\chapterstyle

article \setbeforesubsecskip1ex \setaftersubsecskip-0.25em \setsubsecheadstyle

In this note, we discuss the properties of a function and a differential equation on a smooth manifold. If we speak about a manifold $\mathcal{M}$ of dimension $m$ we always mean a smooth manifold in the sense of [1], i.e. the subset of some $\mathbb{R}^{k}$ with $k\geq{m}$ and $\mathcal{M}$ is locally diffeomorphic to $\mathbb{R}^{m}$. In the context of this note, we need the notions of measure zero and dense. A set $A\subset\mathcal{M}$ of a manifold $\mathcal{M}$ is a set of measure zero if there is a collection of smooth charts $\{U_{l},\phi_{l}\}$ whose domains cover $A$ and such that $\phi_{l}(A\cap{U_{l}})$ have measure zero in $\mathbb{R}^{n}$, i.e. the $\phi_{l}(A\cap{U_{l}})$ can be covered for any $\varepsilon$ by a countable collection of open balls whose volumes sum up to less than $\varepsilon$, for details see e.g. [2, Chapter 10]. To define dense, we need the topological closure $\overline{A}$ of a set $A\subset\mathcal{M}$, i.e. the intersection of all closed sets in $\mathcal{M}$ that contain $A$. A dense subset of a smooth manifold $\mathcal{M}$ is a set $A\subset\mathcal{M}$ such that the topological closure fulfills $\overline{A}=\mathcal{M}$, see e.g. [2, Appendix, Topology]. $A$ is dense if and only if every nonempty open subset of $X$ has non-empty intersection with $A$. The complement $\mathbb{R}^{n}\setminus{A}$ of a set of measure zero $A\subset\mathbb{R}^{n}$ is dense in $\mathbb{R}^{n}$, since if there is a point $x\in\mathbb{R}^{n}$ such that there is an open $U\subset\mathbb{R}^{n}$ with $x\in{U}$ and $U\cap{(\mathbb{R}^{n}\setminus{A})}=\emptyset$, then $A$ contains an open set and cannot have measure zero, see also [3, Chapter 2].

Here, we consider a function and a differential equation on the set of special orthogonal matrices ${SO(n)}=\{\Theta\in\mathbb{R}^{n\times{n}}\rvert{\Theta^{-1}=\Theta^{\mathsf{T}},~\det(\Theta)=1}\}$. $SO(n)$ is a smooth manifold of dimension $\tfrac{n(n-1)}{2}$ with the subspace topology induced by $\mathbb{R}^{n\times{n}}$. The tangent space $T_{\Theta}SO(n)$ at $\Theta$ is given by

(1)

$$\begin{aligned} T_{\Theta}SO(n)&=\{X\in\mathbb{R}^{n\times{n}}\rvert{X=\Omega\Theta,~\Omega\in\mathbb{R}^{n\times{n}},~\Omega=-\Omega^{\mathsf{T}}}\}\end{aligned}\text{.}$$

The Riemannian metric $g:T_{\Theta}SO(n)\times{T_{\Theta}SO(n)}\rightarrow\mathbb{R}$ induced by the standard Euclidean metric on $SO(n)$ is given by

(2)

$$g(\Omega_{1}\Theta,\Omega_{2}\Theta)=\mathop{\mathrm{tr}}\left({(\Omega_{1}\Theta)^{\mathsf{T}}\Omega_{2}\Theta}\right)=\mathop{\mathrm{tr}}\left({\Omega_{1}^{\mathsf{T}}\Omega_{2}}\right)\text{.}$$

In the following, we define the differential and the Hessian of a function $f:SO(n)\rightarrow\mathbb{R}$ at a point $\Theta_{0}\in{SO(n)}$. Let $\Gamma:(-\varepsilon,\varepsilon)\rightarrow{SO(n)}$ be a smooth curve with $\dot{\Gamma}(t)=\Omega(t)\Gamma(t)$, $\Gamma(0)=\Theta_{0}$ and $\dot{\Gamma}(0)=\Omega_{0}\Theta_{0}$ with $\Omega_{0}\in\mathbb{R}^{n\times{n}}$ and $\Omega_{0}=-\Omega_{0}^{\mathsf{T}}$. The differential $d{f}_{\Theta_{0}}:T_{\Theta_{0}}SO(n)\rightarrow{T_{f(\Theta_{0})}\mathbb{R}\cong\mathbb{R}}$ of a function $f:SO(n)\rightarrow\mathbb{R}$ at a point $\Theta_{0}$ evaluated at $\Omega_{0}\Theta_{0}\in{T_{\Theta_{0}}SO(n)}$ is defined by

