Realization of the Quantum Superalgebra Uq(gl(1|1)) via q-Differential Operators

Realization of the Quantum Superalgebra $U_q(\mathfrak{gl}(1|1))$ via $q$-Differential Operators

Hi, I'm akaghef. I talked with Gemini and found the following representation.
I've checked the expression, but the definitions of the algebras have not been checked.
Definitions of algebras are extracted from AS.

1. The Lie Superalgebra $\mathfrak{gl}(1|1)$

We begin by recalling the definition of the Lie superalgebra $\mathfrak{gl}(1|1)$. Let $\mathbb{C}^{1|1}$ be a $(1|1)$-dimensional complex super vector space. $\mathfrak{gl}(1|1)$ is generated by the even elements $h_1, h_2$ (degree 0) and the odd elements $e, f$ (degree 1).
The defining relations are given by the supercommutator $[a, b] = ab - (-1)^{|a||b|}ba$:
$$ \begin{aligned} {}[ h_1,e]&= e, & [h_2, e] &= -e, \\ [h_2, f] &= f, & [h_1, f] &= -f, \\ [h_1, h_2] &= 0, & [e, f] &= h_1 + h_2, \\ [e, e] &= 0, & [f, f] &= 0. \end{aligned} $$

2. The Quantum Enveloping Superalgebra $U_q(\mathfrak{gl}(1|1))$

The quantum enveloping superalgebra $U_q = U_q(\mathfrak{gl}(1|1))$ is the unital superalgebra over $\mathbb{C}(q)$ generated by the elements $E, F$ (odd, degree 1) and invertible elements of the form $q^h$ for $h \in P^*$ (even, degree 0).
Let $K = q^{h_1 + h_2}$. The defining relations are:
$$ \begin{gathered} q^0 = 1, \quad q^h q^{h'} = q^{h+h'}, \\ q^h E = q^{\langle h, \alpha \rangle} E q^h, \qquad q^h F = q^{-\langle h, \alpha \rangle} F q^h, \\ EF + FE = \frac{K - K^{-1}}{q - q^{-1}}, \\ E^2 = F^2 = 0. \end{gathered} $$
Here, $\alpha = \epsilon_1 - \epsilon_2$ is the simple root.

3. Differential Calculus on Superspace

To construct a representation of these algebras, we consider the polynomial superalgebra $\mathbb{C}[x, \eta]$ generated by one even variable $x$ and one odd (Grassmann) variable $\eta$.

Variables and Derivatives

• Even variable: $x$ ($|x|=0$)
• Odd variable: $\eta$ ($|\eta|=1$), satisfying $\eta^2 = 0$.
  We introduce the derivatives $\partial_x$ (even) and $\partial_\eta$ (odd), satisfying the super-Leibniz rules:
  $$ \begin{aligned} {}[\partial_x, x] &=\p_xx-x\p_x= 1, & [\partial_x, \eta] &= 0, \\ [\partial_\eta, \eta] &= \partial_\eta\eta+\eta\partial_\eta= 1, & [\partial_\eta, x] &= 0. \end{aligned} $$

Euler Operators (Number Operators)

We define the Euler operators $\theta_x$ and $\theta_\eta$ as follows:
$$ \theta_x = x \partial_x, \qquad \theta_\eta = \eta \partial_\eta. $$
Since $\eta$ is a Grassmann variable, the operator $\theta_\eta$ is idempotent, i.e., $\theta_\eta^2 = \theta_\eta$. This property allows for a simple expansion of the exponential (or $q$-power) operator:
$$ q^{\theta_\eta} = 1 + (q-1)\theta_\eta. $$
This identity is crucial for calculations involving the quantum Cartan generator.

