Use implicit function theorem to show $O(n)$ is a manifold In class today our teacher mentioned that one can use the implicit function theorem to show that $O(n) \subseteq \mathbb{R}^{n^2}$ is a submanifold...that is, map $A \mapsto A^* A$, and set it equal to the identity matrix. There should be $n(n+1)/2$ equations in $n^2$ variables ($A^*A$ is always symmetric, so we only need $n(n+1)/2$ equations). When I asked how to show that the jacobian matrix of this transformation had full row rank (the condition of the implicit function theorem), my teacher said "take the derivative near the identity, and you see that the derivative is just a matrix plus its transpose ."
I wasn't sure what he meant. Can someone clarify?
 A: There's a standard trick for computing total derivatives of maps between vector spaces. Basically, in this situation, we have
$$
df_x(v) = \lim_{h \to 0} \frac{f(x + h v) - f(x)}{h}
$$
when $v$ is a tangent vector at $x$. The reason this works is because in a vector space, vectorial addition allows you to quickly identify all tangent spaces as just another copy of the original vector space. Outside this specific situation, what I wrote above is nonsense and you have to work in coordinates to regain this formula.
In your situation, you have a map $A \mapsto \phi(A) = A^* A$, where the domain is $M_{n \times n}$, the set of $n \times n$ matrices, and the range is the subspace of symmetric matrices.
A tangent vector to $A \in M_{n \times n}$ is just another $n \times n$ matrix $B \in M_{n \times n}$. So think of $B \in T_{I}M_{n \times n}$, where $I$ is the identity, and compute
$$
d \phi_I(B) = \lim_{h \to 0} \frac{1}{h} (\phi(I + h B) - \phi(I)) = \lim_{h \to 0} \frac{1}{h} ((I + h B)^*(I + h B) - I^* I)
$$
Now think about the rank of $d \phi_I$ and compare it to the dimensions of the domain and range (note: it matters very much that the range is the space of symmetric matrices!).
A: Let $f:\mathbb R^{n^2}\to Sym$ denote the map $A\mapsto AA^*$. Pick $A\in f^{-1}(I)$, $$Df_A(B)=\lim_{t\to 0} \frac{f(A+tB)-f(A)}{t}=\lim_{t\to 0} \frac{(A+tB)(A^*+tB^*)-I}{t}=\lim_{t\to 0} AB^*+BA^*+tBB^*$$
$$ \lim_{t\to 0} AB^*+BA^*+tBB^* = AB^*+BA^*$$
Thus, $Df_A(B)=AB^*+BA^*$.
