I Can I visualize O(3) \ SO(3) in some way?

[BerriesAndCream]

Hello.

I know that a 3×3 orthogonal matrix with determinant = 1 (so a 3×3 special orthogonal matrix) is a rotation in 3D.
I was wondering if there is a 3×3 orthogonal matrix with determinant = –1 could be visualised in some way.

Thank you!

 

[Hill]

Hello.

I know that a 3×3 orthogonal matrix with determinant = 1 (so a 3×3 special orthogonal matrix) is a rotation in 3D.
I was wondering if there is a 3×3 orthogonal matrix with determinant = –1 could be visualised in some way.

Thank you!

It can be visualized as a rotation of reflection.

 

[pines-demon]

_

Hello.

I know that a 3×3 orthogonal matrix with determinant = 1 (so a 3×3 special orthogonal matrix) is a rotation in 3D.
I was wondering if there is a 3×3 orthogonal matrix with determinant = –1 could be visualised in some way.

Thank you!

SO(3) is the group of rotations in 3D (no complex numbers). O(3) is the group of rotations and reflections, you can decompose any matrix O(3) as the product of a matrix in SO(3) and a reflection matrix.

 

[BerriesAndCream]

thanks!

 

[jbergman]

Hello.

I know that a 3×3 orthogonal matrix with determinant = 1 (so a 3×3 special orthogonal matrix) is a rotation in 3D.
I was wondering if there is a 3×3 orthogonal matrix with determinant = –1 could be visualised in some way.

Thank you!

$$O(n) \cong SO(n) \rtimes O(1)$$
$O(1) \in \{1, -1\}$ and $\rtimes$ is the semidirect product. So in some sense $O(n)$ can just be thought of like a direct product of $SO(n)$ and $O(1)$. Geometrically as a manifold, Lie Group, $O(n)$ breaks into two components. $SO(n)$ is the connected component containing the Identity with $det(A) = 1$ and there is an isomorphic component that doesn't contain the identity with $det(A) = -1$. So $O(n)$ can also be thought of as two copies of $SO(n)$.