Let $\mathfrak{g}$ be a finite dimensional Lie algebra and $B:\mathfrak{g} \times \mathfrak{g} \rightarrow \mathbb{K}$ be a non-degenerate invariant symmetric bilinear form. Then there exists a unique $C_{B} \in \mathfrak{U(g)}$, called the Casimir element of $\mathfrak{g}$ corresponding to $B$, such that $$\sum_{i=1}^{n} e_{i}(B^{\sharp}e_{i}^{*}), $$ for any basis $e_{1},...,e_{n}$ of $\mathfrak{g}$, with corresponding dual basis $e_{1}^{*},...,e_{n}^{*}$, where $B^{\sharp}:\mathfrak{g^{*}} \rightarrow \mathfrak{g}$ denotes the inverse of the linear isomorphism $B^{\flat}: \mathfrak{g} \rightarrow \mathfrak{g^{*}}, X \mapsto B(X,-)$. Additionally, $C_{B}$ is a quadratic central element of $\mathfrak{U(g)}$, i.e. $$C_{B} \in \mathfrak{U_{2}(g)} \setminus \mathfrak{U_{1}(g)}$$ and $$C_{B} \in Z(\mathfrak{U(g)}). $$
I'm studying this proposition on Casimir element but i don't understand why to show that $C_{B}$ is a quadratic central element of $\mathfrak{U(g)}$ i have to prove that $C_{B} \in \mathfrak{U_{2}(g)} \setminus \mathfrak{U_{1}(g)}$. What means that an element is "quadratic"? Can i have a strict definition?
Is there anyone who can help me?