Comparing the common matrix groups over $\mathbb{C}$

Question

I am studying Lie groups and the book I am using starts off with matrix Lie groups. It states that the following matrix groups are of considerable importance (all over $\mathbb{C}$):

• $GL(n)$, the set of matrices with nonzero determinant
• $SL(n)$, the set of matrices with unit determinant
• $U(n)$, the set of unitary matrices
• $SU(n)$, the set of unitary matrices with unit determinant
• $O(n)$, the set of orthogonal matrices
• $SO(n)$, the set of orthogonal matrices with unit determinant

I am trying to organize these groups in order of inclusion. Clearly we have that $SO(n)$ and $SU(n)$ are both subgroups of $SL(n) \leqslant GL(n)$. We also have that $U(n)$ contains $SU(n)$ and similarly $O(n)$ contains $SO(n)$.

I have now two questions:

1. Over $\mathbb{R}$, it automatically follows that an orthogonal and unitary mean the same thing, but this is not true over $\mathbb{C}$ as there are orthogonal matrices that are not unitary ($AA^T = I$ but $AA^* \neq I$) and vice versa. With that said, is there any relation that can be said between $U(n)$ and $O(n)$ over $\mathbb{C}$? In the book I am reading it says that we may also define $O(n)$ to be the set of square complex matrices that preserve the bilinear form: $$(x,y) = \sum_j x_jy_j$$ whereas a unitary matrix preserves the inner product on $\mathbb{C}$. Is this the only difference between the two groups?

2. Based on the definitions of these groups I get a feeling that some of them make up the others. For example, since $\det{A} = \pm 1$ for all $A \in U(n)$, does this mean that the only two components of $U(n)$ are $SL(n)$ and those with determinant $-1$? Can the components of any other of these groups be given purely in terms of the others?

Also, though slightly unrelated to my two questions above, what is the geometric difference between orthogonal matrices and unitary matrices? I know that orthogonal matrices represent a rotation or a reflection, but wouldn't a unitary matrix also represent the same thing?

Answer 1

Regarding question 1: You've mostly addressed this in your edits. Over $\Bbb C$, it is true that the orthogonal matrices preserve the bilinear form $(\cdot,\cdot)$ that you have defined. Note that this bilnear form does not satisfy the usual definition of an "inner-product" for a vector space over $\Bbb C$; for one, it is not guaranteed that $(x,x) \neq 0$ if $x \neq 0$.

On the other hand the unitary matrices do not preserve a bilinear form. Instead, $U(n)$ consists of the matrices that preserve the sesquilinear form. This sesquilinear form $\langle \cdot, \cdot \rangle$ is defined by $$ \langle x,y \rangle = \sum_{j}x_j \overline{y_j} $$ (notation varies as to which vector is conjugated). This form does satisfy the definition of an inner product. A matrix will be both orthogonal and unitary if and only if its entries are real.

Regarding question 2: Note that the determinant of an element of $U(n)$ can be any complex number with magnitude $1$, not just $\pm 1$. As a consequence, it turns out that $O(n)$ consists of two disconnected "copies" of $SO(n)$ (corresponding to the possible determinants $\pm 1$, whereas $SU(n)$ consists of an entire contiguous circle's worth of (hence infinitely many) "copies" of $U(n)$.

Regarding the relationship between $SO$ and $SU$, there is not always a nice way to think of one in terms of the other. However, it turns out that there is a nice way to think about each element of $SO(3)$ as corresponding to one of two possible matrices from $SU(2)$.