Center of a linear algebraic group

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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?
 

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  • #2
martinbn
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There are plenty of groups with trivial or finite centre, hence zero dimensional. So ##c(m)## can be zero for nonzero ##m##.
 
  • #3
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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.
 
  • #4
martinbn
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The identity is always in the center.
Of course, but it is a single point and has dimension 0.
 
  • #5
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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?
 
  • #6
Office_Shredder
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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)
 
  • #7
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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}##.
 
  • #8
74
31
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?
 
  • #9
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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?
 
  • #10
martinbn
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Aren't the maximal abelian subgroups conjugate to the diagonal subgroup?
 

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