Simple Witt algebra

Definition

Suppose $p$ is a prime number greater than 3. Suppose $F$ is a field of characteristic $p$. The $p$-dimensional simple Witt algebra over $F$, denoted $W(1;{\underline {1}})$, is a Lie algebra over $F$defined as follows.

• The additive group has basis $e_{-1},e_{0},\dots ,e_{p-2}$. Explicitly, it is $\bigoplus _{i=-1}^{p-2}Fe_{i}$.
• The Lie bracket is defined as follows on the basis:

$[e_{i},e_{j}]:=\left\lbrace {\begin{array}{rl}(j-i)e_{i+j},&-1\leq i+j\leq p-2\\0,&{\mbox{otherwise}}\\\end{array}}\right.$

Note in particular that we can take $F$ to be the prime field $\mathbb {F} _{p}$, or we can take $F$ to be the finite field $\mathbb {F} _{q}$ where $q$ is a power of the prime $p$, or we can take $F$ as an infinite algebraically closed field of characteristic $p$.

However, although we can define a bracket of this sort in characteristics other than $p$, we do not get a Lie algebra in those characteristics, because the Jacobi identity fails.

Verification of alternation

This is obvious from the definition:

• $[e_{i},e_{i}]=0$ both in the case $2i\notin \{-1,\dots ,p-2\}$ and in the case $2i\in \{-1,\dots ,p-2\}$.
• $[e_{i},e_{j}]=-[e_{j},e_{i}]$ for $i\neq j$ both in the case $i+j\notin \{-1,\dots ,p-2\}$ and in the case $i+j\in \{-1,\dots ,p-2\}$.

Verification of Jacobi identity

Note that one of the cases in the verification crucially uses that the characteristic is $p$.

It suffices to verify the Jacobi identity on triples of the form $(e_{i},e_{j},e_{k})$. Explicitly, it suffices to check that for all $i,j,k\in \{-1,0,\dots ,p-2\}$, we have:

$[[e_{i},e_{j}],e_{k}]+[[e_{j},e_{k}],e_{i}]+[[e_{k},e_{i}],e_{j}]=0$

We make cases:

No.	Case description	Details
1	$i+j+k,i+j,j+k,i+k$ are all between -1 and $p-2$.	The expression becomes $((j-i)(k-(i+j))+(k-j)(i-(j+k))+(i-k)(j-(k+i)))e_{i+j+k}$, so showing it is zero reduces to showing that $(j-i)(k-(i+j))+(k-j)(i-(j+k))+(i-k)(j-(k+i))=0$. This can be broken down into two verifications: $(j-i)k+(k-j)i+(i-k)j=0$ $(j-i)(i+j)+(k-j)(j+k)+(i-k)(k+i)=(j^{2}-i^{2})+(k^{2}-j^{2})+(i^{2}-k^{2})=0$
2	$i+j+k$ is not between -1 and $p-2$	In this case, all the summands $[[e_{i},e_{j}],e_{k}],[[e_{j},e_{k}],e_{i}],[[e_{k},e_{i}],e_{j}]$ are zero. Note that the stage of the computation where each becomes zero depends on whether $i+j,j+k,i+k$ are in $\{-1,\dots ,p-2\}$, but they all eventually become zero.
3	Case that $i+j+k$ is between $-1$ and $p-2$ but not all the sums $i+j,j+k,k+i$ are	This can happen only if one of $i,j,k$ is -1. Assume that $k=-1$ (without loss of generality). There are two subcases.
3.1	THIS CASE USES THE CHARACTERISTIC: $i+j=p-1,k=-1$	We know that $[[e_{i},e_{j}],e_{k}]=0$, but if we weren't truncating, it would have been $(j-i)(k-(i+j))e_{i+j+k}=(j-i)(-1-(p-1))e_{p-2}=(j-i)(-p)e_{p-2}=0$ because of characteristic $p$. In other words, the "would have been" answer and the actual answer don't differ. Thus, we can revert to the calculation of Case (1).
3.2	$i=1,j=k=-1$	We know that $[[e_{j},e_{k}],e_{i}]=0$, but if we weren't truncating, it would have been $(k-j)(i-(j+k))e_{i+j+k}=0$. In other words, the "would have been" answer and the actual answer don't differ. Thus, we can revert to the calculation of Step (1).

Related notions

• Zassenhaus algebra is a subalgebra of the simple Witt algebra corresponding to an additive subgroup $G$ of $F$.
• Panferov Lie algebra is defined similarly, except that the indices used to label the algebras are somewhat shifted.

References

• here -- convert to formal reference