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In odd dimensions, we can represent any improper "rotation" G as $-\mathbb{1}\cdot R$ where $R\in SO(d)$. In even dimensions, $-\mathbb{1} \in SO(d)$ and we cant do this. Is there a way of writing an improper rotation in terms of proper rotations in even dimensions? Also, if we have improper rotation G (det = -1) in even dimensions, is there anything general that we can conclude about the order of the cyclic subgroup that it generates? For example, in even dimensions the order of the cyclic group generated by G must be even.

jdizzle
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  • How about simply use reflection (over some axis of choice)? – Gina Jul 02 '14 at 15:20
  • Use a block matrix $\begin{pmatrix}0 & I_n\I_n&0\end{pmatrix}$. – i. m. soloveichik Jul 02 '14 at 15:28
  • Thank you for the replies. Gina - I want to write a general form for G, as I can do in odd dimensions. Could I then do this as G = F$\cdot$R where F is a reflection matrix around one specified axis? – jdizzle Jul 02 '14 at 16:18
  • I.M Soloveichik - thank you for your reply. Could you please explain to me how I can use the matrix you have described to write a general improper rotation matrix? – jdizzle Jul 02 '14 at 16:19
  • @jdizzle Since the rotations $SO(n)$ have index 2 in the group $O(n)$ then any improper is the block matrix identified above multiplied by any rotation. – i. m. soloveichik Jul 03 '14 at 03:19
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    @soloveichik - thank you for your reply. I dont understand how this can be the case, as the block matrix you have provided has a determinant of 1 for even dimensions greater than 2, whereas the improper rotations have a determinant of -1. To clarify, Im looking for matrix X such that G=X.R where G is any improper rotation and R $\in$ SO(n) – jdizzle Jul 08 '14 at 15:00

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Instead take the identity matrix $I$ and change the (1,1) entry to -1 to give the matrix $J$. Certainly has $\det(J)=-1$ and also $JJ^t=I$, so it is orthogonal.

The kernel of the homomorphism $O(n)\rightarrow \{+1,-1\}$ is $SO(n)$ so $SO(n)$ has index 2 in $O(n)$. Thus $O(n)$ is a disjoint union of two left cosets, $O(n)=SO(n) \cup J\cdot SO(n)$.

i. m. soloveichik
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    The first paragraph of the answer was very helpful. The second paragraph is way beyond my pay grade (i.e. explanation is suitable only to the initiated :-))) ) – András Aszódi Sep 15 '20 at 15:43