# Definition Given a [[Lie algebra]] $\mathfrak{g}$, let $\{X_i\}_{i=1\ldots n}$ be a [[basis]]. Then the [[commutator|commutation]] relations take the form: $ [X_i, X_j] = \sum_{k=1}^n c_{ij}^{\,\,\,\,k}X_k $ The $c_{ij}^{\,\,\,\,k}$ are the structure constants. "Structure constants" is a misnomer since they are basis dependent. ## As components of a Tensor Define a (2,1) tensor $T$ on the Lie algebra $\mathfrak{g}$ by: $ T(X, Y, f) \equiv f([X,Y]) \quad \forall X,Y \in \mathfrak{g}, \,\, f \in \mathfrak{g}^* $ Then, let $\{f^k\}_{k=1,\ldots,n}$ be the basis dual to $\{X_i\}_{i=1\ldots n}$, T has components: $ T_{ij}^{\,\,\,\,k} = f^k([X_i,X_j]) = f^k(c_{ij}^{\,\,\,\,l}X_l) = f^k(X_l)c_{ij}^{\,\,\,\,l} = \delta^k_{\,\,l} c_{ij}^{\,\,\,\,l} = c_{ij}^{\,\,\,\,k} $ # Lie Algebra Isomorphisms Sometimes, to prove that two [[Lie Algebra|Lie algebras]] are [[Lie Algebra Isomorphism|isomorphic]], the easiest way is to make a smart choice of bases. Let $\{X_i\}_{i=1\ldots n}$ be a [[basis]] for $\mathfrak{g}$ and $\{Y_i\}_{i=1\ldots n}$ a basis for $\mathfrak{h}$. Then the commutation relations take the form: $ \begin{align} [X_i, X_j] &= \sum_{k=1}^n c_{ij}^{\,\,\,\,k}X_k\\ [Y_i, Y_j] &= \sum_{k=1}^n d_{ij}^{\,\,\,\,k}Y_k \end{align} $ If we pick our basis intelligently so that $c_{ij}^{\,\,\,\,k} = d_{ij}^{\,\,\,\,k} \,\, \forall \,\, i, j, k$, then the map: $ \begin{align} \phi: \quad \mathfrak{g} &\rightarrow \mathfrak{h}\\ v^iX_i &\mapsto v^iY_i \end{align} $ is an isomorphism.