(3)

$$d{f}_{\Theta_{0}}(\Omega_{0}\Theta_{0})=\tfrac{d}{d{t}}\rvert_{t=0}(f\circ\Gamma)(t)\text{.}$$

The critical points of $f$ are the points $\Theta_{0}$ where $d{f}_{\Theta_{0}}$ is not surjective. Because of $\dim(T_{f(\Theta_{0})}\mathbb{R})=1$, this means that these are the points $\Theta_{0}$ where $d{f}_{\Theta_{0}}=0$. The gradient of $f$ is defined as the unique vector field $\mathop{\mathrm{grad}}{f}$ with

(4)

$$df_{\Theta_{0}}(\Omega_{0}\Theta_{0})=g(\mathop{\mathrm{grad}}{f}(\Theta_{0}),\Omega_{0}\Theta_{0})\text{,}$$

see e.g. [2, Chapter 11]. The Hessian $H_{f}(\Theta_{0})$ of $f$ at a critical point $\Theta_{0}$ evaluated at $(\Omega_{0}\Theta_{0},\Omega_{0}\Theta_{0})$ is defined by

(5)

$$H_{f}(\Theta_{0})(\Omega_{0}\Theta_{0},\Omega_{0}\Theta_{0})=\tfrac{d^{2}}{d{t}^{2}}\rvert_{t=0}(f\circ\Gamma)(t)\text{.}$$

Since the Hessian at a critical point is bilinear and symmetric, we have for $(\Omega_{1}+\Omega_{2})\Theta_{0}\in{T_{\Theta_{0}}SO(n)}$ with $\Omega_{1,2}=-\Omega_{1,2}^{\mathsf{T}}$ the equality

(6)

$$H_{f}(\Theta_{0})((\Omega_{1}+\Omega_{2})\Theta_{0},(\Omega_{1}+\Omega_{2})\Theta_{0})\\ =H_{f}(\Theta_{0})(\Omega_{1}\Theta_{0},\Omega_{1}\Theta_{0})+2H_{f}(\Theta_{0})(\Omega_{1}\Theta_{0},\Omega_{2}\Theta_{0})+H_{f}(\Theta_{0})(\Omega_{2}\Theta_{0},\Omega_{2}\Theta_{0})\text{.}$$

As a consequence, the value $H_{f}(\Theta_{0})(\Omega_{1}\Theta_{0},\Omega_{2}\Theta_{0})$ can be computed utilizing the values $H_{f}(\Theta_{0})(\Omega_{1}\Theta_{0},\Omega_{1}\Theta_{0})$, $H_{f}(\Theta_{0})(\Omega_{2}\Theta_{0},\Omega_{2}\Theta_{0})$, $H_{f}(\Theta_{0})((\Omega_{1}+\Omega_{2})\Theta_{0},(\Omega_{1}+\Omega_{2})\Theta_{0})$ and (6). For details on the Hessian at a critical point, see [4, Appendix C.5].

In the following we prove Lemma 1.
Proof. a) As stated above, $\Gamma:(-\varepsilon,\varepsilon)\rightarrow{SO(n)}$ is a differentiable curve with $\dot{\Gamma}(t)=\Omega(t)\Gamma(t)$, $\Gamma(0)=\Theta_{0}$ and $\dot{\Gamma}(0)=\Omega_{0}\Theta_{0}$ with $\Omega_{0}\in\mathbb{R}^{n\times{n}}$ and $\Omega_{0}=-\Omega_{0}^{\mathsf{T}}$. Then

