# What do the closures of cyclic groups in $\textrm{GL}_n$ look like?

Let $k$ be algebraically closed, $G = \textrm{GL}_n$ in the Zariski topology, and let $g \in G$. Let $H$ be the subgroup generated by $g$. Assume that $g$ does not have finite order.

Question: What are the possible dimensions of the subgroup $\overline{H}$?

If $g$ is diagonalizable, then $H$ sits inside the group of diagonal matrices, hence $\overline{H}$ is of dimension $\leq n$. In general, we can assume $H$ consists of upper triangular matrices, so the dimension of $\overline{H}$ is bounded above by the dimension of the group of upper triangular matrices.

Another thing I noticed is that if $n > 1$, and $k$ is uncountable, then $H$ and hence $\overline{H}$ are never irreducible. This is because $H$ is countable, and a countable set in Zariski $n$-space is never irreducible unless it is a singleton set (projection onto affine coordinates is a morphism of varieties, allowing you to reduce to the case of a countable irreducible subset of $\mathbb{A}^1(k)$).

• you are asking what are the subgroups of the group of invertible upper triangular matrices, and which ones are cyclic ? – reuns Mar 28 '16 at 20:39
• Note that without loss of generality you can assume $g$ is in Jordan form. Then by basically the same reasoning as the diagonal case you can see that $n$ is always an upper bound for the dimension of the closure of $H$. As for your last paragraph, $H$ countable does not imply $\bar{H}$ countable. – Nate Mar 28 '16 at 20:55
• @Nate I know, I meant $H$ cannot be irreducible. – D_S Mar 28 '16 at 22:51
• $H$ isn't even an algebraic set in general, it doesn't really make sense to say it is or is not irreducible. $\bar{H}$ on the other hand is algebraic, and can (and often will) be irreducible. – Nate Mar 29 '16 at 15:54
• As an example, take $k = \mathbb{C}$, $n=1$, and $g = 2$. The powers of 2 are a Zariski dense subset of $GL_1(\mathbb{C}) \cong \mathbb{C}^\times$. – Nate Mar 29 '16 at 15:58

I think that I have an answer for the question if you assume that the field is (algebraically closed) of characteristic 0.

I would say that the possible dimensions are $\{1,...,n\}$.

Claim: If $g=g_s g_u$ is the Jordan decomposition of $g$ with $g_s$ semisimple and $g_u$ unipotent, let $T':=\overline{<g_s>}$ and $U':=\overline{<g_u>}$. Then $\overline{H}=U'\times T'$.

From theorem 2.4.8 of Springer Linear algebraic groups both $g_s$ and $g_u$ are in $K:=\overline{H}$. Then \begin{align}T'\subseteq K\text{ and } U' \subseteq K\end{align}

Consider now $GL(n) \times GL(n) \to GL(n)$, $(A,B) \mapsto ABA^{-1}B^{-1}$. We can restrict it to $T' \times U'$, this is continuous and on a dense subset (namely $\{(g_s^a,g_u^b)\}_{a,b \in \mathbb{Z}}$) it is $1$, so it is always 1. Then any element of $T'$ commutes with any element of $U'$ so $T' \times U'$ is a subgroup of $K$. It is closed since it is an algebraic subgroup, and contains $g$, which proves the claim.

Then we need to understand just the case in which $g=g_u$ or $g=g_s$.

If $g=g_u \neq 1$ then $H=\mathbb{G}_a$: in fact $g$ is an upper triangular matrix with 1 on the diagonal, then it makes sense to write $log(g)$ using the power series, and if $h$ is an upper triangular matrix with 0 on the diagonal it makes sense to write $exp(g)$, again using the power series; and these maps are one the inverse of the other.

More explicitly I am saying that if $U$ are the upper triangular matrices with 1 on the diagonal and $\mathbb{A}^N$ are the upper triangular matrices with 0 on the diagonal \begin{align}log:U \to \mathbb{A}^N \text{is an isomorphism with inverse } exp \end{align}

Then we are interested on the closure of $log(g^n)=nlog(g)$ which is $\mathbb{A}^1$.

If $g=g_s$: I think that the dimension could be any number between 1 and $n$. It is enough to prove that it can be $n$ since for $n'<n$, $\mathbb{G}_m^{n'}$ is a closed subgroup of $\mathbb{G}_m^n$. Then what we need to see is that, given a torus $T$, we can find an element $g \in T$ such that the subgroup it generates is dense. In fact pick in $\mathbb{G}_m^n$ the element $(2,3,5,...,p_n)$ where $p_i$ is the $i$-th prime number. Then it generates the whole torus: otherwise there would be a non-0 character which is one on it, which means that there are integers $m_1,...,m_n$ not all 0 such that $\prod p_i^{m_i}=1$, which is impossible.

On the other hand we can't have that $T'$ has dimension $n$ and $U'$ has dimension 1: if $T'$ has dimension $n$ then $g$ has $n$ distinct eigenvalues, i.e. it is semisimple.