^{-\lambda /\kappa }(v^{(i)}(\ln (1-x))^{i}+ \kappa ^{-1}v^{(i-1)}(\ln (1-x))^{i-1}+o^{(i)}(x)), \\&\lim _{x\rightarrow 1^{-}}o^{(i)}(x) =0,\quad i=0,\ldots ,n-1 \end{aligned}$$

Proof is the same as for Lemmas 1, 2.

Corollary 1

If the above restriction conditions on the parameters of typical modules are satisfied, an equivalent form of asymptotic solutions of (3.5) around $x=0$ is

$$\begin{aligned} f(x)=x^{{\varOmega }_{12}/\kappa }(v_{b}+o(v)) \end{aligned}$$

(7.11)

where $v_{t}$ is the same as v in the case when there are no Jordan block structure for the action of ${\varOmega }_{12}$, and $v_{b}$ is the bottom vector $v^{(n-1)}$ when there is a Jordan block of size n for the action of ${\varOmega }_{12}$.

Proof

In the case without Jordan block this is just change of notations. In the case when there is Jordan block of size n, we split ${\varOmega }_{12}=$ ${\varOmega }_{12}^{d}+{\varOmega }_{12}^\mathrm{nil}$ into diagonal and nilpotent parts and write $x^{{\varOmega }_{12}/\kappa }=x^{{\varOmega }_{12}^{d}/\kappa }\sum _{i}\frac{1}{i!}\left( \frac{{\varOmega }_{12}^\mathrm{nil}}{ \kappa }\ln x\right) ^{i}$. The action of it on the bottom vector of the set of generalized eigenvectors of ${\varOmega }_{12}$ will generate the sum of vectors proportional to $(\ln x)^{i}v^{(i)}$ where $v^{(i)}$ are the same as in (7.6). Therefore the representation (7.6) is related to the expansion (7.11) by a change of basis of solutions of KZ equation. $\square$

This corollary enables to use without changes the standard proofs of BTC structure of category of ${\mathfrak {gl}}(1|1)$-modules with associator and braiding defined through the KZ solutions and their monodromies.

1.2 Basis for ${\mathfrak {gl}}(1|1)$ and its modules

The ${\mathfrak {gl}}(1|1)$ generators are E, N,  $\psi ^{\pm }$ with commutation relations $[N,\psi ^{\pm }]=\pm \psi ^{\pm }$, $\{\psi ^{+},\psi ^{-}\}=E$ and E is central. (Maybe some other choice of basis will be more convenient?) Chevalley involution can be chosen as $\omega (E)=-E,$ $\omega (N)=-N,$ $\omega (\psi ^{\pm })=\pm \psi ^{\mp }$ and produces the dual representation. The basis for typical representation ${\mathcal {T}}_{e,n}$ of gl(1|1) can be chosen as

$$\begin{aligned} N=\left( \begin{array}{cc} n+1/2 &{} 0 \\ 0 &{} n-1/2 \end{array} \right) ,\;E=\left( \begin{array}{cc} e &{} 0 \\ 0 &{} e \end{array} \right) ,\;\psi ^{+}=\left( \begin{array}{cc} 0 &{} e \\ 0 &{} 0 \end{array} \right) ,\;\psi ^{-}=\left( \begin{array}{cc} 0 &{} 0 \\ 1 &{} 0 \end{array} \right) \qquad \end{aligned}$$

(7.12)

The basis for weights of module ${\mathcal {T}}_{e,n}$ is $u=\uparrow =\left( {\begin{array}{c}1 \\ 0\end{array}}\right)$ (even highest weight), and $v=\downarrow =\left( {\begin{array}{c}0\\ 1\end{array}}\right)$ (odd), and for dual module ${\mathcal {T}}_{e,n}^{*}$—$u^{*}=\left( {\begin{array}{c}0\\ 1\end{array}}\right)$ (odd lowest weight), and $v^{*}=\left( {\begin{array}{c}-1\\ 0\end{array}}\right)$ (even). For one-dimensional atypical representation ${\mathcal {A}}_{n}$ there is one vector $v_{0}$ with the action of the algebra generators $\psi ^{+}v_{0}=\psi ^{-}v_{0}=Ev_{0}=0$, $Nv_{0}=nv_{0}$. The algebra action on it explicitly:

$$\begin{aligned}&N\cdot \uparrow =(n+1/2)\uparrow ,\;N\cdot \downarrow =(n-1/2)\uparrow ,\;\psi ^{+}\cdot \uparrow =\psi ^{-}\cdot \downarrow =0, \nonumber \\&\psi ^{-}\cdot \uparrow =\downarrow ,\;\psi ^{+}\cdot \downarrow =e\uparrow \end{aligned}$$