(11)		$\displaystyle d{f}_{\Theta_{0}}(\Omega_{0}\Theta_{0})$	$\displaystyle=\frac{d}{d{t}}\rvert_{t=0}\left(n-\mathop{\mathrm{tr}}\left({\Gamma(t)}\right)\right)=-\mathop{\mathrm{tr}}\left({\Omega_{0}\Theta_{0}}\right)$
			$\displaystyle=-\frac{1}{2}\mathop{\mathrm{tr}}\left({\Theta_{0}^{\mathsf{T}}\Omega_{0}^{\mathsf{T}}+\Omega_{0}\Theta_{0}}\right)$
			$\displaystyle=-\frac{1}{2}\mathop{\mathrm{tr}}\left({(\Theta_{0}-\Theta_{0}^{\mathsf{T}})\Omega_{0}}\right)$
			$\displaystyle=-\frac{1}{2}\mathop{\mathrm{tr}}\left({\Theta_{0}^{\mathsf{T}}(\Theta_{0}-\Theta_{0}^{\mathsf{T}})\Omega_{0}\Theta_{0}}\right)\text{.}$

Therefore, the critical points of $f$ are given by

(12)

$$\{\Theta_{0}\in{SO(n)}\rvert{\Theta_{0}=\Theta_{0}^{\mathsf{T}}}\}\text{.}$$

With the definition of the Riemannian metric by (2), the gradient $\mathop{\mathrm{grad}}{f}_{\Theta}$ at $\Theta_{0}$ is given by

(13)

$$\mathop{\mathrm{grad}}{f}(\Theta_{0})=\frac{1}{2}(\Theta_{0}-\Theta_{0}^{\mathsf{T}})\Theta_{0}^{\mathsf{T}}\text{.}$$

b) Let $\Theta_{0}$ denote a critical point of $f$. As stated above, $\Gamma:(-\varepsilon,\varepsilon)\rightarrow{SO(n)}$ is a differentiable curve with $\dot{\Gamma}(t)=\Omega(t)\Gamma(t)$, $\Gamma(0)=\Theta_{0}$ and $\dot{\Gamma}(0)=\Omega_{0}\Theta_{0}$ with $\Omega_{0}\in\mathbb{R}^{n\times{n}}$ and $\Omega_{0}=-\Omega_{0}^{\mathsf{T}}$. Then

(14)		$\displaystyle H_{f}(\Theta_{0})(\Omega_{0}\Theta_{0},\Omega_{0}\Theta_{0})$	$\displaystyle=\frac{d^{2}}{d{t^{2}}}\rvert_{t=0}(f(\Gamma(t))=-\mathop{\mathrm{tr}}\left({\ddot{\Gamma}(t)}\right)\rvert_{t=0}$
			$\displaystyle=-\mathop{\mathrm{tr}}\left({\dot{\Omega}(t)\Gamma(t)+\Omega(t)\dot{\Gamma}(t)}\right)\rvert_{t=0}$
			$\displaystyle=-\mathop{\mathrm{tr}}\left({\Omega_{0}^{2}\Theta_{0}}\right)=\mathop{\mathrm{tr}}\left({\Omega_{0}^{\mathsf{T}}\Theta_{0}\Omega}\right)=\mathop{\mathrm{tr}}\left({\Theta_{0}^{\mathsf{T}}\Omega_{0}^{\mathsf{T}}\Theta_{0}\Omega_{0}\Theta_{0}}\right)\text{,}$

where we utilized that $\mathop{\mathrm{tr}}\left({\dot{\Omega}(0)\Theta_{0}}\right)=0$ since $\dot{\Omega}(t)$ is skew symmetric for all $t$ and $\Theta_{0}=\Theta_{0}^{\mathsf{T}}$ since $\Theta_{0}$ is a critical point. Utilizing (6) we get

(15)		$\displaystyle H_{f}(\Theta_{0})(\Omega_{1}\Theta_{0},\Omega_{2}\Theta_{0})$	$\displaystyle=\frac{1}{2}\biggl{(}H_{f}(\Theta_{0})((\Omega_{1}+\Omega_{2})\Theta_{0},(\Omega_{1}+\Omega_{2})\Theta_{0})$
			$\displaystyle~~~~-H_{f}(\Theta_{0})(\Omega_{1}\Theta_{0},\Omega_{1}\Theta_{0})-H_{f}(\Theta_{0})(\Omega_{2}\Theta_{0},\Omega_{2}\Theta_{0})\biggr{)}$
			$\displaystyle=\frac{1}{2}\mathop{\mathrm{tr}}\left({\!\!(\Omega_{1}+\Omega_{2})^{\mathsf{T}}\Theta_{0}(\Omega_{1}+\Omega_{2})\!-\!\Omega_{1}^{\mathsf{T}}\Theta_{0}\Omega_{1}\!-\!\Omega_{2}^{\mathsf{T}}\Theta_{0}\Omega_{2}\!\!}\right)$
			$\displaystyle=\frac{1}{2}\mathop{\mathrm{tr}}\left({\Omega_{1}^{\mathsf{T}}\Theta_{0}\Omega_{2}+\Omega_{2}^{\mathsf{T}}\Theta_{0}\Omega_{1}}\right)\text{.}$

