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I am working in a group called $$\mathrm{UI}_n=\{A\in \mathrm{GL}_n(\mathbb{F}_q[[t]]) \mid A_{ij}(0) \neq 0 \text{ implies } i\leq j, i=j \text{ implies } A_{ij}(0)=1\}$$

Which can be thought of as the inverse limit indexed by k of $$\mathrm{UI}_n^{(k)}=\{A\in \mathrm{GL}_n(\mathbb{F}_q[t]/t^k\mathbb{F}_q[t]) \mid A_{ij}(0) \neq 0 \text{ implies } i\leq j, i=j \text{ implies } A_{ij}(0)=1\}$$

I am wondering if knowing the or commutator of each finite part has any relation to the center of commutator of the limit.

Update: I have proven the center of $\mathrm{UI}_n$ are the scalar matrices, which seems to track as those project to some of the center (the center of $\mathrm{UI}_n^{(k)}$ also includes the top right entry) of the finite part, but the commutator seems harder to deal with in $\mathrm{UI}_n.$ Thus, it would be helpful to know if knowing anything about the commutator in each finite part says anything about the commutator of the infinite group.

Mattan Feldman
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2 Answers2

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You can calculate the commutator of an element from the centraliser of its reductions in a very general case. Let $R$ be a ring that is adically separated with respect to an ideal $I$ (so $\bigcap_{i=k}^\infty I^k = 0$). Given two matrices $A, B \in \mathrm{M}_n(R)$, we denote their reductions to $\mathrm{M}_n(R / I^k)$ by $A_k$ and $B_k$. Then $AB = BA$ if and only if $A_kB_k = B_kA_k$ for all $k$. The implication $\Rightarrow$ is clear, since $R \to R / I^k$ is a ring homomorphism. For $\Leftarrow$, note that $AB - BA$ reduces to $A_kB_k - B_kA_k = 0 \in \mathrm{M}_n(R)$ for all $k$. This means $AB - BA \in \mathrm{M}_n(I^k)$ for all $k$. Since $R$ is $I$-adically complete, this means $AB = BA$.

From this we can deduce that for every subgroup $G \leq \operatorname{Gl}_n(R)$ and every element $x \in G$, the centraliser $\mathrm{Z}_G(x)$ is the intersection of all preimages $\pi_k^{-1}(\mathrm{Z}_{G_k}(x_k))$. Here, $pi_k : \operatorname{Gl}_n(R) \to \operatorname{Gl}_n(R / I^k)$ is the reduction map, $G_k = \pi_k(G)$, and $x_k = \pi_k(x)$. To check this statement, let $y \in G$. Then $y \in \mathrm{Z}_G(x)$ means $xy = yx$, which is equivalent to $x_ky_k = y_kx_k$ for all $k$, i.e. $y_k \in \mathrm{Z}_{G_k}(x_k)$ for all $k$. In other words, $y \in \pi_k^{-1}(\mathrm{Z}_{G_k}(x_k))$.

Now for your concrete case: We can pick $R = \mathbb{F}_q[[t]]$, $I = (t)$, and $G = \mathrm{UI}_n$. Then $G_k = \mathrm{UI}_n^{(k)}$. Then for $x \in \mathrm{UI}_n$, we get $$ \mathrm{Z}_{\mathrm{UI}_n}(x) = \bigcap_{k = 1}^\infty\pi_k^{-1}(\mathrm{Z}_{\mathrm{UI}_n^{(k)}}(x_k)) = \varprojlim_k \mathrm{Z}_{\mathrm{UI}_n^{(k)}}(x_k) \leq \mathrm{UI_n}. $$

Tzimmo
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  • You showed this for the center, will the same argument work for the commutator subgroup aka the derived subgroup? I believe it should but just making sure – Mattan Feldman Jun 22 '25 at 14:23
  • I guess I mixed up commutators with centralisers in my head. Sorry. The argument can be salvaged if the ring is $I$-adically complete, I suppose. – Tzimmo Jun 22 '25 at 15:38
  • I’m not very informed on projective limits, but I believe my group should be? Could you edit the answer to reflect this? And explain being $I$-adically complete makes it work? – Mattan Feldman Jun 22 '25 at 22:21
  • The only reason I’m not sure about the completion is that I am qouteinting entries of the elements but not the whole group in my construction. – Mattan Feldman Jun 22 '25 at 22:26
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Using the ideas put forth by @Tzimmo I have come up with an answer.

Since $\mathrm{UI}_n-1$ is $I$-adically complete with the ideal $I=(t^k\mathrm{M}_n),$ if a subgroup $H$ of $\mathrm{UI}_n$ satisfies $H/I^k$ normal in $\mathrm{UI}_n/I^k$ for all $k$, then $H$ is normal in $\mathrm{UI}_n$.

Important note: the converse is not true in general, for example, the normal subgroup of $\mathrm{UI}_n$, of elements of determinant 1, has quotients that are not normal in their respective projections.

Mattan Feldman
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