(7.13)

For four-dimensional atypical representation ${\mathcal {P}}_{n}$ one can choose

$$\begin{aligned} N= & {} \left( \begin{array}{cccc} n+1 &{} 0 &{} 0 &{} 0 \\ 0 &{} n &{} 0 &{} 0 \\ 0 &{} 0 &{} n &{} 0 \\ 0 &{} 0 &{} 0 &{} n-1 \end{array} \right) ,\;\psi ^{+}=\frac{1}{2}\left( \begin{array}{cccc} 0 &{} 1 &{} 1 &{} 0 \\ 0 &{} 0 &{} 0 &{} 1 \\ 0 &{} 0 &{} 0 &{} -1 \\ 0 &{} 0 &{} 0 &{} 0 \end{array} \right) ,\;\psi ^{-}=\frac{1}{2}\left( \begin{array}{cccc} 0 &{} 0 &{} 0 &{} 0 \\ -1 &{} 0 &{} 0 &{} 0 \\ 1 &{} 0 &{} 0 &{} 0 \\ 0 &{} 1 &{} 1 &{} 0 \end{array} \right) , \nonumber \\ E= & {} 0\times Id_{4} \end{aligned}$$

(7.14)

And the weights of the module

$$\begin{aligned} u_{1}=t=\left( \begin{array}{c} 0 \\ 1 \\ 1 \\ 0 \end{array} \right) ,\;v_{1}=r=\left( \begin{array}{c} 1 \\ 0 \\ 0 \\ 0 \end{array} \right) ,\;v_{2}=l=\left( \begin{array}{c} 0 \\ 0 \\ 0 \\ 1 \end{array} \right) ,\;u_{2}=b=\frac{1}{2}\left( \begin{array}{c} 0 \\ 1 \\ -1 \\ 0 \end{array} \right) \qquad \end{aligned}$$

(7.15)

the even vectors are $u_{1,2}$, the odd $v_{1,2}$. This module is self-dual. The algebra action on it

$$\begin{aligned} N\cdot t&=nt,\;N\cdot r=(n+1)r,\;N\cdot l=(n-1)l,\;N\cdot b=nb, \nonumber \\ \psi ^{+}\cdot t&=r,\;\psi ^{+}\cdot l=b,\;\psi ^{+}\cdot r=\psi ^{+}\cdot b=0,\; \nonumber \\ \psi ^{-}\cdot t&=l,\;\psi ^{-}\cdot r=-b,\;\psi ^{-}\cdot l=\psi ^{-}\cdot b=0,\; \end{aligned}$$

(7.16)

We will use the following choice of Casimir element

$$\begin{aligned} {\varOmega }=NE+EN+\psi ^{-}\psi ^{+}-\psi ^{+}\psi ^{-}+E^{2} \end{aligned}$$

(7.17)

and its tensor analog

$$\begin{aligned} {\varOmega }_{ij}=N_{i}\otimes E_{j}+E_{i}\otimes N_{j}+\psi _{i}^{-}\otimes \psi _{j}^{+}-\psi _{i}^{+}\otimes \psi _{j}^{-}+E_{i}\otimes E_{j} \end{aligned}$$

(7.18)

where the lower indices denote the spaces where the generator acts.

$\widehat{{\mathfrak {gl}}}(1|1)$ commutation relations

$$\begin{aligned}{}[N_{r},E_{s}]=rk\delta _{r+s},\;[N_{r},\psi _{s}^{\pm }]=\pm \psi _{r+s}^{\pm },\;\{\psi _{r}^{+},\psi _{s}^{-}\}=E_{r+s}+rk\delta _{r+s} \end{aligned}$$

(7.19)

One can rescale generators in such a way that k will become 1 (if it is not 0), but we will keep it. The generic k will mean $e/k\notin {\mathbb {Z}}$ for all the modules involved into correlation function, as well as for all the modules appearing in tensor product decomposition. A remark: the structure of all modules for non-generic k for $\widehat{{\mathfrak {gl}}} (1|1)$ and their tensor product decomposition is of course well known, but the KZ for this case and its solutions is another (next...) problem.

