PBW for countably presented Kac Moody Lie algebras

Replace an adjacent inversion $x_j{x}_i$, $j>i$, by $x_i{x}_j+[x_j,x_i]$, expanding the bracket into basis vectors. On each term order the measure lexicographically by word length and number of inversions. The switched term decreases inversions; every bracket term decreases length. Each replacement has finitely many terms. The finitely branching reduction tree has finite depth: otherwise recursively selecting its first child with arbitrarily deep descendants gives an infinite descending sequence of measures. Thus reduction terminates in ordered words.

Two reductions on disjoint pairs commute, including their lower-length terms. The only overlapping pair is $zyx$ with $z>y>x$. Reducing the length-three terms in the two orders gives respectively $xyz+[y,x]z+y[z,x]+[z,y]x$ and $xyz+x[z,y]+[z,x]y+z[y,x]$. Their difference is $[[y,x],z]+[y,[z,x]]+[[z,y],x]$ after reducing length-two commutators. This is zero by Jacobi. Those length-two reductions are already unambiguous by induction on length, and this reasoning also holds inside a fixed word context.

Induct on the reduction measure to compare any two first reductions: the disjoint case joins exactly, and the overlap difference has zero normal form by step 2.1; all subsequent comparisons involve smaller measures. Linearity then gives a unique normal form $N$ for every finite polynomial. For any words $u,v$, $N(u(x_j{x}_i-x_i{x}_j-[x_j,x_i])v)=0$; the relations with the other order follow by antisymmetry and those with equal indices are zero. Hence $N$ kills the two-sided defining ideal. Conversely each reduction changes a polynomial by an element of that ideal. The ordered-word inclusion and $N$ are inverse maps after taking the quotient. In particular distinct length-one basis vectors remain independent.

Fix one homogeneous component $V$. Its spanning finite bracket words inherit a finite or countable enumeration; retaining each first word outside the span of its predecessors gives an ordered basis $b_0,b_1,\dots$ of $V$. For a specified subspace $W\le V$, put $E_r=\mathrm{span}(b_0,\dots ,b_r)$ and $W_r=W\cap E_r$. These finite-dimensional spaces exhaust $W$, and $\dim W_r-\dim W_{r-1}$ is zero or one. Starting with the empty basis, do nothing when the dimension is unchanged. When it rises at $r$, finite row reduction gives the unique $v_r\in W_r$ whose $b_r$-coefficient is $1$ and whose coefficients in the previous pivot columns are zero; append $v_r$. Indeed, existence comes from normalizing any element of $W_r\setminus W_{r-1}$ and eliminating its old pivots, while two such vectors differ by an element of $W_{r-1}$ with every pivot coefficient zero and hence are equal. Induction now shows that the vectors obtained through stage $r$ form a basis of $W_r$, so their union is a basis of $W$. Extend it to a basis of $V$ by scanning the $b_r$ and retaining the first vectors outside the span already obtained; the images of the added vectors form a quotient basis of $V\mathrm{/}W$. Applying this fixed construction to the supplied countable list of degrees gives compatible homogeneous bases without a family of choices. The same finite-coordinate exhaustion handles a countably spanned ungraded algebra. No basis for an arbitrary unbased vector space is asserted.