$\gamma_{u}(T+\varepsilon)$. Observe that,

$\displaystyle 2\varepsilon$
$\displaystyle=d\left(\gamma_{u}(T+\varepsilon),\gamma_{u}(T)\right)+d\left(\gamma_{v}(T),\gamma_{v}(T-\varepsilon)\right)$
$\displaystyle>d\left(\gamma_{u}(T+\varepsilon),\gamma_{v}(T-\varepsilon)\right).$

If $p,q,r\in M$ such that $d(p,q)+d(q,r)=d(p,r)$ and there exist the shortest
normal geodesic $\gamma_{1}$ and $\gamma_{2}$ joining $p$ to $q$ and $q$ to
$r$, respectively, then $\gamma_{1}\cup\gamma_{2}$ is smooth at $q$ and
defines a shortest normal geodesic joining $p$ to $r$. Therefore, we have

$\displaystyle L(\gamma_{v}|_{[0,T-\varepsilon]}\cup\beta)$
$\displaystyle=T-\varepsilon+d(\gamma_{v}(T-\varepsilon),\gamma_{u}(T+\varepsilon))$
$\displaystyle<T+\varepsilon=L(\gamma_{u}|_{[0,T+\varepsilon]}).$

This contradiction establishes that $\gamma_{u}|_{[0,T+\varepsilon]}$ is not
minimal.

For the converse, set $T=\rho(u)$ and observe that $\gamma_{u}|_{[0,T]}$ is an
$N$-geodesic. Assuming that $q\vcentcolon=\gamma_{u}(T)$ is not the first
focal point of $N$ along $\gamma_{u}$, we will prove that (ii) holds. Let
$p=\gamma_{u}(0)$ and choose a neighbourhood $\tilde{U}$ of $Tu$ in $\nu$ such
that $\exp_{\nu}|_{\tilde{U}}$ is a diffeomorphism. For sufficiently large
$n$, $q_{n}\vcentcolon=\gamma_{u}\left(T+1/n\right)\in\exp_{\nu}(\tilde{U})$.
Take $N$-geodesics $\gamma_{n}$ parametrized by arc-length joining $p_{n}$ to
$q_{n}$ and set $u_{n}\vcentcolon=\dot{\gamma}_{n}(0)\in
S((T_{p_{n}}N)^{\perp})$. Since $S((T_{p_{n}}N)^{\perp})$ is compact, by
passing to a subsequence, we may assume that $u_{n}$ converges to $v\in
S(N_{p})$. By Observation B,

$\gamma_{v}(T)=\lim_{n\to\infty}\gamma_{u_{n}}\left(T+\textstyle{\frac{1}{n}}\right)=\gamma_{u}(T).$

If $v=u$, then for sufficiently large $n$, $d(p,q_{n})u_{n}\in\tilde{U}$,
whence

$\left(T+\textstyle{\frac{1}{n}}\right)u=d(p,q_{n})u_{n}.$

Taking absolute values on both sides imply $T+1/n>d(p,q_{n})$. This
contradiction implies $v\neq u$. ∎

We now will prove that the map $\rho$ defined in 4.4 is a continuous function.

######  Proposition 4.2.2.

The map $\rho:S(\nu)\to[0,\infty)$, as defined in (4.4), is continuous.

###### Proof.

We will prove that $\rho(u_{n})\to\rho(u)$ whenever $(p_{n},u_{n})\to(p,u)$ in
the unit normal bundle $S(\nu)$. Let $T$ be any accumulation point of the
sequence $\\{\rho(u_{n})\\}$ including $\infty$. By Observation B,
$\gamma_{u}|_{[0,T]}$ is an $N$-geodesic and hence $T\leq\rho(u)$. If
$T=+\infty$, we are done. So let us assume that $T<+\infty$. From 4.2.1, at
least one of the following holds for infinitely many $n$.

  1. (i)

The sequence $\rho(u_{n})$ is the first focal point to $N$ along
$\gamma_{u_{n}}$

  2. (ii)

there exist $v_{n}\in S(\nu)$, $v_{n}\neq u_{n}$ with
$\gamma_{u_{n}}\left(\rho(u_{n})\right)=\gamma_{v_{n}}\left(\rho(u_{n})\right)$.

If (i) is true for infinitely many $n$, then choose infinitely many unit
vectors $\\{w_{n}\\}$ which belong to the kernel
$\ker\left(D\exp_{\nu}(\rho(u_{n})u_{n})\right)$ and are contained in a
compact subset of $S(\nu)$. Choose a convergent subsequence whose limit $w$ is
contained in $\ker\left(D\exp_{\nu}{(Tu)}\right)$. Since $w\neq 0$, the rank
of $D\exp_{\nu}{(Tu)}$ is less than $\dim M$. Thus, $\gamma_{u}(T)$ is the
first focal point of $N$ along $\gamma_{u}$ and $T=\rho(u)$.

If (ii) is true for infinitely many $n$, then we may assume that $v_{n}\to
v\in S(\nu)$. If $v\neq u$, then 4.2.1 (ii) holds for $T$, whence $T=\rho(u)$.
If $v=u$, we claim that $\gamma_{u}(T)$ is the first focal point of $N$ along
$\gamma_{u}$. If not, then the map $\exp_{\nu}$ is regular at $Tu\in\nu$ and
hence the map

$\Phi:\nu\to M\times M,~{}(p,u)\mapsto(p,\exp_{\nu}(p,u))$

is regular at $Tu$. Therefore, $\Phi$ is a diffeomorphism if restricted to an
open neighbourhood $\tilde{U}$ of $Tu$ in $\nu$. Since $v=u$, which implies
for sufficiently large $n$, $(p_{n},\rho(u_{n})u_{n})$ and
$(p_{n},\rho(u_{n})v_{n})$ belong to $\tilde{U}$ and are different. On the
other hand, by assumption $\Phi(\rho(u_{n})u_{n})=\Phi(\rho(u_{n})v_{n})$,
which is a contradiction. Therefore, $\gamma_{u}(T)$ is the first focal point
and $T=\rho(u)$. ∎

####  4.2.2 Characterization in terms of separating set

Recall that the separating set of $N$, $\mathrm{Se}(N)$, consists of all
points $q\in M$ such that at least two distance minimal geodesics from $N$ to
$q$ exist. If $q\in\mathrm{Se}(N)$ but $q\not\in\mathrm{Cu}(N)$, then we have
Figure 4.2, i.e., $\gamma_{1}$ is an $N$-geodesic beyond $q$ while
$\gamma_{2}$ is another $N$-geodesic for $q$. The triangle inequality applied
to $\gamma_{1}(0)$, $q=\gamma_{1}(l)$ and $\gamma_{2}(l+\tau)$ implies that

$d(\gamma_{2}(l+\tau),N)<l+\tau$

while for $\tau$ small enough $d(\gamma_{2}(l+\tau),N)=l+\tau$ as $\gamma_{2}$
is an $N$-geodesic beyond $q$. This contradiction establishes the well-known
fact $\mathrm{Se}(N)\subseteq\mathrm{Cu}(N)$. In quite a few examples, these
two sets are equal (see 3.1.6 where these two are not same). In the case of
$M=\mathbb{S}^{n}$ with $N=\\{p\\}$, the set $\mathrm{Se}(N)$ consists of
$-p$. There is an infinite family of minimal geodesics joining $p$ to $-p$. An
appropriate choice of a pair of such minimal geodesics would create a loop,
which is permissible in the definition of $\mathrm{Se}(N)$. According to
4.2.1, a cut point is either a first focal point of $N$ along a geodesic or it
is a separating point. We will now prove our one of the observations in the
last chapter, that cut locus of a submanifold $N$ is the closure of separating
set of $N$.

######  Theorem 4.2.2.

Let $\mathrm{Cu}(N)$ be the cut locus of a compact submanifold $N$ of a
complete Riemannian manifold $M$. The subset $\mathrm{Se}(N)$ of
$\mathrm{Cu}(N)$ is dense in $\mathrm{Cu}(N).$

###### Proof.

Let $q\in\mathrm{Cu}(N)$ but not in $\mathrm{Se}(N)$. Choose an $N$-geodesic
$\gamma$ joining $N$ to $q$ such that any extension of $\gamma$ is not an
$N$-geodesic. This geodesic $\gamma$ is unique as $q\notin\mathrm{Se}(N)$. We
may write $\gamma(t)=\exp_{\nu}(tx)$, where $\gamma(0)=p\in N$ and
$\gamma^{\prime}(0)=x_{0}\in S(\nu_{p})$. It follows from the definition of
$\rho$ that $q=\exp_{\nu}\left(\rho(x_{0})x_{0}\right)$. We need to show that
every neighborhood of $q$ in $\mathrm{Cu}(N)$ must intersect $\mathrm{Se}(N)$.
Suppose it is false. Let $\delta>0$ and consider
$\overline{B(x_{0},\delta)}~{}$, the closed ball with center $x_{0}$ and
radius $\delta$. Define the cone

$\mathrm{Co}(x_{0},\delta)~{}\\!\vcentcolon=\left\\{tx:0\leq t\leq
1,~{}x\in\overline{B(x_{0},\delta)}~{}\cap S(\nu)\right\\}.$

Since $\overline{B(x_{0},\delta)}~{}\cap S(\nu)$ is homeomorphic to a closed
$(n-1)$-ball for sufficiently small $\delta$, the cone will be homeomorphic to
a closed Euclidean $n$-ball.

Figure 4.5: $\mathrm{Co}(x_{0},\delta)~{}$

Similarly, define another cone

$\mathrm{Co}^{\star}(x_{0},\delta)~{}\\!\vcentcolon=\big{\\{}\rho\left(x/\left\|x\right\|\right)x\,|\,x\in\mathrm{Co}(x_{0},\delta)~{}\\!,~{}x\neq
0\big{\\}}\cup\\{0\\}.$

Note that $\rho(x_{0})$ is finite. As $\rho$ is continuous, due to 4.2.2, for
sufficiently small $\delta$ the term $\rho\left(x/\left\|x\right\|\right)$ is
still finite, whence $\mathrm{Co}^{\star}(x_{0},\delta)~{}$is well defined. We
claim that $\mathrm{Co}^{\star}(x_{0},\delta)~{}$is also homeomorphic to a
closed Euclidean $(n-k)$-ball. Indeed, a non-zero
$x\in\mathrm{Co}(x_{0},\delta)~{}\\!$ implies $x=\lambda\hat{x}$, for some
$\lambda\in(0,1]$ and $\hat{x}\in\overline{B(x_{0},\delta)}~{}\cap S(\nu)$.
Since $\rho(\hat{x})x=\lambda\rho(\hat{x})\hat{x}$, it follows that
$\mathrm{Co}^{\star}(x_{0},\delta)~{}$is the cone of the set

$\\{\rho(\hat{x})\hat{x}\,|\,\hat{x}\in\overline{B(x_{0},\delta)}~{}\cap
S(\nu)\\},$

which is homeomorphic to $\overline{B(x_{0},\delta)}~{}\cap S(\nu)$. Now we
have a dichotomy:

  1. (a)

For a fixed small $\delta>0$, the restriction of $\exp_{\nu}$ to
$\mathrm{Co}^{\star}(x_{0},\delta)~{}$is a homeomorphism to its image because
it is injective, or

  2. (b)

For any $\delta>0$, the restriction of $\exp_{\nu}$ to
$\mathrm{Co}^{\star}(x_{0},\delta)~{}$is not injective.

If (b) holds, choose $v_{n}\neq
w_{n}\in\mathrm{Co}^{\star}(x_{0},\frac{1}{n})$ such that these map to $q_{n}$
under $\exp_{\nu}$. Thus, $q_{n}\in\mathrm{Se}(N)$ and compactness of $S(\nu)$
ensures that $q_{n}$ converges to $q$. If (a) holds, then let
$B(q,\varepsilon)$ denote the open ball in $M$ centered at $q$ with radius
$\varepsilon>0$. We claim that it intersects the complement of
$\exp_{\nu}(\mathrm{Co}^{\star}(x_{0},\delta)~{}\\!\\!)$ in $M$. But it is
true as $\rho(x_{0})x_{0}$ lies on the boundary of
$\mathrm{Co}^{\star}(x_{0},\delta)~{}$and hence it has a neighborhood in
$\mathrm{Co}^{\star}(x_{0},\delta)~{}$which is homeomorphic to a closed
$n$-dimensional Euclidean half plane. Since $\exp_{\nu}$ restricted to
$\mathrm{Co}^{\star}(x_{0},\delta)~{}$was a homeomorphism, the open ball
$B(q,\varepsilon)$ must intersect the points outside the image of
$\exp_{\nu}(\mathrm{Co}^{\star}(x_{0},\delta)~{}\\!\\!)$.

Now take $\varepsilon=\frac{1}{n}$. For each $n$, there exists $q_{n}\in
B\left(q,\frac{1}{n}\right)$ such that
$q_{n}\notin\exp_{\nu}(\mathrm{Co}^{\star}(x_{0},\delta)~{}\\!\\!)$. Since $M$
is complete, for each point $q_{n}$ let $\gamma_{n}$ be a $N$-geodesics
joining $p_{n}\in N$ to $q_{n}$. We may invoke the following result from
Buseman’s book [Busemann, 1955, Theorem 5.16, page 24]. Let
$\left\\{\gamma_{n}\right\\}$ be a sequence of rectifiable curves in a
finitely compact set $X$ and the lengths $\ell(\gamma_{n})$ are bounded. If
the initial points $p_{n}$ of $\gamma_{n}$ forms a bounded set, then
$\left\\{\gamma_{n}\right\\}$ contains a subsequence $\gamma_{n_{k}}$ which
converges uniformly to a rectifiable curve $\tilde{\gamma}$ in $X$ and

$\ell(\tilde{\gamma})\leq\lim\inf\ell\left(\gamma_{n_{k}}\right)$

Since $\\{p_{n}\\}$ lie in the compact set $N$, we obtain a rectifiable curve
$\tilde{\gamma}$ such that

$\ell(\tilde{\gamma})\leq\lim\inf\ell\left(\gamma_{n_{k}}\right)=\lim_{k}\ell(\gamma_{n_{k}})=\lim_{k}d(q_{n_{k}},N)=d(q,N).$

Thus, $\tilde{\gamma}$ is actually an $N$-geodesic joining
$p^{\prime}=\lim_{k}p_{n_{k}}$ to $q$ and the unit tangent vectors
$x_{n_{k}}=\gamma_{n_{k}}^{\prime}(0)$ at $p_{n_{k}}$ converges to the unit
tangent vector $\tilde{x}=\tilde{\gamma}^{\prime}(0)$ at $p^{\prime}$. Since
$x_{0}$ is an interior point of the set $\overline{B(x_{0},\delta)}~{}\cap
S(\nu)$, any sequence in $S(\nu)$ converging to $x_{0}$ must eventually lie in
$\mathrm{Co}(x_{0},\delta)~{}$. According to our choice,
$q_{n_{k}}\notin\exp_{\nu}(\mathrm{Co}^{\star}(x_{0},\delta)~{}\\!\\!)$ and
$x_{n_{k}}$ all lie outside of $\mathrm{Co}(x_{0},\delta)~{}$. Hence
$x_{0}\neq\tilde{x}$ and $\gamma\neq\tilde{\gamma}$. Thus, there are two
distinct $N$-geodesics $\gamma$ and $\tilde{\gamma}$ joining $N$ to $q$, a
contradiction to $q\notin\mathrm{Se}(N)$. This completes the proof. ∎

###  4.3 Topological properties

In this section we will study the structure of the cut locus and the relation
of cut locus to the Thom space. We will also see various applications of this
relation.

####  4.3.1 Structure of the cut locus

On a complete Riemannian manifold it is very difficult to analyze the
structure of cut locus of a point or a submanifold. The main problem is that
cut locus is not $C^{1}$-smooth. For example, the authors in [Gluck and
Singer, 1978] showed that cut locus of a point is not always triangulable.
Regarding the question of cut loci being triangulable, we recall the result
[Buchner, 1977] that the cut locus (of a point) of a real analytic Riemannian
manifold (of dimension $d$) is a simplicial complex of dimension at most
$d-1$. It follows, without much changes, that the result holds for cut loci of
submanifolds as well. Hence, we attribute the following result to Buchner.

######  Theorem 4.3.1 (Buchner 1977).

Let $N$ be an analytic submanifold of a real analytic manifold $M$. If $M$ is
of dimension $d$, then the cut locus $\mathrm{Cu}(N)$ is a simplicial complex
of dimension at most $d-1$.

The obvious modifications to the proof by Buchner are the following:

  1. (i)

Choose $\varepsilon$ to be such that there is a unique geodesic from $p$ to
$q$ if $d(p,q)<\varepsilon$ and if $d(N,q)<\varepsilon$, then there is a
unique $N$-geodesic to $q$;

  2. (ii)

Consider the set $\Omega_{N}(t_{0},t_{1},\ldots,t_{k})$, the space of
piecewise broken geodesics starting at $N$, and define
$\Omega_{N}(t_{0},t_{1},\ldots,t_{k})^{s}$ analogously;

  3. (iii)

The map

$\Omega_{N}(t_{0},t_{1},\ldots,t_{k})^{s}\to N\times M\times\cdots\times
M,\,\,\omega\mapsto(\omega(t_{0}),\omega(t_{1}),\ldots,\omega(t_{k}))$

determines an analytic structure on
$\Omega_{N}(t_{0},t_{1},\ldots,t_{k})^{s}$.

The remainder of the proof works essentially verbatim.

######  Remark 4.3.1.

As we have seen in 3.1.5, the dimension of the cut locus of a $k$-dimensional
submanifold can be of dimension $d-k-1$. However, generically, we may not
expect this to be true. In fact, for real analytic knots (except the unknot)
in $\mathbb{S}^{3}$, it is always the case that the cut locus cannot be
homotopic to a (connected) $1$-dimensional simplicial complex (cf 4.3.3).

####  4.3.2 Thom space via cut locus

Recall that the Thom space $\textup{Th}(E)$ of a real vector bundle $E\to B$
of rank $k$ is $D(E)/S(E)$, where it is understood that we have chosen a
Euclidean metric on $E$. If $B$ is compact, then the Thom space
$\textup{Th}(E)$ is the one-point compactification of $E$. In general, we
compactify the fibres and then collapse the section at infinity to a point to
obtain $\textup{Th}(E)$. Thus, Thom spaces obtained via two different metrics
are homeomorphic. We start with a similar exponential map which is obtained by
the $\rho$ map, and we named it rescaled exponential map.

######  Definition 4.3.1 (Rescaled Exponential).

The rescaled exponential or $\rho$-exponential map is defined to be

$\widetilde{\textup{exp}}:D(\nu)\to
M,\,\,(p,v)\mapsto\left\\{\begin{array}[]{rl}\textup{exp}_{p}(\rho(\hat{v})v)&\textup{if
$v=\|v\|\hat{v}\neq 0$}\\\ p&\textup{if $v=0$.}\end{array}\right.$

We are now ready to prove the main result of this section.

######  Theorem 4.3.2.

Let $N$ be an embedded submanifold inside a closed, connected Riemannian
manifold $M$. If $\nu$ denotes the normal bundle of $N$ in $M$, then there is
a homeomorphism

$\widetilde{\textup{exp}}:D(\nu)/S(\nu)\stackrel{{\scriptstyle\cong}}{{\longrightarrow}}M/\mathrm{Cu}(N).$

###### Proof.

It follows from 4.2.2 that the rescaled exponential is continuous. Moreover,
$\widetilde{\textup{exp}}$ is surjective and
$\widetilde{\textup{exp}}(S(\nu))=\mathrm{Cu}(N)$. If there exists
$(p,v)\neq(q,w)\in D(\nu)$ such that

$\widetilde{\textup{exp}}(p,v)=\widetilde{\textup{exp}}(q,w)=p^{\prime},$

then $d(p^{\prime},N)$ can be computed in two ways to obtain

$d(p^{\prime},N)=\rho(\hat{v})\|v\|=\rho(\hat{w})\|w\|.$

Thus, $T=d(p^{\prime},N)$ is a number such that $\gamma_{v}:[0,T]$ is an
$N$-geodesic and $\gamma_{v}(T)=\gamma_{w}(T)=p^{\prime}$. By 4.2.1, we
conclude that $T=\rho(\hat{v})=\rho(\hat{w})$, whence $\|v\|=\|w\|=1$.
Therefore, $\widetilde{\textup{exp}}$ is injective on the interior of
$D(\nu)$.
As $\mathrm{Cu}(N)$ is closed and $M$ is a compact metric space, the quotient
space $M/\mathrm{Cu}(N)$ is Hausdorff. As the quotient $D(\nu)/S(\nu)$ is
compact, standard topological arguments imply the map induced by the rescaled
exponential is a homeomorphism. ∎

We will now revisit a basic property of Thom space via its connection to the
cut locus. It can be seen that

$\mathrm{Cu}(N_{1}\times N_{2})=(\mathrm{Cu}(N_{1})\times
M_{2})\cup(M_{1}\times\mathrm{Cu}(N_{2}))$ (4.5)

for an embedding $N_{1}\times N_{2}$ inside $M_{1}\times M_{2}$. If $\nu_{j}$
is the normal bundle of $N_{j}$ inside $M_{j}$, then 4.3.2 along with (4.5)
implies that

$\displaystyle\textup{Th}(\nu_{1}\oplus\nu_{2})$
$\displaystyle\cong\frac{M_{1}\times
M_{2}}{(M_{1}\times\mathrm{Cu}(N_{2}))\cup(\mathrm{Cu}(N_{1})\times M_{2})}$
$\displaystyle\cong\frac{M_{1}/\mathrm{Cu}(N_{1})\times
M_{2}/\mathrm{Cu}(N_{2})}{M_{1}/\mathrm{Cu}(N_{1})\vee
M_{2}/\mathrm{Cu}(N_{2})}$
$\displaystyle\cong\textup{Th}(\nu_{1})\wedge\textup{Th}(\nu_{2}).$

Let $N=N_{1}\sqcup N_{2}$ be a disjoint union of connected manifolds of the
same dimension. If $N\hookrightarrow M$, then let $\nu_{j}$ denote the normal
bundle of $N_{j}$ in $M$. If $\nu$ is the normal bundle of $N$ in $M$, then

$\textup{Th}(\nu)\cong\textup{Th}(\nu_{1})\vee\textup{Th}(\nu_{2}).$ (4.6)

