I am reading Serre's "Lie algebras and Lie groups" p.103. Let $k$ be a complete valued field(for example $\mathbb{R}$, $\mathbb{C}$, or $\mathbb{Q}_p$) and $R$ be a finite dimensional associative $k$-algebra. Surely $R$ is an additive Lie group. The book asserts that the unit group $G_m(R)$ is a multiplicative Lie group and also contains the proof, but I cannot understand it. I copy the text here.

"We contend that $G_m(R)$ is an analytic group which is open as a subset of $R$. To show that $G_m(R)$ is open in $R$ it suffices to show that there is a neighborhood of $1$ contained in $G_m(R)$. Now, there exists an open neighborhood $U$ of $0$ in $R$ such that for $x \in U$ the series $\sum x^n$ converges. It follows $V=\{1-x:x \in U\} \subset G_m(R)$ and $V$ is a neighborhood of $1$. To show that $G_m(R)$ is an analytic group it remains to show that multiplication is a morphism. This follows since multiplication in $R$ is bilinear."

I cannot understand the first step and the final step:

  1. Why does there exist an open set $U$ which satisfies $\sum x^n$ converges?
  2. Why is multiplication a manifold morphism? (Also, It seems that we need $x\mapsto x^{-1}$ is a morphism.)

From googling, I've found (ex1) of http://www.math.cornell.edu/~sjamaar/classes/6520/problems/2016-10-26.pdf , but still I cannot solve it.

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    $\begingroup$ I thought strategy. Embed $R⊂End_k(R)$ via left multiplication. Then $G_m(R)⊂Aut_k(R)≃GL_n(k)$. The problem reduces to the case of general linear group and it is well-known. Is it correct? $\endgroup$ – MiRi_NaE Jun 1 '18 at 3:08
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    $\begingroup$ I like this idea, it sounds really plausible, but I have two questions (i) to embed, we require $R$ is unital? (ii) where have we used the fact that $R$ is associative? $\endgroup$ – CL. Jan 19 at 22:07
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    $\begingroup$ @CL. There is no group of units if $R$ is not associative or not unital, so the question wouldn't make sense : these assumptions are implicit in the formulation of the question. In the strategy, $R\subset End_k(R)$ is not even a subalgebra if $R$ is not associative, and $G_m(R)$ not defined $\endgroup$ – Max Jan 20 at 0:15
  • $\begingroup$ The problem with the strategy is that you need to show that the embedding $G_m(R) \to GL_n(k)$ either has an open image, or more generally has a submanifold of $GL_n(k)$ as its image. Neither of these seem obvious to me $\endgroup$ – Max Jan 20 at 0:18
  • $\begingroup$ @Max Sps $R$ is $n$-dimensional. Then $R$ is a closed submanifold of $n^2$-dimensional manifold $End_k(R)$ via the left multiplication as above. Obviously we have $G_m(R) \subset Aut_k(R) \cap R$. We prove the opposite direction. For $z\in Aut_k(R) \cap R$, $z:R\longrightarrow R, x\mapsto zx$ is an isomorphism. Therefore there exists $x\in R$ s.t. $zx=1_R$ and we get $x$ also is in $Aut_k(R) \cap R$, and hence, $xz=1$. In particular $z\in G_m(R)$. $\endgroup$ – MiRi_NaE Jan 20 at 6:11
  1. For $x$ of norm $<1$, $(\displaystyle\sum_{k=0}^n x^k)_n$ is a Cauchy sequence (by the triangle inequality) , thus by completeness of $k$ and finite dimensionality of $R$, it converges in $R$. The open neighbourhood is $||x||<1$.

  2. Multiplication $R\times R\to R$ is bilinear so it must be smooth, thus so is its restriction to $G_m(R)$.

It remains the question of why $x\mapsto x^{-1}$ is smooth. First of all note that for $y=1-x\in V$, $y^{-1}= \sum_n x^n$, so for $y\in V$, $y^{-1}=\sum_n (1-y)^n$, thus the inversion is smooth on a neighbourhood of $1$.

If $x\in G_m(R)$, and $y\in xV$, then $y=xv$ for some $v\in V$ and $y^{-1}= v^{-1}x^{-1}= (\sum_n (1-v)^n)x^n$, so $y^{-1}= (\sum_n (1-x^{-1}y)^n)x^{-1}$ so it is smooth as well

  • $\begingroup$ Can you guarantee that we have an algebra norm $|| \cdot ||$ on $R$? I mean, is it possible to find a norm which satisfies $|| x^2|| \le ||x||^2$? For saying that $\sum_{0\le k\le n} x^k $ is a Cauchy sequence, this is required, I think. $\endgroup$ – MiRi_NaE Jan 20 at 5:30
  • $\begingroup$ It is required indeed; but the embedding $R\to End_k(R)\simeq M_n(k)$, and the fact that norms are equivalent over $R$ ($k$ is complete) provides such a norm by picking one in $M_n(k)$. $\endgroup$ – Max Jan 20 at 11:23

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