Since a basis for a Free Lie Algebra $\cal{FL}$ with n generators $x_1,\ldots x_r$ can be described by the elements $\{ [\ldots[x_{i_1},x_{i_2}]x_{i_2}\ldots]x_{i_m}] , 1 \leq x_{i_j} \leq r , 1 \leq j \leq m , m \in \mathbb{N} \}$ , what is the point of using the Hall basis for $\cal{FL}$? Both basis generate the same graded algebra and the first basis is simpler than the Hall basis .
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closed as offtopic by Ricardo Andrade, Daniel Moskovich, Jack Huizenga, Andrey Rekalo, Stefan Kohl Dec 1 '13 at 9:22This question appears to be offtopic. The users who voted to close gave this specific reason:



Your elements don't form a basis. You haven't given any conditions on the indices $i_j$, so $[x_1,x_1]$ is in your list, but it is zero. Now consider words of length three. Among these are $[[x_1,x_2],x_3]$, $[[x_2,x_3],x_1]$ and $[[x_3,x_1],x_2]$. The sum of these three is zero by the Jacobi identity, so they are not linearly independent. You might want to insist that the list of indices $i_1,\dotsc,i_r$ is strictly increasing, but then $[[x_2,x_3],x_1]$ would not be in the span. 