This implies that

$M/\mathrm{Cu}(N)\cong M/\mathrm{Cu}(N_{1})\vee M/\mathrm{Cu}(N_{2}).$

######  Example 4.3.1.

Consider the two circles

$N_{1}=\\{(\cos t,\sin t,0,0)\,|\,t\in\mathbb{R}\\},\,\,N_{2}=\\{(0,0,\cos
t,\sin t)\,|\,t\in\mathbb{R}\\}$

in $\mathbb{S}^{3}$. The link $N:=N_{1}\sqcup N_{2}$ has linking number $1$.
We claim that the cut locus

$\mathrm{Cu}(N)=\left\\{\frac{1}{\sqrt{2}}(\cos s,\sin s,\cos t,\sin
t)\,|\,s,t\in\mathbb{R}\right\\}$

is a torus. We will prove the claim by showing that the above set is
separating set of $N$. As it is closed, the claim will follow by using 4.2.2.
Let $P=(a,b,0,0)\in N$ with $a^{2}+b^{2}=1$. Note that

$T_{P}\mathbb{S}^{3}\cong
T_{P}N\oplus\left(T_{P}N\right)^{\perp}\cong\operatorname{span}\\{(-b,a,0,0)\\}\oplus\operatorname{span}\\{\mathbf{e}_{3},\mathbf{e}_{4}\\},$

where $\mathbf{e}_{3}=(0,0,1,0)$ and $\mathbf{e}_{4}=(0,0,0,1)$. We take any
unit vector at $P$ which is perpendicular to $T_{P}N$, say
$\mathbf{v}=(0,0,\cos\theta,\sin\theta)$. An $N$-geodesic starting at $P$ in
the direction of $\mathbf{v}$ will be

$\gamma(t)=P\cos t+\mathbf{v}\sin t,~{}=(a\cos t,b\cos t,\cos\theta\sin
t,\sin\theta\sin t),~{}~{}0\leq t\leq\pi.$

We have

$d(\gamma(t),N)=\inf_{X\in N}d(\gamma(t),X)=\min\left\\{\inf_{X\in
N_{1}}d(\gamma(t),X),\inf_{X\in N_{2}}d(\gamma(t),X),\right\\}.$

Look at Figure 4.6 and note that

$d(X,\gamma(t))=\cos^{-1}(X\cdot\gamma(t)).$

$X$
---
$\gamma(t)$
---
$\theta$
---
Figure 4.6: Distance of $X$ to $\gamma(t)$

Therefore, the problem of finding the distance of $N$ to $\gamma(t)$ is
equivalent to maximizing the dot product $X\cdot\gamma(t)$. Let $X\in N_{1}$
and $X=(x,y,0,0),~{}x^{2}+y^{2}=1$. Then

$X\cdot\gamma(t)=ax\cos t+by\cos t.$

Maximizing the above such that $x^{2}+y^{2}=1$ by the method of Lagrange
multiplier, we have

$x=\dfrac{a\cos t}{|\cos t|},\text{ and }y=\dfrac{b\cos t}{|\cos t|}.$

Note that the above expression is well-defined, as if $\cos t=0$, then
$X\cdot\gamma(t)=0$. The maximum value of $X\cdot\gamma(t)$ will be $|\cos t|$
and this is achieved by only one point of $N_{1}$. Similarly, if we maximize
the dot product over $N_{2}$, we get the maximum value $|\sin t|$, which is
also obtained by a single maxima. Thus, $\gamma(t)$ will be a separating point
if and only if

$|\cos t|=|\sin t|\implies t=\dfrac{\pi}{4},\dfrac{3\pi}{4}.$

Therefore, the separating points will be

$\left\\{\frac{1}{\sqrt{2}}(\cos s,\sin
s,\cos\theta,\sin\theta):s,\theta\in\mathbb{R}\right\\}.$

Note that $\mathrm{Cu}(N_{1})=N_{2}$ and vice-versa as well as

$\mathbb{S}^{3}/\mathrm{Cu}(N_{j})\cong(S^{1}\times
S^{2})/(S^{1}\times\infty)$

where $S^{1}\times S^{2}$ is the fibrewise compatification of the normal
bundle of $N_{j}$. We conclude that

$\mathbb{S}^{3}/\mathrm{Cu}(N)\cong\Big{(}\frac{S^{1}\times
S^{2}}{S^{1}\times\infty}\Big{)}\vee\Big{(}\frac{S^{1}\times
S^{2}}{S^{1}\times\infty}\Big{)}.$

There are some topological similarities between $\mathrm{Cu}(N)$ and $M-N$. We
recall that a topological pair $(X,A)$ is called a good pair if $A$ is closed
in $X$ and there is an open subset $U\subseteq X$ with $A\subseteq U$ such
that $A$ is a strong deformation retract in $U$.

######  Lemma 4.3.1.

The cut locus $\mathrm{Cu}(N)$ is a strong deformation retract of $M-N$. In
particular, $(M,\mathrm{Cu}(N))$ is a good pair and the number of path
components of $\mathrm{Cu}(N)$ equals that of $M-N$.

###### Proof.

Consider the map $H:(M-N)\times[0,1]\to M-N$ defined via the normal
exponential map

$\displaystyle
H(q,t)=\begin{cases}\exp_{\nu}\left[\left\\{t\cdot\rho\left(\frac{\exp_{\nu}^{-1}(q)}{\left\|\exp_{\nu}^{-1}(q)\right\|}\right)+(1-t)\left\|\exp_{\nu}^{-1}(q)\right\|\right\\}\frac{\exp_{\nu}^{-1}(q)}{\left\|\exp_{\nu}^{-1}(q)\right\|}\right]&\text{
if }q\notin\mathrm{Cu}(N)\\\\[4.30554pt] q&\text{ if
}q\in\mathrm{Cu}(N).\end{cases}$

If $q\in M-(\mathrm{Cu}(N)\cup N)$, then let $\gamma$ be the unique
$N$-geodesic joining $N$ to $q$. The path $H(q,t)$ is the image of this
geodesic from $q$ to the first cut point along $\gamma$. The continuity of
$\rho$ implies that $H$ is continuous. It also satisfies $H(q,0)=q$ and
$H(q,1)\in\mathrm{Cu}(N)$. The claims about good pair and path components are
clear. ∎

######  Remark 4.3.2.

The above lemma, in particular, shows that the homotopy type of the cut locus
of a submanifold is independent of the choice of the Riemannian metric.

######  Corollary 4.3.1.

If two embeddings $f,g:N\to M$ are ambient isotopic, then $\mathrm{Cu}(f(N))$
and $\mathrm{Cu}(g(N))$ are homotopy equivalent.

###### Proof.

The hypothesis implies that there is a diffeomorphism $\varphi:M\to M$ such
that $\varphi(f(N))=g(N)$. Thus, $M-\mathrm{Cu}(f(N))$ is homeomorphic to
$M-\mathrm{Cu}(g(N))$ and the claim follows from the lemma above. Note that in
the smooth category, the notion of isotopic and ambient isotopic are
equivalent (refer to §8.1 of the book [Hirsch, 1976]). Thus, the same
conclusion holds if we assume that the embeddings are isotopic. ∎

######  Remark 4.3.3.

Without the assumption of $M$ being closed, the above result fails to be true.
One may consider $M=S^{1}\times\mathbb{R}$ with the natural product metric and
$N=S^{1}$. In fact, the universal cover of $M$ is $\mathbb{R}\times\mathbb{R}$
while that of $N$ is $\mathbb{R}$. If we choose a periodic curve in
$\mathbb{R}^{2}$ which is isotopic to the $x$-axis and has non-empty cut locus
in $\mathbb{R}^{2}$, then we may pass via the covering map to obtain an
embedding $g$ of $N$ isotopic to the embedding $f$ identifying $N$ with
$S^{1}\times\\{0\\}$. For this pair, $\mathrm{Cu}(f(N))=\varnothing$ while
$\mathrm{Cu}(g(N))\neq\varnothing$.

Several other identifications between topological invariants can be explored.
For instance, if $\iota:N^{k}\hookrightarrow M^{d}$ is as before such that
$M-N$ is path connected, then

$\iota_{\ast}:\pi_{j}(\mathrm{Cu}(N))\stackrel{{\scriptstyle\cong}}{{\longrightarrow}}\pi_{j}(M)$
(4.7)

if $0\leq j\leq d-k-2$ while $\iota_{\ast}$ is a surjection for $j=d-k-1$. The
proof of this relies on a general position argument, i.e., being able to find
a homotopy of the sphere that avoids $N$, followed by Lemma 4.3.1.
Surjectivity of $\iota_{\ast}$ if $j\leq d-k-1$ is imposed by the requirement
that a sphere $S^{j}$ in general position must not intersect $N^{k}$.
Injectivity of $\iota$ for $j\leq d-k-2$ is imposed by the condition that a
homotopy $S^{j}\times[0,1]$ in general position must not intersect $N^{k}$.
This observation (4.7) generalizes a result in [Sakai, 1996, Proposition 4.5
(1)].

The inclusion $i:\mathrm{Cu}(N)\hookrightarrow M$ induces a long exact
sequence in homology

$\cdots\to H_{j}(\mathrm{Cu}(N))\stackrel{{\scriptstyle
i_{\ast}}}{{\longrightarrow}}H_{j}(M)\to
H_{j}(M,\mathrm{Cu}(N))\stackrel{{\scriptstyle\partial}}{{\longrightarrow}}H_{j-1}(\mathrm{Cu}(N))\to\cdots$

As $(M,\mathrm{Cu}(N))$ is a good pair (cf Lemma 4.3.1), we replace the
relative homology of $(M,\mathrm{Cu}(N))$ with reduced homology of
$M/\mathrm{Cu}(N)\cong\textup{Th}(\nu)$. This results in the following long
exact sequence

$\cdots\to H_{j}(\mathrm{Cu}(N))\stackrel{{\scriptstyle
i_{\ast}}}{{\longrightarrow}}H_{j}(M)\stackrel{{\scriptstyle
q}}{{\longrightarrow}}\widetilde{H}_{j}(\textup{Th}(\nu))\stackrel{{\scriptstyle\partial}}{{\longrightarrow}}H_{j-1}(\mathrm{Cu}(N))\to\cdots$
(4.8)

If $N=\\{p\\}$ is a point, then $\textup{Th}(\nu)=S^{d}$ and (4.8) imply
isomorphisms

$i_{\ast}:H_{j}(\textup{Cu}(p))\stackrel{{\scriptstyle\cong}}{{\longrightarrow}}H_{j}(M),\,\,i^{\ast}:H^{j}(M)\stackrel{{\scriptstyle\cong}}{{\longrightarrow}}H^{j}(\textup{Cu}(p))$

for $j\neq d,d-1$ (cf [Sakai, 1996, Proposition 4.5 (2)]).

######  Remark 4.3.4.

The long exact sequence (4.8) can be interpreted as the dual to the long exact
sequence in cohomology of the pair $(M,N)$. If $N=N_{1}\sqcup\cdots\sqcup
N_{l}$ is a disjoint union of submanifolds of dimension $k_{1},\ldots,k_{l}$
respectively, then the Thom isomorphism implies that

$\widetilde{H}_{j}(\textup{Th}(\nu))\cong\widetilde{H}_{j}(\textup{Th}(\nu_{1}))\oplus\cdots\oplus\widetilde{H}_{j}(\textup{Th}(\nu_{l}))\cong
H_{j-(d-k_{1})}(N_{1})\oplus\cdots\oplus H_{j-(d-k_{l})}(N_{l}),$

where $\nu_{j}$ is the normal bundle of $N_{j}$. Applying Poincaré duality to
each $N_{j}$, we obtain isomorphisms

$\widetilde{H}_{j}(\textup{Th}(\nu))\cong\oplus_{i=1}^{l}H^{d-j}(N_{i})=H^{d-j}(N).$

Poincaré-Lefschetz duality applied to the pair $(M,N)$ provides isomorphisms

$\check{H}^{j}(M,N)\cong H_{d-j}(M-N).$ (4.9)

As $M$ and $N$ are triangulable, $\check{\textup{C}}$ech cohomology may be
replaced by singular cohomology. Since $M-N$ deforms to $\mathrm{Cu}(N)$ by
Lemma 4.3.1, we have isomorphisms

$H^{j}(M,N)\cong H_{d-j}(\mathrm{Cu}(N)).$ (4.10)

Combining all these isomorphisms, we obtain the long exact sequence in
cohomology for $(M,N)$ from (4.8).

######  Lemma 4.3.2.

Let $N$ be a closed submanifold of $M$ with $l$ components. If $M$ has
dimension $d$, then $H_{d-1}(\mathrm{Cu}(N))$ is free abelian of rank $l-1$
and $H_{d-j}(\mathrm{Cu}(N))\cong H^{j}(M)$ if $j-2\geq k$, where $k$ is the
maximum of the dimension of the components of $N$.

###### Proof.

It follows from (4.9) that

$H_{d-1}(\mathrm{Cu}(N))\cong H^{1}(M,N).$

Consider the long exact sequence associated to the pair $(M,N)$

$0\to H^{0}(M,N)\to H^{0}(M)\stackrel{{\scriptstyle
i^{\ast}}}{{\rightarrow}}H^{0}(N)\to H^{1}(M,N)\to H^{1}(M)\to
H^{1}(N)\to\cdots$

If $N$ has $l$ components, i.e., $N=N_{1}\sqcup\cdots\sqcup N_{l}$ where
$N_{j}$ has dimension $k_{j}$, then $H^{1}(M,N)$ is torsion-free. This follows
from the fact that $i^{\ast}(1)=(1,\ldots,1)$ and $H^{1}(M)$ is free abelian.

######  Remark 4.3.5.

In the above Lemma, if in particular, $H^{1}(M)=0$, then
$H_{d-1}(\mathrm{Cu}(N))\cong\mathbb{Z}^{l-1}$.

The long exact sequence for the pair $(M,N)$ imply that there are isomorphisms

$H_{d-j}(\mathrm{Cu}(N))\cong
H^{j}(M,N)\stackrel{{\scriptstyle\cong}}{{\longrightarrow}}H^{j}(M)$ (4.11)

if $j\geq k+2$, where $k=\max\\{k_{1},\ldots,k_{l}\\}$. ∎

######  Remark 4.3.6.

Cut locus can be very hard to compute. For a general space, we have the notion
of topological dimension. This notion coincides with the usual notion if the
space is triangulable. However, in [Barratt and Milnor, 1962], the authors
proved that the singular homology of a space may be non-zero beyond its
topological dimension. Čech (co)homology is better equipped to detect
topological dimension and is the reason why one may prefer it over singular
homology due to the generic fractal like nature of cut loci (see the remarks
following C in Chapter 1). Although the topological dimension of
$\mathrm{Cu}(N)$ is at most $d-1$, it is not apparent that
$H_{d-1}(\mathrm{Cu}(N))$ is a free abelian group.

There are several applications of this discussion.

######  Theorem 4.3.3.

Let $N$ be a smooth homology $k$-sphere embedded in a Riemannian manifold
homeomorphic to $S^{d}$. If $d\geq k+3$, then the cut locus $\mathrm{Cu}(N)$
is homotopy equivalent to $S^{d-k-1}$.

###### Proof.

As $N$ has codimension at least $3$, its complement is path-connected. It
follows from (4.7) and Lemma 4.3.1 that $M-N$ is $(d-k-2)$-connected. In
particular, $M-N$ is simply-connected and by Hurewicz isomorphism,
$H_{j}(M-N)=0$ if $j\leq d-k-2$. Note that $H_{d}(M-N)=0$ as $M-N$ is a non-
compact manifold of dimension $d$.

If $k>0$, then by Lemma 4.3.2, $H_{d-1}(M-N)=0$. Moreover, by Poincaré-
Lefschetz duality (4.9), we infer that the only non-zero higher homology of
$M-N$ is $H_{d-k-1}(M-N)\cong\mathbb{Z}$. By Hurewicz Theorem there is an
isomorphism, $\pi_{d-k-1}(M-N)\cong\mathbb{Z}$. Let

$\alpha:S^{d-k-1}\to M-N$

be a generator. The map $\alpha_{\ast}$ induces an isomorphism on all homology
groups between two simply-connected CW complexes. It follows from Whitehead’s
Theorem that $\alpha$ is a homotopy equivalence. Using Lemma 4.3.1, we obtain
our homotopy equivalence
$H_{1}\mathbin{\mathchoice{\vbox{\hbox{$\scriptstyle\circ$}}}{\vbox{\hbox{$\scriptstyle\circ$}}}{\vbox{\hbox{$\scriptscriptstyle\circ$}}}{\vbox{\hbox{$\scriptscriptstyle\circ$}}}}\alpha:S^{d-k-1}\to\mathrm{Cu}(N)$.

If $k=0$, then by Lemma 4.3.2, $H_{d-1}(M-N)\cong\mathbb{Z}$. Arguments
similar to the $k>0$ case now applies to obtain a homotopy equivalence with
$S^{d-1}$. ∎

The above result is foreshadowed by 3.1.5 where we showed that the cut locus
of $N=\mathbb{S}^{k}_{i}$ inside $M=\mathbb{S}^{d}$ is
$\mathbb{S}^{d-k-1}_{l}$. It also differs from Poincaré-Lefschetz duality in
that we are able to detect the exact homotopy type of the cut locus. In fact,
when $M$ and $N$ are real analytic and the embedding is also real analytic,
then by 4.3.1 we infer that $\mathrm{Cu}(N)$ is a simplicial complex of
dimension at most $d-1$. Towards this direction, 4.3.3 can be pushed further.

######  Proposition 4.3.1.

Let $N$ be a real analytic homology $k$-sphere embedded in a real analytic
homology $d$-sphere $M$. If $d\geq k+3$, then the cut locus $\mathrm{Cu}(N)$
is a simplicial complex of dimension at most $(d-1)$, having the homology of
$(d-k-1)$-sphere with fundamental group isomorphic to that of $M$.

The proof of this is a combination of ideas used in the proof of 4.3.3. The
homotopy type cannot be deduced here due to the presence of a non-trivial
fundamental group. An intriguing example can be obtained by combining 4.3.1
and Poincaré homology sphere.

######  Example 4.3.2 (Cut locus of $0$-sphere in Poincaré sphere).

Let $\tilde{I}$ be the binary icosahedral group. It is a double cover of $I$,
the icosahedral group, and can be realized a subgroup of $SU(2)$. It is known
that $H_{1}(\tilde{I};\mathbb{Z})=H_{2}(\tilde{I};\mathbb{Z})=0$, i.e., it is
perfect and the second homology of the classifying space $B\tilde{I}$ is zero.
A presentation of $\tilde{I}$ is given by

$\tilde{I}=\langle s,t\,|\,(st)^{2}=s^{3}=t^{5}\rangle.$

In fact, if we construct a cell complex $X$ of dimension $2$ using the
presentation above, then $X$ has one $0$-cell, two $1$-cells and two
$2$-cells. The cellular chain complex, as computed from the presentation, is
given by

$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathbb{Z}^{2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\left(\begin{smallmatrix}-1&2\\\
3&-5\end{smallmatrix}\right)}$$\textstyle{\mathbb{Z}^{2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{0}$$\textstyle{\mathbb{Z}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0}$

Therefore, $H_{1}(X)=H_{2}(X)=0$ while $\pi_{1}(X)=\tilde{I}$.

In contrast, consider the cut locus $C$ of the $0$-sphere in
$SU(2)/\tilde{I}$, the Poincaré homology sphere. As $SU(2)$ is real analytic,
so is the homology sphere. By 4.3.1, $C$ is a finite, connected simplicial
complex of dimension $2$ such that $\pi_{1}(C)\cong\tilde{I}$ and
$H_{\bullet}(C;\mathbb{Z})\cong H_{\bullet}(S^{2};\mathbb{Z})$. The existence
of this space is interesting for the following reason: although $X\vee S^{2}$
has the same topological invariants as $C$, we are unable to determine whether
$X\vee S^{2}$ is homotopy equivalent to $C$.

In the codimension two case, we have two results.

######  Theorem 4.3.4.

Let $\Sigma$ be a closed, orientable, real analytic surface of genus $g$ and
$N$ a non-empty, finite subset. Then $\mathrm{Cu}(N)$ is a connected graph,
homotopy equivalent to a wedge product of $|N|+2g-1$ circles.

###### Proof.

As $\Sigma-N$ is connected, Lemma 4.3.1 implies that $\mathrm{Cu}(N)$ is
connected. It follows from 4.3.1 that $\mathrm{Cu}(N)$ is a finite
$1$-dimensional simplicial complex, i.e., a finite graph. In this case,
$\textup{Th}(\nu)$ is a wedge product of $|N|$ copies of $S^{2}$ (cf (4.6)).
We consider (4.8) with $j=2$:

$0\stackrel{{\scriptstyle
i_{\ast}}}{{\longrightarrow}}\mathbb{Z}\stackrel{{\scriptstyle
q}}{{\longrightarrow}}\widetilde{H}_{2}(\vee_{|N|}S^{2})\stackrel{{\scriptstyle\partial}}{{\longrightarrow}}H_{1}(\mathrm{Cu}(N))\stackrel{{\scriptstyle
i_{\ast}}}{{\longrightarrow}}H_{1}(\Sigma)\to 0$

Note that $H_{d-1}(\Sigma)$ is torsion-free, whence all the groups appearing
in the long exact sequence are free abelian groups. This implies that

$\dim_{\mathbb{Z}}H_{1}(\mathrm{Cu}(N))=2g+|N|-1.$

As $\mathrm{Cu}(N)$ is connected finite graph, collapsing a maximal tree $T$
results in a quotient space $\mathrm{Cu}(N)/T$ which is homotopic to
$\mathrm{Cu}(N)$ as well being a wedge product of $|N|+2g-1$ circles. ∎

######  Remark 4.3.7.

The authors in [Itoh and Vîlcu, 2015] proved that every finite, connected
graph can be realized as the cut locus (of a point) of some surface. There
remains the question of orientability of the surface. As noted in the proof of
4.3.4, if the surface is orientable and $|N|=1$, then the graph has an even
number of generating cycles. If $\Sigma$ is non-orientable, then
$\Sigma\cong(\mathbb{RP}^{2})^{\\#k}$ has non-orientable genus $k$ and the
oriented double cover of $\Sigma$ has genus $g=k-1$. Recall that
$H_{1}(\Sigma)\cong\mathbb{Z}^{k-1}\oplus\mathbb{Z}_{2}$ and
$H_{2}(\Sigma)=0$. Looking at (4.8) with $j=2$ we obtain

$0\to\mathbb{Z}\to
H_{1}(\mathrm{Cu}(p))\to\mathbb{Z}^{k-1}\oplus\mathbb{Z}_{2}\to 0.$

Thus, $H_{1}(\mathrm{Cu}(p))\cong\mathbb{Z}^{k}$ as homology of graphs are
free abelian. Let $B_{\varepsilon}(\mathrm{Cu}(p))$ denote the
$\varepsilon$-neighbourhood of $\mathrm{Cu}(p)$ in $\Sigma$. For $\varepsilon$
sufficiently small, this is a surface such that
$\overline{B_{\varepsilon}(\mathrm{Cu}(p))}$ has one boundary component. The
compact surface $B_{\varepsilon}(\mathrm{Cu}(p))$ is reminiscent of ribbon
graphs. The surface $\Sigma$ can be obtained as the connect sum of a disk
centered at $p$ and the closure of $B_{\varepsilon}(\mathrm{Cu}(p))$.
Therefore, non-orientability of $\Sigma$ is equivalent to non-orientability of
$B_{\varepsilon}(\mathrm{Cu}(p))$. A similar observation appears in the
unpublished article [Itoh and Vîlcu, 2011, Theorem 3.7].