Conformal dimension of Virasoro primary field $h=e\left( n+\frac{e}{2} \right)$.

We are going to find basis for invariants of level zero KZ equations for $N=2,3,4$. Recall that level zero equations in the case of ${\mathfrak {gl}} (1|1)$ mean that $\sum e_{i}=0$, if typical representations are involved in correlation function. In addition the invariants can be classified according to the N-grading of the space of states V of correlation function.

1.3 Examples of solutions of KZ equation for correlation functions

In this section we collect examples of explicit form of KZ $N=2,3$ solutions on the space of ${\mathfrak {gl}}(1|1)$ invariant functions. This class of solutions is the most interesting in the context of KZ equations for correlation functions of intertwining operators of affine Lie superalgebra ${\mathfrak {gl}}(1|1)^{\vee }$. Similar calculations have been done in the paper [39].

1. $N=2$

There is one invariant for ${{\mathcal {TT}}}$ correlation function in the basis described above $I_{0}^{{{\mathcal {TT}}}}=\uparrow \downarrow +\downarrow \uparrow$, and the list of invariants for ${\mathcal {PP}}$ correlation function is

$$\begin{aligned} I_{-1}^{{\mathcal {PP}}}&=rb-br \nonumber \\ I_{0,1}^{{\mathcal {PP}}}&=tb+rl-lr+bt \nonumber \\ I_{0,2}^{{\mathcal {PP}}}&=bb \nonumber \\ I_{1}^{{\mathcal {PP}}}&=lb-bl \end{aligned}$$

(7.20)

The first subindex denotes the value of $n_{1}+n_{2}$. (Recall that it is not an eigenvalue of N acting on the tensor product state. The latter is 0 for ${\mathfrak {g}}$-invariant correlation function.) Projection of KZ $N=2$ equation onto this basis gives an ODE with solutions

$$\begin{aligned} f(z_{1},z_{2})=[A(z_{1}-z_{2})^{\delta _{12}/k}]I_{0}^{{\mathcal {TT}}}, \;\delta _{ij}=n_{i}e_{j}+n_{j}e_{i}+e_{i}e_{j} \end{aligned}$$

(7.21)

for ${\mathcal {T}}_{e_{1},n_{1}}{\mathcal {T}}_{-e_{1},n_{2}}$ correlation function (A is a constant), and solutions

$$\begin{aligned} f(z_{1},z_{2})= & {} const\times I_{\pm 1}^{{\mathcal {PP}}},\text { for } n_{1}+n_{2}=\pm 1 \nonumber \\ f(z_{1},z_{2})= & {} AI_{0,2}^{{\mathcal {PP}}}+(2A\kappa ^{-1}\ln (z_{1}-z_{2})+B)I_{0,1}^{{\mathcal {PP}}}\text { \ for }n_{1}+n_{2}=0 \end{aligned}$$

(7.22)

where A, B are constants. This is an example of logarithms in correlation functions of logarithmic vertex operator algebras.

2. $N=3$

There are two invariants for ${\mathcal {TTT}}$ correlation in the same notations as above

$$\begin{aligned} I_{-1/2}^{{\mathcal {TTT}}}&=(\uparrow \uparrow \downarrow +\uparrow \downarrow \uparrow +\downarrow \uparrow \uparrow ) \nonumber \\ I_{+1/2}^{{\mathcal {TTT}}}&=(e_{1}\uparrow \downarrow \downarrow -e_{2}\downarrow \uparrow \downarrow +e_{3}\downarrow \downarrow \uparrow ), \end{aligned}$$

(7.23)

(Of course $e_{1}+e_{2}+e_{3}=0$.) Invariants of ${\mathcal {TTP}}$ correlations are

$$\begin{aligned} I_{-1}^{{\mathcal {TTP}}}&=\uparrow \uparrow b-\uparrow \downarrow r-\downarrow \uparrow r \nonumber \\ I_{0,1}^{{\mathcal {TTP}}}&=e_{1}(\uparrow \uparrow l+\uparrow \downarrow t+\downarrow \uparrow t)+\uparrow \downarrow b+\downarrow \downarrow r \nonumber \\ I_{0,2}^{{\mathcal {TTP}}}&=\uparrow \downarrow b+\downarrow \uparrow b \nonumber \\ I_{1}^{{\mathcal {TTP}}}&=e_{1}(\uparrow \downarrow l+\downarrow \uparrow l)+\downarrow \downarrow b \end{aligned}$$

