# Wednesday October 23

Recall from last time:

For $\lieg$ a lie algebra, we define $T(\lieg)$ the tensor algebra, and the universal enveloping algebra $U(\lieg) = T(\lieg)/\sim$ where $x\tensor y - y\tensor x \sim [x, y]$.

We also described the *PBW Theorem*, which provides a basis for $U(\lieg)$.

## Proof of PBW Theorem

*Proof of PBW Theorem:*

We have $T(\lieg) = \spanof\theset{x_{j_1} \tensor \cdots \tensor x_{j_k} \suchthat j_1, \cdots, j_k \in I}$, where we note that there are not required to be ordered.
Thus $U(\lieg) = \spanof\theset{y_{j_1} \tensor \cdots \tensor y_{j_k} \suchthat j_1, \cdots, j_k \in I}$, where which are again not required to be ordered.
We would thus like to express every term here as some linear combination of monomials in the $y_{i_j}$ with increasing indices.
We proceed by inducting on $k$, the number of tensor factors occurring.
The base case is clear.

For $k> 1$, supposing that the element is *not* a PBW monomial, then there is some inversion in the indices $(j_1, \cdots, j_k)$, i.e. there is at least one $i$ such that $j_{i+1} < j_i$.
Now for any two indices $a,b \in I$, we have
$$
\iota(x_b \tensor x_a) = \iota(x_a \tensor x_b + [x_b, x_a]) \implies  y_b y_a = y_a y_b + \iota([x_b, x_a])
$$
Since $[x_b, x_a] = \sum_t F x_t$ and $\iota[x_b, x_a] = \sum_t F y_t$.

But then $y_{j_1} \cdots y_{j_k} = y_{i_1} y_{i_2} \cdots y_{j_k} + \text{ lower degree terms}$ where $i_1 \leq i_2 \cdots i_k$ is a non-decreasing rearrangement of the $j_i$.
By the inductive hypothesis, the lower degree terms are spanned by PBW monomials, so we're done.

*Proof of linear independence:*

*Claim:* 
Let $\vector x \coloneqq x_{j_1} \tensor \cdots \tensor x_{j_n}$ for an arbitrary indexing sequence, and $\vector x_{(k)}$ be this tensor with the $j_k$ and $j_{k+1}$ terms swapped, and $\vector x_{[k]}$ be this tensor with $x_{j_k}, x_{j_{k+1}}$ replaced by their bracket.

Then there exists a linear map 

\begin{align*}
f: T(\lieg) \to R \coloneqq F[\theset{z_i}_{i\in I}] \\
f(x_{i_1} \tensor \cdots \tensor x_{i_n}) = z_{i_1} \cdots z_{i_n} \\
f(\vector x - \vector x_{(k)}) = f(\vector x_{[k]})
.\end{align*}

By collecting terms, we can write
$$
\vector x - \vector x_{(k)} - \vector x_{[k]} = x_{j_1} \tensor \cdots \tensor x_{j_{k-1}} \tensor \left( (x_{j_k} \tensor x_{j_{k+1}}) - (x_{j_{k+1}} \tensor x_{j_k}) - [x_{j_k}, x_{j_{k+}}]   \right) \tensor \cdots
$$

So we can take $J$ to be the ideal generated by all elements of this form, and we find that $J \subset \ker f$, and thus $f$ descends to a map $\overline f$ on $U(\lieg)$.
We then know that if $\overline f$ applied to any PBW monomial is $z_{i_1}^{r_1} \cdots z_{i_n}^{r_n}$, which are linearly independent in $R$, then any PBW monomial will be linearly independent in $U(\lieg)$.

*Proof of claim:*

For each $\vector x$, define an *index*
$$
\lambda(\vector x) = \#\theset{ (a,b) \in \theset{1, \cdots, n}^2 \suchthat a<b, j_a < j_b  }.
$$

Then 
$$
\theset{\vector x \suchthat \lambda(\vector x) = 0} = \theset{ x_{i_1} \tensor \cdots \tensor x_{i_n} \suchthat i_1 \leq \cdots \leq i_n }.  
$$

So set $T^{n, k} = \theset{\vector x \in T^n(\lieg) \suchthat \lambda(\vector x) \leq k}$; we then have a filtration $T^{n, 0} \injects T^{n, 1} \injects \cdots \injects T^n(\lieg)$.

**Step 1:** 
We'll construct $f$ by induction on $n$.

For $n> 0$, set $f(\vector x) = z_{j_1} \cdots z_{j_n}$ if $\lambda(\vector x) = 0$.
We now induct on the index $k$ at a fixed power $n > 0$. 
The base case is clear.

For $k>0$, there exists an inversion $(\ell, \ell+1)$, i.e. some indices $i_{\ell} > \i_{\ell+1}$.
Set $f(\vector x) = f(\vector x_{(\ell)}) - f(\vector x_{[\ell]})$, where the LHS is in $T^{n, k}$ and the RHS terms are in $T^{n, k-1}$ and $T^{n-1}(\lieg$ respectively.

**Step 2:**
We'll check that $f$ is well-defined.

In the above definition, note that $f(\vector x)$ can be defined using different inversions of the indices, we'd like to show that these yield the same map.

Let $(\ell, \ell+1)$ and $(\ell', \ell'+1)$ be two distinct inversions. Then set

\begin{align*}
a = x_{j_\ell} \\
b = x_{j_{\ell+1} }\\
c = x_{j_\ell'} \\
d = x_{j_{\ell'+1}} \\
.\end{align*}

Then we have several cases:

**Case 1: $\ell + 1 < \ell'$.** 

Then

\begin{align*}
f(\vector x_{(\ell)}) + f(\vector x_{[\ell]}) 
&= f( \cdots b\tensor a \cdots c\tensor d \cdots ) \\
+ f( \cdots \tensor [a, b] \tensor \cdots c\tensor d \cdots ) \\
&= f( \cdots b\tensor a \cdots d\tensor c \cdots )
+ f( \cdots b\tensor a \cdots [c, d] \cdots )
+ f( \cdots \tensor [a, b] \tensor \cdots d\tensor c \cdots ) 
+ f( \cdots \tensor [a, b] \tensor \cdots [c, d] \cdots ) \\
&= f(\vector x_{(\ell']}) + f(\vector x_{[\ell']} )
.\end{align*}

**Case 2: $\ell+1 = \ell'$**

Then

\begin{align*}
f(\vector x_{(\ell)}) + f(\vector x_{[\ell]}) 
&= f(\cdots b\tensor a\tensor x) + f(\cdots [a,b] \tensor c) \\
&= f(b\tensor c \tensor a)
+ f(c\tensor [a,b])
+ f(b\tensor [a,c]) +
f([[a,b], c]) \\
&= f(c\tensor b \tensor a)
+ f(c\tensor [a,b])
+ f(b \tensor [a,c])
+ f([[a,b], c]) 
+ f(b\tensor [a,c])
+ f(a \tensor [b, c])
+ f([[b,c], a])\\
&= f(\vector x_{(\ell')}) + f(\vector x_{[\ell']})
.\end{align*}

where the last equality is found by expanding the expression backwards.