######  Example 4.3.3 (Homology spheres of codimension two).

In continuation of 4.3.3, let $N\hookrightarrow S^{k+2}$ be a homology sphere
of dimension $k\geq 1$. Since $N$ has codimension two, $S^{k+2}-N$ is path
connected and so is $\mathrm{Cu}(N)$. We are not assuming that the metric on
$S^{k+2}$ is real analytic. Using (4.10) and the long exact sequence in
cohomology of $(S^{k+2},N)$, we infer that
$H_{1}(\mathrm{Cu}(N))\cong\mathbb{Z}$ and all higher homology groups vanish.
However, the Hurewicz Theorem cannot be used here to establish that
$\pi_{1}(\mathrm{Cu}(N))\cong\mathbb{Z}$.

In particular cases, we may conclude that $\mathrm{Cu}(N)$ is homotopic to a
circle. It was proved in [Plotnick, 1982] that certain homology $3$-spheres
$N$, obtained by a Dehn surgery of type $\frac{1}{2a}$ on a knot, smoothly
embed in $S^{5}$ with complement a homotopy circle. Since $M-N$ deforms to
$\mathrm{Cu}(N)$, it follows that there is a map
$\alpha:S^{1}\to\mathrm{Cu}(N)$ inducing isomorphisms on homotopy and homology
groups.

If $k=1$, then a homology $1$-sphere is just a knot $K$ in $S^{3}$. Since
$S^{3}-K$ deforms to $\mathrm{Cu}(K)$, the fundamental group of the cut locus
is the knot group. Moreover, in the case of real analytic knots in
$\mathbb{S}^{3}$, the cut locus is a finite simplicial complex of dimension at
most $2$ (cf 4.3.1). Except for the unknot, the knot group is never a free
group while the fundamental group of a connected, finite graph is free. This
observation establishes that $\mathrm{Cu}(K)$ is always a $2$-dimensional
simplicial complex, whenever $K$ is a non-trivial (real analytic) knot in
$\mathbb{S}^{3}$.

Finally, we will end this chapter by proving that the complement of cut locus
deforms to the submanifold.

######  Theorem 4.3.5.

Let $N$ be a closed embedded submanifold of a complete Riemannian manifold
$M$. Let $d:M\to\mathbb{R}$ be the distance function with respect to $N$. If
$f=d^{2}$, then its restriction to $M-\mathrm{Cu}(N)$ is a Morse-Bott
function, with $N$ as the critical submanifold. Moreover, the gradient flow of
$f$ deforms $M-\mathrm{Cu}(N)$ to $N$.

###### Proof.

It follows from 4.2.1 the map $\exp_{\nu}^{-1}:M-\left(\mathrm{Cu}(N)\cup
N\right)\to\nu-\\{0\\}$ is an (into) diffeomorphism and
$\operatorname{dist}(N,q)=\left\|\exp^{-1}_{\nu}(q)\right\|$ and hence the
distance function is of class $C^{\infty}$ at $q\in M-\left(\mathrm{Cu}(N)\cup
N\right)$. Using Fermi coordinates (cf 4.1.1), we have seen that the distance
squared function is smooth around $N$ and therefore it is smooth on
$M-\mathrm{Cu}(N)$. By 4.1.1, the Hessian of this function at $N$ is non-
degenerate in the normal direction. It is well-known [Sakai, 1996, Proposition
4.8] that $\|\nabla d(q)\|=1$ if $d$ is differentiable at $q\in M$. Thus, for
$q\in M-(\mathrm{Cu}(N)\cup N)$ we have

$\|\nabla f(q)\|=2d(q)\|\nabla d(q)\|=2d(q).$ (4.12)

Let $\gamma$ be the unique unit speed $N$-geodesic that joins $N$ to $q$,
i.e.,

$\gamma:[0,d(q)]\to
M,\,\,\gamma(0)=p,\,\gamma(d(q))=q,\,\|\gamma^{\prime}\|=1.$

We may write $\nabla f(q)=\lambda\gamma^{\prime}(d(q))+w$, where $w$ is
orthogonal to $\gamma^{\prime}(d(q))$. But

$\left\langle\nabla
f\big{|}_{q},\gamma^{\prime}(d(q))\right\rangle=\frac{d}{dt}f(\gamma(d(q)+t))\Big{|}_{t=0}=\frac{d}{dt}(d(q)^{2}+2d(q)t+t^{2})\Big{|}_{t=0}=2d(q).$

Thus, $\lambda=2d(q)$ and combined with (4.12), we conclude that $\nabla
f(q)=2d(q)\gamma^{\prime}(d(q))$. Therefore, the negative gradient flow line
initialized at $q\in M-\mathrm{Cu}(N)$ is given by

$\eta(t)=\gamma(d(q)e^{-2t}).$

These flow lines define a flow which deform $M-\mathrm{Cu}(N)$ to $N$ in
infinite time. ∎

The reader may choose to revisit the example of $GL(n,\mathbb{R})$ discussed
in Section 3.2 and treat it as a concrete illustration of the Theorem above.

##  Chapter 5 Application to Lie groups

Due to the classical results of Cartan, Iwasawa and others, we know that any
connected Lie group $G$ is diffeomorphic to the product of a maximally compact
subgroup $K$ and the Euclidean space. In particular, $G$ deforms to $K$. For
semisimple groups, this decomposition is stronger and is attributed to
Iwasawa. The Killing form on the Lie algebra $\mathfrak{g}$ is non-degenerate
and negative definite for compact semi-simple Lie algebras. For such a Lie
group $G$, consider the Levi-Civita connection associated to the bi-invariant
metric obtained from negative of the Killing form. This connection coincides
with the Cartan connection.

We will consider two examples, both of which are non-compact and non
semisimple. We prove that these Lie groups $G$ deformation retract to
maximally compact subgroups $K$ via gradient flows of appropriate Morse-Bott
functions. This requires a choice of a left-invariant metric which is
right-$K$-invariant, and a careful analysis of the geodesics associated with
the metric. In particular, we provide a possibly new proof of the surjectivity
of the exponential map for $U(p,q)$. The results of this chapter is based on
joint work with Basu [Basu and Prasad, 2021, §4].

###  5.1 Matrices with positive determinant

Let $g$ be a left-invariant metric on $GL(n,\mathbb{R})$, the set of all
invertible matrices. Recall that a left-invariant metric $g$ on a Lie group is
determined by its restriction at the identity. For $A\in GL(n,\mathbb{R})$,
consider the left multiplication map $l_{A}:GL(n,\mathbb{R})\to
GL(n,\mathbb{R}),~{}B\mapsto AB$. This extends to a linear isomorphism from
$M(n,\mathbb{R})$ to itself. Thus, the differential
$(Dl_{A})_{I}:T_{I}GL(n,\mathbb{R})\to T_{A}GL(n,\mathbb{R})$ is an
isomorphism and given by $l_{A}$ itself. For $X,Y\in T_{I}GL(n,\mathbb{R})$,

$g_{I}(X,Y)=g_{A}((Dl_{A})_{I}X,(Dl_{A})_{I}Y)=g_{A}(AX,AY).$

We choose the left-invariant metric on $GL(n,\mathbb{R})$ generated by the
Euclidean metric at $I$. Therefore,

$g_{A^{-1}}(X,Y)=\left\langle
AX,AY\right\rangle_{I}:=\mathrm{tr}\left((AX)^{T}\\!AY\right)=\mathrm{tr}\left(X^{T}\\!A^{T}\\!AY\right).$

Note that this metric is right-$O(n,\mathbb{R})$-invariant. We are interested
in the distance between an invertible matrix $A$ (with $\det(A)>0$) and
$SO(n,\mathbb{R})$. Since $SO(n,\mathbb{R})$ is compact, there exists $B\in
SO(n,\mathbb{R})$ such that $d(A,B)=\operatorname{dist}(A,SO(n,\mathbb{R}))$.

######  Lemma 5.1.1.

If $D$ is a diagonal matrix with positive diagonal entries
$\lambda_{1},\cdots,\lambda_{n}$, then

$\operatorname{dist}(D,SO(n,\mathbb{R}))=d(D,I).$

Moreover, $I$ is the unique minimizer and the associated minimal geodesic is
given by $\gamma(t)=e^{t\log D}$.

###### Proof.

Let $B\in SO(n,\mathbb{R})$ satisfying
$d(D,B)=\operatorname{dist}(D,SO(n,\mathbb{R}))$. Since with respect to the
left-invariant metric $GL^{+}(n,\mathbb{R})$ is complete, there exists a
minimal geodesic $\gamma:[0,1]\to GL^{+}(n,\mathbb{R})$ joining $B$ to $D$,
i.e.,

$\gamma(0)=B,~{}~{}\gamma(1)=D,~{}~{}\text{ and }~{}~{}l(\gamma)=d(D,B).$

The first variational principle implies that $\gamma^{\prime}(0)$ is
orthogonal to $T_{B}SO(n,\mathbb{R})$. It follows from [Martin and Neff, 2016,
§2.1] that $\eta(t)=e^{tW}$ is a geodesic if $W$ is a symmetric matrix.
Moreover, $\eta^{\prime}(0)=W$ is orthogonal to $T_{I}SO(n,\mathbb{R})$. As
left translation is an isometry and isometry preserves geodesic, it follows
that $\gamma(t)=Be^{tW}$ is a geodesic with $\gamma^{\prime}(0)$ orthogonal to
$T_{B}SO(n,\mathbb{R})$. By the defining properties of $\gamma$,
$D=\gamma(1)=Be^{W}$. Since $e^{W}$ is symmetric positive definite, we obtain
two polar decompositions of $D$, i.e., $D=ID$ and $D=Be^{W}$. By the
uniqueness of the polar decomposition for invertible matrices, $B=I$ and
$D=e^{W}$.

In order to compute $d(I,D)$, note that

$e^{W}=D=e^{\log D},$

where $\log D$ denotes the diagonal matrix with entries
$\log\lambda_{1},\cdots,\log\lambda_{n}$. As $W$ and $\log D$ are symmetric,
and matrix exponential is injective on the space of symmetric matrices, we
conclude that $W=\log D$. The geodesic is given by $\gamma(t)=e^{t\log D}$ and

$\operatorname{dist}(D,SO(n,\mathbb{R}))=\left\|\gamma^{\prime}(0)\right\|_{I}=\left\|\log
D\right\|_{I}=\left(\sum_{i=1}^{n}(\log\lambda_{i})^{2}\right)^{\frac{1}{2}}.$
(5.1)

Thus, the distance squared function will be given by
$\sum_{i=1}^{n}(\log\lambda_{i})^{2}$. ∎

Now for any $A\in GL^{+}(n,\mathbb{R})$ we can apply the SVD decomposition,
i.e., $A=UDV^{T}$ with $\sqrt{A^{T}A}=VDV^{T}$ and $\log\sqrt{A^{T}A}=V(\log
D)V^{T}$. Note that $U,V\in SO(n,\mathbb{R})$ and $D$ is a diagonal matrix
with positive entries. The left-invariant metric is right-invariant with
respect to orthogonal matrices. Thus,

$\operatorname{dist}(A,SO(n,\mathbb{R}))=\operatorname{dist}(D,SO(n,\mathbb{R}))=\left\|\log
D\right\|_{I},$

where the last equality follows from the lemma (see (5.1)). As

$\left\|\log D\right\|_{I}=\left\|V(\log
D)V^{T}\right\|_{I}=\left\|\log\sqrt{A^{T}A}\right\|_{I},$

it follows from the arguments of the lemma and the metric being
bi-$O(n,\mathbb{R})$-invariant that

$\gamma(t)=Ue^{t\log D}V^{T}$

is a minimal geodesic joining $UV^{T}$ to $A$, realizing
$\operatorname{dist}(A,SO(n,\mathbb{R}))$. As the minimizer $UV^{T}$ is
unique, $\mathrm{Se}(SO(n,\mathbb{R}))$ is empty, implying that
$\mathrm{Cu}(SO(n,\mathbb{R}))$ is empty as well. In fact,
$UV^{T}=A\sqrt{A^{T}A}^{-1}$ and

$\gamma(t)=Ue^{t\log D}V^{T}=UV^{T}Ve^{t\log
D}V^{T}=A\sqrt{A^{T}A}^{-1}e^{t\log\sqrt{A^{T}A}}.$ (5.2)

If we compare (3.10), the deformation of $GL(n,\mathbb{R})$ to
$O(n,\mathbb{R})$ inside $M(n,\mathbb{R})$, with (5.2), then in both of the
cases, an invertible matrix $A$ deforms to $A\sqrt{A^{T}A}^{-1}$. Finally,
observe that the normal bundle of $SO(n,\mathbb{R})$ is diffeomorphic to
$GL^{+}(n,\mathbb{R})$.

###  5.2 Indefinite unitary groups

Let $n$ be a positive integer with $n=p+q$. Consider the inner product on
$\mathbb{C}^{n}$ given by

$\left\langle(w_{1},\ldots,w_{n}),(z_{1},\ldots,z_{n})\right\rangle=z_{1}\overline{w_{1}}+\cdots+z_{p}\overline{w_{p}}-z_{p+1}\overline{w_{p+1}}-\cdots-
z_{n}\overline{w_{n}}.$

This is given by the matrix $I_{p,q}$ in the following way:

$\left\langle\mathbf{w},\mathbf{z}\right\rangle=\overline{\mathbf{w}}^{t}I_{p,q}\mathbf{z}=\left(\begin{array}[]{ccc}\overline{w}_{1}&\cdots&\overline{w}_{n}\end{array}\right)\left(\begin{array}[]{cc}I_{p}&0\\\
0&-I_{q}\end{array}\right)\left(\begin{array}[]{c}z_{1}\\\ \vdots\\\
z_{n}\end{array}\right)$

Let $U(p,q)$ denote the subgroup of $GL(n,\mathbb{C})$ preserving this
indefinite form, i.e., $\mathcal{A}\in U(p,q)$ if and only if
$\mathcal{A}^{\ast}I_{p,q}\mathcal{A}=I_{p,q}$. In particular,
$\det\mathcal{A}$ is a complex number of unit length. By convention,
$I_{n,0}=I_{n}$ and $I_{0,n}=-I_{n}$, both of which corresponds to
$U(n,0)=U(n)=U(0,n)$, the unitary group. In all other cases, the inner product
is indefinite.

The group $U(1,1)$ is given by matrices of the form

$\mathcal{A}=\left(\begin{array}[]{cc}\alpha&\beta\\\
\lambda\overline{\beta}&\lambda\overline{\alpha}\end{array}\right),\,\,\lambda\in
S^{1},\,\,|\alpha|^{2}-|\beta|^{2}=1.$

More generally, we shall use

$\mathcal{A}=\left(\begin{array}[]{cc}A&B\\\ C&D\end{array}\right)$

to denote an element of $U(p,q)$. It follows from the definition that
$\mathcal{A}\in U(p,q)$ if and only if

$\displaystyle A^{\ast}A-C^{\ast}C$ $\displaystyle=$ $\displaystyle I_{p}$
$\displaystyle A^{\ast}B-C^{\ast}D$ $\displaystyle=$ $\displaystyle 0_{p\times
q}$ $\displaystyle B^{\ast}B-D^{\ast}D$ $\displaystyle=$ $\displaystyle-
I_{q}.$

Observe that if $Av=0$, then

$0=A^{\ast}Av=C^{\ast}Cv+v,$

which implies that $C^{\ast}C$, a positive semi-definite matrix, has $-1$ as
an eigenvalue unless $v=0$. Therefore, $A$ is invertible, and the same
argument works for $D$.

######  Lemma 5.2.1.

The intersection of $U(p+q)$ with $U(p,q)$ is $U(p)\times U(q)$. Moreover, if
$\mathcal{A}\in U(p,q)$, then
$\mathcal{A}^{\ast},\sqrt{\mathcal{A}^{\ast}\mathcal{A}}\in U(p,q)$.

###### Proof.

If $\mathcal{A}\in U(p)\times U(q)$, then

$\displaystyle A^{\ast}A+C^{\ast}C$ $\displaystyle=$ $\displaystyle I_{p}$
$\displaystyle B^{\ast}B+D^{\ast}D$ $\displaystyle=$ $\displaystyle I_{q}.$

This implies that both $B$ and $C$ are zero matrices. If $\mathcal{A}\in
U(p,q)$, then $\mathcal{A}^{\ast}=I_{p,q}\mathcal{A}^{-1}I_{p,q}$ and

$\displaystyle(\mathcal{A}^{\ast}\mathcal{A})^{\ast}I_{p,q}(\mathcal{A}^{\ast}\mathcal{A})$
$\displaystyle=(\mathcal{A}^{\ast}\mathcal{A})I_{p,q}(\mathcal{A}^{\ast}\mathcal{A})$
$\displaystyle=I_{p,q}\mathcal{A}^{-1}I_{p,q}\mathcal{A}I_{p,q}I_{p,q}\mathcal{A}^{-1}I_{p,q}\mathcal{A}$
$\displaystyle=I_{p,q}=\mathcal{A}^{\ast}I_{p,q}\mathcal{A}.$

This also implies that $\mathcal{A}I_{p,q}\mathcal{A}^{\ast}=I_{p,q}$.

All the eigenvalues of $\mathcal{A}^{\ast}\mathcal{A}$ are positive. Moreover,
if $\lambda$ is an eigenvalue of $\mathcal{A}^{\ast}\mathcal{A}$ with
eigenvector $\mathbf{v}=(v_{1},\ldots,v_{p},v_{p+1},\ldots,v_{n})$, then

$I_{p,q}\mathbf{v}=\mathcal{A}^{\ast}\mathcal{A}\,I_{p,q}\mathcal{A}^{\ast}\mathcal{A}\mathbf{v}=\lambda(\mathcal{A}^{\ast}\mathcal{A}\,I_{p,q}\mathbf{v}),$

which implies that $\lambda^{-1}$ is also an eigenvalue with eigenvector
$\mathbf{v}^{\prime}=(v_{1},\ldots,v_{p},-v_{p+1},\ldots,-v_{n})$. If
$\\{\mathbf{v}_{1},\ldots,\mathbf{v}_{n}\\}$ is an eigenbasis of
$\mathcal{A}^{\ast}\mathcal{A}$ with (possibly repeated) eigenvalues
$\lambda_{1},\ldots,\lambda_{n}$, then

$\sqrt{\mathcal{A}^{\ast}\mathcal{A}}\,I_{p,q}\sqrt{\mathcal{A}^{\ast}\mathcal{A}}\mathbf{v}_{j}=\sqrt{\mathcal{A}^{\ast}\mathcal{A}}\,I_{p,q}\sqrt{\lambda_{j}}\mathbf{v}_{j}=\sqrt{\lambda_{j}}\sqrt{\mathcal{A}^{\ast}\mathcal{A}}\mathbf{v}_{j}^{\prime}=\mathbf{v}_{j}^{\prime}=I_{p,q}\mathbf{v}_{j}.$

Thus, $\sqrt{\mathcal{A}^{\ast}\mathcal{A}}$ satisfies the defining relation
for a matrix to be in $U(p,q)$. ∎

We may use the polar decomposition (for matrices in $GL(n,\mathbb{C})$) to
write

$\mathcal{A}=U|\mathcal{A}|,\,\,\textup{where}\,\,U=\mathcal{A}\left(\sqrt{\mathcal{A}^{\ast}\mathcal{A}}\right)^{-1},|\mathcal{A}|=\sqrt{\mathcal{A}^{\ast}\mathcal{A}},$

where $U,|\mathcal{A}|\in U(p,q)$. For $\textup{U}(1,1)$ this decomposition
takes the form

$\left(\begin{array}[]{cc}\alpha&\beta\\\
\lambda\overline{\beta}&\lambda\overline{\alpha}\end{array}\right)=\left(\begin{array}[]{cc}\frac{\alpha}{|\alpha|}&0\\\
0&\lambda\frac{\overline{\alpha}}{|\alpha|}\end{array}\right)\left(\begin{array}[]{cc}|\alpha|&\frac{|\alpha|\beta}{\alpha}\\\
\frac{|\alpha|\overline{\beta}}{\overline{\alpha}}&|\alpha|\end{array}\right)$

The Lie algebra $\mathfrak{u}_{p,q}$ is given by matrices $X\in
M_{n}(\mathbb{C})$ such that

$X^{\ast}I_{p,q}+I_{p,q}X=0.$

This is real Lie subalgebra of $M_{p+q}(\mathbb{C})$. It contains the
subalgebras $\mathfrak{u}_{p},\mathfrak{u}_{q}$ as Lie algebras of the
subgroups $U(p)\times I_{q}$ and $I_{p}\times U(q)$. Consider the inner
product

$\left\langle\cdot,\cdot\right\rangle:\mathfrak{u}_{p,q}\times\mathfrak{u}_{p,q}\to\mathbb{R},\,\,\,\left\langle
X,Y\right\rangle:=\textup{trace}(X^{\ast}Y).$

######  Lemma 5.2.2.

The inner product is symmetric and positive-definite.