c)i) Since $\Theta_{0}$ is symmetric, $\Theta_{0}$ is orthonormally diagonalizable, i.e. $\Theta_{0}=\Pi^{\mathsf{T}}D\Pi$ for some diagonal $D$ and orthonormal $\Pi$ where the columns of $\Pi$ are eigenvectors of $\Theta_{0}$. Since $\Theta_{0}^{\mathsf{T}}\Theta_{0}=I$, we get $D^{2}=I$ and consequently the eigenvalues are $\pm{1}$. Since $\Theta_{0}\neq{I}$ and $\det(\Theta_{0})=1$, we always have an even number of negative eigenvalues. A similarity transformation leaves the trace invariant, hence a critical point $\Theta_{0}$ fulfills

(16)

$$\mathop{\mathrm{tr}}\left({\Theta_{0}}\right)=n-4k\text{,}$$

where $k\in\{0,\ldots,\lfloor{\frac{n}{2}}\rfloor\}$ is the number of eigenvalue pairs which are $-1$.  
c)ii) We start by showing that each $\mathcal{F}_{k}$ is path connected and thus connected. Let $k\in\{0,\ldots,\lfloor{\frac{n}{2}}\rfloor\}$ be arbitrary but fixed and let $\Theta_{1},\Theta_{2}\in\mathcal{F}_{k}$. Then there are orthogonal $\Pi_{1},\Pi_{2}$ such that $\Theta_{1}=\Pi_{1}^{\mathsf{T}}D\Pi_{1}$ and $\Theta_{2}=\Pi_{2}^{\mathsf{T}}D\Pi_{2}$. Furthermore, there are real skew-symmetric matrices $\Omega_{1},\Omega_{2}$ such that $\Pi_{1}=\exp(\Omega_{1})$ and $\Pi_{2}=\exp(\Omega_{2})$ with $\exp$ denoting the matrix exponential. Then $\alpha:[0,1]\rightarrow{SO(n)}$ defined by

(17)

$$t\mapsto\exp\left(\Omega_{1}^{\mathsf{T}}(1-t)\right)\exp\left(\Omega_{2}^{\mathsf{T}}t\right)D\exp\left(\Omega_{2}t\right)\exp\left(\Omega_{1}(1-t)\right)$$

is a smooth curve in $\mathcal{F}_{k}$ which connects $\Theta_{1}$ and $\Theta_{2}$. Since $\Theta_{1},\Theta_{2}\in\mathcal{F}_{k}$ were arbitrary, this implies the path-connectedness of $\mathcal{F}_{k}$. To show that $\mathcal{F}_{k}$ is isolated, we utilize that $n-4l=f\rvert_{\mathcal{F}_{l}}\neq{f\rvert_{\mathcal{F}_{k}}}=n-4k$ for $l\neq{k}$. Then there is a $\varepsilon(l)$ with $(n-4l-\varepsilon(l),n-4l+\varepsilon(l))\cap(n-4k-\varepsilon(l),n-4k+\varepsilon(l))=\emptyset$ for $k\neq{l}$. As a consequence, the intersection of the preimage of these sets under $f$ is empty. Since $f$ is continuous and both, $(n-4l-\varepsilon(l),n-4l+\varepsilon(l))$ and $(n-4k-\varepsilon(l),n-4k+\varepsilon(l))$ are open, their preimages are open and contain $\mathcal{F}_{l}$ and $\mathcal{F}_{k}$ respectively. With $U_{k}(l)=f^{-1}\bigl{(}(n-4l-\varepsilon(l),n-4l+\varepsilon(l))\bigr{)}$ we thus have $U_{k}(l)\cap{\mathcal{F}_{l}}=\emptyset$. Since this is possible for every $l\in\{0,\ldots,\lfloor{\tfrac{n}{2}}\rfloor\}$ and since a finite intersection of open sets is an open set, we find an open neighborhood $U$ of $\mathcal{F}_{k}$ such that $U\cap{\mathcal{F}_{l}}=\emptyset$ for all $l\in\{0,\ldots,\lfloor{\tfrac{n}{2}}\rfloor\}$.  
c)iii) The property that the $\mathcal{F}_{k}$ are submanifolds is given in [5]. The tangent space follows from (7).  
c)iv) Let $k\in\{0,\ldots,\lfloor{\frac{n}{2}}\rfloor\}$ be arbitrary but fixed and $\Theta_{0}\in\mathcal{F}_{k}$. Since $\ker{H}_{f}(\Theta_{0})\supset{T_{\Theta_{0}}\mathcal{F}_{k}}$ is always true, we have to check $\ker{H}_{f}(\Theta_{0})\subset{T_{\Theta_{0}}\mathcal{F}_{k}}$. Since every critical point is symmetric, there is an orthogonal $\Pi$ such that $\Pi^{\mathsf{T}}\Theta_{0}\Pi=D$ where $D$ is a diagonal matrix with non-zero diagonal elements. Furthermore, we know that

