fix(riemannian-geometry): revise do Carmo astrolabe OCR text

projects/riemannian-geometry/.astrolabe/atoms/049b5985dbae.md

@@ -8,13 +8,13 @@ source: tex
 src: docarmo
 title: covering transformations are translations
 ---
-Let $M$ be a compact Riemannian manifold and $\alpha$ a covering transformation
+Let $M$ be a compact Riemannian manifold and let $\alpha\neq\mathrm{ident}$ be a covering transformation
 of $\tilde{M}$, considered with the covering metric. Then $\alpha$ is a
 translation of $\tilde{M}$.
 
 *Proof.* Let $\tilde{p}\in\tilde{M}$ and let $g\in\pi_1(M;p)$, $p=\pi(\tilde{p})$,
 the element corresponding to $\alpha$ under the isomorphism mentioned in the
-introduction of this Section. We can assume $\alpha\neq\mathrm{ident}$. By
+introduction of this Section. By
 Cartan's Theorem (2.2), there exists a closed geodesic $\gamma$ of $M$ in the
 free homotopy class determined by $g$. Choose a point $q\in\gamma$. Then $\gamma$
 is homotopic to the closed path $\sigma g\sigma^{-1}$, where $\sigma$ is a path
@@ -40,8 +40,8 @@ It follows that if $\tilde{\gamma}(s)$ is a point of the lift of $\gamma$ starti
 from $\tilde{q}$, we have, by uniqueness of lifting,
 
 $$
-\alpha(\tilde{\gamma}(s))=\alpha_{\tilde{q}}(\tilde{\gamma}(s))\in\gamma,
+\alpha(\tilde{\gamma}(s))=\alpha_{\tilde{q}}(\tilde{\gamma}(s))\in\tilde{\gamma},
 $$
 
 which shows that $\tilde{\gamma}$ is invariant by $\alpha$, and proves the
-Proposition. $\blacksquare$
+Proposition. $\blacksquare$

projects/riemannian-geometry/.astrolabe/atoms/081d380c8cea.md

@@ -14,10 +14,12 @@ $c(t_0)$, $t_0\in I$ (i.e. $V_0\in T_{c(t_0)}M$). Then there exists a unique
 parallel vector field $V$ along $c$ with $V(t_0)=V_0$; $V(t)$ is called the
 *parallel transport* of $V(t_0)$ along $c$.
 
-*Proof.* By compactness $c([t_0,t_1])$ can be covered by finitely many coordinate
-neighborhoods, and uniqueness makes the local definitions coincide on overlaps,
-so it suffices to prove the theorem when $c(I)$ lies in one coordinate
-neighborhood $\mathbf{x}(U)$. Writing $V=\sum v^j X_j$, parallelism gives
+*Proof.* Suppose first that the theorem is proved when $c(I)$ is contained in a
+local coordinate neighborhood. By compactness, for any $t_1\in I$,
+$c([t_0,t_1])$ can be covered by finitely many coordinate neighborhoods, and
+uniqueness makes the local definitions coincide on overlaps, so it suffices to
+prove the theorem when $c(I)$ lies in one coordinate neighborhood
+$\mathbf{x}(U)$. Writing $V=\sum v^j X_j$, parallelism gives
 $0=\frac{DV}{dt}=\sum_j\frac{dv^j}{dt}X_j+\sum_{i,j}\frac{dx_i}{dt}v^j\nabla_{X_i}X_j$.
 Putting $\nabla_{X_i}X_j=\sum_k\Gamma_{ij}^k X_k$ yields the system of $n$ linear
 ODEs in $v^k(t)$,
@@ -28,4 +30,4 @@ $$
 
 which possesses a unique solution with $v^k(t_0)=v_0^k$. Since the system is
 linear, the solution is defined for all $t\in I$, proving existence and
-uniqueness. $\blacksquare$
+uniqueness. $\blacksquare$

projects/riemannian-geometry/.astrolabe/atoms/093b05290da4.md

@@ -17,7 +17,7 @@ M=B_\rho(p)\cup B_\rho(q),
 $$
 
 where $B_r(p)$ denotes the open geodesic ball of radius $r$ and center $p$, and
-$\rho$ is such that $\pi/2\sqrt\delta<\rho<\pi$.
+$\rho$ is such that $\pi/(2\sqrt\delta)<\rho<\pi$.
 
