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Why is the special orthogonal group considered the rotation group?

  1. Sep 3, 2012 #1
    I understand that the special orthogonal group consists of matrices x such that [itex]x\cdot x=I[/itex] and [itex]detx=1[/itex] where I is the identity matrix and det x means the determinant of x. I get why the matrices following the rule [itex]x\cdot x=I[/itex] 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.
  2. jcsd
  3. Sep 3, 2012 #2
    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.
  4. Sep 3, 2012 #3
    Did you mean [itex]x\cdot x^T=I[/itex]??

    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.
  5. Sep 4, 2012 #4


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    To prove that rotations have determinant 1, you must first define the term "rotation". It's perfectly OK to just take "R is said to be a rotation if R is a member of SO(3)" as the definition, but then there's nothing to prove. Another option is "R is said to be a rotation if R is a member of the largest connected subgroup of O(3)". Then your task would be to prove that the largest connected subgroup is SO(3).

    (I actually prefer the terminology that calls members of O(3) rotations, and members of SO(3) proper rotations).
  6. Sep 4, 2012 #5
    Ya, I forgot the transpose part.

    Thanks to all those who replied, I'm pretty sure I get it now.
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