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An exercise in a book I'm reading is to show that $O(2)$ is not isomorphic to $SO(2)\times \{-1,1\}$.

The problem is, I don't believe the statement. Let me elaborate why:

$O(2)$ consists of orthogonal matrices of either determinant $1$ or $-1$. The subset $SO(2) \subseteq O(2)$ is precisely the set of matrices of determinant $1$. So one should be able to define an isomorphism as follows:

$A \in O(2)$ maps to $(A,1)$ if $\det A = 1$

and

$A \in O(2)$ maps to $(-A, -1)$ if $\det A = -1$.

Why is this not an isomorphism? But even if it was clear to me that this is not an isomorphism, it is not clear to me why it should be impossible to find one.

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    $\begingroup$ $-A$ has the same determinant as $A$, so it isn't guaranteed to itself be an element of $SO(2)$. $\endgroup$ Dec 7 '14 at 7:04
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While this isn't true, it's close enough to true that the confusion is quite believable. In fact, for $n$ odd the map $SO(n)\times \mathbb{Z}_2\to O(n)$ given by $(A,m)\mapsto mA$ is an isomorphism. It's clear that this is a bijection, since in odd dimensions the map $A\mapsto -A$ reverses orientation/determinant. And it's a group homomorphism: $(A,m)(A',m')\mapsto mAm'A'=mm'AA',$ the image of $(AA',mm')$. This basically is just because$-SO(n)$ is a coset $SO(n)*-I$ of a central element $-I$.

But for $n$ even this can't happen: central elements are scalar matrices, and scalar matrices of even dimension have nonnegative determinant. We do still get a bijection (even a diffeomorphism) $SO(n)\times \mathbb{Z}_2\to O(n)$, but here instead of sending $(A,-1)$ to $-A$ we must send it, among other possible choices, to $A$ with the sign of its first column flipped: this exhibits $-SO(n)$ as the coset of the block matrix $\begin{pmatrix} -1&0\\0&I_{n-1}\end{pmatrix}=B$. But again, since $B$ is not central, this is not a group homomorphism.

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  • $\begingroup$ Very nicely put! $\endgroup$ Dec 7 '14 at 7:24
  • $\begingroup$ @Travis thank you kindly! $\endgroup$ Dec 7 '14 at 7:25
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    $\begingroup$ So, in other words, $O(n)\cong SO(n)\rtimes Z_2$, and for n odd, the semidirect product is actually a direct product? $\endgroup$
    – AG learner
    Aug 16 '15 at 17:12
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    $\begingroup$ @YilongZhang yep. $\endgroup$ Aug 17 '15 at 17:19
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As Qiaochu points out, your proposed map isn't an isomorphism.

But these two groups are nonisomorphic for the simple reason that $SO(2) \times \{-1, 1\}$ is abelian (it is a product of abelian groups) whereas $O(2)$ is not: For example, the orthogonal matrices $$\begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix}, \begin{pmatrix}-1 & 0 \\ 0 & 1\end{pmatrix} \in O(2)$$ do not commute.

You can think of this perhaps usefully as an infinite version of the fact that $\mathbb{Z}_n \times \mathbb{Z}_2$ is not isomorphic to $D_{2n}$ (for large enough $n$). Put another way, all rotations of the plane commute, but in general reflections of the plane do not.

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