# Set of orthogonal matrices is a compact manifold

The following is a problem from Munkres's Analysis in Manifolds.

Problem: Let $$\mathcal O(3)$$ denote the set of all orthogonal 3 by 3 matrices, considered as a subspace of $$\mathbf R^9$$.

(a) Define a $$C^\infty$$ function $$f:\mathbf R^9\to\mathbf R^6$$ such that $$\mathcal O(3)$$ is the solution set of the equation $$f(\mathbf x)=\mathbf0$$.

(b) Show that $$\mathcal O(3)$$ is a compact 3-manifold in $$\mathbf R^9$$ without boundary. [Hint: Show the rows of $$Df(\mathbf x)$$ are independent if $$\mathbf x\in\mathcal O(3)$$.]

In Euclidean space of course compactness is equivalent to being closed and bounded.

So far, I can only show that $$\mathcal O(3)$$ is closed: Considering $$\mathcal O(3)$$ as a subspace of $$\mathbf R^9$$, if we define a map $$g:\mathbf R^9\to \mathbf R^9$$ by $$g(A)=A^TA$$, then $$g(A)=I$$ if and only if $$A$$ is orthogonal. Hence, $$\mathcal O(3)=g^{-1}(\{I\})$$.Further, $$g$$ is polynomial in the entries of a matrix, so it is continuous. As $$\{I\}$$ is closed and $$g$$ is continuous, $$\mathcal O(3)$$ is closed.

But I'm a bit confused on how to show $$\mathcal O(3)$$ is bounded. I've seen other responses using the fact that $$\|v\|=\|Av\|$$ whenever $$A$$ is orthogonal; I don't understand why this shows that $$\mathcal O(3)$$ is bounded, though. There must be something conceptual that I'm missing here.

I'm also unable to come up with the function $$f$$ required in part (a). If I can find such an $$f$$, then I imagine it just takes a bit of linear algebra to show that $$Df(\mathbf x)$$ has linearly independent rows — in which case, the result for (b) follows.

Any hints on how to construct $$f$$ or show that $$\mathcal O(3)$$ is bounded would be greatly appreciated.

• (1) Your $g$ is not linear. (2) The function you want is essentially $g$ (the range of $g$ is not all of $\mathbf{R}^9$; can you figure out how to identify it with $\mathbf{R}^6$? (3) look at the diagonal entries of $g$: their values provide bounds for the entries of $A$. – Willie Wong May 29 at 2:02
• Thanks, I didn't know what I was thinking. I've fixed that part. I know that the range of $g$ is a subset of the set of symmetric matrices. How should I associate the range of $g$ with $\mathbf R^6$, though? – buffle May 29 at 2:05

Let's use the operator norm on the space $$M_{3}(\Bbb{R})$$, which is defined as: for any $$A \in M_3(\Bbb{R})$$, \begin{align} \lVert A \rVert_{\text{op}}:= \sup\{\lVert Ax\rVert: \, \, \lVert x \rVert = 1\}, \end{align} where the norms $$\lVert \cdot\rVert$$ are the Euclidean norms on $$\Bbb{R}^3$$ (the one induced by the "standard" inner product). Then, for an orthogonal $$A$$, we have $$\lVert Av \rVert = \lVert v\rVert$$, so it follows that they have operator norm $$1$$: \begin{align} \lVert A \rVert_{\text{op}} &= 1. \end{align} Thus, $$\mathcal{O}(3)$$ lies on the unit sphere of $$M_3(\Bbb{R})$$ (relative to the operator norm). It is a relatively simple theorem to show that in finite-dimensions, every norm on a vector space generates the same topology. Thus, $$\mathcal{O}(3)$$, being a closed subset of a compact space (the unit sphere) is in fact compact.

Note that the map $$g$$ you defined can be written as follows: $$g: M_3(\Bbb{R}) \to \text{Sym}_3(\Bbb{R})$$, $$g(A) := A^tA$$. Then, $$\mathcal{O}(3) = g^{-1}(\{I\})$$. Here, $$M_3(\Bbb{R})$$ is a $$9$$-dimensional vector space, and $$\text{Sym}_3(\Bbb{R})$$ is a $$6$$-dimensional vector space. Now, you have to verify whether the hypothesis of the regular-value theorem is satisfied (this theorem is really a consequence of the inverse/implicit function theorem).

So, if you manage to show that for every $$A \in \mathcal{O}(3) = g^{-1}(\{I\})$$, the derivative $$Dg_A: M_3(\Bbb{R}) \to \text{Sym}_3(\Bbb{R})$$ is surjective, then the regular value theorem tells you that $$\mathcal{O}(3)$$ is a $$9-6=3$$ dimensional submanifold (without boundary) of $$M_3(\Bbb{R})$$. So, really all you have to do is calculate the derivative $$Dg_A$$ and show it is surjective.

If you want to think in terms of "$$f$$", then you look at $$f: M_3(\Bbb{R}) \to \text{Sym}_3(\Bbb{R})$$ defined by$$f(A) = g(A) - I$$, so that $$\mathcal{O}(3) = f^{-1}(\{0\})$$.

By the way, there is nothing special about $$3$$-dimensions. By considering the map $$G: M_n(\Bbb{R}) \to \text{Sym}_n(\Bbb{R})$$ given as $$G(A):= A^tA$$, and reasoning very similarly you can show that $$\mathcal{O}(n)= G^{-1}(\{I\})$$ is a compact, $$\dfrac{n(n-1)}{2}$$ dimensional submanifold of $$M_n(\Bbb{R})$$.

• Thank you for the help! I used your steps and was able to solve the problem. – buffle May 29 at 5:41
• Also, your last remark about there being nothing special about $3\times3$ matrices was illuminating. I thought about it more closely and it makes sense to me now. Thanks again! – buffle May 29 at 8:00
• @buffle I'm glad it was helpful. Also, as you may have noticed, I purposely avoided talking about $\Bbb{R}^9$ and $\Bbb{R}^6$, and just directly referred to the vector spaces, without choosing an isomorphism. THe reason is that almost all of the differential calculus you learn in $\Bbb{R}^n$ (for example, as presented in Munkres or Spivak) can be generalized with almost zero extra effort to the setting of Banach spaces (for example take a look at Loomis and Sternberg's Advanced Calculus, or Henri Cartan's Differential Calculus) – peek-a-boo May 29 at 8:12