###### Proof.

Note that

$\left\langle
X,Y\right\rangle=\textup{trace}(-I_{p,q}XI_{p,q}Y)=\textup{trace}(-I_{p,q}YI_{p,q}X)=\left\langle
Y,X\right\rangle.$

Since $\overline{\left\langle X,Y\right\rangle}=\left\langle Y,X\right\rangle$
due to the invariance of trace under transpose, we conclude that the inner
product is real and symmetric. It is positive-definite as $\left\langle
X,X\right\rangle=\textup{trace}(X^{\ast}X)\geq 0$ and equality holds if and
only if $X$ is the zero matrix. ∎

The Riemannian metric obtained by left translations of
$\left\langle\cdot,\cdot\right\rangle$ will also be denoted by
$\left\langle\cdot,\cdot\right\rangle$. We shall analyze the geodesics for
this metric. The Lie algebra $\mathfrak{u}_{p}\oplus\mathfrak{u}_{q}$ of
$U(p)\times U(q)$ consists of

$\left(\begin{array}[]{cc}A&0\\\
0&D\end{array}\right),\,\,A+A^{\ast}=0,\,D+D^{\ast}=0.$

Let $\mathfrak{n}$ denote the orthogonal complement of
$\mathfrak{u}_{p}\oplus\mathfrak{u}_{q}$ inside $\mathfrak{u}_{p,q}$. As
$\mathfrak{n}$ is of (complex) dimension $pq$, and

$\left\\{\left(\begin{array}[]{cc}0&B\\\
B^{\ast}&0\end{array}\right)\,\Big{|}\,B\in M_{p,q}(\mathbb{C})\right\\}$

is contained in $\mathfrak{n}$, this is all of it. We may verify that

$\displaystyle\left[\left(\begin{array}[]{cc}A&0\\\
0&D\end{array}\right),\left(\begin{array}[]{cc}0&B\\\
B^{\ast}&0\end{array}\right)\right]$ $\displaystyle=$
$\displaystyle\left(\begin{array}[]{cc}0&AB-BD\\\
DB^{\ast}-B^{\ast}A&0\end{array}\right)\in\mathfrak{n}$
$\displaystyle\left[\left(\begin{array}[]{cc}0&B\\\
B^{\ast}&0\end{array}\right),\left(\begin{array}[]{cc}0&C\\\
C^{\ast}&0\end{array}\right)\right]$ $\displaystyle=$
$\displaystyle\left(\begin{array}[]{cc}BC^{\ast}-CB^{\ast}&0\\\
0A&B^{\ast}C-C^{\ast}B\end{array}\right)\in\mathfrak{u}_{p}\oplus\mathfrak{u}_{q}.$

######  Lemma 5.2.3.

Let $\gamma$ be the integral curve, initialized at $e$, for a left-invariant
vector field $Y$. This curve is a geodesic if $Y(e)$ either belongs to
$\mathfrak{n}$ or to $\mathfrak{u}_{p}\oplus\mathfrak{u}_{q}$.

###### Proof.

The Levi-Civita connection $\nabla$ is given by the Koszul formula

$2\left\langle X,\nabla_{Z}Y\right\rangle=Z\left\langle
X,Y\right\rangle+Y\left\langle X,Z\right\rangle-X\left\langle
Y,Z\right\rangle+\left\langle Z,[X,Y]\right\rangle+\left\langle
Y,[X,Z]\right\rangle-\left\langle X,[Y,Z]\right\rangle.$

Putting $Z=Y$ and $X$, two left-invariant vector fields, in the above, we
obtain

$\left\langle X,\nabla_{Y}Y\right\rangle=\left\langle Y,[X,Y]\right\rangle.$

To prove our claim, it suffices to show that $\nabla_{Y}Y=0$, ie,
$\left\langle Y,[X,Y]\right\rangle=0$ for any $X$. Let us assume that
$Y(e)\in\mathfrak{n}$. If $X(e)\in\mathfrak{n}$, then
$[X(e),Y(e)]\in\mathfrak{u}_{p}\oplus\mathfrak{u}_{q}$, which implies that
$\left\langle Y(e),[X(e),Y(e)]\right\rangle=0$. If
$X(e)\in\mathfrak{u}_{p}\oplus\mathfrak{u}_{q}$, then

$\displaystyle\left\langle Y,[X,Y]\right\rangle$ $\displaystyle=$
$\displaystyle\left\langle\left(\begin{array}[]{cc}0&B\\\
B^{\ast}&0\end{array}\right),\left(\begin{array}[]{cc}0&AB-BD\\\
DB^{\ast}-B^{\ast}A&0\end{array}\right)\right\rangle$ $\displaystyle=$
$\displaystyle\textup{trace}\left(\begin{array}[]{cc}B(DB^{\ast}-B^{\ast}A)&0\\\
0&B^{\ast}(AB-BD)\end{array}\right)$ $\displaystyle=$
$\displaystyle\textup{trace}(BDB^{\ast}-BB^{\ast}A)+\textup{trace}(B^{\ast}AB-B^{\ast}BD)$
$\displaystyle=$ $\displaystyle 0$

by the cyclic property of trace. Thus, $\nabla_{Y}Y=0$ if
$Y(e)\in\mathfrak{n}$; similar proof works if
$Y(e)\in\mathfrak{u}_{p}\oplus\mathfrak{u}_{q}$. ∎

######  Remark 5.2.1.

An integral curve of a left-invariant vector field (also called $1$-parameter
subgroups) need not be a geodesic in $U(p,q)$. For instance, if $X+Y$ is a
left-invariant vector field given by
$X(e)\in\mathfrak{u}_{p}\oplus\mathfrak{u}_{q}$ and $Y(e)\in\mathfrak{n}$,
then $\nabla_{X+Y}(X+Y)=0$ if and only if $\nabla_{X}Y=\frac{1}{2}[X,Y]$ and
$\nabla_{Y}X=\frac{1}{2}[Y,X]$. This happens if and only if the metric is bi-
invariant, i.e.,

$\left\langle[X,Z],Y\right\rangle=\left\langle X,[Z,Y]\right\rangle.$

This is not true in general; for instance, with
$X(e)\in\mathfrak{u}_{p}\oplus\mathfrak{u}_{q}$ and linearly independent
$Y(e),Z(e)\in\mathfrak{n}$, we get
$\left\langle[X,Z],Y\right\rangle-\left\langle X,[Z,Y]\right\rangle\neq 0$.

Consider the matrix

$Y=\left(\begin{array}[]{cc}0&B\\\
B^{\ast}&0\end{array}\right)\in\mathfrak{n}.$

Let $B=U\sqrt{B^{\ast}B}$ and $B^{\ast}=\sqrt{B^{\ast}B}\,U^{\ast}$ be the
polar decompositions for the rectangular matrices. It follows from direct
computation that

$\displaystyle e^{Y}$ $\displaystyle=$
$\displaystyle\left(\begin{array}[]{cc}I_{p}+\frac{BB^{\ast}}{2!}+\frac{(BB^{\ast})^{2}}{4!}+\cdots&\frac{B}{1!}+\frac{B(B^{\ast}B)}{3!}+\frac{B(B^{\ast}B)^{2}}{5!}+\cdots\\\
\frac{B^{\ast}}{1!}+\frac{(B^{\ast}B)B^{\ast}}{3!}+\frac{(B^{\ast}B)^{2}B^{\ast}}{5!}+\cdots&I_{q}+\frac{B^{\ast}B}{2!}+\frac{(B^{\ast}B)^{2}}{4!}+\cdots\end{array}\right)$
$\displaystyle=$
$\displaystyle\left(\begin{array}[]{cc}\cosh(\sqrt{BB^{\ast}})&U\sinh(\sqrt{B^{\ast}B})\\\
\sinh(\sqrt{B^{\ast}B})U^{\ast}&\cosh(\sqrt{B^{\ast}B})\end{array}\right).$

It can be checked that

$e^{\mathfrak{n}}\cap\left(U(p)\times U(q)\right)=\\{I_{n}\\}.$

It is known that the non-zero eigenvalues of $Y$ are the non-zero eigenvalues
of $\sqrt{BB^{\ast}}$ and their negatives.

######  Theorem 5.2.1.

For any element $\mathcal{A}\in U(p,q)$, the associated matrix
$\sqrt{\mathcal{A}^{\ast}\mathcal{A}}$ can be expressed uniquely as $e^{Y}$
for $Y\in\mathfrak{n}$. Moreover, there is a unique way to express
$\mathcal{A}$ as a product of a unitary matrix and an element of
$e^{\mathfrak{n}}$, and it is given by the polar decomposition.

In order to prove the result, we discuss some preliminaries on logarithm of
complex matrices. In general, there is no unique logarithm. However, the
Gregory series

$\log A=-\sum_{m=0}^{\infty}\frac{2}{2m+1}\left[(I-A)(I+A)^{-1}\right]^{2m+1}$

converges if all the eigenvalues of $A\in M_{n}(\mathbb{C})$ have positive
real part, see [Higham, 2008, §11.3, page 273]. In particular, $\log A$ is
well-defined for Hermitian positive-definite matrix. This is often called the
principal logarithm of $A$. This logarithm satisfies $e^{\log A}=A$. There is
an integral form of logarithm that applies to matrices without real or zero
eigenvalues; it is given by

$\log A=(A-I)\int_{0}^{1}\left[s(A-I)+I\right]^{-1}ds.$

######  Lemma 5.2.4.

The inverse of $\mathcal{A}^{\ast}\mathcal{A}+I_{n}$ for $\mathcal{A}\in
U(p,q)$ is given by

$\left[\mathcal{A}^{\ast}\mathcal{A}+I_{n}\right]^{-1}=\frac{1}{2}\left(\begin{array}[]{cc}I_{p}&-A^{-1}B\\\
-B^{\ast}(A^{\ast})^{-1}&I_{q}\end{array}\right).$

###### Proof.

Since $\mathcal{A}^{\ast}\mathcal{A}$ has only positive eigenvalues,
$\mathcal{A}^{\ast}\mathcal{A}+I_{n}$ has no kernel. We note that

$\mathcal{A}^{\ast}\mathcal{A}+I_{n}=\left(\begin{array}[]{cc}2C^{\ast}C+2I_{p}&2A^{\ast}B\\\
2B^{\ast}A&2B^{\ast}B+2I_{q}\end{array}\right)=\left(\begin{array}[]{cc}2A^{\ast}A&2A^{\ast}B\\\
2B^{\ast}A&2D^{\ast}D\end{array}\right).$

The inverse matrix satisfies

$\left(\begin{array}[]{cc}2A^{\ast}A&2A^{\ast}B\\\
2B^{\ast}A&2D^{\ast}D\end{array}\right)\left(\begin{array}[]{cc}E&F\\\
F^{\ast}&G\end{array}\right)=\left(\begin{array}[]{cc}I_{p}&0\\\
0&I_{q}\end{array}\right).$

As the matrices are Hermitian, the three constraints that $E,F,G$ must satisfy
(and are uniquely determined by) are

$\displaystyle E$ $\displaystyle=$
$\displaystyle\textstyle{\frac{1}{2}}(A^{\ast}A)^{-1}-A^{-1}BF^{\ast}$
$\displaystyle G$ $\displaystyle=$
$\displaystyle\textstyle{\frac{1}{2}}(D^{\ast}D)^{-1}-D^{-1}CF$ $\displaystyle
F$ $\displaystyle=$ $\displaystyle-A^{-1}BG.$

We note that $E=\frac{1}{2}I_{p}$, $G=\frac{1}{2}I_{q}$ and
$F=-\frac{1}{2}A^{-1}B$ satisfy the above equations. For instance,

$\displaystyle\frac{1}{2}(A^{\ast}A)^{-1}-A^{-1}BF^{\ast}$
$\displaystyle=\frac{1}{2}(A^{\ast}A)^{-1}+\frac{1}{2}A^{-1}BB^{\ast}(A^{\ast})^{-1}$
$\displaystyle=\frac{1}{2}(A^{\ast}A)^{-1}+\frac{1}{2}A^{-1}(AA^{\ast}-I_{p})(A^{\ast})^{-1}=\frac{1}{2}I_{p},$

where $BB^{\ast}=AA^{\ast}-I_{p}$ is a consequence of $\mathcal{A}^{\ast}\in
U(p,q)$. Yet another consequence is $AC^{\ast}=BD^{\ast}$, which is equivalent
to

$A^{-1}B=(D^{-1}C)^{\ast}.$

In a similar vein,

$\displaystyle\frac{1}{2}(D^{\ast}D)^{-1}-D^{-1}CF$
$\displaystyle==\frac{1}{2}(D^{\ast}D)^{-1}+\frac{1}{2}D^{-1}CC^{\ast}(D^{\ast})^{-1}$
$\displaystyle=\frac{1}{2}(D^{\ast}D)^{-1}+\frac{1}{2}D^{-1}(DD^{\ast}-I_{q})(D^{\ast})^{-1}$
$\displaystyle=\frac{1}{2}I_{q},$

where $CC^{\ast}=DD^{\ast}-I_{q}$ is due to $\mathcal{A}^{\ast}\in U(p,q)$. ∎

###### Proof of 5.2.1.

We use Gregory series expansion for computing the principal logarithm of
$\mathcal{A}^{\ast}\mathcal{A}$ along with Lemma 5.2.4:

$\displaystyle\log(\mathcal{A}^{\ast}\mathcal{A})$
$\displaystyle=\sum_{m=0}^{\infty}{\frac{2}{2m+1}}\left[2\left(\begin{array}[]{cc}A^{\ast}A-I_{p}&A^{\ast}B\\\
B^{\ast}A&D^{\ast}D-I_{q}\end{array}\right)\frac{1}{2}\left(\begin{array}[]{cc}I_{p}&-A^{-1}B\\\
-B^{\ast}(A^{\ast})^{-1}&I_{q}\end{array}\right)\right]^{2m+1}$
$\displaystyle=\sum_{m=0}^{\infty}{\frac{2}{2m+1}}\left(\begin{array}[]{cc}0&A^{-1}B\\\
B^{\ast}(A^{\ast})^{-1}&0\end{array}\right)^{2m+1}.$

We set $Y=\frac{1}{2}\log(\mathcal{A}^{\ast}\mathcal{A})$. It is clear that
$Y\in\mathfrak{n}$ and $e^{Y}=\sqrt{\mathcal{A}^{\ast}\mathcal{A}}$. It is
known that the exponential map is injective on Hermitian matrices. This
implies the uniqueness of $Y$.

If $U_{1}e^{Y_{1}}=U_{2}e^{Y_{2}}$ are two decompositions of $\mathcal{A}\in
U(p,q)$ with $U_{i}\in\textup{U}(p)\times\textup{U}(q)$ and
$Y_{i}\in\mathfrak{n}$, then

$e^{2Y_{1}}=e^{Y_{1}}U_{1}^{\ast}U_{1}e^{Y_{1}}=e^{Y_{2}}U_{2}^{\ast}U_{2}e^{Y_{2}}=e^{2Y_{2}}.$

By the injectivity of the exponential map (on Hermitian matrices), we obtain
$Y_{1}=Y_{2}$, which implies that $U_{1}=U_{2}$. ∎

We infer the following (see [Yakubovich and Starzhinskii, 1975, Lemma 1, page
211] for a different proof) result.

######  Corollary 5.2.1.

The exponential map $\textup{exp}:\mathfrak{u}_{p,q}\to U(p,q)$ is surjective.

###### Proof.

Using the polar decomposition and 5.2.1,

$\mathcal{A}=\mathcal{A}\big{(}\sqrt{\mathcal{A}^{\ast}\mathcal{A}}\big{)}^{-1}\sqrt{\mathcal{A}^{\ast}\mathcal{A}}=\mathcal{A}\big{(}\sqrt{\mathcal{A}^{\ast}\mathcal{A}}\big{)}^{-1}e^{Y}.$

Since the matrix exponential is surjective for $U(p)\times U(q)$, choose
$Z\in\mathfrak{u}_{p}\oplus\mathfrak{u}_{q}$ such that
$e^{Z}=\mathcal{A}(\sqrt{\mathcal{A}^{\ast}\mathcal{A}})^{-1}$. By Baker-
Campbell-Hausdorff formula, we may express $e^{Z}e^{Y}$ as exponential of an
element in $\mathfrak{u}_{p,q}$. ∎

The distance from any matrix $\mathcal{A}\in U(p,q)$ to $U(p)\times U(q)$ is
given by the length of the curve

$\gamma(t)=\mathcal{A}\big{(}\sqrt{\mathcal{A}^{\ast}\mathcal{A}}\big{)}^{-1}e^{tY},$

which can be computed (and simplified via left-invariance) as follows

$\ell(\gamma)=\int_{0}^{1}\|\gamma^{\prime}(t)\|_{\gamma(t)}\,dt=\int_{0}^{1}\|Y\|\,dt=\|Y\|.$

Note that

$\|Y\|^{2}=\textup{trace}(Y^{\ast}Y)=\textup{trace}\left[\textstyle{\frac{1}{4}}(\log(\mathcal{A}^{\ast}\mathcal{A}))^{2}\right].$

Thus, the distance squared function is given by

$d^{2}:U(p,q)\to\mathbb{R},\,\,\mathcal{A}\mapsto\textstyle{\frac{1}{4}}\textup{trace}\left[\left(\log(\mathcal{A}^{\ast}\mathcal{A})\right)^{2}\right].$

##  Chapter 6 Equivariant cut locus

Let $M$ be a smooth manifold on which a compact Lie group $G$ acts freely. It
is known that $M/G$ is a smooth manifold. Moreover, if $M$ has a Riemannian
metric and $G$ acts isometrically, then there is an induced Riemannian metric
on $M/G$. Let $N$ be a $G$-invariant submanifold of $M$. We want to find the
cut locus of $N/G$ in $M/G$. In this chapter we will discuss the equality
between $\mathrm{Cu}(N)/G$ and $\mathrm{Cu}(N/G)$. We will start the chapter
by motivating with an example, then state the theorem and then recall
Riemannian submersion which will be useful for the proof of the theorem. At
the end we will discuss an application of this result to complex projective
hypersurfaces.

###  6.1 Statement of the theorem

Let us discuss an example which will be helpful to arrive at the statement of
the main theorem. Let $\mathbb{RP}^{n}$ denotes the $n$-dimensional real
projective space which is obtained from $\mathbb{S}^{n}$ by identifying points
$p$ and $-p$. Equivalently, this space can be obtained from the action of
$\mathbb{Z}_{2}$ on $\mathbb{S}^{n}$. We know from 2.3.4 that for a point
$p\in\mathbb{RP}^{n}$, the cut locus is $\mathbb{RP}^{n-1}$. In 3.1.5, we
showed that
$\mathrm{Cu}\left(\mathbb{S}_{i}^{k}\right)=\mathbb{S}_{l}^{n-k-1}$, where
$\mathbb{S}_{i}^{k}\hookrightarrow\mathbb{S}^{n}$ denote the embedding of the
$k$-sphere in the first $k+1$ coordinates while $\mathbb{S}^{n-k-1}_{l}$
denote the embedding of the $(n-k-1)$-sphere in the last $n-k$ coordinates. So
we have

${\mathbb{S}^{n}\supset\left\\{p,-p\right\\}}$${\mathbb{S}^{n-1}}$${\mathbb{RP}^{n}\supset\\{p\\}}$${\mathbb{RP}^{n-1}}$$\scriptstyle{\mathbb{Z}_{2}}$$\scriptstyle{\mathrm{Cu}}$$\scriptstyle{\mathbb{Z}_{2}}$$\scriptstyle{\mathrm{Cu}}$

Similarly, one can see a similar diagram for the complex projective space
$\mathbb{CP}^{n}$ which is obtained by taking $\mathbb{S}^{1}$ action on
$\mathbb{S}^{{2n+1}}$. We have

${\mathbb{S}^{2n+1}\supset\mathbb{S}_{i}^{1}}$${\mathbb{S}^{2n-1}_{f}}$${\mathbb{CP}^{n}\supset\\{p\\}}$${\mathbb{CP}^{n-1}}$$\scriptstyle{\mathbb{S}^{1}}$$\scriptstyle{\mathrm{Cu}}$$\scriptstyle{\mathbb{S}^{1}}$$\scriptstyle{\mathrm{Cu}}$

Thus, it is natural to ask whether the following diagram commutes for a
compact Lie group $G$.

${M\supset N}$${\mathrm{Cu}(N)}$${M/G\supset
N/G}$${\mathrm{Cu}(N)/G}$$\scriptstyle{G}$$\scriptstyle{\mathrm{Cu}}$$\scriptstyle{G}$$\scriptstyle{\mathrm{Cu}}$

The above diagram make sense if $N$ and $\mathrm{Cu}(N)$ are $G$-invariant
subsets. Note that if the action is free and isometric, i.e., length of the
curves $\gamma$ and $g\cdot\gamma$ are the same, then $\mathrm{Se}(N)$ is
$G$-invariant.

$N$
---
$M$
---
$\gamma$
---
$\eta$
---
$p$
---

$N$
---
$M$
---
$g\cdot\gamma$
---
$g\cdot\eta$
---
$g\cdot p$
---

Figure 6.1: $\mathrm{Se}(N)$ is $G$-invariant

Now to show that the cut locus of $N$, which is the closure of
$\mathrm{Se}(N)$, is $G$-invariant we take $x\in\overline{\mathrm{Se}(N)}$. So
there exists a sequence $\left(x_{n}\right)\subset\mathrm{Se}(N)$ such that
$x_{n}\to x$ (this convergence is with respect to the Riemannian metric). This
implies $g\cdot x_{n}\to g\cdot x$ as the action is continuous. Hence, $g\cdot
x\in\mathrm{Se}(N).$

The following theorem tells us that the above diagram commutes if $G$ is a
compact Lie group and the action is free and isometric.

######  Theorem 6.1.1 (Equivariant cut locus theorem).

Let $M$ be a closed and connected Riemannian manifold and $G$ be any compact
Lie group which acts on $M$ freely and isometrically. Let $N$ be any
$G$-invariant closed submanifold of $M$, then we have an equality

$\mathrm{Cu}(N)/G=\mathrm{Cu}(N/G).$

######  Remark 6.1.1.

  1. (i)

If the action of $G$ is not isometric, then we can construct a $G$-invariant
metric on $M$ by averaging any metric on $M$ over $G$. In fact, for any $p\in
M$ and any vectors $\mathbf{v}_{1},\mathbf{v}_{2}\in T_{p}M$, we can define

$\left\langle\mathbf{v}_{1},\mathbf{v}_{2}\right\rangle\vcentcolon=\int_{G}\left\langle\mathbf{v}_{1},\mathbf{v}_{2}\right\rangle_{g\cdot
p}~{}dg,$

where the integral is taken with respect to the Haar measure. Then the theorem
is valid with respect to the new metric.