	$\displaystyle H_{f}(\Theta_{0})(\Omega_{1}\Theta_{0},\Omega_{2}\Theta_{0})$	$\displaystyle=-\frac{1}{2}\mathop{\mathrm{tr}}\left({\Omega_{1}^{\mathsf{T}}\Theta_{0}\Omega_{2}+\Omega_{2}^{\mathsf{T}}\Theta_{0}\Omega_{1}}\right)$
		$\displaystyle=-\frac{1}{2}\mathop{\mathrm{tr}}\left({\Theta_{0}^{\mathsf{T}}\Omega_{1}^{\mathsf{T}}\Theta_{0}\Omega_{2}\Theta_{0}+\Theta_{0}^{\mathsf{T}}\Omega_{2}^{\mathsf{T}}\Theta_{0}\Omega_{1}\Theta_{0}}\right)$
		$\displaystyle=-\frac{1}{2}\mathop{\mathrm{tr}}\left({(\Omega_{1}\Pi^{\mathsf{T}}D\Pi)^{\mathsf{T}}\Pi^{\mathsf{T}}D\Pi(\Omega_{2}\Pi^{\mathsf{T}}D\Pi)}\right)$
		$\displaystyle~~~-\frac{1}{2}\mathop{\mathrm{tr}}\left({(\Omega_{2}\Pi^{\mathsf{T}}D\Pi)^{\mathsf{T}}\Pi^{\mathsf{T}}D\Pi(\Omega_{1}\Pi^{\mathsf{T}}D\Pi)}\right)$
		$\displaystyle=-\frac{1}{2}\mathop{\mathrm{tr}}\left({(\Pi\Omega_{1}\Pi^{\mathsf{T}}D)^{\mathsf{T}}D(\Pi\Omega_{2}\Pi^{\mathsf{T}}D)}\right)$
		$\displaystyle~~~-\frac{1}{2}\mathop{\mathrm{tr}}\left({(\Pi\Omega_{2}\Pi^{\mathsf{T}}D)^{\mathsf{T}}D(\Pi\Omega_{1}\Pi^{\mathsf{T}}D)}\right)$
		$\displaystyle=-\frac{1}{2}\mathop{\mathrm{tr}}\left({(\tilde{\Omega}_{1}D)^{\mathsf{T}}D(\tilde{\Omega}_{2}{D})+(\tilde{\Omega}_{2}D)^{\mathsf{T}}D(\tilde{\Omega}_{1}{D})}\right)$
		$\displaystyle=H_{f}(D)(\tilde{\Omega}_{1}D,\tilde{\Omega}_{2}D)$

where $\tilde{\Omega}_{1}=\Pi\Omega_{1}\Pi^{\mathsf{T}}$ and $\tilde{\Omega}_{2}=\Pi\Omega_{2}\Pi^{\mathsf{T}}$. As a consequence

