Need to show that $I$ is a regular value of $A^tA$.

Question

I'm working through a proof in geometry and I am stuck trying to show that on the set $SO(n)$ the map $F(A) =A^tA$ is a submersion. I guess I'm not sure what it means to take the derivative of a matrix operation. I have looked at several pdf's on the subject and am not quite sure how to take the partials of $F$ or how to show that the associated Jacobian is surjective. Any feedback would be greatly appreciated. Regards.

Answer 1

So you should view your $F$ as a map from $M=M_n$, the space of all $n$ by $n$ matrices, to $S=\text{Sym}_n$, the set of $n$ by $n$ symmetric matrices, and $F(A)=A^tA$. To show that $F$ is a submersion at some $A\in SO(n)$ (thus $A^tA=I$), we need to show that the derivative map at $A$, denoted $DF_A: TM_A\to TS_I$, where $TM_A$ is the tangent space of $M$ at $A$ and $TS_I$ is the tangent space of $S$ at $I$, is onto.

Now $M=M_n$ is the space of all $n$ by $n$ matrices, which is diffeomorphic to $\mathbb{R}^{n\times n}$, thus the tangent space of $M$ at $A$, $TM_A$, can be identified with $M_n$ itself. For $S=\text{Sym}_n$, it is embedded in $M_n\cong\mathbb{R}^{n\times n}$, and the tangent space of $S$ at $I$ is defined to be $$TS_I:=\{\gamma'(0)\in M_n|\gamma:(-\varepsilon,\varepsilon)\to \text{Sym}_n\subset M_n, \gamma(0)=I\}.$$ Here $\gamma$ are smooth curves. Then it is not hard to show that $TS_I= \text{Sym}_n$.

Lastly we look at the definition of $DF_A:TM_A= M_n\to TS_I=\text{Sym}_n$: $$DF_A(Q):=\frac{d}{ds}|_{s=0}F(\gamma(s)), \gamma:(-\varepsilon,\varepsilon) \to M_n, \gamma(0)=A,\gamma'(0)=Q,\gamma\text{ smooth}.$$ Thus $DF_A(Q)=\frac{d}{ds}|_{s=0}(\gamma(s)(\gamma(s))^t) =\gamma'(0)(\gamma(0))^t+\gamma(0)(\gamma'(0))^t =QA^t+AQ^t$. For any $P\in TS_I=\text{Sym}_n$, the equation $$QA^t+AQ^t=P$$ always has a solution for $Q$, for example, we may let $Q=\frac{P(A^{t})^{-1}}{2}.$

This shows that $DF_A: TM_A\to TS_I$ is onto for each $A$ with $A^tA=I$, in other words, $I$ is a regular value of $F:M\to S$.