Morse function on $\mathrm{SO}(n)$ I would like to prove that the following function is a Morse function,
$F : \mathrm{SO}(n)\to\mathbb{R}$ 
$$A=(a_{ij})\mapsto\sum_{i=1}^na_{ii}\lambda_i$$
with $0<\lambda_1<...<\lambda_n$. 
If one use the following local parametrization $e : \mathrm{Ant}(n)\to\mathrm{SO}(n)$
$$e(A)=\exp A=\sum_{i=1}^n\frac{A^n}{n!}$$
which is a local diffeomorphism from a neighborhood of $0$ (in the Lie algebra of antisymmetric matrix) to a neighborhood of the identity matrix in $\mathrm{SO}(n)$. We can check that the identity is a critical point of $F$ by the fact that the $0$ is a critical point of $F\circ e$.


*

*How to use $e$ to parametrize the whole of $\mathrm{SO}(n)$, is it useful to compose with a left translation?

*How to prove that the critical points of $F$ are precisely the diagonal matrix of $\mathrm{SO}(n)$ and that they are non degenerate?
Many thanks for your help!
 A: I will use $A^i{}_j$ to denote the $(i,j)^\text{th}$ entry of a matrix $A$. I will not be using the Einstein summation convention, so no sigma means no sum.
This intersects vaguely with some of my research from a few months ago. I developed this after a long week of trying to find an elegant way to prove a similar function was Morse. Eventually, I broke down and used this. It's messy, but it does the job.
A basic theorem from Lie group theory says that, for a Lie group $G$, the exponential map $\exp$ is a local diffeomorphism near $0\in\mathfrak{g}$. As a consequence, there is a neighborhood $U$ of the identity $\mathrm{e}$ in $G$ such that $\exp^{-1}|_U$ is a chart. Since left multiplication by any $g\in G$ is a diffeomorphism, it follows that the map $\exp^{-1}|_U\circ(g^{-1})^L$, where $(\cdot)^L$ indicates left multiplication, is a chart around $g\in G$.
Define $\mathfrak{T}(i,j)$ as the matrix $T\in\mathfrak{so}(n)$ with $T^i{}_j=t=-T^j{}_i$ and $T^k{}_l=0$ for all $k,l\not\in\{i,j\}$. Then, $\exp(\mathfrak{T}(i,j))$ is the special orthogonal matrix $A$ with $A^i{}_i=A^j{}_j=\cos(t)$ and $A^i{}_j=\sin(t)=-A^j{}_i$, and $A^k{}_l=0$ for all $k,l\not\in\{i,j\}$. Since we are differentiating at $0$, we may define $T(i,j)$ as the matrix $T\in SO(n)$ such that $T^i{}_i=T^j{}_j=\sqrt{1-t^2}$ and $T^i{}_j=t=-T^j{}_i$ and use $T(i,j)$ for $\exp(\mathfrak{T}(i,j))$.
If $A\in SO(n)$, we may consider $f^{(i,j)}_A(t)=F(A\,T(i,j))$ and differentiate at $t=0$. An easy calculation reveals that $$f^{(i,j)}_A(t)=(\lambda_iA^i{}_i+\lambda_jA^j{}_j)\sqrt{1-t^2}-(\lambda_iA^i{}_j-\lambda_jA^j{}_i)t.$$ Differentiating leaves us with $\lambda_jA^j{}_i-\lambda_iA^i{}_j$. Since this must equal $0$ for all appropriate pairs $(i,j)$, it follows that $A^i{}_j=A^j{}_i=0$.
Showing that the critical points are nondegenerate is simply a matter of differentiating again for each case to compute the Hessian, and then proving that the Hessian is nondegenerate. I'll leave it as an exercise. 
Hint: 

 Utilize the fact that, if $\mathfrak{S}(i,j)$ is defined similarly to $\mathfrak{T}(i,j)$ above, except $t$ is replaced with $s$, and $t$ and $s$ are close to $0\in\mathfrak{so}(n)$, then $\exp(\mathfrak{T}(i,j)+\mathfrak{S}(k,l))\approx\exp(\mathfrak{T}(i,j))\exp(\mathfrak{S}(k,l))$.

As an expert tip, for doing this in the future, always check that the possibly nondegenerate critical points are isolated before doing these calculations. It saves a LOT of trouble.
A: F is a linear map on $\mathbb{R}^{n^2}$ so its differential on any tangent plane to an embedded sub-manifold is the restriction of the same linear map. We know the tangent plane at identity of $SO(n)$ is the space of all skew-symmetry matrix: $T_I=\{X\ |\ X+X^T=0\}$. So the tangent plane at any $q\in SO(n)$ is just $dL_q(T_I)=\{ qX\ |\ X+X^T=0\}$ where $L_q$ is the left translation by $q$.
Let $F(qX)\equiv 0$, we have $$\sum_{i=1}^n \sum_{j=1}^n q_{ij}X_{ji}\lambda_i=0, \forall X\in T_I$$
 Set $X$ be the matrix with only $X_{kl}=-X_{lk}=1$ and all other entries $0$, we get $-q_{kl}\lambda_k+q_{lk}\lambda_l=0$. Let k=1 that is $q_{1l}=q_{l1}\lambda_l/\lambda_1$ so $1=\sum_{l=1}^nq_{1l}^2\ge\sum_{l=1}^n q_{l1}^2=1$, equality holds if and only if all $q_{l1}=0, \forall l>1$. This way we can show that all elements except the diagonal ones are zero.