	$\displaystyle\ker{H}_{f}(\Theta_{0})$	$\displaystyle=\{X\in{T_{\Theta_{0}}SO(n)}\rvert{\text{ $H_{f}(\Theta_{0})(X,Y)=0$ for all $Y\in{T_{\Theta_{0}}SO(n)}$ }}\}$
		$\displaystyle=\{X\in{T_{D}SO(n)}\rvert{\text{ $\mathop{\mathrm{tr}}\left({X^{\mathsf{T}}D\tilde{\Omega}_{2}D-D\tilde{\Omega}_{2}DX}\right)=0$ for all $\tilde{\Omega}_{2}{D}\in{T_{D}SO(n)}$ }}\}$
		$\displaystyle=\{X\in{T_{D}SO(n)}\rvert{\text{ $\mathop{\mathrm{tr}}\left({\tilde{\Omega}_{2}D(X^{\mathsf{T}}-X)D}\right)=0$ for all $\tilde{\Omega}_{2}{D}\in{T_{D}SO(n)}$ }}\}\text{.}$

Observe that $\tilde{\Omega}_{2}$ is skew symmetric. Furthermore, $X^{\mathsf{T}}-X$ is skew symmetric and since $D$ is diagonal with non-zero entries, $D(X^{\mathsf{T}}-X)D$ is skew symmetric. Because the equation $\mathop{\mathrm{tr}}\left({D\tilde{\Omega}_{2}D(X^{\mathsf{T}}-X)}\right)=0$ has to hold for all skew-symmetric $\tilde{\Omega}_{2}$, we obtain $D(X^{\mathsf{T}}-X)D=0$. With the non-singular $D$ this implies $X^{\mathsf{T}}-X=0$ which is equivalent to $X=X^{\mathsf{T}}$. In c)ii) we showed $T_{\Theta_{0}}\mathcal{F}_{k}$ is $\{\Sigma\in\mathbb{R}^{n\times{n}}\rvert{\Sigma=\Sigma^{\mathsf{T}}}\}\cap{T_{\Theta_{0}}SO(n)}$, thus the previous calculation shows $\ker{H}_{f}(\Theta_{0})\subset{T_{\Theta_{0}}\mathcal{F}_{k}}$.  
d) Since $\Theta_{0}=\Theta_{0}^{\mathsf{T}}$, $\Theta_{0}$ is orthogonally diagonalizable, i.e. $\Theta_{0}=\Pi^{\mathsf{T}}D\Pi$ for some diagonal $D$ and orthogonal $\Pi$. Therefore

(18)

$\displaystyle H_{f}((\Omega_{0}\Theta_{0}),(\Omega_{0}\Theta_{0}))$

$\displaystyle=\mathop{\mathrm{tr}}\left({\Omega_{0}^{\mathsf{T}}\Theta_{0}\Omega_{0}}\right)=-\mathop{\mathrm{tr}}\left({\tilde{\Omega}_{0}^{2}D}\right)\text{,}$

where $\tilde{\Omega_{0}}=\Pi\Omega_{0}\Pi^{\mathsf{T}}$ is skew symmetric. Consequently, $\mathop{\mathrm{tr}}\left({\Omega_{0}^{\mathsf{T}}\Theta_{0}\Omega_{0}}\right)$ is definite for all skew symmetric $\Omega_{0}$ at a critical point $\Theta_{0}$ if and only if $\mathop{\mathrm{tr}}\left({\tilde{\Omega_{0}}^{2}D}\right)$ is definite for all skew symmetric $\tilde{\Omega_{0}}$ where $D=\Pi\Theta_{0}\Pi^{\mathsf{T}}$ is diagonal. Thus, we have to consider $H_{f}$ only for diagonal $D$, i.e.

(19)		$\displaystyle H_{f}((\Omega_{0}\Theta_{0}),(\Omega_{0}\Theta_{0}))$	$\displaystyle=-\mathop{\mathrm{tr}}\left({\Omega_{0}^{2}{D}}\right)=-\sum_{1\leq{k}\leq{n}}{\left(\Omega_{0}^{2}D\right)_{kk}}$
			$\displaystyle=-\sum_{1\leq{k}\leq{n}}{(\Omega_{0}^{2})_{kk}D_{kk}}=-\sum_{1\leq{k}\leq{n}}{\sum_{1\leq{l}\leq{n}}{(\Omega_{0})_{kl}(\Omega_{0})_{lk}D_{kk}}}$
			$\displaystyle=\sum_{1\leq{k,l}\leq{n}}{((\Omega_{0})_{kl})^{2}D_{kk}}=\sum_{\begin{subarray}{c}1\leq{k,l}\leq{n}\\ k\neq{l}\end{subarray}}{((\Omega_{0})_{kl})^{2}D_{kk}}$
			$\displaystyle=\sum_{1\leq{k}<{l}\leq{n}}{(D_{kk}+D_{ll})((\Omega_{0})_{kl})^{2}}\text{.}$

