Difference between the Lie groups $O(2,2)$ and $O(2)\times O(2)$

Question

I'm interested in the differences in the groups but also in the Lie algebra associated. I know that two groups can have the same lie algebra if they differ from discrete elements, for instance: $SO(n)$ and $O(n)$ should have the same algebra. But then if I have a group $O(2,2)$, what is the associated Lie algebra? Does $O(2)\times O(2)$ have the same associated Lie algebra that $SO(2)\times SO(2)$ does?

Answer 1

No, the groups $O(2, 2)$ and $O(2) \times O(2)$ (and the corresponding algebras) are different---in fact, they have different dimensions as Lie groups.

The group $O(2, 2)$ is the group preserving an inner product of signature $(2, 2)$ on a $4$-dimensional real vector space $\Bbb V$. Concretely, if $g$ is the usual (definite-signature) inner product on $\Bbb R^2$, then we may identify $O(2, 2)$ is the group preserving $g \oplus -g$ on $\Bbb R^2 \oplus \Bbb R^2$. (The matrix representation of this bilinear form w.r.t. the standard basis is the matrix given in Tsemo Aristide's excellent answer. Thus, the explicit matrix representations of $O(2, 2)$ and $\mathfrak{so}(2, 2)$ for this choice are the ones given there.) It has dimension $\dim O(2, 2) = \frac{1}{2} (4) (4 - 1) = 6$.

The Lie algebra $\mathfrak{so}(2, 2)$ of $O(2, 2)$ is actually decomposable: $\mathfrak{so}(2, 2) \cong \mathfrak{sl}(2, \Bbb R) \oplus \mathfrak{sl}(2, \Bbb R)$. Better yet, this lifts to an isomorphism $SO_0(2, 2) \cong SL(2, \Bbb R) \times SL(2, \Bbb R)$ (see this Physics SE answer); here, $SO_0(2, 2)$ just denotes the identity component of $O(2, 2)$.

On the other hand, we may identify $O(2)$ with the group preserving the inner product $g$ itself, and it has dimension $1$, so it has abelian Lie algebra $\mathfrak{so}(2, \Bbb R) \cong \Bbb R$. Thus, $O(2) \times O(2)$ has dimension $2$ and abelian Lie algebra $\mathfrak{so}(2, \Bbb R) \oplus \mathfrak{so}(2, \Bbb R) \cong \Bbb R^2$.

We can view $O(2) \times O(2)$ naturally, however, as a subgroup of $O(2, 2)$ in terms of our above realization: It is exactly the subgroup that preserves the vector space decomposition $\Bbb R^2 \oplus \Bbb R^2$.

Alternatively, the inclusion $O(2) \hookrightarrow SL(2, \Bbb R)$ gives a different way to view $O(2) \times O(2)$ as a subgroup of $O(2, 2)$: $$O(2) \times O(2) \hookrightarrow SL(2, \Bbb R) \times SL(2, \Bbb R) \cong SO_0(2, 2) \leq O(2, 2) .$$

Answer 2

Consider $J=\pmatrix{1&0&0&0\cr 0&1&0&0\cr 0&0&-1&0\cr 0&0&0&-1}$ $M$ in $SO(2,2)$ i.e $M^t J M=I$. The Lie algebra of $SO(2,2)$ is the set of $4\times 4$ matrices such that $A^t J+JA=0$.

The Lie algebra $so(2)$ of $SO(2)$ is the set of $2\times 2$ matrices such that $A^t A=I$ and the Lie algebra of $SO(2)\times SO(2)$ is the product $so(2)\times so(2)$.