# Property of group of $k \times k$ orthogonal matrices

Does the group of $k \times k$ orthogonal matrices lie on an $(k^2 - 1)$-sphere? If so, of what radius? If not, does it lie on some sort of other object?

• The group itself is an "object" like a sphere. It's an algebraic variety. Do you have some reason for asking this? – Matt Samuel Apr 29 '16 at 2:47
• – Will Jagy Apr 29 '16 at 2:49
• anyway, look for books with the title "Matrix Groups." – Will Jagy Apr 29 '16 at 2:56
• oh; new and unregistered.. – Will Jagy Apr 29 '16 at 2:58

## 2 Answers

First you need a definition of what a sphere is in the space of $n \times n$ matrices. For example, you may want to do this using a norm. But then the answer highly depends on the choice of norm.

For example, if you consider the following norm on the space of $n \times n$ matrices: $$||A|| = \sqrt{\mbox{tr} ( A^t A )}$$

Then since orthogonal matrices satisfy $A^tA = I$, we have $|| A || = \sqrt{n}$. So orthogonal matrices do lie on a sphere in the space of all matrices with respect to this norm. Notice that scaling the norm changes the radius of the sphere.

However, if you consider this second norm: $$||A||_{\max} = \max_{ij} |A_{ij}|$$ Then for example the identity matrix has norm $1$, while the orthogonal matrix $$\frac{1}{2}\begin{pmatrix} \sqrt{2} & \sqrt{2} \\ \sqrt{2} & -\sqrt{2} \end{pmatrix}$$ has norm $\sqrt{2}/2$. So the orthogonal matrices do not lie on a sphere with respect to this norm.

The columns in an $n\times n$ orthogonal matrix $A$ are all unit vectors, hence $\|A\|^2=n$ where the square norm is defined by the trace form $\|A\|^2=\mathrm{tr}(A^TA)=\sum_i\|\vec{a}_i\|^2=\sum_{i,j}a_{ij}^2$.

This means not only is $\mathrm{O}(n)$ contained in $\mathbb{S}^{n^2-1},$ it's contained in $\underbrace{\mathbb{S}^{n-1}\times\cdots\times\mathbb{S}^{n-1}}_n$.

The same can be said for the rows of $\mathrm{O}(n)$, so it's inside a different copy of $(\mathbb{S}^{n-1})^n$ too.

To get a more narrow view of the topology of $\mathrm{O}(n)$, one needs to understand the orbit-stabilizer theorems for group actions in terms of fiber bundles. Since $\mathrm{O}(n)$ acts transitively on $\mathbb{S}^{n-1}$ with a point-stabilizer a copy of $\mathrm{O}(n-1)$, we have fiber bundles $\mathrm{O}(n-1)\to\mathrm{O}(n)\to\mathbb{S}^{n-1}$. So that means inductively $\mathrm{O}(n)$ is kind of like $((\cdots(\mathbb{S}^0\times\mathbb{S}^1)\times\cdots)\times\mathbb{S}^{n-2})\times\mathbb{S}^{n-1}$, except with very "twisted" products making it look very different.

For example, $\mathrm{O}(3)\simeq\mathbb{S}^0\times\mathbb{RP}^3$, where $\mathbb{RP}^3\simeq\mathbb{S}^3/\{\pm1\}$ is like $\mathbb{S}^3$ divided by two, and the so-called Hopf fibration says that $\mathbb{S}^3$ is essentially a bunch of $\mathbb{S}^1$s arranged in the shape of $\mathbb{S}^2$, the same way a Mobius band is a bunch of $[0,1]$s arranged in the shape of a $\mathbb{S}^1$.

Using the complex trace form $\|A\|^2=\mathrm{tr}(A^*A)$, one sees that $\mathrm{U}(n)$ similarly exists inside two different copies of $(\mathbb{S}^{2n-1})^n$ sitting inside $\mathbb{C}^n$, and more narrowly the orbit-stabilizer theorem gives fiber bundles $\mathrm{U}(n-1)\to\mathrm{U}(n)\to\mathbb{S}^{2n-1}$.