# Is this unipotent group, over characteristic 2, connected?

Let $$E_{ij}(x)\in \mathrm{Mat}_{7\times7}(\overline{\mathbb{F}}_2)$$ be the matrix with zeros everywhere, except for the value $$x$$ at $$ij$$. Set $$a(x)=1+E_{12}(x)+E_{34}(x)+E_{56}(x),\quad b(y)=1+E_{23}(y)+E_{45}(y)+E_{67}(y),$$ for two indeterminates $$x,y$$. Set $$G_8=\langle a(x), b(y) \mid x, y\in\overline{\mathbb{F}}_2\rangle.$$ Is $$G_8$$ Zariski closed and connected?

For easier visualization, here are the generating matrices:

$$a(x)=\begin{pmatrix} 1&x& & & & & \\ &1&0& & & & \\ & &1&x& & & \\ & & &1&0& & \\ & & & &1&x& \\ & & & & &1&0\\ & & & & & &1 \end{pmatrix} ,\ b(y)=\begin{pmatrix} 1&0& & & & & \\ &1&y& & & & \\ & &1&0& & & \\ & & &1&y& & \\ & & & &1&0& \\ & & & & &1&y\\ & & & & & &1 \end{pmatrix}$$

To put things on a more general footing, given such a description of a subgroup of an algebraic group, how does one show that it is connected?

• It's generated by the union of two connected (1-parameter) subgroups so is clearly Zariski-connected. – YCor Mar 27 at 8:56
• To see whether it's Zariski-closed... probably one can describe the group, at the price of a few computations (I'd compute various commutators such as $[a(x),b(y)]$, $[a(x),[a(y),b(z)]]$, etc.) – YCor Mar 27 at 8:59
• ... and I'd start beforehand solving the analogous problem in size $<7$. It gives the projections to the northwest and southeast blocks, so is certainly useful. – YCor Mar 27 at 9:46
• Springer LAG Prop. 2.2.6 answers both questions: The union of irreducible subvarieties, each containing 1, generates a closed connected subgroup. – Friedrich Knop Mar 27 at 19:35

Good news and bad news here: This group is connected and Zariski closed for nice general reasons. However, I assume that Dror considered this group because $$a(1)$$ and $$b(1)$$ generate a subgroup isomorphic to the dihedral group of order $$16$$, and are thus relevant to my question. It turns out that the group generated by $$a(x)$$ and $$b(y)$$ has dimension at least $$9$$. The issue is that "the $$\mathbb{F}_2$$ points of the group generated by $$X$$" can be much larger than "the group generated by the $$\mathbb{F}_2$$ points of $$X$$".

Good news: See Section 2.h of Milne, Algebraic Groups for the following results. Let $$G$$ be an algebraic group scheme over $$k$$, let $$X$$ be a geometrically connected, geometrically reduced, subvariety of $$G$$ over $$k$$ and let $$\phi: X \to G$$ be a map defined over $$k$$ such that $$\phi(X)$$ is closed under inversion.

Then there is a smallest algebraic sub-group-scheme $$H$$ of $$G$$ such that $$\phi$$ factors through $$H$$, and this group is connected. Define $$\phi_n : X^n \to G$$ by $$(x_1, \ldots, x_n) \mapsto \phi(x_1) \cdots \phi(x_n)$$, then $$\phi^n : X^n \to H$$ is dominant for $$n$$ sufficiently large.

Furthermore (this part isn't in Milne, but seems right to me), let $$n$$ be large enough that $$\phi^n : X^n \to H$$ is dominant. Then I claim that $$\phi^{2n} : X(k^{\mathrm{alg}})^{2n} \to H(k^{\mathrm{alg}})$$ is surjective. Proof: Let $$U$$ be an nonempty open set within the image of $$\phi^n$$. Since $$H$$ is irreducible, for any $$h \in H(k^{\mathrm{alg}})$$, the open sets $$U$$ and $$U^{-1} h$$ have nonempty intersection, so we can find some point $$u_1 \in H(k^{\mathrm{alg}})$$ which is both in $$U(k^{\mathrm{alg}})$$ and of the form $$u_2^{-1} h$$ for $$u_2 \in U(k^{\mathrm{alg}})$$, and thus $$h = u_1 u_2$$.