  2. (ii)

Recall that, in 4.3.5, we proved that the gradient flow of distance squared
function from the submanifold $N$ deforms $M-\mathrm{Cu}(N)$ to $N$. By
construction, the flow lines are $G$-invariant. So we also have that
$M/G-\mathrm{Cu}(N)/G$ deforms to $N/G$.

To prove the theorem we have to encounter mainly two problems.

  1. a.

Whether a distance minimal geodesic in $M$ projects down to a distance minimal
geodesic in $M/G$?

  2. b.

Whether a distance minimal geodesic in $M/G$ lifts to a distance minimal
geodesic in $M$?

###  6.2 Proof of the equivariant cut locus theorem

In this section we will recall some results on Riemannian submersion and most
of the results can be found in the book [Michor, 2008, Chapter V, section 26].

######  Definition 6.2.1.

Let $\pi:E\to B$ be a smooth principal $G$-bundle. An Ehresmann connection on
$E$ is a smooth subbundle $\mathcal{H}$ of $TE$, called the horizontal bundle
of the connection, such that $TE=\mathcal{H}\oplus\mathcal{V}$, where
$\mathcal{V}_{p}=\ker\left(d\pi_{p}:T_{p}E\to T_{\pi(p)}B\right)$.

The bundle $\mathcal{V}$ is called the vertical bundle, and it is independent
of the connection chosen. It follows from the definition that
$\mathcal{H}_{p}$ depends smoothly on $p$ and
$\mathcal{H}_{p}\cap\mathcal{V}_{p}=\\{0\\}$. Moreover, the map $d\pi_{p}$
restricts to $\mathcal{H}_{p}$ is an isomorphism on $T_{\pi(p)}B$. If $E$ is a
Riemannian manifold with metric $g$, then by choosing a horizontal bundle
$\mathcal{H}$ we have $\mathcal{H}_{p}=\mathcal{V}_{p}^{\perp}$.

######  Definition 6.2.2.

A smooth submersion $\pi:(E,g)\to(B,g^{\prime})$ is called a Riemannian
submersion if the linear map $d\pi_{p}$ preserves the length of the horizontal
vectors for each point $p\in E$. Equivalently, $d\pi_{p}$ is a linear isometry
between $\mathcal{H}_{p}$ and $T_{\pi(p)}B$.

Using the above definitions we can define the following type of vectors.

######  Definition 6.2.3.

Let $\pi:E\to B$ be a Riemannian submersion. A vector field
$X\in\mathfrak{X}(E)$ is called

  * •

vertical if for any $p\in E$, $X_{p}\in\mathcal{V}_{p}$, denoted by
$X^{\text{ver}}$, and

  * •

horizontal if for any $p\in E$, $X_{p}\in\mathcal{H}_{p}$, denoted by
$X^{\text{hor}}$.

We can uniquely decompose any vector field $X\in\mathfrak{X}(E)$ as

$X=X^{\text{ver}}+X^{\text{hor}}$

into its horizontal and vertical components.

Once we have a connection, we now have a preferred way of lifting vectors from
$TB$ to $TE$. Recall that a vector $\tilde{X}\in T_{e}E$ is a lift of $X\in
T_{\pi(e)B}$ if $T_{e}\pi(\tilde{X})=X$. In absence of a connection, there are
many different choices of lifts of a vector, and any two choices differ by a
vertical vector. That is, if $\tilde{X},\tilde{X}^{\prime}$ are lifts of $X$,
then $\tilde{X}-\tilde{X}^{\prime}$ is vertical. Once we have a connection, we
can define the horizontal lift (with respect to a connection $\mathcal{H}$) of
$X$ as the horizontal component of any lift of $X$. This definition is, of
course, independent of the choice of lift, since any two differ by a vertical
vector, whose horizontal component vanishes. Similarly, we can lift vector
fields by lifting them in a pointwise fashion.

######  Definition 6.2.4 (Horizontal lift of vector fields).

Let $X\in\mathfrak{X}(B)$ be a vector field and $\mathcal{H}\subset TE$ an
Ehresmann connection on $E$. We define the horizontal lift of $X$ as the
vector field $\tilde{X}\in\mathfrak{X}(E)$ which satisfies $d\pi(\tilde{X})=X$
and $\tilde{X}_{e}\in\mathcal{H}_{e}$ for all $e\in E$.

Suppose that we have a curve $\gamma:[0,1]\to B$. At each point over the
curve, we have a vector $\gamma^{\prime}(t)\in T_{\gamma(t)}B$, which we can
lift to the fiber above $\gamma(t)$. So if we choose a starting point
$e_{0}\in\pi^{-1}(\gamma(0))$, we can find an integral curve along all these
lifted vectors on the fibers over the curve $\gamma$. In the end we obtain a
curve $\tilde{\gamma}:[0,1]\to E$ satisfying
$\pi\circ\tilde{\gamma}=\gamma,~{}\tilde{\gamma}(0)=e_{0}$, and
$\tilde{\gamma}^{\prime}(t)\in\mathcal{H}_{\tilde{\gamma}(t)}$ for all $t$. We
call it a horizontal lift of $\gamma$.

######  Definition 6.2.5 (Horizontal lift of a curve).

Let $\pi:E\to B$ be a fiber bundle with a connection $\mathcal{H}$. Let
$\gamma$ be a smooth curve in $B$ through $\gamma(0)=b$. Let $e\in E$ be such
that $\pi(e)=b$. A horizontal lift of $\gamma$ through $e$ is a curve
$\tilde{\gamma}$ in $E$ such that
$\pi\circ\tilde{\gamma}=\gamma,~{}\tilde{\gamma}(0)=e$, and
$\tilde{\gamma}\kern 0.56917pt^{\prime}(t)\in\mathcal{H}_{\tilde{\gamma}(t)}$.

For every point $t_{0}\in(0,1)$, we can find $\varepsilon>0$ such that the
vector field $\gamma^{\prime}$ can be extended to a vector field over
$\gamma\big{|}_{(t_{0}-\varepsilon,t_{0}+\varepsilon)}$. Then we look at the
horizontal vector field $\tilde{X}$ defined on the bundle $E\big{|}_{U}$,
where $U\supseteq\gamma((t_{0}-\varepsilon,t_{0}+\varepsilon))$. Then
$\tilde{\gamma}$ is the integral curve of $\tilde{X}$ starting at the
prescribed point $e_{0}\in\pi^{-1}(\gamma(0))$. Since $[0,1]$ is compact, a
usual gluing argument will help us to construct a horizontal lift of $\gamma$.
Hence, we have the following proposition. For a detailed proof of the
proposition we refer the reader to [Lang, 1999, Chapter XIV, Proposition
3.5(i)].

######  Proposition 6.2.1.

Given a smooth path $\gamma:[0,1]\to B$ such that $\gamma(0)=b$ and
$e_{0}\in\pi^{-1}(b)$, there is a unique horizontal lift $\tilde{\gamma}$ of
$\gamma$ through $e_{0}\in E$.

Recall the Quotient manifold theorem [Lee, 2013, Theorem 21.10].

######  Theorem 6.2.1 (Quotient Manifold Theorem).

Suppose a Lie group $G$ acting smoothly, freely, and properly on a smooth
manifold $M$. Then the orbit space $M/G$ is a topological manifold of
dimension equal to $\dim M-\dim G$, and has a unique smooth structure with the
property that the quotient map $\pi:M\to M/G$ is a smooth submersion.

Using the above theorem, we can define a unique metric on $M/G$ such that
$\pi$ is a Riemannian submersion. In fact, for any $x\in M/G$ and
$p\in\pi^{-1}(x)$, we take the vertical space $\mathcal{V}_{p}\vcentcolon=\ker
d\pi_{p}$ and the horizontal space
$\mathcal{H}_{p}\vcentcolon=\mathcal{V}_{p}^{\perp}$ so that
$T_{p}M=\mathcal{V}_{p}\oplus\mathcal{H}_{p}$. Since $d\pi_{p}$ is surjective,
then $d\pi_{p}\big{|}_{\mathcal{H}_{p}}:\mathcal{H}_{p}\to T_{\pi(p)}M/G$ is a
bijection. Define

$h_{x}(v,w)\vcentcolon=\left\langle
d\pi^{-1}\big{|}_{\mathcal{H}_{p}}(v),d\pi^{-1}\big{|}_{\mathcal{H}_{p}}(w)\right\rangle.$

Since the action is isometric, the metric is independent of the choice of
point $p$. Thus, $h$ defines a well-defined metric on $M/G$ and $\pi$ is a
Riemannian submersion.

The following is the key lemma for proving 6.1.1 and the proof of the same can
be found in [Michor, 2008, Lemma 26.11].

######  Lemma 6.2.1.

Let $\left(E,g_{E}\right)$ and $\left(B,g_{B}\right)$ be two Riemannian
manifolds and $\pi:E\to B$ be a Riemannian submersion. Let $\gamma$ be a
geodesic in $B$ and $\tilde{\gamma}$ be the horizontal lift of $\gamma$. Then
we have:

  1. 1.

The length of $\gamma$ and $\tilde{\gamma}$ are same.

  2. 2.

$\tilde{\gamma}^{\prime}(t)$ is perpendicular to each fiber
$E_{\tilde{\gamma}(t)}$.

  3. 3.

$\tilde{\gamma}$ is a geodesic in $E$.

We also have the following result.

######  Theorem 6.2.2 (O’Neill).

Let $\pi:E\to B$ be a Riemannian submersion. If $\tilde{\gamma}$ is a geodesic
in $E$ and $\tilde{\gamma}^{\prime}(0)\in\mathcal{H}_{\tilde{\gamma}(0)}$,
then $\tilde{\gamma}^{\prime}(t)\in\mathcal{H}_{\tilde{\gamma}(t)}$ for all
$t$. Moreover, $\pi\circ\tilde{\gamma}$ is a geodesic in $B$ and the length is
preserved.

The above theorem can also be proved using Lemma 6.2.1, see [Michor, 2008,
Corollary 26.12].

Combining Lemma 6.2.1 and 6.2.2, we have the following correspondence.

######  Theorem 6.2.3.

There is a one-to-one correspondence between the geodesics on $M/G$ and
geodesics on $M$ which are horizontal.

We now ready to prove 6.1.1.

###### Proof of 6.1.1.

Note that if $\tilde{\gamma}$ is an $N$-geodesic, then
$\tilde{\gamma}^{\prime}(1)\in\mathcal{H}_{\tilde{\gamma}(1)}$ and from 6.2.2
$\tilde{\gamma}^{\prime}(t)\in\mathcal{H}_{\tilde{\gamma}(t)}$ and hence
$\tilde{\gamma}$ is a horizontal geodesic which implies
$\gamma=\pi\circ\tilde{\gamma}$ is a geodesic in $M/G$.

Let $\tilde{p}\in\mathrm{Se}(N)$. This implies there exists at least two
$N$-geodesic, say $\tilde{\gamma}$ and $\tilde{\eta}$ such that
$l(\tilde{\gamma})=l(\tilde{\eta})=d(\tilde{p},N)$. Let us denote
$\gamma=\pi\circ\tilde{\gamma}$ and $\eta=\pi\circ\tilde{\eta}$. Due to
uniqueness of horizontal lift (6.2.1), $\gamma$ and $\eta$ will be two
different geodesics and the lengths will be same. Also, note that as
$\tilde{\gamma}$ is an $N$-geodesic, $\gamma$ will be an $N/G$ geodesic.
Otherwise, there exists an $N/G$-geodesic $\delta$ joining $p$ to $N/G$ which
gives a horizontal lift $\tilde{\delta}$ whose length is strictly less than
$\tilde{\gamma}$, a contradiction (see Figure 6.2). Hence,
$\pi(\tilde{p})\in\mathrm{Se}(N/G)$.

Figure 6.2: $N$-geodesics maps to $N/G$-geodesics

On the other hand, if $\gamma$ is an $N/G$ geodesic starting from $p$, then
its horizontal lift $\tilde{\gamma}$ will be a geodesic. In fact, it will be
an $N$-geodesic. If not, let $\tilde{\eta}$ be such that
$l\left(\tilde{\eta}\right)=d\left(\tilde{p},N\right)$ which implies
$\tilde{\eta}$ is horizontal. Hence, $\eta$ will be a geodesic and

$\displaystyle l(\gamma)$ $\displaystyle=d\left(p,N/G\right)=l(\eta)$
$\displaystyle=l\left(\tilde{\eta}\right)<l\left(\tilde{\gamma}\right)=l(\gamma),$

a contradiction. Thus $\tilde{p}\in\mathrm{Se}(N)$. This proves that
$\mathrm{Se}(N)/G=\mathrm{Se}(N/G)$. In order to prove the theorem, note that
we have the following relation.

$\mathrm{Se}(N/G)\subseteq\overline{\mathrm{Se}(N)}/G\subseteq\overline{\mathrm{Se}(N/G)}.$

As $\overline{\mathrm{Se}(N)}$ is a closed set and $G$ is a compact Lie group,
so $\overline{\mathrm{Se}(N)}/G$ is closed. Thus, we have

$\overline{\mathrm{Se}(N)}/G=\overline{\mathrm{Se}(N/G)}\implies\mathrm{Cu}(N)/G=\mathrm{Cu}(N/G).$

∎

We will discuss some examples based on the above result. Recall from 3.1.5,

$\mathrm{Cu}\left(\mathbb{S}^{k}_{i}\right)=\mathbb{S}^{n-k-1}_{l}.$

######  Example 6.2.1 (Real projective spaces).

Take $M=\mathbb{S}^{n},~{}N=\mathbb{S}^{k}_{i}$ and
$G=\mathbb{S}^{0}\cong\mathbb{Z}_{2}$. Applying 6.1.1, we get

$\mathrm{Cu}\left(\mathbb{RP}^{k}_{i}\right)\cong\mathbb{RP}^{n-k-1}_{l}.$

######  Example 6.2.2.

Take $M=\mathbb{S}^{2n+1},~{}N=\mathbb{S}^{2k+1}_{i}$ and $G=\mathbb{S}^{1}$.
Applying 6.1.1, we get

$\mathrm{Cu}\left(\mathbb{CP}^{k}_{i}\right)\cong\mathbb{CP}^{n-k-1}_{l}.$

######  Example 6.2.3 (Lens space).

Consider $\mathbb{S}^{2n+1}\subset\mathbb{C}^{n+1}$. Let $p$ be a prime number
and $\xi=e^{\frac{2\iota\pi}{p}}$ be a primitive $p^{\text{th}}$ root of unity
and let $q_{1},\cdots,q_{n+1}$ be integers coprime to $p$. Consider
$\mathbb{Z}_{p}=\left\\{1,\xi,\xi^{2},\cdots,\xi^{p-1}\right\\}$ and let it
acts on $\mathbb{S}^{2n+1}$ by
$\xi\left(z_{1},\cdots,z_{n+1}\right)\vcentcolon=\left(\xi^{q_{1}}z_{1},\cdots,\xi^{q_{n+1}}z_{n+1}\right)$.
The orbit space is denoted by $L(p;q_{1},\ldots,q_{n+1})$ and is called a
_lens space_. Since
$\mathrm{Cu}\left(\mathbb{S}^{2n-1}\subset\mathbb{S}^{2n+1}\right)=\mathbb{S}^{1}$,
so taking the $\mathbb{Z}_{p}$ action, gives that the cut locus of
$L\left(p;q_{1},\ldots,q_{n}\right)$ in $L\left(p;q_{1},\ldots,q_{n+1}\right)$
is $\mathbb{S}^{1}$. In general, we have

$\mathrm{Cu}\left(L\left(p;q_{1},\ldots,q_{k+1}\right)\right)\cong
L\left(p;q_{k+2},\ldots,q_{n+1}\right).$

###  6.3 Cut locus of hypersurface in complex projective space

Let $\mathbf{z}=\left(z_{0},\cdots,z_{n}\right)\in\mathbb{C}^{n+1}$ and
$[\mathbf{z}]\in\mathbb{CP}^{n}$, then the Fermat hypersurface of degree $d$
is given by the polynomial $f(\mathbf{z})=z_{0}^{d}+\cdots+z_{n}^{d}$,

$X(d)\vcentcolon=\\{[\mathbf{z}]:f(\mathbf{z})=0\\}.$

The homotopy type of the complement of the above hypersurface is well studied
in the article [Kulkarni and Wood, 1980]. Since the partial derivatives
$\frac{\partial f}{\partial z_{j}}$ do not vanish simultaneously on
$\mathbb{C}^{n+1}-\\{0\\}$, the hypersurface is nonsingular. We wish to find
its cut locus and want to compare our result with the result in [Kulkarni and
Wood, 1980, Proposition 3.1]. In that paper, the authors found the homotopy
type of the complement of the hypersurface.

######  Proposition 6.3.1 ([Kulkarni and Wood, 1980]).

The complement of $X(d)$ is homotopic to the base space of the $n$ universal
principal $\mathbb{Z}_{d}$-bundle constructed by Milnor from the join of $n+1$
copies of $\mathbb{Z}_{d}$.

Using 4.3.5, the same can be studied by looking at the cut locus of $X(d)$ in
$\mathbb{CP}^{n}$. We will use 6.1.1 to find the cut locus of $X(d)$. Recall
that there is a principal $\mathbb{S}^{1}$-bundle given by

$\pi:\mathbb{S}^{2n+1}\to\mathbb{CP}^{n},~{}(z_{0},\cdots,z_{n})\mapsto[z_{0}:\cdots:z_{n}].$

Therefore, using 6.1.1 it is enough to find the cut locus of
$\pi^{-1}(X(d))\vcentcolon=\tilde{X}(d)$. We propose the following conjecture.

######  Conjecture.

The cut locus of $\tilde{X}(d)\subseteq\mathbb{S}^{2n+1}$ is
$\mathbb{Z}_{d}^{\star(n+1)}\times_{\mathbb{Z}_{d}}\mathbb{S}^{1}$, where
$\times_{\mathbb{Z}_{d}}$ is the diagonal action of $\mathbb{Z}_{d}$ and
$\star$ denotes the topological join of spaces.

Using 6.1.1, the cut locus of $X(d)$ will be $\mathbb{Z}_{d}^{\star(n+1)}$ and
hence we are recovering the $n^{\text{th}}$-stage of Milnor join construction
of classifying space for $\mathbb{Z}_{d}$. Note that we can see
$\mathbb{Z}_{d}\subset\mathbb{S}^{1}$ and $\mathbb{S}^{2n+1}$ is the join of
$n$ circles. Hence, $\mathbb{Z}_{d}^{\star(n+1)}\subset\mathbb{S}^{2n+1}$ and
also note that

$\mathbb{Z}_{d}^{\star(n+1)}\times\mathbb{S}^{1}\hookrightarrow\mathbb{S}^{2n+1},~{}\left(\mathbf{v},e^{\iota\theta}\right)\mapsto\mathbf{v}e^{-\iota\theta}$

gives a well-defined map from
$\mathbb{Z}_{d}^{\star(n+1)}\times_{\mathbb{Z}_{d}}\mathbb{S}^{1}$ to
$\mathbb{S}^{2n+1}$. We will prove some particular cases of the above
conjecture. More precisely, we will prove the above conjecture holds for $d=2$
and arbitrary $n$ (6.3.1) and $n=1$ and arbitrary $d$ (6.3.2). Let us denote

$\widetilde{\mathrm{Cu}}=\mathrm{Cu}(\tilde{X}(d)),\text{ and
}\widetilde{\mathrm{Se}}=\mathrm{Se}(\tilde{X}(d)).$

######  Theorem 6.3.1 (Cut locus of $X(2)$).

The cut locus of $\tilde{X}(2)\subseteq\mathbb{S}^{2n+1}$ is
$\mathbb{S}^{n}\times_{\mathbb{Z}_{2}}\mathbb{S}^{1}\cong\\{(\mathbf{v}\cos\theta,\mathbf{v}\sin\theta):\mathbf{v}\in\mathbb{S}^{n},\theta\in[0,2\pi]\\}$.
Hence, the cut locus of $X(2)$ in $\mathbb{CP}^{n}$ will be
$\mathbb{RP}^{n+1}$.

###### Proof.

We will show that
$\\{(\mathbf{v}\cos\theta,\mathbf{v}\sin\theta):\mathbf{v}\in\mathbb{S}^{n},\theta\in\mathbb{R}\\}=\mathrm{Se}(\tilde{X}(2))$.
Let $\mathbf{v}\in\mathbb{S}^{n}$ and $\theta\in\mathbb{R}$. Let us write
$z_{j}=x_{j}+\iota y_{j}$. We can write $\tilde{X}(2)$ as

$\displaystyle\tilde{X}(2)$
$\displaystyle=\left\\{\left(z_{0},\cdots,z_{n}\right)\in\mathbb{C}^{n+1}:\sum_{i=0}^{n}z_{i}^{2}=0,\text{
and }\sum_{i=0}^{n}\left|z_{i}\right|^{2}=1\right\\}$
$\displaystyle=\left\\{\left(x_{0},y_{0},\cdots,x_{n},y_{n}\right):\sum_{i=0}^{n}x_{i}^{2}=\frac{1}{2}=\sum_{i=0}^{n}y_{i}^{2},\text{
and }\sum_{i=0}^{n}x_{i}y_{i}=0\right\\}$
$\displaystyle=\left\\{\left(x_{0},x_{1},\cdots,x_{n},y_{0},y_{1},\cdots,y_{n}\right):\sum_{i=0}^{n}x_{i}^{2}=\frac{1}{2}=\sum_{i=0}^{n}y_{i}^{2},\text{
and }\sum_{i=0}^{n}x_{i}y_{i}=0\right\\}.$

If $A\in O(n+1)$, then $\tilde{A}=\left(\begin{array}[]{c|c}A&\mathbf{0}\\\
\hline\cr\mathbf{0}&A\end{array}\right)\in SO(2n+2)$. Note that

  1. i

$\tilde{A}\in\mathrm{Iso}\left(\mathbb{S}^{2n+1}\right)$, where
$\mathrm{Iso}(M)$ denotes the set of all isometries of $M$.

  2. ii

$\tilde{A}$ maps $\tilde{X}(2)$ to itself, and $\widetilde{\mathrm{Cu}}$ to
itself.

