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This question is my doubt from Kac's book on Infinite dimensional Lie algebras.

We start with an arbitrary matrix A, and we define the realization of A and using the generators $\{e_i,f_i : 1 \le i \le n\}$ and $h$ we define the auxiliary lie algebra $g^{-}(A)$. We define $n_-$ to be the sub algebra generated by $\{f_i : 1 \le i \le n\}$. Later in theorem 1.2 he claims that $n_-$ is freely generated by $\{f_i : 1 \le i \le n\}$.

My question is, by defintion $n_-$ is generated by $\{f_i : 1 \le i \le n\}$. Then why we have to again prove it is freely generated by $\{f_i : 1 \le i \le n\}$. What is the difference between "generated by" and "freely generated" by ?

Thanks for your valuable timing.

GA316
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A free algebra $A$ on a set $S$ of generators is one with no relations among the generators. Informally, this means that the only expression of the form $\sum_{n \in \mathbb Z_{\ge 0}^S} a_n S^n = 0$ is one where all coefficients $a_n$ are $0$. Formally, it means that any set map $S \to B$, where $B$ is an algebra, extends to an algebra map $A \to B$. Thus, showing that the algebra is freely generated by a certain set is, loosely speaking, showing that that set of generators is minimal.

Compare to the situation for a vector space, where "$V$ is generated by $S$" means "$S$ spans $V$", but "$V$ is freely generated by $S$" means "$S$ is a basis for $V$". That is, 'free' here corresponds to 'linearly independent'. (Of course, in an algebra, the concept 'linearly independent' is replaced by 'algebraically independent'.)

LSpice
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  • can you please tell me how the arbitrary element looks like inside $n_-$? – GA316 Apr 29 '15 at 17:45
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    @GA316, by definition, an element of $\mathfrak n_-$ is an algebraic combination of $f_-$'s. The claim is that every element can be uniquely represented as such a combination. – LSpice Apr 29 '15 at 18:48