Compactness of the Classical Groups

MOM Alok’s thinking about conic sections again.

The Classical Groups

The classical groups (that I was thinking about) are:

1. Real Orthogonal Group: $O(n,\mathbb{R})$. Real-valued matrices $A$ such that $A^{T}A=I$ where $A^T$ is the transpose of $A$, and $I$ is the identity matrix.
2. Complex Orthogonal Group: $O(n,\mathbb{C})$. Complex-valued matrices $A$ such that $A^{\ast }A=I$ where $A^{\ast }$ is the conjugate transpose of $A$.
3. Unitary Group: $U(n)$. Complex-valued matrices $U$ such that $U^{\ast }U=I$ where $U^{\ast }$ is the conjugate transpose of $U$.
4. Real Indefinite Orthogonal Group: $O(p,q)$. Real matrices $A$ such that $A^{T}JA=J$ where $J$ is the matrix with $p$ positive 1s and $q$ negative 1s on the diagonal and 0s elsewhere.
5. Complex Indefinite Orthogonal Group: $O(p,q,\mathbb{C})$. Complex-valued matrices $A$ such that $A^{\ast }JA=J$ where $J$ is the matrix with $p$ positive 1s and $q$ negative 1s on the diagonal and 0s elsewhere.

There’s also the special versions of these groups, which have the additional requirement that each element has determinant 1.

We’ll look at the differences between the real and complex groups, in particular if they are compact or not. I’m using the nonstandard characterization of compactness because it made it really easy to figure out the answer in my head for each case. Which is why I’m writing this.

Compactness

This definition is equivalent to the open cover definition over arbitary topological spaces.

Definition: A set $S$ in a topological space $X$ ($S\subseteq X$) is compact if and only if every point $s$ in its nonstandard extension ${}^{\ast }S$ is nearstandard (infinitely close to some standard point $p$ in the original set)¹.

In symbols:

So $\mathbb{R}$ is not compact because its nonstandard extension is ${\mathbb{R}}^{\ast }$, which includes infinite numbers. The open interval $(0,1)$ is not compact because its nonstandard extension is $(0,1{)}^{\ast }$ which includes infinitesimal numbers like $\epsilon$, which is infinitely close to 0, except that 0 is missing because it’s an OPEN interval.

This is the part about compactness generalizing “closed and bounded” in Euclidean space. In fact, it even explains why the “closed and bounded” definition works in the first place. The only nonstandard elements in Euclidean space that aren’t nearstandard are infinite numbers, so as long as the set is closed (contains all its limit points), then boundedness guarantees compactness.

Now let’s use this.

The Real Orthogonal Group: $O(n,\mathbb{R})$

$n=1$

This is the group of all 1x1 matrices with determinant 1. Which is just $\{1\}$, a single point. Finite sets are compact because their nonstandard extension is the same set, so every point is nearstandard. Plus intuitively, compactness is a generalization of finiteness, so every finite set should be compact.

$n\ge 2$

Imagine a matrix $\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ in $O(2,\mathbb{R})$. For it to be orthogonal, we need:

This gives us:

1. $$a^2+b^2=1$$
2. $$c^2+d^2=1$$
3. $$ac+bd=0$$

The first two equations tell us that the columns are unit vectors, and the third equation tells us that they are orthogonal.

I realized it’s compact because the unit length condition prevents any coordinates from being infinitely large. Like if you plug in an infinite number $\mathrm{\infty }$ for $a$, then ${\mathrm{\infty }}^2+b^2=1$, so $b^2=1-{\mathrm{\infty }}^2$, which is impossible (in the reals).

Same holds for higher $n$ for the same reason. So $O(n,\mathbb{R})$ is compact for all $n$.

Complex Orthogonal Group: $O(n,\mathbb{C})$

This is a funny group to think about since its defining condition is the same as the real case, but whenever you look at complex numbers, you should be multiplying them by their conjugates. But we don’t do that here.

$n=1$

This is the same as $O(1,\mathbb{R})$, so it’s compact.

$n\ge 2$

This is not compact. The reason is that now we can satisfy the equation $a^2+b^2=1$ with infinite $a$. Let $a=\mathrm{\infty }$. Then $b=-\sqrt{1-{\mathrm{\infty }}^2}$ which is negative infinite, which is impossible.

The matrix $\left(\begin{array}{cc}\mathrm{\infty }& -\sqrt{1-{\mathrm{\infty }}^2}\\ \sqrt{1-{\mathrm{\infty }}^2}& \mathrm{\infty }\end{array}\right)$ is in $O(2,\mathbb{C}{)}^{\ast }$, but is not nearstandard.

So for $n\ge 2$, $O(n,\mathbb{C})$ is not compact.

Unitary Group $U(n)$

The defining condition is $U^{\ast }U=I$, where $U^{\ast }$ is the conjugate transpose of $U$. And it’s that conjugate transpose that makes all the difference.

This is compact for all $n$.

$n=1$

Now instead of a single point, the defining condition reduces to $\overline{a}=1$, which is the unit circle. So it’s compact because the unit circle is closed and bounded.

$n\ge 2$

For a 2x2 unitary matrix $U=\left(\begin{array}{cc}a+bi& c+di\\ e+fi& g+hi\end{array}\right)$, the condition $U^{\ast }U=I$ gives:

1. $a^2+b^2+e^2+f^2=1$ (first column has unit length)
2. $c^2+d^2+g^2+h^2=1$ (second column has unit length)
3. $(ac+bd)+(eg+fh)+(bc-ad+fg-eh)i=0$ (columns are orthogonal)

All the coefficients are real numbers, so this is like $O(n,\mathbb{R})$. It’s compact. Ditto for higher $n$.

Indefinite Real Orthogonal Group $O(p,q)$

Over the reals

This is basically a group where $p$ unit basis vectors square to positive 1 and $q$ minus unit basis vectors square to negative 1. As long as both $p$ and $q$ are positive, the group is not compact because positive infinite numbers can be “compensated” for by negative infinite numbers, breaking compactness.

For example, consider $O(1,1)$. A matrix in this group satisfies:

This gives us:

1. $a^2-c^2=1$ (first diagonal)
2. $b^2-d^2=-1$ (second diagonal)
3. $ab-cd=0$ (off-diagonal)

We can satisfy these with infinite values. For example: $\left(\begin{array}{cc}\cosh \mathrm{\infty }& \sinh \mathrm{\infty }\\ \sinh \mathrm{\infty }& \cosh \mathrm{\infty }\end{array}\right)$ gives a matrix in ${}^{\ast }O(1,1)$ that is not nearstandard.

Let’s just look at the top left element to get the idea across:

Over the complex numbers

For the complex case, $O(p,q,\mathbb{C})$ is actually equal to $O(p+q,\mathbb{C})$, so it’s not compact by the reasoning in the previous section. The reason it’s equal is that you can multiply any basis vector $e_j$ by $i$ and get a basis vector of the opposite signature. This is basically the same as how hyperbolas, ellipses, and parabolas are all the same shape over the (projective) complex numbers.

Oh and for completeness, the symplectic group $Sp(2n,\mathbb{R})$ is not compact either. The general linear groups aren’t because they have no constraints on their entries. The special groups are compact because they’re closed subsets of compact groups. Although the indefinite orthogonal group isn’t compact, it has compact subgroups $O(p)$ and $O(q)$, at least over the reals.