4. Construction of Representations

4.1 Classical Limit ($q \to 1$)

First, we recover the fundamental representation of the classical Lie superalgebra $\mathfrak{gl}(1|1)$ on the space spanned by $\{x, \eta\}$. Using the notation of Euler operators, we identify the Cartan generators directly with $\theta$:
$$ \begin{aligned} \rho(h_1) &= \theta_x, \\ \rho(h_2) &= \theta_\eta, \\ \rho(e) &= x \partial_\eta, \\ \rho(f) &= \eta \partial_x. \end{aligned} $$
It is straightforward to verify that these satisfy the relations in Section 1. For instance, the supercommutator of $e$ and $f$ is:
$$\begin{aligned} {}[\rho(e), \rho(f)] &= x \partial_\eta (\eta \partial_x) + \eta \partial_x (x \partial_\eta) \\ &= x(1 - \eta \partial_\eta)\partial_x + \eta (1 + x \partial_x) \partial_\eta \\ &= x \partial_x - x \eta \partial_\eta \partial_x + \eta \partial_\eta + \eta x \partial_x \partial_\eta. \end{aligned}$$
Since $x, \eta$ commute and $\partial_x, \partial_\eta$ commute, the cross terms cancel, yielding:
$$[\rho(e), \rho(f)] = x \partial_x + \eta \partial_\eta = \rho(h_1) + \rho(h_2).$$

4.2 Quantum Deformation

To obtain the representation of $U_q(\mathfrak{gl}(1|1))$, we deform the differential operators using the $q$-number definition:
$$ [A]_q = \frac{q^A - q^{-A}}{q - q^{-1}}. $$
Using the Euler operators defined above, we propose the following representation $\rho$ on the superspace $\mathbb{C}[x, \eta]$:
$$ \begin{aligned} \rho(K) &= q^{\theta_x + \theta_\eta} = q^{\theta_x}q^{\theta_\eta}, \\ \rho(E) &= x \partial_\eta, \\ \rho(F) &= \frac{\eta}{x} [\theta_x]_q. \end{aligned} $$
Here, $\rho(F)$ involves the $q$-deformation of the even Euler operator $\theta_x$, while the odd part remains simple due to the properties of $\theta_\eta$.

The vector space $V$ spanned by the variables $x$ and $\eta$ corresponds to the fundamental representation (or vector representation) of $\mathfrak{gl}(1|1)$. By restricting the action of the differential operators to the linear subspace $V = \text{span}\{x, \eta\}$, we recover the standard defining representation:
\begin{align} \rho(e) \eta &= x, & \rho(e) x &= 0, \\ \rho(f) x &= \eta, & \rho(f) \eta &= 0. \end{align}
This corresponds exactly to the matrix form where $e=E_{12}$ and $f=E_{21}$ under the basis identification $v_1=x$ and $v_2=\eta$.

Verification of the Quantum Relation

We verify the defining relation $EF + FE = [K]_q = [\theta_x + \theta_\eta]_q$.
Recall the properties:

• $\theta_\eta$ is idempotent, since $[\theta_\eta]_q = \theta_\eta$ and $q^{\theta_\eta} = 1 + (q-1)\theta_\eta$.
• Shift property: $f(\theta_x) x = x f(\theta_x + 1)$.
• addition of q-bracket: $[A+B]_q=q^A[B]_q+q^{-B}[A]_q$
  \begin{align} \rho(E)\rho(F) + \rho(F)\rho(E) &= x \partial_\eta \frac{\eta}{x} [\theta_x]_q + \frac{\eta}{x} [\theta_x]_q x \partial_\eta \\ &= (1 - \theta_\eta) [\theta_x]_q + \eta [\theta_x + 1]_q \partial_\eta \\ &= [\theta_x]_q + \left( [\theta_x + 1]_q - [\theta_x]_q \right) \theta_\eta\\ &= [\theta_x]_q \left( 1 + (q-1)\theta_\eta \right) + q^{-\theta_x} \theta_\eta \\ &= [\theta_x]_q q^{\theta_\eta} + q^{-\theta_x} [\theta_\eta]_q\\ &= [\theta_x + \theta_\eta]_q \\ &= \rho\left(\frac{K - K^{-1}}{q - q^{-1}}\right) \end{align}
  Other relations are easier to check.