Lie group of orthogonal matrix group - what would this mean?

In Lie group terms, this means that the Lie algebra of an orthogonal matrix group consists of skew-symmetric matrices. Going the other direction, the matrix exponential of any skew-symmetric matrix is an orthogonal matrix (in fact, special orthogonal).

I am not sure what this would mean. So, the elements that go into "lie bracket" consists of only skew-symmetric matrices? I am very confused here, and can anyone explain this?

• exponentiate $X+X^T=0$ to get $e^Xe^{X^T}=1$ – yoyo Apr 10 '13 at 0:13
• differentiate $X(t)X(t)^T=I$ (with $X(0)=I$) at $t=0$ to get $X'(0)+X'(0)^T=0$ – yoyo Apr 10 '13 at 0:33

The orthogonal group is a set of matrices satisfying a particular condition.

$$O^TO = \mathbb{I}.$$

Using matrix multiplication as the group multiplication this set satisfies the axioms for a Lie group.

All Lie groups have an associated space called a Lie algebra. A Lie algebra is a vector space with an extra structure which is the Lie bracket. The exponential map provides a map from the Lie algebra to the Lie group. This map is some times surjective, sometimes injective and sometimes neither depending on the properties of the group.

In this case the Lie algebra consists of the set of all skew-symmetric matrices. It is the Lie algebra shared by both the orthogonal group and the special orthogonal group. The exponential map maps every skew-symmetric matrix to a special orthogonal matrix (which form a subgroup of the orthogonal group). Just FYI the reason for this lack of surjectivity comes from the fact that the orthogonal group is not connected.

The Lie bracket is an operator that can act between any two matrices in the Lie algebra. Here it is just the commutator.