(7.24)

and the list of invariants of ${\mathcal {PPP}}$ correlations are

$$\begin{aligned} I_{-2}^{{\mathcal {PPP}}}&=rrb-rbr+brr \nonumber \\ I_{-1,1}^{{\mathcal {PPP}}}&=trb-tbr-rrl-rbt+lrr+brt \nonumber \\ I_{-1,2}^{{\mathcal {PPP}}}&=rtb+rrl-rlr+rbt-btr-brt \nonumber \\ I_{-1,3}^{{\mathcal {PPP}}}&=rbb-brb \nonumber \\ I_{-1,4}^{{\mathcal {PPP}}}&=rbb-bbr \nonumber \\ I_{0,1}^{{\mathcal {PPP}}}&=btb+brl-blr+bbt \nonumber \\ I_{0,2}^{{\mathcal {PPP}}}&=bbb \nonumber \\ I_{1,1}^{{\mathcal {PPP}}}&=tlb-tbl-rll+llr-lbt+blt \nonumber \\ I_{1,2}^{{\mathcal {PPP}}}&=ltb+lrl-llr+lbt-btl-blt \nonumber \\ I_{1,3}^{{\mathcal {PPP}}}&=lbb-bbl \nonumber \\ I_{1,4}^{{\mathcal {PPP}}}&=blb-bbl \nonumber \\ I_{2}^{{\mathcal {PPP}}}&=llb-lbl+bll \end{aligned}$$

(7.25)

Projection of KZ equation in the form (3.5) onto these bases gives systems of ODEs with the following solutions. If the space of invariants with fixed first subindex, i.e., fixed sum of $n_{1}+n_{2}+n_{3}$, is one-dimensional equal to I, then the solution for correlation function in all three cases can be written as

$$\begin{aligned} f(x)=Ax^{\alpha /\kappa }(1-x)^{\beta /\kappa }I \end{aligned}$$

(7.26)

where $A\in {\mathbb {C}}$ is a constant, and $\alpha ,\beta$ are eigenvalues of ${\varOmega }_{12},{\varOmega }_{23}$ acting on I, respectively.

In the ${\mathcal {TTP}}$ case with $n_{1}+n_{2}+n_{3}=0$ solution contains logarithms:

$$\begin{aligned} f(x)= & {} Ax^{\delta _{12}/\kappa }(1-x)^{\delta _{23}/\kappa } \nonumber \\&\left[ I_{0,1}^{{\mathcal {TTP}}}+ \left( B+\frac{e_{1}}{\kappa }(\ln (1-x)-\ln x)\right) I_{0,2}^{{\mathcal {TTP}}}\right] \end{aligned}$$

(7.27)

In the ${\mathcal {PPP}}$ case with $n_{1}+n_{2}+n_{3}=0$ the solution is trivial

$$\begin{aligned} f(x)=AI_{0,1}^{{\mathcal {PPP}}}+BI_{0,2}^{{\mathcal {PPP}}},\;A,B\in {\mathbb {C}} \end{aligned}$$

(7.28)

But in the case $n_{1}+n_{2}+n_{3}=\pm 1$ there are logarithms in the solutions:

$$\begin{aligned} f^{\pm }(x)= & {} A^{\pm }I_{\pm 1,1}^{{\mathcal {PPP}}}+B^{\pm }I_{\pm 1,2}^{ {\mathcal {PPP}}} \nonumber \\&+\left( C_{3}^{\pm }+\frac{A^{\pm }-B^{\pm }}{\kappa }\ln x+ \frac{B^{\pm }-2A^{\pm }}{\kappa }\ln (1-x)\right) I_{\pm 1,3}^{{\mathcal {PPP}}} \nonumber \\&+\left( C_{4}^{\pm }+\frac{B^{\pm }}{\kappa }\ln x+\frac{A^{\pm }-B^{\pm } }{\kappa }\ln (1-x)\right) I_{\pm 1,3}^{{\mathcal {PPP}}} \end{aligned}$$

(7.29)

where $A^{\pm },B^{\pm },C_{3,4}^{\pm }$ are constants.