 *Proof.* By the injectivity-radius estimate of Section 3, $B_\rho(p)$ contains no
 points of $C_m(p)$, so it is diffeomorphic via $\exp$ to a Euclidean ball, and
@@ -32,14 +32,14 @@ $\partial B_\rho(q)$ is path connected, $\partial B_\rho(p)\cap\partial B_\rho(q
 giving $r_o$ with $d(r_o,p)=d(r_o,q)=\rho$. Take $\lambda$ minimizing from $p$ to
 $r_o$; by Berger's Lemma there is $\gamma$ minimizing from $p$ to $q$ with
 $\langle\gamma'(0),\lambda'(0)\rangle\geq0$. Let $s\in\gamma$ with $d(p,s)=\rho$.
-Applying Rauch's Theorem (\entryref{18ec1e38faa8} of Chap. 10), comparing $M^n$ with a
+Applying Rauch's Theorem (\entryref{0a8a561b495c} of Chap. 10), comparing $M^n$ with a
 sphere $S^n$ of curvature $\delta$: since the angle
-$\sphericalangle r_o p s\leq\pi/2$, $d(r_o,s)\leq\pi/2\sqrt\delta$. As
+$\sphericalangle r_o p s\leq\pi/2$, $d(r_o,s)\leq\pi/(2\sqrt\delta)$. As
 $d(r_o,p)=d(r_o,q)=\rho$ and some $s$ has $d(r_o,s)<\rho$, the distance from $r_o$ to
 $\gamma$ is realized by an interior point $s_o$, with the minimizing geodesic from
 $r_o$ to $s_o$ orthogonal to $\gamma$ and
-$d(r_o,\gamma)=d(r_o,s_o)\leq\pi/2\sqrt\delta$. Since $d(p,q)\leq\pi/\sqrt\delta$,
-either $d(p,s_o)\leq\pi/2\sqrt\delta$ or $d(q,s_o)\leq\pi/2\sqrt\delta$; in the
-former case, since $d(r_o,s_o)\leq\pi/2\sqrt\delta$ and
-$\sphericalangle p s_o r_o=\pi/2$, Rauch gives $d(p,r_o)\leq\pi/2\sqrt\delta<\rho$,
-contradicting $d(p,r_o)=\rho$ (the other case is analogous). $\square$
+$d(r_o,\gamma)=d(r_o,s_o)\leq\pi/(2\sqrt\delta)$. Since $d(p,q)\leq\pi/\sqrt\delta$,
+either $d(p,s_o)\leq\pi/(2\sqrt\delta)$ or $d(q,s_o)\leq\pi/(2\sqrt\delta)$; in the
+former case, since $d(r_o,s_o)\leq\pi/(2\sqrt\delta)$ and
+$\sphericalangle p s_o r_o=\pi/2$, Rauch gives $d(p,r_o)\leq\pi/(2\sqrt\delta)<\rho$,
+contradicting $d(p,r_o)=\rho$ (the other case is analogous). $\square$

projects/riemannian-geometry/.astrolabe/atoms/096001bed96a.md

@@ -13,7 +13,7 @@ manifold $M^n$ satisfies $0<K\leq 1$, then $i(M)\geq\pi$.
 
 *Proof.* By compactness there are $p,q\in M$ with $q\in C_m(p)$ and $d(p,q)=i(M)$.
 Suppose $d(p,q)<\pi$. If $p$ is conjugate to $q$, then $d(p,q)\geq\pi$ by
-\entryref{be0d62658bc0} of Chap. 10, a contradiction. So $p$ is not conjugate to $q$; by
+\entryref{ba45ce517a2a} of Chap. 10, a contradiction. So $p$ is not conjugate to $q$; by
 Propositions 2.12 and 2.13 there is a closed geodesic $\gamma$ through $p=\gamma(0)$
 and $q$ with $\ell(\gamma)<2\pi$. Parallel transport along $\gamma$ leaves invariant
 a vector $V$ orthogonal to $\gamma$ (Lemma 3.8 of Chap. 9); computing the second
@@ -25,4 +25,4 @@ $q_s=\sigma_s(0)$ to $\gamma_s(0)$ has $\sigma_s(0)\to q$, and an accumulation p
 $w$ of $\sigma_s'(0)$ gives a minimizing geodesic $\sigma(t)=\exp_q tw$ from $q$ to
 $p$. By the first-variation formula $\sigma_s'(0)\perp\gamma_s'$ at $q_s$, so
 $\sigma'(0)\perp\gamma'$ at $q$. This yields three minimizing geodesics from $p$ to
-$q$; since $p$ is not conjugate to $q$, this contradicts \entryref{d8300a62544d}. $\square$
+$q$; since $p$ is not conjugate to $q$, this contradicts \entryref{d8300a62544d}. $\square$