Thus, if $p\in\widetilde{\mathrm{Se}}\subseteq\widetilde{\mathrm{Cu}}$ and let
$\gamma$ and $\eta$ be two distance minimal geodesics joining $p$ to
$\tilde{X}(2)$, then $\tilde{A}\gamma$ and $\tilde{A}\eta$ will be two minimal
geodesics joining $\tilde{A}p$ to $\tilde{X}(2)$. As the action of $O(n+1)$ on
$\mathbb{S}^{n}$ is transitive, it suffices to check if
$\mathbf{e}_{1}\cos\theta+\mathbf{e}_{n+2}\sin\theta\in\widetilde{\mathrm{Se}}$.
We can further reduce our work by looking at the matrix

$B=\begin{pmatrix}\begin{array}[]{cccc|cccc}\cos\theta&0&\cdots&0&\sin\theta&0&\cdots&0\\\
0&&&&0&&&\\\ \vdots&&I_{n}&&\vdots&&\mbox{\Large 0}_{n}&\\\ 0&&&&0&&&\\\
\hline\cr-\sin\theta&0&\cdots&0&\cos\theta&0&\cdots&0\\\ 0&&&&0&&&\\\
\vdots&&\mbox{\Large 0}_{n}&&\vdots&&I_{n}&\\\ 0&&&&0&&&\\\
\end{array}\end{pmatrix},$

which is again an isometry of $\mathbb{S}^{2n+1}$ and sends
$\mathbf{e}_{1}\cos\theta+\mathbf{e}_{n+2}\sin\theta$ to $\mathbf{e}_{1}$.
Hence, it is enough to prove that $\mathbf{e}_{1}\in\widetilde{\mathrm{Se}}$.
Note that

$\operatorname{dist}\left(\mathbf{e}_{1},\tilde{X}(2)\right)=\frac{\pi}{4}.$

Let

$\mathbf{v}_{1}=\mathbf{e}_{2n+2},~{}\mathbf{v}_{2}=-\mathbf{e}_{2n+2}.$

Consider the geodesics

$\gamma(t)=\mathbf{e}_{1}\cos t+\mathbf{e}_{2n+2}\sin t,\text{ and
}\eta(t)=\mathbf{e}_{1}\cos t-\mathbf{e}_{2n+2}\sin t,~{}t\in\mathbb{R}.$

Note that $\gamma$ and $\eta$ intersect the set $\tilde{X(2)}$ at
$t=\frac{\pi}{4}$, so their lengths are same, and it is equal to the distance
between $\mathbf{e}_{1}$ and the submanifold. This proves that
$\mathbf{e}_{1}\in\widetilde{\mathrm{Se}}$.

Conversely, if $p=(\mathbf{v},\mathbf{w})\in\widetilde{\mathrm{Se}}$, then we
will show that there exists $\mathbf{u}\in\mathbb{S}^{n}$ and
$\theta\in\mathbb{R}$ such that $\mathbf{v}=\mathbf{u}\cos\theta$ and
$\mathbf{w}=\mathbf{u}\sin\theta$. Note that if $\\{\mathbf{v},\mathbf{w}\\}$
is linearly dependent, then there exists $\theta\in\mathbb{R}$ such that

$\mathbf{v}=\hat{\mathbf{v}}\cos\theta,~{}~{}\mathbf{w}=\hat{\mathbf{v}}\sin\theta$

and hence $p\in\widetilde{\mathrm{Se}}$. So we assume that $\mathbf{v}$ and
$\mathbf{w}$ are linearly independent. Suppose
$p\notin\mathbb{S}^{n}\times_{\mathbb{Z}_{2}}\mathbb{S}^{1}$. We need to show
that $p\notin\widetilde{\mathrm{Se}}$. To the contrary, let us assume that
$p\in\widetilde{\mathrm{Se}}$. Consider a unit speed geodesic $\gamma(t)$ in
the direction of $\mathcal{v}=(\mathbf{v}_{1},\mathbf{v}_{2})\in
T_{(\mathbf{v},\mathbf{w})}\mathbb{S}^{2n+1}$ which implies

$\displaystyle\left\|\mathbf{v}_{1}\right\|^{2}+\left\|\mathbf{v}_{2}\right\|^{2}=1$
(6.1)
$\displaystyle\left\langle\mathbf{v}_{1},\mathbf{v}\right\rangle+\left\langle\mathbf{v}_{2},\mathbf{w}\right\rangle=0$
(6.2)

Consider the curve $\gamma(t)=\left(\mathbf{v}\cos t+\mathbf{v}_{1}\sin
t,\mathbf{w}\cos t+\mathbf{v}_{2}\sin t\right)$ for $t\in\mathbb{R}$. Note
that
$2\leq\operatorname{rank}[\mathbf{v},\mathbf{w},\mathbf{v}_{1},\mathbf{v}_{2}]\leq
4$. We will prove that none of the cases is possible. Since
$p\in\widetilde{\mathrm{Se}}$, so $p\in\widetilde{\mathrm{Cu}}$ which implies
there exists $t\in\mathbb{R}$ such that
$\gamma(t)\in\tilde{X}(2),\gamma^{\prime}(t)\in\left(T_{\gamma(t)}\tilde{X}(2)\right)^{\perp}$
and $t$ will be minimum among all such values. So we have

$\displaystyle\left\|\mathbf{v}\right\|^{2}\cos^{2}t+\left\|\mathbf{v}_{1}\right\|^{2}\sin^{2}t+\left\langle\mathbf{v},\mathbf{v}_{1}\right\rangle\sin
2t=\frac{1}{2},$ (6.3)
$\displaystyle\left\|\mathbf{w}\right\|^{2}\cos^{2}t+\left\|\mathbf{v}_{2}\right\|^{2}\sin^{2}t+\left\langle\mathbf{w},\mathbf{v}_{2}\right\rangle\sin
2t=\frac{1}{2},$
$\displaystyle\left\langle\mathbf{v},\mathbf{w}\right\rangle\cos^{2}t+\left\langle\mathbf{v}_{1},\mathbf{v}_{2}\right\rangle\sin^{2}t+\frac{1}{2}\left(\left\langle\mathbf{v},\mathbf{v}_{2}\right\rangle+\left\langle\mathbf{v}_{1},\mathbf{w}\right\rangle\right)\sin
2t=0.$ (6.4)

Now we will make use of the other condition
$\gamma^{\prime}(t)\in\left(T_{\gamma(t)}\tilde{X}(2)\right)^{\perp}$.
Consider the vector $\mathcal{u}=\left(-\mathbf{w}\cos t+\mathbf{v}_{2}\sin
t,\mathbf{v}\cos t+\mathbf{v}_{1}\sin t\right)$. We claim that $\mathcal{u}\in
T_{\gamma(t)}\tilde{X}(2)$. Note that $(\mathbf{v},\mathbf{w})\in
T_{(\mathbf{p},\mathbf{q})}\tilde{X}(2)$ implies
$\left\langle\mathbf{p},\mathbf{v}\right\rangle=0,\left\langle\mathbf{q},\mathbf{w}\right\rangle=0$
and
$\left\langle\mathbf{p},\mathbf{w}\right\rangle+\left\langle\mathbf{q},\mathbf{v}\right\rangle=0$.

$\displaystyle\left\langle\mathbf{p},\mathbf{v}\right\rangle$
$\displaystyle=\left\langle\left(\mathbf{v}\cos t+\mathbf{v}_{1}\sin
t\right),\left(-\mathbf{w}\cos t-\mathbf{v}_{2}\sin t\right)\right\rangle$
$\displaystyle=-\left\langle\mathbf{v},\mathbf{w}\right\rangle\cos^{2}t-\left\langle\left(\mathbf{v},\mathbf{v}_{2}\right\rangle+\left\langle\mathbf{v}_{1},\mathbf{w}\right\rangle\right)\cos
t\sin t-\left\langle\mathbf{v}_{1},\mathbf{v}_{2}\right\rangle\sin^{2}t$
$\displaystyle=0\qquad(\text{from \eqref{eq:geodesic-on-X(2)-3}}).$

Similarly, $\left\langle\mathbf{q},\mathbf{w}\right\rangle=0$, and

$\displaystyle\kern
14.22636pt\left\langle\mathbf{p},\mathbf{w}\right\rangle+\left\langle\mathbf{q},\mathbf{v}\right\rangle$
$\displaystyle=\left\langle\mathbf{v}\cos t+\mathbf{v}_{1}\sin
t,\mathbf{v}\cos t+\mathbf{v}_{1}\sin
t\right\rangle+\left\langle\mathbf{w}\cos t+\mathbf{v}_{2}\sin
t,-\mathbf{w}\cos t-\mathbf{v}_{2}\sin t\right\rangle$
$\displaystyle=\left\|\mathbf{v}\right\|^{2}\cos^{2}t+\left\|\mathbf{v}_{1}\right\|^{2}\sin^{2}t+2\left\langle\mathbf{v},\mathbf{v}_{1}\right\rangle\cos
t\sin
t-\left\|\mathbf{w}\right\|^{2}\cos^{2}t-\left\|\mathbf{v}_{2}\right\|^{2}\sin^{2}t$
$\displaystyle\kern
56.9055pt-2\left\langle\mathbf{w},\mathbf{v}_{2}\right\rangle\cos t\sin t$
$\displaystyle=\cos^{2}t\left(\left\|\mathbf{v}\right\|^{2}-\left\|\mathbf{w}\right\|^{2}\right)+\sin^{2}t\left(\left\|\mathbf{v}_{1}\right\|^{2}-\left\|\mathbf{v}_{2}\right\|^{2}\right)+\sin
2t\left(\left\langle\mathbf{v},\mathbf{v}_{1}\right\rangle-\left\langle\mathbf{w},\mathbf{v}_{2}\right\rangle\right)$
$\displaystyle=0\qquad(\text{from \eqref{eq:tangent_space_of_sphere_cond-1}
and \eqref{eq:tangent_space_of_sphere_cond-2}}).$

Therefore, $\mathcal{u}\in T_{\gamma(t)}\tilde{X}(2)$ and hence
$\left\langle\mathcal{u},\gamma^{\prime}(t)\right\rangle=0$ which implies

$\left\langle\mathbf{v},\mathbf{v}_{2}\right\rangle-\left\langle\mathbf{v}_{1},\mathbf{w}\right\rangle=0.$
(6.5)

Define

$\tilde{\mathbf{u}}_{1}=\sqrt{2}\left(\mathbf{v}\cos t+\mathbf{v}_{1}\sin
t\right),\text{ and }\tilde{\mathbf{u}}_{2}=\sqrt{2}\left(\mathbf{w}\cos
t+\mathbf{v}_{2}\sin t\right).$

Note that $\tilde{\mathbf{u}}_{1}\perp\tilde{\mathbf{u}}_{2}$ and both are
vectors in $\mathbb{R}^{n+1}$. We extend
$\left\\{\tilde{\mathbf{u}}_{1},\tilde{\mathbf{u}}_{2}\right\\}$ to an
orthonormal basis of $\mathbb{R}^{n+1}$, say
$\left\\{\tilde{\mathbf{u}}_{1},\tilde{\mathbf{u}}_{2},\tilde{\mathbf{u}}_{3},\ldots,\tilde{\mathbf{u}}_{n+1}\right\\}$.
If $\left(\mathbf{w}_{1},\mathbf{w}_{2}\right)\in T_{\gamma(t)}\tilde{X}(2)$,
then $\left\langle\mathbf{w}_{1},\tilde{\mathbf{u}}_{1}\right\rangle=0$,
$\left\langle\mathbf{w}_{2},\tilde{\mathbf{u}}_{2}\right\rangle=0$ and
$\left\langle\mathbf{w}_{1},\tilde{\mathbf{u}}_{2}\right\rangle+\left\langle\mathbf{w}_{2},\tilde{\mathbf{u}}_{1}\right\rangle=0$
as
$\gamma(t)=\frac{1}{\sqrt{2}}(\tilde{\mathbf{u}}_{1},\tilde{\mathbf{u}}_{2})$.
This implies
$\mathbf{w}_{1},\mathbf{w}_{2}\in\operatorname{Span}\left\\{\tilde{\mathbf{u}}_{3},\ldots,\tilde{\mathbf{u}}_{n+1}\right\\}$
or

$\mathbf{w}_{1}=-\tilde{\mathbf{u}}_{2}+\sum_{j\geq
3}c_{j}\tilde{\mathbf{u}}_{j}\text{ and
}\mathbf{w}_{2}=\tilde{\mathbf{u}}_{1}+\sum_{j\geq
3}d_{j}\tilde{\mathbf{u}}_{j}.$

Since $\mathbf{w}_{i}\in T_{\gamma(t)}\tilde{X}(2)$,

$\displaystyle\left\langle\gamma^{\prime}(t),\left(\mathbf{w}_{1},\mathbf{w}_{2}\right)\right\rangle=0$
$\displaystyle\implies\text{ for }j\geq
3,~{}\left\langle\gamma^{\prime}(t),\left(\tilde{\mathbf{u}}_{j},0\right)\right\rangle=0,\text{
and
}\left\langle\gamma^{\prime}(t),\left(0,\tilde{\mathbf{u}}_{j}\right)\right\rangle=0.$

The above implies

$-\mathbf{v}\sin t+\mathbf{v}_{1}\cos t,-\mathbf{w}\sin t+\mathbf{v}_{2}\cos
t\in\operatorname{Span}\left\\{\tilde{\mathbf{u}}_{1},\tilde{\mathbf{u}}_{2}\right\\}.$
(6.6)

Since $-\mathbf{v}\sin t+\mathbf{v}_{1}\cos
t\in\operatorname{Span}\left\\{\tilde{\mathbf{u}}_{1},\tilde{\mathbf{u}}_{2}\right\\}$,

$\displaystyle-\mathbf{v}\sin t+\mathbf{v}_{1}\cos
t=\alpha\sqrt{2}\left(\mathbf{v}\cos t+\mathbf{v}_{1}\sin
t\right)+\beta\sqrt{2}\left(\mathbf{w}\cos t+\mathbf{v}_{2}\sin t\right)$
$\displaystyle\implies$ $\displaystyle\mathbf{v}(\sin t-\alpha\sqrt{2}\cos
t)+\mathbf{w}(-\beta\sqrt{2}\cos t)+$ $\displaystyle\kern
42.67912pt\mathbf{v}_{1}(\cos t-\alpha\sqrt{2}\sin
t)+\mathbf{v}_{2}(\beta\sqrt{2}\sin t)=0.$ (6.7)

If
$\operatorname{rank}[\mathbf{v},\mathbf{w},\mathbf{v}_{1},\mathbf{v}_{2}]=4$,
then from equation (6.7)

$\displaystyle\sin t-\alpha\sqrt{2}\cos t=0=\cos t-\alpha\sqrt{2}\sin
t,~{}\text{and }-\beta\cos t=0=\beta\sin t.$

This implies $2\alpha^{2}=-1$, which is absurd. Thus, all four vectors can not
be linearly independent. Now we are remaining with two cases:
$\operatorname{rank}[\mathbf{v},\mathbf{w},\mathbf{v}_{1},\mathbf{v}_{2}]=2$
or
$\operatorname{rank}[\mathbf{v},\mathbf{w},\mathbf{v}_{1},\mathbf{v}_{2}]=3$.

Case 1:
$\operatorname{rank}[\mathbf{v},\mathbf{w},\mathbf{v}_{1},\mathbf{v}_{2}]=2:$
Let $\mathbf{v}_{1}=\alpha\mathbf{v}+\beta\mathbf{w}$ and
$\mathbf{v}_{2}=\gamma\mathbf{v}+\delta\mathbf{w}$ for some
$\alpha,\beta,\gamma,\delta\in\mathbb{R}$. Observe that

$\displaystyle\eqref{eq:perp-cond}\implies$
$\displaystyle\gamma\left\|\mathbf{v}\right\|^{2}-\beta\left\|\mathbf{w}\right\|^{2}=(\alpha-\delta)\left\langle\mathbf{v},\mathbf{w}\right\rangle$
$\displaystyle\implies$
$\displaystyle(\gamma+\beta)\left\|\mathbf{v}\right\|^{2}-(\alpha-\delta)\left\langle\mathbf{v},\mathbf{w}\right\rangle=\beta$
(6.8) $\displaystyle\eqref{eq:tangent_space_of_sphere_cond-2}\implies$
$\displaystyle\alpha\left\|\mathbf{v}\right\|^{2}+\delta\left\|\mathbf{w}\right\|^{2}=-(\beta+\gamma)\left\langle\mathbf{v},\mathbf{w}\right\rangle$
$\displaystyle\implies$
$\displaystyle(\alpha-\delta)\left\|\mathbf{v}\right\|^{2}+(\beta+\gamma)\left\langle\mathbf{v},\mathbf{w}\right\rangle=-\delta.$
(6.9)

Using equations (6.8) and (6.9), we obtain

$\left\|\mathbf{v}\right\|^{2}\left((\alpha-\delta)^{2}+(\beta+\gamma)^{2}\right)=\beta\gamma+\beta^{2}-\alpha\delta+\delta^{2}.$

Note that $(\alpha-\delta)^{2}+(\beta+\gamma)^{2}=0$ implies $\alpha=\delta$
and $\beta=-\gamma$. But from equations (6.8) and (6.9) implies that
$\alpha=\beta=\delta=\gamma=0$, which is not possible. Thus,

$\begin{split}&\left\|\mathbf{v}\right\|^{2}=\frac{\beta\gamma+\beta^{2}-\alpha\delta+\delta^{2}}{(\alpha-\delta)^{2}+(\beta+\gamma)^{2}},\\\
&\left\|\mathbf{w}\right\|^{2}=\frac{\beta\gamma+\delta^{2}-\alpha\delta+\beta^{2}}{(\alpha-\delta)^{2}+(\beta+\gamma)^{2}},\text{
and }\\\
&\left\langle\mathbf{v},\mathbf{w}\right\rangle=\frac{-(\alpha\beta+\gamma\delta)}{(\alpha-\delta)^{2}+(\beta+\gamma)^{2}}.\end{split}$

From equation (6.3), we have

$\displaystyle\left(\alpha^{2}+\gamma^{2}\right)\left\|\mathbf{v}\right\|^{2}+2\left(\alpha\beta+\gamma\delta\right)\left\langle\mathbf{v},\mathbf{w}\right\rangle+\left(\beta^{2}+\delta^{2}\right)\left\|\mathbf{w}\right\|^{2}=1$
$\displaystyle\begin{split}\implies&\left(\alpha^{2}+\gamma^{2}\right)\left(\beta\gamma+\beta^{2}-\alpha\delta+\delta^{2}\right)-2(\alpha\beta+\gamma\delta)^{2}\\\
&\kern
28.45274pt+\left(\beta^{2}+\delta^{2}\right)\left(\beta\gamma+\delta^{2}-\alpha\delta+\beta^{2}\right)-(\alpha-\delta)^{2}-(\beta+\gamma)^{2}=0\end{split}$
$\displaystyle\begin{split}\implies&(\beta\gamma-\alpha\delta)\left(\sum\alpha^{2}\right)+2\left(\alpha^{2}+\gamma^{2}\right)\left(\beta^{2}+\delta^{2}\right)-2(\alpha\beta+\gamma\delta)^{2}\\\
&\kern
56.9055pt-\left(\sum\alpha^{2}\right)+2(\alpha\delta-\beta\gamma)=0\end{split}$
$\displaystyle\implies$
$\displaystyle(\beta\gamma-\alpha\delta-1)\left(\sum\alpha^{2}\right)+2(\beta\gamma-\alpha\delta)^{2}-2(\beta\gamma-\alpha\delta)=0$
$\displaystyle\implies$
$\displaystyle(\beta\gamma-\alpha\delta-1)\left(\sum\alpha^{2}\right)+2(\beta\gamma-\alpha\delta)(\beta\gamma-\alpha\delta-1)=0$
$\displaystyle\implies$
$\displaystyle(\beta\gamma-\alpha\delta-1)\left((\alpha-\delta)^{2}+(\beta+\delta)^{2}\right)=0,$

which implies

$\beta\gamma-\alpha\delta=1.$ (6.10)

Now we will use the condition that $\gamma(t)\in\tilde{X}(2)$ which was given
by equations (6.3) and (6.4). From (6.3) we have

$\displaystyle\left\|\mathbf{v}\right\|^{2}\cos^{2}+\left(\alpha^{2}\left\|\mathbf{v}\right\|^{2}+\beta^{2}\left\|\mathbf{w}\right\|^{2}+2\alpha\beta\left\langle\mathbf{v},\mathbf{w}\right\rangle\right)\sin^{2}t$
$\displaystyle\kern
56.9055pt+\left(\alpha\left\|\mathbf{v}\right\|^{2}+\beta\left\langle\mathbf{v},\mathbf{w}\right\rangle\right)\sin
2t=\frac{1}{2}$
$\displaystyle\begin{split}\implies&\sin^{2}t\left(\alpha^{2}\left(1+\beta^{2}+\delta^{2}\right)+\beta^{2}\left(1+\alpha^{2}+\gamma^{2}\right)-2\alpha\beta(\alpha\beta+\gamma\delta)\right)\\\
&\kern 42.67912pt+\cos^{2}t\left(1+\beta^{2}+\delta^{2}\right)\\\ &\kern
42.67912pt+\sin
2t\left(\alpha\left(1+\beta^{2}+\delta^{2}\right)-\beta(\alpha\beta+\gamma\delta)\right)=\frac{2+\sum\alpha^{2}}{2}\end{split}$
$\displaystyle\implies$
$\displaystyle\left(\alpha^{2}+\beta^{2}+1\right)\sin^{2}t+\left(\beta^{2}+\delta^{2}+1\right)\cos^{2}t+(\alpha-\delta)\sin
2t=\frac{2+\sum\alpha^{2}}{2}$ $\displaystyle\implies$
$\displaystyle\alpha^{2}+\beta^{2}+\cos^{2}t\left(\beta^{2}+\delta^{2}+1-\alpha^{2}-\beta^{2}+1\right)+(\alpha-\delta)\sin
2t=\frac{\sum\alpha^{2}}{2}$ $\displaystyle\implies$
$\displaystyle\left(\delta^{2}-\alpha^{2}\right)\cos^{2}t+(\alpha-\delta)\sin
2t-\frac{\gamma^{2}+\delta^{2}-\alpha^{2}-b^{2}}{2}=0,$

which simplifies to

$\left(\delta^{2}-\alpha^{2}\right)\cos 2t+2(\alpha-\delta)\sin
2t=\gamma^{2}-\beta^{2}.$

Similarly, using (6.4) we have

$-(\alpha+\delta)(\beta+\gamma)\cos 2t+2(\beta+\gamma)\sin
2t=(\alpha-\delta)(\beta-\gamma).$

Writing the last two relation into matrix form we have

$\displaystyle\begin{bmatrix}-(\alpha-\delta)(\alpha+\delta)&2(\alpha-\delta)\\\
-(\alpha+\delta)(\beta+\gamma)&2(\beta+\gamma)\end{bmatrix}\begin{bmatrix}\cos
2t\\\ \sin 2t\end{bmatrix}=\begin{bmatrix}\gamma^{2}-\beta^{2}\\\
(\alpha-\delta)(\beta-\gamma)\end{bmatrix}.$ (6.11)

Note that the rank of the coefficient matrix is $1$ and this can occur if one
row is linear multiple of the other.

