# Why is the special orthogonal group considered the rotation group?

tensor33
I understand that the special orthogonal group consists of matrices x such that $x\cdot x=I$ and $detx=1$ where I is the identity matrix and det x means the determinant of x. I get why the matrices following the rule $x\cdot x=I$ are matrices involved with rotations because they preserve the dot products of vectors. The part I dont get is why the matrices involved with rotation must have determinant 1.

Muphrid
A pure rotation should not dilate or shrink any volume. The determinant of a linear operator is the scale factor by which the volume element increases. Hence, the determinant of a rotation operator must be 1.

Staff Emeritus
Homework Helper
I understand that the special orthogonal group consists of matrices x such that $x\cdot x=I$ and $detx=1$ where I is the identity matrix and det x means the determinant of x. I get why the matrices following the rule $x\cdot x=I$ are matrices involved with rotations because they preserve the dot products of vectors. The part I dont get is why the matrices involved with rotation must have determinant 1.

Did you mean $x\cdot x^T=I$??

A pure rotation should not dilate or shrink any volume. The determinant of a linear operator is the scale factor by which the volume element increases. Hence, the determinant of a rotation operator must be 1.

That doesn't really explain why the determinant can't be -1, which is what the OP is asking.
The reason is that rotations preserves the orientation of a basis. If we have a right-handed basis, then rotations of this will be right-handed as well.
As an example of an orthogonal matrix that does not preserve the orientation, you can probably take a reflection.

Staff Emeritus
I understand that the special orthogonal group consists of matrices x such that $x\cdot x=I$ and $detx=1$ where I is the identity matrix and det x means the determinant of x. I get why the matrices following the rule $x\cdot x=I$ are matrices involved with rotations because they preserve the dot products of vectors. The part I dont get is why the matrices involved with rotation must have determinant 1.
Did you mean $x\cdot x^T=I$??