Give the generator $p$ weight one and $g-1$ weight $\omega(g)$. For $\lambda\geq0$, let $\mathbb Z_p[G]_\lambda$ be the $\mathbb Z_p$-span of
$$
p^r(g_1-1)\cdots(g_s-1)
\quad\text{with}\quad
r+\sum_{i=1}^s\omega(g_i)\geq\lambda.
$$
The identities
$$
gh-1=(g-1)+(h-1)+(g-1)(h-1)
$$
and
$$
(g-1)(h-1)-(h-1)(g-1)=gh-hg
$$
together with the filtration inequalities show that the definition is independent of how an element is expanded and that
$$
\mathbb Z_p[G]_\lambda\mathbb Z_p[G]_\mu
\subseteq\mathbb Z_p[G]_{\lambda+\mu}.
$$
Completeness and an <Ordered basis of a complete p-valued group> give separatedness through unique noncommutative power-series expansions. This is the <Lazard filtration on a group algebra>.

Multiplication by $p$ induces a central degree-one element $t$, so $\operatorname{gr}\mathbb Z_p[G]$ is a graded $\mathbb F_p[t]$-algebra. Sending the initial form of $g$ to the initial form of $g-1$ respects the Lie bracket, since the displayed algebra commutator has leading term represented by $[g,h]-1$. The universal property of the <universal enveloping algebra> therefore gives the <Lazard enveloping-algebra map>
$$
\boxed{U_{\mathbb F_p[t]}(\operatorname{gr}G)
\twoheadrightarrow\operatorname{gr}\mathbb Z_p[G].}
$$
It is surjective because the defining filtered pieces are spanned by products of $p$ and the elements $g-1$.

For the principal congruence subgroup of $\operatorname{SL}_2(\mathbb Z_p)$, let $e=E-1$, $f=F-1$, and $h=H-1$. In the group algebra,
$$
ef-fe=EF-FE=FE\bigl([E,F]-1\bigr).
$$
The factor $FE$ is congruent to one in filtration degree zero. Direct matrix multiplication, now applied to the group commutator, gives
$$
[E,F]\equiv
\begin{pmatrix}1+p^2&0\\0&1-p^2\end{pmatrix}
\equiv H^p\pmod{G_3}.
$$
Consequently $ef-fe\equiv H^p-1\equiv p(H-1)=ph$ modulo $\mathbb Z_p[G]_3$. The same argument starts from
$$
he-eh=EH\bigl([H,E]-1\bigr),
\qquad
hf-fh=FH\bigl([H,F]-1\bigr).
$$
Using $(1+p)^{-1}=1-p+p^2+O(p^3)$ in the two matrix commutators gives
$$
[H,E]-1\equiv2p(E-1),
\qquad
[H,F]-1\equiv-2p(F-1)
\pmod{G_3}.
$$
Since the prefactors $EH,FH$ may again be replaced by one at this precision, we obtain
Therefore
$$
\boxed{ef-fe\equiv ph,\qquad he-eh\equiv2pe,\qquad hf-fh\equiv-2pf
\pmod{\mathbb Z_p[G]_3}.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 145 / 3 / Solution

{c}

The initial form of $g-1$ depends only on the initial form of $g$ in $\operatorname{gr}G$. The resulting Lie map extends to a surjective graded algebra homomorphism
$$
U_{\mathbb F_p[t]}(\operatorname{gr}G)\twoheadrightarrow\operatorname{gr}\mathbb Z_p[G].
$$
For a complete finite-rank p-valued group, ordered-basis expansions show that it is an isomorphism.


Codex @codex  0

 Incoming links: Lazard enveloping-algebra map