  * •

Let $\alpha=\delta$ and $\beta\neq-\gamma$. The system (6.11) has a solution,
so $\beta-\gamma=0$. Note that

$\displaystyle\eqref{eq:rk-2_cond-1}\implies\beta\left(\left\|\mathbf{v}\right\|^{2}-\left\|\mathbf{w}\right\|^{2}\right)=0\implies\beta=0\text{
or }\left\|\mathbf{v}\right\|^{2}=\left\|\mathbf{w}\right\|^{2}.$

But, note that $\beta=0$ can not be possible because if $\beta=0=\gamma$, then
using (6.10) $\alpha^{2}=-1$, a contradiction. Therefore,
$\left\|\mathbf{v}\right\|^{2}=\frac{1}{2}=\left\|\mathbf{w}\right\|^{2}$.
Again using (6.9),

$\displaystyle\left\langle\mathbf{v},\mathbf{w}\right\rangle=\frac{-\alpha}{2\beta}.$

Since

$\displaystyle\left\|\mathbf{v}_{1}\right\|^{2}+\left\|\mathbf{v}_{2}\right\|^{2}=1$
$\displaystyle\implies\frac{\left(\alpha^{2}+\beta^{2}\right)}{2}+4\alpha\beta\left(\frac{-\alpha}{2\beta}\right)+\frac{\left(\alpha^{2}+\beta^{2}\right)}{2}=1$
$\displaystyle\implies\beta^{2}-\alpha^{2}=1.$

Consider

$\displaystyle\left\langle\mathbf{v},\mathbf{w}\right\rangle=-\frac{\alpha}{2\beta}$
$\displaystyle\implies
4\beta^{2}\left\langle\mathbf{v},\mathbf{w}\right\rangle^{2}=\alpha^{2}=\beta^{2}-1$
$\displaystyle\implies\beta^{2}=\frac{1}{1-4\left\langle\mathbf{v},\mathbf{w}\right\rangle^{2}},~{}\alpha^{2}=\frac{4\left\langle\mathbf{v},\mathbf{w}\right\rangle^{2}}{1-4\left\langle\mathbf{v},\mathbf{w}\right\rangle^{2}}.$

Note that the above expression is valid as $\mathbf{v}$ and $\mathbf{w}$ are
linearly independent and each has norm $\frac{1}{\sqrt{2}}$. Thus, if
$p\in\widetilde{\mathrm{Se}}$, then the only two possible directions are
$\mathcal{v}_{1}=(\mathbf{v}_{1},\mathbf{v}_{2})$ and
$-\mathcal{v}=(-\mathbf{v}_{1},-\mathbf{v}_{2})$. If $\gamma$ and $\eta$ be
two geodesics in $\mathcal{v}_{1}$ and $-\mathcal{v}_{1}$ direction
respectively, then they intersect $\tilde{X}(2)$ at $t$ and $\pi-t$
respectively. As their lengths are same, and they are unit speed geodesics, so
$t=\pi-t$ which implies $t=\frac{\pi}{2}$. This implies
$\left(\mathbf{v}_{1},\mathbf{v}_{2}\right)\in\tilde{X}(2)$ which means

$\displaystyle\left\langle\mathbf{v}_{1},\mathbf{v}_{2}\right\rangle=0$
$\displaystyle\implies\frac{\alpha\beta}{2}+\frac{\alpha\beta}{2}+2\alpha\beta\left\langle\mathbf{v},\mathbf{w}\right\rangle=0$
$\displaystyle\implies\alpha\beta(2+\left\langle\mathbf{v},\mathbf{w}\right\rangle)=0$
$\displaystyle\implies\alpha=0.$

If $\alpha=0=\delta$, then $\beta^{2}=1$ which implies $\beta=\pm 1$, and
hence $(\mathbf{v},\mathbf{w})\in\tilde{X}(2)$, which is a contradiction.

  * •

Let $\beta+\gamma=0$ or $\alpha-\delta\neq 0$. But $\alpha-\delta\neq 0$, so
$\beta=\gamma=0$ which implies $\mathbf{v}_{1}=\alpha\mathbf{v}$ and
$\mathbf{v}_{2}=\delta\mathbf{w}$. Observe that

$\displaystyle\eqref{eq:rk-2_cond-1}$
$\displaystyle\implies\left\langle\mathbf{v},\mathbf{w}\right\rangle=0\implies\left\langle\mathbf{v}_{1},\mathbf{v}_{2}\right\rangle=0.$
$\displaystyle\eqref{eq:tangent_space_of_sphere_cond-1}$
$\displaystyle\implies\alpha^{2}\left\|\mathbf{v}\right\|^{2}+\delta^{2}\left\|\mathbf{w}\right\|^{2}=1$
$\displaystyle\implies\left\|\mathbf{w}\right\|^{2}\left(\alpha^{2}+\delta^{2}\right)=1-\alpha^{2}$
$\displaystyle\implies\left\|\mathbf{w}\right\|^{2}=\frac{1-\alpha^{2}}{\alpha^{2}+\delta^{2}},\text{
and
}\left\|\mathbf{v}\right\|^{2}=\frac{\delta^{2}-1}{\alpha^{2}+\delta^{2}}.$

Now we use the condition that $\alpha\delta=-1$ to obtain,

$\alpha^{2}=\frac{\left\|\mathbf{w}\right\|^{2}}{\left\|\mathbf{v}\right\|^{2}},\text{
and
}\delta^{2}=\frac{\left\|\mathbf{v}\right\|^{2}}{\left\|\mathbf{w}\right\|^{2}},$

which is again fixed and there are only two possible directions
$\mathcal{v}_{1}$ and $-\mathcal{v}_{1}$ and hence this case is also not
possible.

  * •

Finally, both row are non-zero and let $\alpha+\delta=\lambda(\beta+\gamma)$.
So (6.11) become

$\displaystyle\begin{bmatrix}-(\alpha+\delta)\lambda(\beta+\gamma)&2\lambda(\beta+\gamma)\\\
-(\alpha+\delta)(\beta+\gamma)&2(\beta+\gamma)\end{bmatrix}\begin{bmatrix}\cos
2t\\\ \sin
2t\end{bmatrix}=\begin{bmatrix}-\left(\beta^{2}-\gamma^{2}\right)\\\
\lambda\left(\beta^{2}-\gamma^{2}\right)\end{bmatrix}$ $\displaystyle\implies$
$\displaystyle\begin{bmatrix}-(\alpha+\delta)\lambda&2\lambda\\\
-(\alpha+\delta)&2\end{bmatrix}\begin{bmatrix}\cos 2t\\\ \sin
2t\end{bmatrix}=\begin{bmatrix}-(\beta+\gamma)\\\
\lambda(\beta+\gamma)\end{bmatrix},$

which implies $\lambda^{2}=-1$, a contradiction.

Thus, we have proved that
$\operatorname{rank}[\mathbf{v},\mathbf{w},\mathbf{v}_{1},\mathbf{v}_{2}]\neq
2$.

Case 2:
$\operatorname{rank}[\mathbf{v},\mathbf{w},\mathbf{v}_{1},\mathbf{v}_{2}]=3:$
Since rank is $3$, without loss of generality we assume that
$\mathbf{v}_{2}\in\operatorname{Span}\left\\{\mathbf{v},\mathbf{w},\mathbf{v}_{1}\right\\}$.
Let us write $\mathbf{v}_{2}=a\mathbf{v}+b\mathbf{w}+c\mathbf{v}_{1}$ for some
$a,b,c\in\mathbb{R}$. Since the rank is three we can assume that all the
vectors are in $\mathbb{R}^{3}$ and let $\times$ denote the vector cross
product. Let

$\displaystyle\mathcal{v}_{1}$ $\displaystyle=\left(\mathbf{v}\cos
t+\mathbf{v}_{1}\sin t)\right)\times\left(\mathbf{w}\cos t+\mathbf{v}_{2}\sin
t\right)$
$\displaystyle=(\mathbf{v}\times\mathbf{w})\cos^{2}t+\left(\mathbf{v}\times\mathbf{v}_{2}\right)\cos
t\sin t+\left(\mathbf{v}_{1}\times\mathbf{w}\cos t\sin
t\right)+\left(\mathbf{v}_{1}\times\mathbf{v}_{2}\right)\sin^{2}t.$

Using (6.6), we have

$\displaystyle\mathcal{v}_{1}\cdot\left(-\mathbf{v}\sin t+\mathbf{v}_{1}\cos
t\right)=0$
$\displaystyle\implies\left[\mathbf{v},\mathbf{w},\mathbf{v}_{1}\right]\cos
t+\left[\mathbf{v}_{1},\mathbf{v},\mathbf{v}_{2}\right]\sin t=0$ (6.12)
$\displaystyle\mathcal{v}_{1}\cdot\left(-\mathbf{v}\sin t+\mathbf{v}_{2}\cos
t\right)=0$
$\displaystyle\implies\left[\mathbf{v},\mathbf{w},\mathbf{v}_{2}\right]\cos
t+\left[\mathbf{v}_{1},\mathbf{w},\mathbf{v}_{2}\right]\sin t=0.$ (6.13)

Note that

$\displaystyle\left[\mathbf{v}_{1},\mathbf{v},\mathbf{v}_{2}\right]=\left[\mathbf{v}_{1},\mathbf{v},a\mathbf{v}+b\mathbf{w}+c\mathbf{v}_{1}\right]=b\left(\mathbf{v},\mathbf{w},\mathbf{v}_{1}\right).$

So from (6.12),

$\left[\mathbf{v},\mathbf{w},\mathbf{v}_{1}\right]\cos
t+b\left[\mathbf{v},\mathbf{w},\mathbf{v}_{1}\right]\sin t=0\implies\cos
t+b\sin t=0.$ (6.14)

Above implies,

$\cos t=\frac{-b}{1+b^{2}},\text{ and }\sin t=\frac{1}{1+b^{2}}.$

Similarly, using (6.13),

$\displaystyle\left[\mathbf{v},\mathbf{w},a\mathbf{v}+b\mathbf{w}+c\mathbf{v}_{1}\right]\cos
t+\left[\mathbf{v}_{1},\mathbf{w},a\mathbf{v}+b\mathbf{w}+c\mathbf{v}_{1}\right]\sin
t=0$ $\displaystyle\implies$ $\displaystyle
c\left(\mathbf{v},\mathbf{w},\mathbf{v}_{1}\right)\cos
t+a\left[\mathbf{v}_{1},\mathbf{w},\mathbf{v}\right]\sin t=0$
$\displaystyle\implies$
$\displaystyle\left[\mathbf{v},\mathbf{w},\mathbf{v}_{1}\right](c\cos t-a\sin
t)=0$ $\displaystyle\implies$ $\displaystyle c\cos t=a\sin t.$ (6.15)

Using (6.14) and (6.15) we obtain

$a+bc=0.$

Now we collect some more conditions using previous conditions.

$\displaystyle\eqref{eq:perp-cond}$ $\displaystyle\implies
a\left\|\mathbf{v}\right\|^{2}+b\left\langle\mathbf{v},\mathbf{w}\right\rangle+c\left\langle\mathbf{v},\mathbf{v}_{1}\right\rangle-\left\langle\mathbf{v}_{1},\mathbf{w}\right\rangle=0$
(6.16) $\displaystyle\implies-
bc\left\|\mathbf{v}\right\|^{2}+b\left\langle\mathbf{v},\mathbf{w}\right\rangle+c\left\langle\mathbf{v}_{1},\mathbf{v}\right\rangle-\left\langle\mathbf{v}_{1},\mathbf{w}\right\rangle=0$
$\displaystyle\implies\left(\mathbf{v}_{1}-b\mathbf{v}\right)\cdot(\mathbf{w}-c\mathbf{v})=0$
$\displaystyle\eqref{eq:tangent_space_of_sphere_cond-2}$
$\displaystyle\implies\left\langle\mathbf{v},\mathbf{v}_{1}\right\rangle+a\left\langle\mathbf{v},\mathbf{w}\right\rangle+b\left\|\mathbf{w}\right\|^{2}+c\left\langle\mathbf{v}_{1},\mathbf{w}\right\rangle=0$
(6.17)
$\displaystyle\implies\left\langle\mathbf{v},\mathbf{v}_{1}\right\rangle-
bc\left\langle\mathbf{v},\mathbf{w}\right\rangle+b\left\|w\right\|^{2}+c\left\langle\mathbf{v},\mathbf{v}_{1}\right\rangle=0$
$\displaystyle\implies\mathbf{v}_{1}\cdot(\mathbf{v}-c\mathbf{w})=b\mathbf{w}\cdot(c\mathbf{v}-\mathbf{w}).$

Multiply (6.16) by $c$ and add to (6.17) to obtain

$\left\langle\mathbf{v}_{1},\mathbf{v}\right\rangle=\frac{bc^{2}}{1+c^{2}}\left\|\mathbf{v}\right\|^{2}-\frac{b}{1+c^{2}}\left\|\mathbf{w}\right\|^{2}].$
(6.18)

Similarly,

$\left\langle\mathbf{v}_{1},\mathbf{w}\right\rangle=\frac{-bc}{1+c^{2}}+b\left\langle\mathbf{v},\mathbf{w}\right\rangle.$
(6.19)

We use (6.3) and substitute the value of $\cos t,\sin t$ and use (6.19) and
(6.18) to obtain

$\left\|\mathbf{v}_{1}\right\|^{2}=\frac{b^{2}\left(c^{2}-1\right)}{1+c^{2}}\left\|\mathbf{v}\right\|^{2}-\frac{2b^{2}}{1+c^{2}}\left\|\mathbf{w}\right\|^{2}+\frac{1+b^{2}}{2}.$

Now use $\left\|\gamma(t)\right\|^{2}=1$,

$\displaystyle\frac{b^{2}\left(c^{2}-1\right)}{1+c^{2}}\left\|\mathbf{v}\right\|^{2}+\frac{b^{2}\left(c^{2}-1\right)}{1+c^{2}}\left\|\mathbf{w}\right\|^{2}+\frac{b^{2}\left(c^{2}-1\right)^{2}+\left(c^{2}+1\right)^{2}}{2\left(1+c^{2}\right)}=1$
$\displaystyle\implies$
$\displaystyle\frac{b^{2}\left(c^{2}-1\right)}{1+c^{2}}\left(\left\|\mathbf{v}\right\|^{2}+\left\|\mathbf{w}\right\|^{2}\right)+\frac{b^{2}\left(c^{2}-1\right)^{2}+\left(c^{2}+1\right)^{2}}{2\left(1+c^{2}\right)}=1$
$\displaystyle\implies$ $\displaystyle
2b^{2}\left(c^{2}-1\right)+b^{2}\left(c^{2}-1\right)^{2}=2\left(1+c^{2}\right)-\left(1+c^{2}\right)^{2}$
$\displaystyle\implies$
$\displaystyle\left(c^{2}-1\right)\left(2b^{2}+b^{2}c^{2}-b^{2}\right)=\left(1+c^{2}\right)\left(1-c^{2}\right)$
$\displaystyle\implies$
$\displaystyle\left(c^{2}-1\right)\left(b^{2}+b^{2}c^{2}+1+c^{2}\right)=0$
$\displaystyle\implies$ $\displaystyle c=\pm 1.$

We also have

$\displaystyle\left(\mathbf{v}\cos t+\mathbf{v}_{1}\sin
t\right)\cdot\left(\mathbf{w}\cos
t+\left(a\mathbf{v}+b\mathbf{w}+c\mathbf{v}_{1}\right)\sin t\right)=0$
$\displaystyle\implies$
$\displaystyle\left(-b\mathbf{v}+\mathbf{v}_{1}\right)\cdot\left(-b\mathbf{w}+a\mathbf{v}+b\mathbf{w}+c\mathbf{v}_{1}\right)=0$
$\displaystyle\implies$ $\displaystyle
b^{2}c\left\|\mathbf{v}\right\|^{2}-2bc\left\langle\mathbf{v},\mathbf{v}_{1}\right\rangle+c\left\|\mathbf{v}_{1}\right\|^{2}=0$
$\displaystyle\implies$ $\displaystyle\frac{c\left(1+b^{2}\right)}{2}=0$
$\displaystyle\implies$ $\displaystyle c=0,$

which is a contradiction. Hence, the rank can not be $3$. Therefore,
$\mathbf{v}$ and $\mathbf{w}$ are linearly dependent and hence,
$(\mathbf{v},\mathbf{w})\in\mathbb{S}^{n}\times_{\mathbb{Z}_{2}}\mathbb{S}^{1}$.
∎

We now prove that the cut locus of $\tilde{X}(d)$ when $n=1$ will be
$\left(\mathbb{Z}_{d}\star\mathbb{Z}_{d}\right)\times_{\mathbb{Z}_{d}}\mathbb{S}^{1}$.

######  Note 6.3.1 (Cut locus of $X(2)$ for $n=1$).

For $n=1$, the cut locus of $\tilde{X}(2)$ is

$\mathrm{Cu}(\tilde{X}(2))=\left\\{\frac{1}{2}\left(\cos s+\sin t,\sin s+\cos
t,\sin s-\cos t,-\cos s+\sin t\right):s,t\in\mathbb{R}\right\\}.$

Moreover, the cut locus of $X(2)$ will be $\mathbb{RP}^{1}$.

###### Proof.

Let us write $z_{j}=x_{j}+\iota y_{j}$. Note that

$\displaystyle\tilde{X}(2)$
$\displaystyle=\left\\{\left(z_{0},z_{1}\right)\in\mathbb{C}^{2}:z_{0}^{2}+z_{1}^{2}=0,\text{
and }\left|z_{0}\right|^{2}+\left|z_{1}\right|^{1}=1\right\\}$
$\displaystyle=\left\\{\left(x_{0},y_{0},x_{1},y_{1}\right):x_{0}^{2}+x_{1}^{2}=\frac{1}{2}=y_{0}^{2}+y_{1}^{2},\text{
and }x_{0}y_{0}+x_{1}y_{1}=0\right\\}.$

Since $\left(x_{0},x_{1}\right)$ and $\left(y_{0},y_{1}\right)$ can not be
zero vectors, so without loss of generality, we assume that $x_{0}y_{1}\neq
0$. Since

$\displaystyle
x_{0}y_{0}+x_{1}y_{1}=0\implies\frac{y_{0}}{y_{1}}+\frac{x_{1}}{x_{0}}=0\implies
y_{0}=-y_{1}\left(\frac{x_{1}}{x_{0}}\right).$

Now as

$\displaystyle y_{0}^{2}+y_{1}^{2}=\frac{1}{2}\implies
y_{1}^{2}\left(\frac{x_{1}}{x_{0}}\right)^{2}+y_{1}^{2}=\frac{1}{2}\implies
y_{1}^{2}\left(\frac{x_{1}^{2}}{x_{0}^{2}}+1\right)=\frac{1}{2}\implies
y_{1}=\pm x_{0}.$

Similarly, we have

$x_{1}=\pm y_{0}.$

Therefore,

$\displaystyle\tilde{X}(2)$
$\displaystyle=\left\\{\left(x_{0},y_{0},y_{0},-x_{0}\right):x_{0}^{2}+y_{0}^{2}=\frac{1}{2}\right\\}\sqcup\left\\{\left(x_{0},-y_{0},y_{0},x_{0}\right):x_{0}^{2}+y_{0}^{2}=\frac{1}{2}\right\\}$
$\displaystyle=S^{1}_{1}\sqcup S^{1}_{2}.$

Define a linear transformation

$T:\mathbb{R}^{4}\to\mathbb{R}^{4},~{}(a,b,c,d)\mapsto\frac{1}{\sqrt{2}}(a-d,b+c,b-c,a+d).$

Note that $T$ maps $\mathbb{S}^{3}$ onto $\mathbb{S}^{3}$ and is an isometry
hence it preserves the cut locus. So

$\displaystyle\mathrm{Cu}(T(\tilde{X}(2)))$
$\displaystyle=\mathrm{Cu}\left(T\left(S^{1}_{1}\right)\sqcup
T\left(S^{1}_{2}\right)\right)$
$\displaystyle=\mathrm{Cu}\left(\left\\{(a,b,0,0):a^{2}+b^{2}=1\right\\}\sqcup\left\\{(0,0,c,d):c^{2}+d^{2}=1\right\\}\right)$
$\displaystyle=\left\\{\frac{1}{\sqrt{2}}(\cos s,\sin s,\cos t,\sin
t):s,t\in\mathbb{R}\right\\}.$

Therefore, the cut locus of $X(2)$ can be found by the inverse transformation
which is

$T^{-1}(x,y,z,w)=\frac{1}{2}(x+2,y+z,y-z,w-x).$

Hence,

$\displaystyle\mathrm{Cu}(\tilde{X}(2))$
$\displaystyle=\left\\{\frac{1}{2}\left(\cos s+\sin t,\sin s+\cos t,\sin
s-\cos t,-\cos s+\sin t\right):s,t\in\mathbb{R}\right\\}$
$\displaystyle\cong\mathbb{S}^{1}\times\mathbb{S}^{1}.$

Quotient with $\mathbb{S}^{1}$ give the required cut locus. ∎

######  Theorem 6.3.2.

For $n=1$, we have

$\widetilde{\mathrm{Cu}}=\left(\mathbb{Z}_{d}\star\mathbb{Z}_{d}\right)\times_{\mathbb{Z}_{d}}\mathbb{S}^{1}.$

###### Proof.

Note that

$\mathbb{Z}_{d}\star\mathbb{Z}_{d}=\left\\{\mathbf{v}=(\xi^{k}\cos\phi,\xi^{l}\sin\phi):0\leq
k,l\leq d-1,~{}\text{and }0\leq\phi\leq\frac{\pi}{2}\right\\},$

where $\xi^{k}$ is a $d^{\text{th}}$ root of unity. Let
$\mathbf{v}\in\mathbb{Z}_{d}\star\mathbb{Z}_{d}$ and $\theta\in(0,2\pi)$. We
will show that $\mathbf{v}e^{\iota\theta}\in\widetilde{\mathrm{Se}}$. Due to
$\mathbb{S}^{1}$-symmetry, it is enough to show that
$\mathbf{v}\in\widetilde{\mathrm{Se}}$. Now consider the matrix