projects/riemannian-geometry/.astrolabe/atoms/0dc5984b8e92.md

@@ -15,11 +15,12 @@ $$
 (\overline{\nabla}_X B)(Y,Z,\eta)=(\overline{\nabla}_Y B)(X,Z,\eta).
 $$
 
-If, in addition, the codimension of the immersion is $1$, then
+If, in addition, the codimension of the immersion is $1$ and $\eta$ is chosen as a
+local unit normal field, then
 $\nabla_X^\perp\eta=0$, hence
 
 $$
-\overline{\nabla}_X B(Y,Z,\eta)=X\langle S_\eta(Y),Z\rangle-\langle S_\eta(\nabla_X Y),Z\rangle-\langle S_\eta(Y),\nabla_X Z\rangle=\langle\nabla_X(S_\eta(Y)),Z\rangle-\langle S_\eta(\nabla_X Y),Z\rangle.
+(\overline{\nabla}_X B)(Y,Z,\eta)=X\langle S_\eta(Y),Z\rangle-\langle S_\eta(\nabla_X Y),Z\rangle-\langle S_\eta(Y),\nabla_X Z\rangle=\langle\nabla_X(S_\eta(Y)),Z\rangle-\langle S_\eta(\nabla_X Y),Z\rangle.
 $$
 
 Therefore, in this case, the Codazzi equation can be written
@@ -34,4 +35,4 @@ compatibility equations in the local theory of surfaces (Cf. M. do Carmo [dC 2],
 pp. 235–236). It is possible to state an analogous theorem to the fundamental
 theorem of local surface theory; see K. Tenenblat, "On isometric immersions of
 Riemannian manifolds", Boletim da Soc. Bras. de Mat. vol. 2 (1971), 23–36, and
-H. Jacobowitz, "The Gauss-Codazzi equations", Tensor N.S., 39 (1982), 15–22.
+H. Jacobowitz, "The Gauss-Codazzi equations", Tensor N.S., 39 (1982), 15–22.

projects/riemannian-geometry/.astrolabe/atoms/13f6b45547fe.md

@@ -8,7 +8,8 @@ source: tex
 src: docarmo
 title: the index form
 ---
-It is often convenient to write (5) using
+It is often convenient to write (5) using, on each interval where $V$ is
+differentiable,
 $\frac{d}{dt}\langle V,\frac{DV}{dt}\rangle=\langle V,\frac{D^2 V}{dt^2}\rangle+\langle\frac{DV}{dt},\frac{DV}{dt}\rangle$.
 Taking a geodesic $\gamma:[0,a]\to M$ and a partition
 $0=t_0<t_1<\cdots<t_{k+1}=a$, this gives
@@ -28,4 +29,4 @@ the general case,
 
 $$
 \frac{1}{2}E''(0)=I_a(V,V)-\Big\langle\frac{D}{ds}\frac{\partial f}{\partial s},\gamma'\Big\rangle(0,0)+\Big\langle\frac{D}{ds}\frac{\partial f}{\partial s},\gamma'\Big\rangle(0,a).\qquad(6)
-$$
+$$

projects/riemannian-geometry/.astrolabe/atoms/167b9abd6797.md

@@ -38,16 +38,18 @@ $$
 Since $K\leq 0$ and $T_c(\tilde{M})$ has zero curvature, we can apply \entryref{0a8a561b495c} (application of Rauch's Theorem) and obtain that
 
 $$
-\ell(\Gamma_C)\leq\ell(\gamma_C)\quad(<,\text{ if }K<0).
+\ell(\Gamma_C)\leq\ell(\gamma_C),
 $$
 
+with strict inequality if $K<0$.
 It follows that
 
 $$
-A^2+B^2-2AB\cos\gamma\leq\ell(\Gamma_C)^2\leq\ell(\gamma_C)^2=C^2\quad(<,\text{ if }K<0),
+A^2+B^2-2AB\cos\gamma\leq\ell(\Gamma_C)^2\leq\ell(\gamma_C)^2=C^2,
 $$
 
-which proves (i).
+where the last inequality is strict if $K<0$.
+This proves (i).
 
