# A question regarding the basis of $\mathfrak{sl_2}$.

The basis of the lie algebra $\mathfrak{sl}_2$ comprises of the matrices $\mathfrak{u}=\begin{pmatrix} 0&1\\0&0\end{pmatrix},\mathfrak{v}= \begin{pmatrix} 0&0\\1&0\end{pmatrix}$ and $\mathfrak{w}=\begin{pmatrix} 1&0\\0&-1\end{pmatrix}$.

Note that the operations in this algebra are usual addition and bracket multiplication.

However, note that $[\mathfrak{u},\mathfrak{v}]=\mathfrak{w}$. Then why do we need $\mathfrak{w}$ to be an element of the basis at all, if it can be generated using other elements in the basis? Are we referring to a vector space basis here, and not the generating set of the Lie Algebra?

• >Are we referring to a vector space basis here, and not the generating set of the Lie Algebra? >> Yes, that is the answer. – Wolfram Feb 1 '17 at 23:47

Yes, the dimension $\dim(L)$ of a Lie algebra $L$ is by definition the dimension of the underlying vector space. And the vector space of traceless $2\times 2$-matrices over a field $K$ is $3$-dimensional. Note that a basis here is not unique. For example, another basis is given by $$\mathfrak{u}=\begin{pmatrix} 0&1\\0&0\end{pmatrix},\mathfrak{v}= \begin{pmatrix} 0&0\\1&0\end{pmatrix},\mathfrak{w}=\begin{pmatrix} 1 & 1\\-1 &-1\end{pmatrix}$$ giving a Lie algebra isomorphic to $\mathfrak{sl}_2(K)$, but where $ad(\mathfrak{w})$ is nilpotent.