# Munkres Problem: Define a $C^{\infty}$ map $f: \mathbb{R}^9 \to \mathbb{R}^6$ such that $O(3)$ is the solution of $f(x)=0$.

On Munkres's book analysis on manifold chap "the boundary of manifold", question 3, says: let $O(3)$ the set of orthogonal matrices, as a subspace of $\mathbb{R}^9$.

a) define a $C^{\infty}$ map $f: \mathbb{R}^9 \to \mathbb{R}^6$ such that $O(3)$ is the solution set of equation $f(x)=0$.

What does it mean?

• Are you saying you don't understand the question? If you say precisely what's causing you trouble, you'll be more likely to get the answer you're looking for. – Mark Jul 12 '16 at 15:25

Identify $\Bbb R^9$ with the set of $3\times 3$ real matrices and $\Bbb R^6$ with the set of symmetric $3\times 3$ real matrices. Then, $$f(A)=AA^{T}-I.$$
• @Jianluca The point is that $f(A)=AA^T-I$ is always symmetric, i.e. $f(A)^T=f(A)$. But the set of symmetric $3\times 3$ matrices has only $6$ independent coordinates. – Spenser Jul 12 '16 at 16:11
A $3 \times 3$ matrix is orthogonal iff its columns $c_1,c_2,c_3$ satisfy the six relations $$c_i \cdot c_j = 0 \quad (i < j), \qquad c_i \cdot c_i = 1 \quad (i=1,2,3).$$
This lets us express an element of $O(3)$ precisely as an element of the kernel of the map $M_3 \cong \mathbb{R}^9 \to \mathbb{R}^6$ defined by $$(c_1,c_2,c_3) \mapsto (c_1 \cdot c_2,\,c_1 \cdot c_3,\, c_2 \cdot c_3,\,c_1 \cdot c_1 -1,\, c_1 \cdot c_1 -1,\, c_2 \cdot c_2 -1,\,c_3 \cdot c_3 - 1).$$
You can express each of the coordinate functions as a map $\mathbb{R}^9 \to \mathbb{R}$ which, as it involves only addition and multiplication, is clearly smooth. For example, writing $c_i = (x_i,y_i,z_i)$, the first coordinate function is $x_1x_2+y_1y_2+z_1z_2$.