Equivalence of Casimir Element Expressions for gl₂(ℝ)

Definition 1 (Standard Basis and Casimir Element). Let the Lie algebra ${\mathfrak{gl}}_2(\mathbb{R})$ be spanned by the standard basis over $\mathbb{R}$:

$$H=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right),E=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right),F=\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right),I=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right).$$

The element $I$ spans the center of ${\mathfrak{gl}}_2(\mathbb{R})$. The quadratic Casimir element derived from the trace form on ${\mathfrak{gl}}_2(\mathbb{R})$ is the element ${\Omega }_{\text{std}}$ in the universal enveloping algebra $U({\mathfrak{gl}}_2(\mathbb{R}))$ given by:

$${\Omega }_{\text{std}}=\frac{1}{2}{H}^2+EF+FE+\frac{1}{2}{I}^2.$$

Definition 2 (Alternative Basis and Casimir Element). Let ${\mathfrak{gl}}_2(\mathbb{R})$ be spanned by the alternative basis:

$$X=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right),Y=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right),K=\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right),I=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right).$$

Within the complexified universal enveloping algebra $U({\mathfrak{gl}}_2(\mathbb{C}))$, we define the raising operator $R$ and lowering operator $L$ as $R=\frac{1}{2}(X+iY)$ and $L=\frac{1}{2}(X-iY)$. An alternative expression for the Casimir element is the element ${\Omega }_{\text{alt}}$ given by:

$${\Omega }_{\text{alt}}=-\frac{1}{2}{K}^2+RL+LR+\frac{1}{2}{I}^2.$$

Theorem 1 (Equivalence of Casimir Formulations for ${\mathfrak{gl}}_2(\mathbb{R})$). The two formulations of the quadratic Casimir element are equal.

$${\Omega }_{\text{std}}={\Omega }_{\text{alt}}.$$

Proof. The elements of the two bases are related by the following linear transformations: $X=H$, $Y=E+F$, and $K=E-F$. The products $E^2$ and $F^2$ are zero in the algebra of matrices, and thus also in $U({\mathfrak{gl}}_2(\mathbb{R}))$. The calculation proceeds by expressing ${\Omega }_{\text{alt}}$ in the standard basis $\{H,E,F,I\}$.

$$\begin{array}{rl}{\Omega }_{\text{alt}}& =-\frac{1}{2}{K}^2+RL+LR+\frac{1}{2}{I}^2\\ & =-\frac{1}{2}{K}^2+\frac{1}{4}(X+iY)(X-iY)+\frac{1}{4}(X-iY)(X+iY)+\frac{1}{2}{I}^2\\ & =-\frac{1}{2}{K}^2+\frac{1}{4}(X^2+Y^2+i[Y,X])+\frac{1}{4}(X^2+Y^2-i[Y,X])+\frac{1}{2}{I}^2\\ & =-\frac{1}{2}{K}^2+\frac{1}{2}(X^2+Y^2)+\frac{1}{2}{I}^2\\ & =-\frac{1}{2}(E-F{)}^2+\frac{1}{2}\left(H^2+(E+F{)}^2\right)+\frac{1}{2}{I}^2\\ & =-\frac{1}{2}(E^2-EF-FE+F^2)+\frac{1}{2}(H^2+E^2+EF+FE+F^2)+\frac{1}{2}{I}^2\\ & =-\frac{1}{2}(-EF-FE)+\frac{1}{2}(H^2+EF+FE)+\frac{1}{2}{I}^2\\ & =\frac{1}{2}(EF+FE)+\frac{1}{2}{H}^2+\frac{1}{2}(EF+FE)+\frac{1}{2}{I}^2\\ & =\frac{1}{2}{H}^2+EF+FE+\frac{1}{2}{I}^2\\ & ={\Omega }_{\text{std}}.\end{array}$$

$\square$