$A=\begin{pmatrix}\lambda_{1}&0\\\ 0&\lambda_{2}\end{pmatrix}\in U(2),\text{
such that }\lambda_{1}^{d}=1=\lambda_{2}^{d}.$

We observe that $A$ maps $\tilde{X}(d)$ to itself and hence it is enough to
show that $p=(\cos\phi,0,\sin\phi,0)\in\widetilde{\mathrm{Se}}$. Note that

$\displaystyle\tilde{X}(d)$
$\displaystyle=\left\\{\left(z_{1},z_{2}\right)\in\mathbb{S}^{3}:z_{1}^{d}+z_{2}^{d}=0\right\\}$
$\displaystyle=\left\\{\left(z_{1},z_{2}\right)\in\mathbb{S}^{3}:z_{1}^{d}=-z_{2}^{d},~{}\left|z_{1}\right|=\frac{1}{\sqrt{2}}=\left|z_{2}\right|\right\\}$
$\displaystyle=\left\\{\left(z,\xi z\right)\in\mathbb{S}^{3}:\xi^{d}=-1,\text{
and }\left|z\right|=\frac{1}{\sqrt{2}}\right\\}$
$\displaystyle=\bigsqcup_{k=0}^{d-1}\left\\{(z,\xi
z):\xi=e^{\frac{(2k+1)\pi}{d}},~{}|z|=\frac{1}{\sqrt{2}}\right\\}=\bigsqcup_{k=0}^{d-1}X_{k}$

We now compute the distance of $p$ from $X_{k}$ and will show that the
distance is same with at least two components, which will show that
$p\in\widetilde{\mathrm{Se}}$. Note that for any point
$Q=(x_{1},y_{1},x_{2},y_{2})\in\tilde{X}(d)$, the distance between $p$ and $Q$
is given by (look at the Figure 6.3)

$\theta$
---
$p$
---
$Q=(x_{1},y_{1},x_{2},y_{2})$
---
Figure 6.3: distance of $p$ to $\tilde{X}(d)$

$\mathrm{dist}(p,Q)=\cos^{-1}(p\cdot
Q)=\cos^{-1}\left(x_{1}\cos\phi+x_{2}\sin\phi\right).$

Therefore, the distance between $p$ and the set $\tilde{X}(d)$ is given by

$\displaystyle\mathrm{dist}(p,\tilde{X}(d))$
$\displaystyle=\inf\left\\{\mathrm{dist}(p,Q):Q\in\tilde{X}(d)\right\\}$
$\displaystyle=\inf\left\\{\cos^{-1}\left(x_{1}\cos\phi+x_{2}\sin\phi\right):\left(x_{1},y_{1},x_{2},y_{2}\right)\in\tilde{X}(d)\right\\}.$

As $\cos^{-1}$ is a decreasing function, it is equivalent to maximize
$x_{1}\cos\phi+x_{2}\sin\phi$ such that
$x_{2}=x_{1}\cos\left(\frac{(2k+1)\pi}{d}\right)-y_{1}\sin\left(\frac{(2k+1)\pi}{d}\right)$,
and $x_{1}^{2}+y_{1}^{2}=\frac{1}{2}$. This maximum value will be

$\sqrt{\frac{1+\sin 2\phi\cos\left(\frac{(2k+1)\pi}{d}\right)}{2}}.$

Therefore, the distance from $X_{k}$ will be

$\cos^{-1}\left(\sqrt{\frac{1+\sin
2\phi\cos\left(\frac{(2k+1)\pi}{d}\right)}{2}}\right).$

Note that if $d$ is even, then the above distance is same from $X_{k}$ and
$X_{d-1-k}$, therefore, the point is a separating point and hence is a cut
point. If $d$ is odd, then the above still holds. The only thing to make sure
that the distance from each component is smaller than the distance from
$X_{\frac{d-1}{2}}$, but this is true as $0\leq\phi\leq\frac{\pi}{2}$.
Therefore, we proved that $p\in\widetilde{\mathrm{Se}}$ and hence it is in
$\widetilde{\mathrm{Cu}}$. The inverse inclusion follows with a similar
argument. In fact, if we take any other point, then there will be only one
distance minimal geodesic, that will occur from one component. Let
$p=\left(r_{1}e^{\iota\theta_{1}},r_{2}e^{\iota\theta_{2}}\right)$ be any
point in $\mathbb{S}^{3}$ not in the above form. Note that if
$\theta_{1}=\theta_{2}$, then due to $\mathbb{S}^{1}$-equivariant,
$p\in\widetilde{\mathrm{Se}}$ if and only if
$(r_{1},0,r_{2},0)\in\widetilde{\mathrm{Se}}$, which is a point in the above
form. So we assume that $\theta_{1}\neq\theta_{2}$ and
$\theta_{1},\theta_{2}\in[0,2\pi/d]$. Then we have to minimize the distance
from $p$ to $\tilde{X}(d)$, that is

$\operatorname{dist}(p,\tilde{X}(d))=\inf\left\\{\cos^{-1}(p\cdot(z_{1},z_{2})):(z_{1},z_{2})\in\tilde{X}(d)\right\\}.$

This is equivalent to maximizing the dot product $p\cdot(z_{1},z_{2})$, and
this maximum will be achieved from one component, say $X_{l}$, of
$\tilde{X}(d)$ as $0<|\theta_{1}-\theta_{2}|<2\pi/d$. A similar computation
shows that there is only one distance minimal geodesic from the component
$X_{l}$, and this shows that $p\notin\widetilde{\mathrm{Se}}$. Hence, the
theorem is proved. ∎

######  Remark 6.3.1.

In [Audin, 2005, §2.2], the author has shown that the function

$f:\mathbb{CP}^{n}\to\mathbb{R},~{}[x+\iota
y]\mapsto\dfrac{\left\|y\right\|^{2}}{\left(1+\left\|y\right\|^{2}\right)^{2}}$

is a Morse-Bott function with two critical submanifolds $-$ $X(2)$ and
$\mathbb{RP}^{n}$. We know that $\mathbb{CP}^{n}\setminus X(2)$ deformation
retracts to $\mathbb{RP}^{n}$ via Morse-Bott flow, whereas Lemma 4.3.1 implies
that $\mathbb{CP}^{n}\setminus X(2)$ deformation retracts to
$\mathrm{Cu}(X(2))$. Hence, $\mathrm{Cu}(X(2))$ and $\mathbb{RP}^{n}$ have the
same homotopy type. However, it’s not a priori clear whether they are equal.
Moreover, our calculation shows a computation of
$\mathrm{Cu}(T_{1}\mathbb{S}^{n})\subseteq\mathbb{S}^{2n+1}$, where
$T_{1}\mathbb{S}^{n}$ is the unit tangent bundle of $\mathbb{S}^{n}$. We
deduce that the cut locus is the unique non-trivial $\mathbb{S}^{1}$-bundle of
$\mathbb{RP}^{n}$ and this is a new computation.

## Bibliography

  * [Angulo Ardoy and Guijarro, 2011] Angulo Ardoy, P. and Guijarro, L. (2011).  Balanced split sets and Hamilton-Jacobi equations.  Calc. Var. Partial Differential Equations, 40(1-2):223–252.
  * [Audin, 2005] Audin, M. (2005).  On the topology of Lagrangian submanifolds. Examples and counter-examples.  Port. Math. (N.S.), 62(4):375–419.
  * [Banyaga and Hurtubise, 2004] Banyaga, A. and Hurtubise, D. (2004).  Lectures on Morse homology, volume 29 of Kluwer Texts in the Mathematical Sciences.  Kluwer Academic Publishers Group, Dordrecht.
  * [Barratt and Milnor, 1962] Barratt, M. G. and Milnor, J. (1962).  An example of anomalous singular homology.  Proc. Amer. Math. Soc., 13:293–297.
  * [Basu and Prasad, 2021] Basu, S. and Prasad, S. (2021).  A connection between cut locus, Thom space and Morse-Bott functions.  available at https://arxiv.org/abs/2011.02972, to appear in Algebraic & Geometric Topology.
  * [Besse, 1978] Besse, A. L. (1978).  Manifolds all of whose geodesics are closed, volume 93 of Ergebnisse der Mathematik und ihrer Grenzgebiete [Results in Mathematics and Related Areas].  Springer-Verlag, Berlin-New York.  With appendices by D. B. A. Epstein, J.-P. Bourguignon, L. Bérard-Bergery, M. Berger and J. L. Kazdan.
  * [Bott, 1954] Bott, R. (1954).  Nondegenerate critical manifolds.  Ann. of Math. (2), 60:248–261.
  * [Buchner, 1977] Buchner, M. A. (1977).  Simplicial structure of the real analytic cut locus.  Proc. Amer. Math. Soc., 64(1):118–121.
  * [Busemann, 1955] Busemann, H. (1955).  The geometry of geodesics.  Academic Press Inc., New York, N. Y.
  * [Cheeger and Ebin, 1975] Cheeger, J. and Ebin, D. G. (1975).  Comparison theorems in Riemannian geometry.  North-Holland Mathematical Library, Vol. 9. North-Holland Publishing Co., Amsterdam-Oxford; American Elsevier Publishing Co., Inc., New York.
  * [Daniilidis et al., 2018] Daniilidis, A., Deville, R., Durand-Cartagena, E., and Rifford, L. (2018).  Self-contracted curves in Riemannian manifolds.  J. Math. Anal. Appl., 457(2):1333–1352.
  * [do Carmo, 1992] do Carmo, M. P. a. (1992).  Riemannian geometry.  Mathematics: Theory & Applications. Birkhäuser Boston, Inc., Boston, MA.  Translated from the second Portuguese edition by Francis Flaherty.
  * [Fermi, 1922] Fermi, E. (1922).  Sopra i fenomeni che avvengono in vicinanza di una linea oraria.  Rend. Lincei, 31:21–23.
  * [Flaherty, 1965] Flaherty, F. J. (1965).  The cut locus of a submanifold.  PhD thesis.  Thesis (Ph.D.)–University of California, Berkeley.
  * [Gluck and Singer, 1978] Gluck, H. and Singer, D. (1978).  Scattering of geodesic fields. I.  Ann. of Math. (2), 108(2):347–372.
  * [Gray, 2004] Gray, A. (2004).  Tubes, volume 221 of Progress in Mathematics.  Birkhäuser Verlag, Basel, second edition.  With a preface by Vicente Miquel.
  * [Hatcher, 2002] Hatcher, A. (2002).  Algebraic topology.  Cambridge University Press, Cambridge.
  * [Hebda, 1983] Hebda, J. J. (1983).  The local homology of cut loci in Riemannian manifolds.  Tohoku Math. J. (2), 35(1):45–52.
  * [Hebda, 1995] Hebda, J. J. (1995).  Cut loci of submanifolds in space forms and in the geometries of Möbius and Lie.  Geom. Dedicata, 55(1):75–93.
  * [Higham, 2008] Higham, N. J. (2008).  Functions of matrices.  Society for Industrial and Applied Mathematics (SIAM), Philadelphia, PA.  Theory and computation.
  * [Hirsch, 1976] Hirsch, M. W. (1976).  Differential topology.  Graduate Texts in Mathematics, No. 33. Springer-Verlag, New York-Heidelberg.
  * [Hopf and Rinow, 1931] Hopf, H. and Rinow, W. (1931).  Ueber den Begriff der vollständigen differentialgeometrischen Fläche.  Comment. Math. Helv., 3(1):209–225.
  * [Itoh and Sabau, 2016] Itoh, J.-i. and Sabau, S. V. (2016).  Riemannian and Finslerian spheres with fractal cut loci.  Differential Geom. Appl., 49:43–64.
  * [Itoh and Vîlcu, 2011] Itoh, J.-i. and Vîlcu, C. (2011).  Orientable cut locus structures on graphs.  available at https://arxiv.org/abs/1103.3136.
  * [Itoh and Vîlcu, 2015] Itoh, J.-i. and Vîlcu, C. (2015).  Every graph is a cut locus.  J. Math. Soc. Japan, 67(3):1227–1238.
  * [Klingenberg, 1959] Klingenberg, W. (1959).  Contributions to Riemannian geometry in the large.  Ann. of Math. (2), 69:654–666.
  * [Kobayashi, 1967] Kobayashi, S. (1967).  On conjugate and cut loci.  In Studies in Global Geometry and Analysis, pages 96–122. Math. Assoc. Amer. (distributed by Prentice-Hall, Englewood Cliffs, N.J.).
  * [Kulkarni and Wood, 1980] Kulkarni, R. S. and Wood, J. W. (1980).  Topology of nonsingular complex hypersurfaces.  Adv. in Math., 35(3):239–263.
  * [Lang, 1999] Lang, S. (1999).  Fundamentals of differential geometry, volume 191 of Graduate Texts in Mathematics.  Springer-Verlag, New York.
  * [Lee, 2013] Lee, J. M. (2013).  Introduction to smooth manifolds, volume 218 of Graduate Texts in Mathematics.  Springer, New York, second edition.
  * [Li and Nirenberg, 2005] Li, Y. and Nirenberg, L. (2005).  The distance function to the boundary, Finsler geometry, and the singular set of viscosity solutions of some Hamilton-Jacobi equations.  Comm. Pure Appl. Math., 58(1):85–146.
  * [Manasse and Misner, 1963] Manasse, F. K. and Misner, C. W. (1963).  Fermi normal coordinates and some basic concepts in differential geometry.  J. Mathematical Phys., 4:735–745.
  * [Mantegazza and Mennucci, 2003] Mantegazza, C. and Mennucci, A. C. (2003).  Hamilton-Jacobi equations and distance functions on Riemannian manifolds.  Appl. Math. Optim., 47(1):1–25.
  * [Martin and Neff, 2016] Martin, R. J. and Neff, P. (2016).  Minimal geodesics on ${\rm GL}(n)$ for left-invariant, right-${\rm O}(n)$-invariant Riemannian metrics.  J. Geom. Mech., 8(3):323–357.
  * [McKay, 2015] McKay, B. (2015).  A summary of progress on the Blaschke conjecture.  ICCM Not., 3(2):33–45.
  * [Michor, 2008] Michor, P. W. (2008).  Topics in differential geometry, volume 93 of Graduate Studies in Mathematics.  American Mathematical Society, Providence, RI.
  * [Milnor, 1963] Milnor, J. (1963).  Morse theory.  Annals of Mathematics Studies, No. 51. Princeton University Press, Princeton, N.J.  Based on lecture notes by M. Spivak and R. Wells.
  * [Myers, 1935] Myers, S. B. (1935).  Connections between differential geometry and topology. I. Simply connected surfaces.  Duke Math. J., 1(3):376–391.
  * [Myers, 1936] Myers, S. B. (1936).  Connections between differential geometry and topology II. Closed surfaces.  Duke Math. J., 2(1):95–102.
  * [Omori, 1968] Omori, H. (1968).  A class of Riemannian metrics on a manifold.  J. Differential Geometry, 2:233–252.
  * [Ozols, 1974] Ozols, V. (1974).  Cut loci in Riemannian manifolds.  Tohoku Math. J. (2), 26:219–227.
  * [Plotnick, 1982] Plotnick, S. (1982).  Embedding homology $3$-spheres in $S^{5}$.  Pacific J. Math., 101(1):147–151.
  * [Poincaré, 1905] Poincaré, H. (1905).  Sur les lignes géodésiques des surfaces convexes.  Trans. Amer. Math. Soc., 6(3):237–274.
  * [Rauch, 1959] Rauch, H. E. (1959).  Geodesics and curvature in differential geometry in the large, volume 1.  Yeshiva University, Graduate School of Mathematical Sciences.
  * [Sabau and Tanaka, 2016] Sabau, S. V. and Tanaka, M. (2016).  The cut locus and distance function from a closed subset of a Finsler manifold.  Houston J. Math., 42(4):1157–1197.
  * [Sakai, 1977] Sakai, T. (1977).  On cut loci of compact symmetric spaces.  Hokkaido Math. J., 6(1):136–161.
  * [Sakai, 1978] Sakai, T. (1978).  On the structure of cut loci in compact Riemannian symmetric spaces.  Math. Ann., 235(2):129–148.
  * [Sakai, 1979] Sakai, T. (1979).  Cut loci of compact symmetric spaces.  In Minimal submanifolds and geodesics (Proc. Japan-United States Sem., Tokyo, 1977), pages 193–207. North-Holland, Amsterdam-New York.
  * [Sakai, 1984] Sakai, T. (1984).  Comparison and finiteness theorems in Riemannian geometry.  In Geometry of geodesics and related topics (Tokyo, 1982), volume 3 of Adv. Stud. Pure Math., pages 125–181. North-Holland, Amsterdam.
  * [Sakai, 1996] Sakai, T. (1996).  Riemannian geometry, volume 149 of Translations of Mathematical Monographs.  American Mathematical Society, Providence, RI.  Translated from the 1992 Japanese original by the author.
  * [Sharafutdinov, 2007] Sharafutdinov, V. A. (2007).  Proof of soul theorem (translated with some modifications).
  * [Singer and Gluck, 1976] Singer, D. and Gluck, H. (1976).  The existence of nontriangulable cut loci.  Bull. Amer. Math. Soc., 82(4):599–602.
  * [Singh, 1987a] Singh, H. (1987a).  On the cut locus and the focal locus of a submanifold in a Riemannian manifold.  Ann. Univ. Sci. Budapest. Eötvös Sect. Math., 30:139–144.
  * [Singh, 1987b] Singh, H. (1987b).  On the cut locus and the focal locus of a submanifold in a Riemannian manifold. II.  Publ. Inst. Math. (Beograd) (N.S.), 41(55):119–124.
  * [Singh, 1988] Singh, H. (1988).  Closest point of the cut locus to submanifold.  Proc. Indian Acad. Sci. Math. Sci., 98(2-3):179–186.
  * [Takeuchi, 1978] Takeuchi, M. (1978).  On conjugate loci and cut loci of compact symmetric spaces. I.  Tsukuba J. Math., 2:35–68.
  * [Takeuchi, 1979] Takeuchi, M. (1979).  On conjugate loci and cut loci of compact symmetric spaces. II.  Tsukuba J. Math., 3(1):1–29.
  * [Thom, 1972] Thom, R. (1972).  Sur le cut-locus d’une variété plongée.  J. Differential Geometry, 6:577–586.
  * [von Mangoldt, 1881] von Mangoldt, H. (1881).  Ueber diejenigen Punkte auf positiv gekrümmten Flächen, welche die Eigenschaft haben, dass die von ihnen ausgehenden geodätischen Linien nie aufhören, kürzeste Linien zu sein.  J. Reine Angew. Math., 91:23–53.
  * [Wall, 1977] Wall, C. T. C. (1977).  Geometric properties of generic differentiable manifolds.  In Geometry and topology (Proc. III Latin Amer. School of Math., Inst. Mat. Pura Aplicada CNPq, Rio de Janeiro, 1976), Lecture Notes in Math., Vol. 597, pages 707–774. Springer, Berlin.
  * [Whitehead, 1935] Whitehead, J. H. C. (1935).  On the covering of a complete space by the geodesics through a point.  Ann. of Math. (2), 36(3):679–704.
  * [Wolter, 1979] Wolter, F.-E. (1979).  Distance function and cut loci on a complete Riemannian manifold.  Arch. Math. (Basel), 32(1):92–96.
  * [Yakubovich and Starzhinskii, 1975] Yakubovich, V. A. and Starzhinskii, V. M. (1975).  Linear differential equations with periodic coefficients. 1, 2.  Halsted Press [John Wiley & Sons] New York-Toronto, Ont.,; Israel Program for Scientific Translations, Jerusalem-London.  Translated from Russian by D. Louvish.

## Index

  * $C^{1}$-smooth §4.3.1
  * Čech cohomology §3.2
  * conjugate locus §2.3.2
  * conjugate points via exponential map Definition 2.3.4
  * conjugate points via Jacobi fields Definition 2.3.3
  * critical points Definition 2.2.1
  * $\mathrm{Cu}(N)$ Chapter 1, Theorem
  * cut locus Chapter 1
  * cut locus of a point Definition 2.3.2
  * cut locus of a subset Definition 3.1.2
  * cut point §2.3
  * degenerate critical point Definition 2.2.3
  * $df_{p}$ Definition 2.2.1
  * distance minimal geodesic Definition 2.3.1, Definition 3.1.1
  * $d_{N}$ Chapter 1
  * Ehresmann connection Definition 6.2.1
  * equidistant set §3.2, Definition 3.2.1
  * equivariant cut locus theorem Theorem 6.1.1
  * exponential map §4.2
  * $\exp_{\nu}$ §2.1.1
  * Fermi coordinates §1.3, §2.1, §2.1.2
  * first conjugate point §2.3.2
  * first focal point §4.2.1
  * focal point Definition 4.2.2
  * geodesic variation §2.3.2
  * geodesically complete §2.3.1
  * good pair §4.3.2
  * Gregory series §5.2
  * height function of sphere Example 2.2.2
  * height function of torus Example 2.2.3
  * Hessian Definition 2.2.2
  * Hessian matrix §3.2
  * $\mathrm{Hess}_{p}(f)$ 2.4
  * horizontal bundle Definition 6.2.1
  * horizontal lift Definition 6.2.5
  * horizontal lift of a curve Definition 6.2.5
  * horizontal lift of a vector field Definition 6.2.4
  * horizontal vector field 2nd item
  * indefinite unitary groups §5.2
  * index Definition 2.2.3
  * Killing form Chapter 5
  * lens spaces Example 6.2.3
  * Levi-Civita connection Chapter 5
  * Lie algebra Chapter 5
  * Lie group Chapter 5
  * link of a point Definition 3.2.1
  * local homology §3.2
  * Morse function Definition 2.2.4
  * Morse-Bott functions §2.2, §3.2, Chapter 5
  * muliplicity of focal point §4.2.1
  * multiplicity of a conjugate point §2.3.2
  * $N$-geodesic Definition 3.1.1, Definition 3.1.1
  * non-degenerate critical point Definition 2.2.3
  * normal bundle §2.1.1, Theorem 4.3.2
  * normal exponential map §2.1.1
  * one-point compactification §4.3.2