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I have an intuition that at least one of the element should be in the form of diagonal matices with diagonal entries being 1 or -1 (e.g I3)

I don't know if there's any other possiblity

Please give a strong proof

(a similar question is that show the centre of the general linear group GL2(C) consisits of all scalar matrices, I can show all scalar matrices are in the centre, but how to show that they are the only kind?)

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    It is not clear to me why this question got down voted. Hence I will +1 it to compensate the down vote.2012-12-28

1 Answers 1

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Every non-identity element of $SO(3)$ has two fix points as it acts on the unit sphere, which are the axis of the rotation. Two elements commute if they have the same two fixed points - that is, if they rotate along the same axis, or if their axes are orthogonal and they are both $180$ degree turns.

This can be simplified to include the identity element by saying that for $g,h\in SO(3)$, $gh=hg$ if and only if either $\exists x\in S^2: gx=x=hx$ or $\exists x,y\in S^2: x\cdot y=0: hx=x, gy=y, hy=-y, gx=-x$

A quick outline of the proof.

First you need to show that every non-identity element of $SO(3)$ has exactly two fixed points, necessarily antipodal.

Second if $g,h\in SO(3)$ and $gh=hg$. Let $x$ be a fixed point of $h$. Then $gx=g(hx)=(gh)x=(hg)x=h(gx)$ So $gx$ is also a fixed point of $h$. Which means that $gx=x$ or $gx=-x$. If $gx=x$ we are done, so assume $gx=-x$, and let $y$ be a fixed point of $g$. Then we get: $y\cdot x = (gy)\cdot(gx) = y\cdot(-x)=-y\cdot x$ (The first equality is the definition of $SO(3)$.) By symmetric, if $h$ doesn't fix $y$, then $hy=-y$.

So we see that $h$ and $g$ either must share a fixed point, or they must both be $180$ degree rotations around orthogonal axes.

This makes the representative examples:

$g=\left(\begin{matrix}1&0&0\\ 0&\cos\alpha&-sin\alpha\\ 0&\sin\alpha&\cos\alpha \end{matrix}\right),h=\left(\begin{matrix}1&0&0\\ 0&\cos\beta&-sin\beta\\ 0&\sin\beta&\cos\beta \end{matrix}\right)$ and $g=\left(\begin{matrix}1&0&0\\ 0&-1&0\\ 0&0&-1 \end{matrix}\right),h=\left(\begin{matrix}-1&0&0\\ 0&1&0\\ 0&0&-1 \end{matrix}\right)$

with all other pairs being group conjugates of such pairs.

That is different from your question about the center of $GL_2(\mathbb C)$ because that question is not about whether two elements commute, but whether a single element commutes with all other elements of the group.