The critical points $\Theta_{0}$ are such that $\Theta_{0}\in{SO(n)}$ are symmetric, therefore all eigenvalues of $\Theta_{0}$ are real. Observe now that $\Theta_{0}=\Pi^{\mathsf{T}}D\Pi$ with orthogonal $\Pi$ implies $D^{2}=I$. Hence, the eigenvalues are $D_{kk}\in\{-1,1\}$ and since $\Theta_{0}\in{SO(n)}$ the number of $-1$-eigenvalues is even. Consequently we have to determine the definiteness of $H_{f}$ by considering (19) for all diagonal matrices $D$ with $\pm{1}$ on the diagonals where the number of $-1$ entries is zero or even.  
Suppose first, that all $D_{kk}$ are equal to $1$, i.e. $D=I$. The associated $\Theta_{0}$ is $\Theta_{0}=\Pi^{\mathsf{T}}D\Pi=\Pi^{\mathsf{T}}\Pi=I$. $\eqref{eqn:traceson:1}$ then implies $H_{f}(\Omega_{0}\Theta_{0},\Omega_{0}\Theta_{0})=\sum_{1\leq{k}<{l}\leq{n}}{2(\Omega_{0})_{kl}^{2}}$, i.e. $H_{f}>0$ for all skew-symmetric $\Omega_{0}$. Thus $H_{f}$ is positive definite if $\Theta_{0}=I$. Suppose now there is an even number of eigenvalues $D_{kk}$ equal to $-1$. Then, there are indices $l,k$ and $l\neq{k}$ such that $D_{kk}=-1$ and $D_{ll}=-1$, and therefore there are skew symmetric $\Omega_{0}$ such that $H_{f}(\Omega_{0}\Theta_{0},\Omega_{0}\Theta_{0})<0$. As consequence, $H_{f}$ is indefinite at a critical point $\Theta_{0}$ where $\Theta_{0}$ has an even number of negative eigenvalues $D_{kk}=-1$. Therefore, $\Theta_{0}=I$ is the only local (global) minimum of $f$. All other critical points are saddle points.    

Proof. We show only Definition 3a) since b) and c) were shown in Lemma 1. $SO(n)$ is compact, hence $f$ attains its minimal and its maximal value on $SO(n)$. The minimal value of $f$ is zero, the maximal value is $2n-2$ for $n$ odd and $2n$ for $n$ even. If $n$ is odd we thus have

$$L_{c}=\{\Theta\in{SO(n)}\rvert{f(\Theta)\leq{c}}\}=\begin{cases}f^{-1}([0,c])&\text{for $c\leq{2n-2}$}\\ f^{-1}([0,2n-2])&\text{for $c>{2n-2}$}\text{.}\end{cases}$$

If $n$ is even we have

$$L_{c}=\{\Theta\in{SO(n)}\rvert{f(\Theta)\leq{c}}\}=\begin{cases}f^{-1}([0,c])&\text{for $c\leq{2n}$}\\ f^{-1}([0,2n])&\text{for $c>{2n}$}\text{.}\end{cases}$$

Since $f$ is continuous, the preimage of a closed set is a closed set and since $SO(n)$ is bounded, its subsets are bounded as well. Since $SO(n)\subset\mathbb{R}^{n\times{n}}$, the boundedness and closedness of the sublevel sets implies their compactness.    
The convergence properties of the gradient flow associated with $\Theta\mapsto{n-\mathop{\mathrm{tr}}\left({\Theta}\right)}$ are thus determined by the following proposition.

To give a more detailed specification the convergence behavior of the gradient flow of $\Theta\mapsto{n-\mathop{\mathrm{tr}}\left({\Theta}\right)}$ we need the following result.