So we can literally think of $$H$$ as the subgroup of $$G$$ generated by $$X$$, as long as we work with points in the algebraic closure.

Bad news: Let $$G$$ be the group of $$7 \times 7$$ upper triangular matrices with $$1$$'s on the diagonal. Define $$G_k = \{ g \in G : g_{ij}=0 \ \mbox{for} \ i < j < i+2^k \}.$$ So $$G = G_0 \supset G_1 \supset G_2 \supset G_3 = \{ \mathrm{Id} \}$$. Each $$G_k$$ is normal in $$G$$, and each $$G_k/G_{k+1}$$ is a product of copies of $$\mathbb{G}_a$$. Let $$H_k = H \cap G_k$$. I will verify that $$\dim H_0/H_1 = 2$$, $$\dim H_1/H_2 = 3$$ and $$\dim H_2/H_3 \geq 2$$. So the group $$H$$ has dimension at least $$7$$. Details follow. Note that we always have $$g_{ij} = g_{(i+2)(j+2)}$$ for any $$g \in H$$.

Computation of $$H_0/H_1$$ Every product in $$H$$ is of the form $$\begin{bmatrix} 1 & x & \ast & \ast & \ast & \ast & \ast \\ 0 & 1 & y & \ast & \ast & \ast & \ast \\ 0 & 0 & 1 & x & \ast & \ast & \ast \\ 0 & 0 & 0 & 1 & y & \ast & \ast \\ 0 & 0 & 0 & 0 & 1 & x & \ast \\ 0 & 0 & 0 & 0 & 0 & 1 & y \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ \end{bmatrix}$$ and every $$(x,y)$$ occurs, so $$H_0/H_1 \cong \mathbb{G}_a^2$$.

Computation of $$H_1/H_2$$ Let $$h=\prod_j \left( a(x_j) b(y_j) \right) \in H_1$$. Then $$\sum x_i = \sum y_j=0$$. Then $$h_{13} = h_{35} = h_{57} = \sum_{i and $$h_{24} = h_{46} = \sum_{i \geq j} x_i y_j$$. We moreover have $$h_{13} + h_{24} = \sum_{i,j} x_i y_j = \left( \sum_i x_i \right) \left( \sum_j y_j \right) = 0$$, so $$h_{13} = h_{24}$$. So we have $$h_{13} = h_{35} = h_{57} = h_{24} = h_{46}$$. Also, $$h_{14} = h_{36}$$ and $$h_{25} = h_{47}$$. So all matrices of $$H_1$$ are of the form $$\begin{bmatrix} 1 & 0 & p & q & \ast & \ast & \ast \\ 0 & 1 & 0 & p & r & \ast & \ast \\ 0 & 0 & 1 & 0 & p & q & \ast \\ 0 & 0 & 0 & 1 & 0 & p & r \\ 0 & 0 & 0 & 0 & 1 & 0 & p \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ \end{bmatrix}$$ and $$H_1/H_2$$ embeds into $$\mathbb{G}_a^3$$. We compute that $$a(x) b(y) a(x) b(y)$$ has $$(p,q,r) = (xy, x^2 y, x y^2)$$ and every vector in $$(\mathbb{F}_2^{\mathrm{alg}})^3$$ is a sum of vectors of the form $$(xy, x^2 y, x y^2)$$, so $$H_1/H_2 \cong \mathbb{G}_a^3$$.

Computation of $$H_2/H_3$$ At this point, I already know that $$\dim H \geq 5$$, so it is not a $$4$$-dimensional unipotent group. I wasn't able to fully compute $$H_2$$. For $$h \in H_2$$, we definitely have $$h_{15}=h_{37}$$, so $$\dim H_2 \leq 5$$. Every element of $$H_2$$ that I have found obeys the additional equation $$h_{15} = h_{26}$$. I can find enough elements to generate the $$4$$-dimensional subspace cut out by these equations. So $$H_2$$ is a subgroup of $$\mathbb{G}_a^5$$ which contains at least $$\mathbb{G}_a^4$$.

• Nice! Thanks for putting lots of details in the computation. Good luck with your question, even if this direction is a dead end :) – Dror Speiser Mar 27 at 22:48
• Thanks. I found enough elements now to get my bound for $\dim H_2$ up to $4$. – DES-SupportsMonicaAndTransfolk Mar 27 at 23:23