Center of a linear algebraic group

[fresh_42]

TL;DR Summary

linear algebra; center; groups

Let $G\leq GL(n)$ be a linear algebraic group of dimension $m,$ and $C$ its $c$-dimensional center. What do we know about lower and upper bounds of $c=c(m)\,\text{?}$

Clearly $c(0)=0, c(1)=1$ and $n^2\geq c(m)\geq 1$ for $m\neq 0.$ By Schur's Lemma we also know $c(n^2)=1$. Did anybody ever investigated what happens in between?

 

[martinbn]

There are plenty of groups with trivial or finite centre, hence zero dimensional. So $c(m)$ can be zero for nonzero $m$.

 

[fresh_42]

There are plenty of groups with trivial or finite centre, hence zero dimensional. So $c(m)$ can be zero for nonzero $m$.

The identity is always in the center.

 

[martinbn]

The identity is always in the center.

Of course, but it is a single point and has dimension 0.

 

[fresh_42]

Of course, but it is a single point and has dimension 0.

Sure, for some simple S-groups, of which we know the centers. That is if $\mathbb{F}\cdot 1 \cap G <\infty ,$ which are very specific cases. My question was more generic when we have $\mathbb{F}\cdot 1 \subseteq G.$ A few cases of finite centers doesn't answer the question: Are there significant upper and lower bounds for $c(m)$ except $n^2$ and $1\,\text{?}$ Or $0$, I don't care. E.g. what is the maximum of $c(m)\,\text{?}$ $n\,\text{?}$ or $n^\eta,$ and what is the maximal value of the exponent?

 

[Office_Shredder]

If the dimension is $n^2$ then it must contain some small open ball around the identity, which is impossible since $GL(n)$ isn't commutative. So $n^2-1$ is an upper bound I guess? (I'm also pretty sure you can't have an $n^2$ dimensional linear algebraic group except for $GL(n)$, but I'm open to the possibility of a strange counterexample)

 

[jbergman]

Summary:: linear algebra; center; groups

Let $G\leq GL(n)$ be a linear algebraic group of dimension $m,$ and $C$ its $c$-dimensional center. What do we know about lower and upper bounds of $c=c(m)\,\text{?}$

Clearly $c(0)=0, c(1)=1$ and $n^2\geq c(m)\geq 1$ for $m\neq 0.$ By Schur's Lemma we also know $c(n^2)=1$. Did anybody ever investigated what happens in between?

I'm not sure I full understand your question, but for a concrete example the center of $O(n)$ is $\{I_n,-I_n\}$ and $O(n)$ has dimension $\frac{n^2-n}{n}$.

 

[jbergman]

Summary:: linear algebra; center; groups

Let $G\leq GL(n)$ be a linear algebraic group of dimension $m,$ and $C$ its $c$-dimensional center. What do we know about lower and upper bounds of $c=c(m)\,\text{?}$

Clearly $c(0)=0, c(1)=1$ and $n^2\geq c(m)\geq 1$ for $m\neq 0.$ By Schur's Lemma we also know $c(n^2)=1$. Did anybody ever investigated what happens in between?

BTW, what are the tools to even investigate such a question?

 

[fresh_42]

Sure, I know the standard (simple) groups here. But how does $d:=d(n,m):=\dim C(G)$ behave generically? Yes, $0$ is a lower bound, and $n^2$ maybe $n^2-1$ an upper bound, but neither are interesting cases. The question is: Can we say something more significant about $d$ given an arbitrary $m-$dimensional subgroup of $\operatorname{GL}(n)$ then $n^2>d\geq 0.$ Diagonal matrices form an abelian group, so there are groups with $d=m.$ If we write $d(n,m)=O(m^\eta),$ what can we say about $\eta$? Is it bounded with $m$ from above, or is $\eta >1?$. What is the biggest abelian subgroup of $\operatorname{GL}(n)?$ $\eta=1$ is a good guess, but is there a proof? How big are typical centers, i.e. of a randomly chosen subgroup?

 

[martinbn]

Aren't the maximal abelian subgroups conjugate to the diagonal subgroup?