0
$\begingroup$

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?

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

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.