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The central series of U_n with additive quotients

Statement

Let $k$ be a field and $n \geq 1$ , and let $U_{n} \subseteq T_{n} \subseteq {G L}_{n}$ be the upper unitriangular and upper triangular group schemes of The upper unitriangular group scheme U_n and its coordinate ring, with $T_{n} = D_{n} ⋉ U_{n}$ .

Order the pairs $(i, j)$ with $1 \leq i < j \leq n$ by increasing $j - i$ (and arbitrarily, say by increasing $i$ , within a fixed difference), and let $m = n (n - 1) / 2$ . For $0 \leq r \leq m$ let $U_{n}^{(r)}$ be the closed subgroup scheme of $U_{n}$ of matrices whose entries $x_{i j}$ vanish on the first $r$ pairs of the ordering; thus $U_{n}^{(0)} = U_{n}$ and $U_{n}^{(m)} = 1$ , and the $U_{n}^{(r)}$ are closed subgroup schemes of $U_{n}$ stable under conjugation by $T_{n}$ .

Then $U_{n} = U_{n}^{(0)} \supseteq U_{n}^{(1)} \supseteq \dots \supseteq U_{n}^{(m)} = 1$ is a central series of closed subgroup schemes of $U_{n}$ stable under $T_{n}$ : $[U_{n}^{(r)}, U_{n}] \subseteq U_{n}^{(r + 1)}$ for $0 \leq r < m$ , and each successive quotient $U_{n}^{(r)} / U_{n}^{(r + 1)}$ is canonically isomorphic to $G_{a}$ , the isomorphism being given by the coordinate $x_{i j}$ of the $(r + 1)$ -st pair. The diagonal torus $D_{n}$ acts on each quotient through the character $d \mapsto d_{i} d_{j}^{- 1}$ .

Facts & Assumptions

Given: A field $k$ , an integer $n \geq 1$ , the group schemes $U_{n} \subseteq T_{n} \subseteq {G L}_{n}$ , and the ordering of pairs $(i, j)$ , $i < j$ , described in the statement.

[F1]

For every commutative unital $k$ -algebra $R$ , $U_{n} (R)$ is the group of upper unitriangular matrices in ${G L}_{n} (R)$ and $T_{n} (R)$ the group of invertible upper triangular matrices, with $T_{n} = D_{n} ⋉ U_{n}$ . (The upper unitriangular group scheme U_n and its coordinate ring, Upper triangular, lower triangular and diagonal square matrices over a commutative ring)

[F2]

Matrix multiplication is associative, the identity matrix is a unit, and the $(i, j)$ -entry of a product $X Y$ is $\sum_{l} x_{i l} y_{l j}$ ; entrywise these identities hold over every commutative ring. (Matrix arithmetic over a commutative ring is associative, unital and distributive, and transpose reverses products)

[F3]

The commutator of two elements of an abstract group is $[x, y] = x y x^{- 1} y^{- 1}$ , and the lower central series of a group is defined by $G^{(0)} = G$ , $G^{(r + 1)} = [G, G^{(r)}]$ ; a series is central when $[G^{(r)}, G] \subseteq G^{(r + 1)}$ in the indexed form used here. (Subgroup commutators and the lower central series)

Proof

Given: A field $k$ , $n \geq 1$ , the group schemes $U_{n} \subseteq T_{n}$ and the pair ordering of the statement.

1.1F1F2

For $0 \leq r < m$ , let $d$ be the difference of the next pair. Every pair among the first $r$ has difference $e \leq d$ and its $(i, j)$ entry vanishes in each $X \in U_{n}^{(r)} (R)$ . In a product, the linear terms of that entry vanish; every cross term has two positive differences strictly smaller than $e \leq d$ , so its factors vanish as well. For an inverse, write $X = I + N$ and use the finite series $X^{- 1} = I - N + N^{2} - \dots$ . The linear entry is zero; every entry of $N^{q}$ for $q \geq 2$ is a sum over strict index chains whose segment differences are positive and strictly smaller than $e \leq d$ , so each factor vanishes. Thus the first $r$ entries remain zero under product and inverse over every commutative $k$ -algebra $R$ . By [F1] these valued-point subgroups define closed subgroup schemes. The endpoints are $U_{n}^{(0)} = U_{n}$ and $U_{n}^{(m)} = 1$ .

2.1F1F2step 1.1

Fix $r < m$ and let the next pair have difference $d$ . Write $X = I + A \in U_{n}^{(r)} (R)$ and $Y = I + B \in U_{n} (R)$ for an arbitrary $k$ -algebra $R$ . All entries of $A$ have difference at least $d$ , while those of $B$ have difference at least one. In $X Y X^{- 1} Y^{- 1} - I$ , expansion using the finite nilpotent inverse series leaves only words involving at least one $A$ and at least one $B$ ; terms involving only one matrix cancel since the commutator is $I$ if either matrix is zero. Every such word has entries of difference at least $d + 1$ . Therefore the commutator vanishes on all pairs of difference at most $d$ , in particular the first $r + 1$ pairs, and lies in $U_{n}^{(r + 1)} (R)$ . The valued-point criterion proves $[U_{n}^{(r)}, U_{n}] \subseteq U_{n}^{(r + 1)}$ as subgroup schemes. By step 1.1, $U_{n}^{(r + 1)}$ is a subgroup contained in $U_{n}^{(r)}$ . Since $[Y, X] = [X, Y]^{- 1}$ , it also lies in $U_{n}^{(r + 1)}$ , and $Y X Y^{- 1} = [Y, X] X$ shows that conjugation by $U_{n}$ preserves $U_{n}^{(r)}$ . Diagonal conjugation multiplies each coordinate $x_{i j}$ by $d_{i} d_{j}^{- 1}$ and preserves the zero conditions. Since $T_{n} = D_{n} ⋉ U_{n}$ , every term is $T_{n}$ -stable.

3.1F1F2step 2.1

Let $(i, j)$ be the next pair, of difference $d$ . The coordinate $x_{i j} : U_{n}^{(r)} \to G_{a}$ is a homomorphism, since the cross terms $x_{i l} (X) x_{l j} (Y)$ in multiplication have factors of differences strictly less than $d$ , and those entries vanish in $U_{n}^{(r)}$ . Its kernel is exactly $U_{n}^{(r + 1)}$ . It is surjective on every algebra-valued point, with section $a \mapsto I + a E_{i j}$ ; the quotient functor is therefore represented by $G_{a}$ . By step 2.1 the action of $U_{n}$ on this quotient is trivial, while diagonal conjugation multiplies the coordinate by $d_{i} d_{j}^{- 1}$ . Thus the quotient and its stated $T_{n}$ -action are as claimed.

4.1step 2.1step 3.1step 1.1F3∎

By [step 2.1] the series is central and normal in $U_{n}$ , with $T_{n}$ -stable terms, and by [step 3.1] its successive quotients are canonically $G_{a}$ with the diagonal characters displayed; the last term is $U_{n}^{(m)} = 1$ by [step 1.1]. This proves all assertions.

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Sources

J. S. Milne, Algebraic Groups (corrected 2022 printing, Cambridge University Press) (standard reference, not scraped)