Another interesting problem is structure of solutions of KZ equations on a wider N-graded spaces, not necessarily invariants of ${\mathfrak {gl}}(1|1)$. We will address this problem elsewhere.

Appendix B

Here we will describe the basis and tensor product decomposition of $U_{h}( {\mathfrak {gl}}(1|1))$-modules and will prove the Proposition 3.

We will choose $i\pi \kappa ^{-1}=h$ and consider real $\kappa$. We use the following matrix basis for the three types of $U_{h}({\mathfrak {gl}}(1|1))$-modules ${\mathcal {T}}_{e,n}^{\kappa },{\mathcal {A}}_{n}^{\kappa },{\mathcal {P}}_{n}^{\kappa }$ included into ${\mathcal {C}}_{\kappa }$, as the basis for construction of tensor ring. For ${\mathcal {T}}_{e,n}^{\kappa }$

$$\begin{aligned} E=\left( \begin{array}{cc} e &{} 0 \\ 0 &{} e \end{array} \right) ,\;N=\left( \begin{array}{cc} n+1/2 &{} 0 \\ 0 &{} n-1/2 \end{array} \right) ,\;\psi ^{+}=\left( \begin{array}{cc} 0 &{} 2\sinh (eh) \\ 0 &{} 0 \end{array} \right) ,\;\;\psi ^{-}=\left( \begin{array}{cc} 0 &{} 0 \\ 1 &{} 0 \end{array} \right) \end{aligned}$$

with the vectors of the module

$$\begin{aligned} |e,n\rangle =\left( \begin{array}{c} 1 \\ 0 \end{array} \right) \text { \ (even)},\;|e,n-1\rangle =\psi ^{-}|e,n\rangle =\left( \begin{array}{c} 0 \\ 1 \end{array} \right) \text { \ (odd)} \end{aligned}$$

and for four-dimensional module we choose

$$\begin{aligned} N= & {} \left( \begin{array}{cccc} n+1 &{} 0 &{} 0 &{} 0 \\ 0 &{} n &{} 0 &{} 0 \\ 0 &{} 0 &{} n &{} 0 \\ 0 &{} 0 &{} 0 &{} n-1 \end{array} \right) ,\;\psi ^{+}=\left( \begin{array}{cccc} 0 &{} 1 &{} -e^{h} &{} 0 \\ 0 &{} 0 &{} 0 &{} e^{h} \\ 0 &{} 0 &{} 0 &{} 1 \\ 0 &{} 0 &{} 0 &{} 0 \end{array} \right) , \\ \psi ^{-}= & {} \left( \begin{array}{cccc} 0 &{} 0 &{} 0 &{} 0 \\ -1 &{} 0 &{} 0 &{} 0 \\ -e^{-h} &{} 0 &{} 0 &{} 0 \\ 0 &{} e^{-h} &{} -1 &{} 0 \end{array} \right) ,\;E=0\times Id_{4}, \end{aligned}$$

The coordinates of the vectors of the four-dimensional vector space of this representation are graded as in (7.15). Let us note that there are many other matrix presentations of ${\mathcal {P}}_{n}^{\kappa }$ module which can contain some more free numerical parameters.

Proof of Proposition 3

With these basis we can consider decomposition of tensor product of this set of three types of modules using the coproduct (4.1) and show that under some suitable assumptions on parameters of modules they form a ring. The cases

$$\begin{aligned} {\mathcal {A}}_{n}^{\kappa }\otimes {\mathcal {A}}_{n^{\prime }}^{\kappa }= {\mathcal {A}}_{n+n^{\prime }}^{\kappa },\;{\mathcal {A}}_{n}^{\kappa }\otimes {\mathcal {T}}_{e,n^{\prime }}^{\kappa }={\mathcal {T}}_{e,n+n^{\prime }}^{\kappa },\;{\mathcal {A}}_{n}^{\kappa }\otimes {\mathcal {P}}_{e,n^{\prime }}^{\kappa }= {\mathcal {P}}_{e,n+n^{\prime }}^{\kappa } \end{aligned}$$

are obvious. More interesting are the remaining three cases.