Proof. The goal of the proof is to show that ${A}$ has measure zero and that $\mathcal{M}\setminus{A}$ is dense. We show this in the following way. First, we consider the set of points lying in a suitable neighborhood of $\mathcal{N}$ and which contains the orbits of the solutions of the gradient flow $\dot{x}=-\mathop{\mathrm{grad}}{f}(x)$ which eventually converge towards $\mathcal{N}$. We utilize a result from [6] to conclude that this set has measure zero and $\mathcal{M}$ without this set is dense. Then, we utilize this set to derive the same result for ${A}$ utilizing the properties of the flow of the gradient vector field on $\mathcal{M}$.

In the following, we apply [6, Proposition 4.1]. This proposition concerns the case of a a three times continuously differentiable vector field $v:\mathbb{R}^{l}\rightarrow\mathbb{R}^{l}$ together with submanifold of equilibria $\overline{\mathcal{N}}$ in $\mathbb{R}^{l}$ under the assumption that $\overline{\mathcal{N}}$ is normally hyperbolic with respect to $v$. Normal hyperbolicity of $\overline{\mathcal{N}}$ means that the linearization of the vector field $v$ at $x\in\overline{\mathcal{N}}$ has $n-\dim{\overline{\mathcal{N}}}$ eigenvalues with real parts different from zero. Under these assumptions, there exists a neighborhood $\overline{\mathcal{U}}$ of $\overline{\mathcal{N}}$ such that any solution $t\mapsto\phi(t,x_{0})$ of $\dot{x}=v(x)$ with initial condition $x_{0}$ and with a forward orbit $\phi([0,\infty);x_{0})$ in $\overline{\mathcal{U}}$ lies on the stable of manifold $W^{s}_{\text{loc}}(p)$ of a point $p\in\overline{\mathcal{N}}$. $W^{s}_{\text{loc}}(p)$ is defined by

(21)

$$W^{s}_{\text{loc}}(p)=\{x\in\mathcal{U}\rvert{\lim_{t\rightarrow\infty}\phi(t,x)=p}\}\text{.}$$

We can always embed $\mathcal{M}$ into $\mathbb{R}^{l}$ for $l$ large enough, see e.g. [2, Chapter 10], therefore we can utilize [6] also for our case of a vector field on a manifold $\mathcal{M}$.

Since $\mathcal{M}$ is compact and $f$ is smooth, we have a global flow $\phi:\mathbb{R}\times\mathcal{M}\rightarrow\mathcal{M}$, which means that $t\mapsto{\phi(t,x_{0})}$ is a solution of the gradient flow $\dot{x}=-\mathop{\mathrm{grad}}{f}(x)$ defined for all $t\in\mathbb{R}$ and with $\phi(0,x_{0})=x_{0}$, see e.g. [2, Chapter 17]. Furthermore $\phi(t,\cdot):\mathcal{M}\rightarrow\mathcal{M}$ is a diffeomorphism for every $t\in\mathbb{R}$. Since $f$ is a Morse-Bott function, $\mathcal{N}$ is normally hyperbolic, i.e. the linearization of the gradient flow at any $x\in\mathcal{N}$ has exactly $m-n$ eigenvalues with real parts different from zero, see [7, p. 183, Morse-Bott functions]. According to [6, Proposition 4.1], we have a neighborhood $\mathcal{U}$ of $\mathcal{N}$ such that for every solution $\phi(\cdot,x)$ with $x\in\mathcal{N}$ and a forward orbit $\phi([0,\infty],x)$ in $\mathcal{U}$, the solution has to lie in one $W^{s}_{\text{loc}}(p)$ with $p\in\mathcal{N}$. We know from [8, Proposition 3.2], that if we choose $\mathcal{U}$ small enough, then the local stable manifold $W^{s}_{\text{loc}}(\mathcal{N})$ of $\mathcal{N}$ given by

(22)

$$W^{s}_{\text{loc}}(\mathcal{N})=\cup_{p\in\mathcal{N}}W^{s}_{\text{loc}}(p)$$

is a smooth submanifold of dimension $m+k$ where $k<m-n$ is the number of eigenvalues with real part smaller than zero. Since the stable manifold $W^{s}_{\text{loc}}(\mathcal{N})$ is a submanifold of $\mathcal{M}$ with smaller dimension than $\mathcal{M}$, the stable manifold has measure zero and $\mathcal{M}\setminus{W^{s}_{\text{loc}}(\mathcal{N})}$ is dense in $\mathcal{M}$, see [2, Theorem 10.5].