3
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 2
    $\begingroup$ @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
3
+50
$\begingroup$
  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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.