Consider ${\mathcal {T}}_{e,n}^{\kappa }\otimes {\mathcal {T}}_{e^{\prime },n^{\prime }}^{\kappa }$. The calculations of tensor product decomposition of two $U_{ih}({\mathfrak {gl}}(1|1))$-modules ${\mathcal {T}}_{e_{1},n_{1}}^{ \kappa }\otimes {\mathcal {T}}_{e_{2},n_{2}}^{\kappa }$ are completely parallel to the same calculations for ${\mathfrak {gl}}(1|1)$-modules. ${\mathcal {T}}_{e,n}^{\kappa }$ has two states—the highest weight $v_{1}=|\uparrow \rangle$ Grassmann even and $v_{2}=\psi ^{-}v_{1}=|\downarrow \rangle$—Grassmann odd. We can start from two vectors $w_{2}=\alpha _{2}|\uparrow \rangle \otimes |\downarrow \rangle +\beta _{2}|\downarrow \rangle \otimes |\uparrow \rangle$ and $u_{1}=\alpha _{1}|\uparrow \rangle \otimes |\downarrow \rangle +\beta _{1}|\downarrow \rangle \otimes |\uparrow \rangle$ with constraint $\alpha _{2}=-\beta _{1}\beta _{2}/\alpha _{1}$ which guarantees their orthogonality. We consider $w_{2}$ as highest weight of a grading reversed module, i.e., ${\varDelta }(\psi ^{+})w_{2}=0$. It gives $\beta _{2}=-2\alpha _{2}e^{-he_{1}}\sinh (e_{2}h)$. And we consider $u_{1}$ as lowest weight module, with Grassmann even highest weight. It means ${\varDelta }(\psi ^{-})u_{1}=0$, which gives $\beta _{1}=\alpha _{1}e^{he_{2}}$. Then one can easily check that corresponding lowest weight module of the first (grading reversed) module is $2\alpha _{1}\sinh ((e_{1}+e_{2})h)|\downarrow \rangle \otimes |\downarrow \rangle$, and highest weight of the second module is $\alpha _{2}\sinh ((e_{1}+e_{2})h)/\sinh (e_{1}h)|\uparrow \rangle \otimes |\uparrow \rangle$. We see that conditions $\sinh ((e_{1}+e_{2})h)\ne 0$, $\sinh (e_{1}h)\ne 0$, which mean $e_{1}/\kappa \notin {\mathbb {Z}}$, $(e_{1}+e_{2})/\kappa \notin {\mathbb {Z}} \backslash \{0\}$ are sufficient for decomposition

$$\begin{aligned} {\mathcal {T}}_{e_{1},n_{1}}^{\kappa }\otimes {\mathcal {T}}_{e_{2},n_{2}}^{\kappa }={\mathcal {T}}_{e_{1}+e_{2},,n_{1}+n_{2}+1/2}^{\kappa }\oplus {\mathcal {T}}_{e_{1}+e_{2},,n_{1}+n_{2}-1/2}^{\kappa \prime } \end{aligned}$$

In the case $e_{1}+e_{2}=0$ one can check that any vector of the form $|t\rangle =\alpha |\uparrow \rangle \otimes$ $|\downarrow \rangle +\beta |\downarrow \rangle \otimes$ $|\uparrow \rangle$ with $\alpha \ne e^{he_{1}}\beta$ serves as the $|t\rangle$-vector in the basis of the ${\mathcal {P}}_{n_{1}+n_{2}}^{\kappa }$ module of four vectors of the tensor product ${\mathcal {T}}_{e_{1},n_{1}}^{\kappa }\otimes {\mathcal {T}} _{-e_{1},n_{2}}^{\kappa }$. We see that the tensor product ring composed of the $U_{h}({\mathfrak {gl}}(1|1))$-modules ${\mathcal {A}}_{n}^{\kappa }, {\mathcal {T}}_{e,n}^{\kappa },{\mathcal {P}}_{n}^{\kappa }$ is the same as the tensor product ring of the category ${\mathcal {C}}_{\kappa }$ composed of ${\mathcal {A}}_{n},{\mathcal {T}}_{e,n},{\mathcal {P}}_{n}$ for restriction on parameters the same as in Proposition 3. $\square$