 To prove (ii), let us observe that
 
@@ -61,11 +63,12 @@ sides have lengths $A$, $B$ and $C$. Denoting the opposite angles of this triang
 by $\alpha'$, $\beta'$ and $\gamma'$, respectively, we obtain from (i),
 
 $$
-\alpha\leq\alpha',\qquad\beta\leq\beta',\qquad\gamma\leq\gamma'\quad(<,\text{ if }K<0).
+\alpha\leq\alpha',\qquad\beta\leq\beta',\qquad\gamma\leq\gamma'.
 $$
 
+All three inequalities are strict if $K<0$.
 Since $\alpha'+\beta'+\gamma'=\pi$, (ii) follows. $\blacksquare$
 
 From now on, $M$ will denote a complete Riemannian manifold with sectional
 curvature $K<0$. As always, $\pi:\tilde{M}\to M$ denotes the universal covering of
-$M$ with the covering metric. Our goal is to prove the following theorem.
+$M$ with the covering metric. Our goal is to prove the following theorem.

projects/riemannian-geometry/.astrolabe/atoms/1a0870f7d0e5.md

@@ -10,4 +10,4 @@ title: C. Tompkins
 ---
 Suppose that $M$ is compact with zero curvature. If $k<n$, there does not exist
 an isometric immersion $f:M^n\to\mathbb{R}^{n+k}$; in particular, the flat torus
-$T^n$ (see \entryref{5ecedf4a250b}, Chap. 1) cannot be immersed isometrically in $\mathbb{R}^{2n-1}$.
+$T^n$ (see \entryref{a47005f1c0b9}, Chap. 1) cannot be immersed isometrically in $\mathbb{R}^{2n-1}$.

projects/riemannian-geometry/.astrolabe/atoms/1ff5911ba53e.md

@@ -21,8 +21,10 @@ $s\in(-\varepsilon,\varepsilon)$. If $f$ is differentiable, the variation is sai
 *differentiable*. For each $s\in(-\varepsilon,\varepsilon)$, the parametrized curve
 $f_s:[0,a]\to M$ given by $f_s(t)=f(s,t)$ is called a *curve in the variation*, so
 a variation determines a family $f_s(t)$ of neighboring curves of $f_0(t)=c(t)$.
+The variation is proper exactly when the curves in this family have the same
+initial point $c(0)$ and endpoint $c(a)$.
 The parametrized differentiable curve given by $f_t(s)=f(s,t)$, $t$ fixed, is a
 *transversal curve of the variation*. The velocity vector of a transversal curve
 at $s=0$, defined by $V(t)=\frac{\partial f}{\partial s}(0,t)$, is a (piecewise
 differentiable) vector field along $c(t)$ and is called the *variational field* of
-$f$.
+$f$.

projects/riemannian-geometry/.astrolabe/atoms/2660f4a30491.md

@@ -16,10 +16,10 @@ $0<K_{\min}\leq K\leq K_{\max}$, then:
 
 *Proof.* By Bonnet–Myers (Theor. 3.1 of Chap. 9), $M$ is compact. Since $T_1M$ is
 compact, \entryref{31ea0ff2d398} gives $p\in M$ with $d(p,C_m(p))=\inf_{r\in M}d(r,C_m(r))$,
-and $C_m(p)$ compact gives $q\in C_m(p)$ assuming this distance. If $q$ is conjugate
+and $C_m(p)$ compact gives $q\in C_m(p)$ realizing this distance. If $q$ is conjugate
 to $p$, then $d(p,q)\geq\pi/\sqrt{K_{\max}}$ by \entryref{ba45ce517a2a} — case
 (a). If $q$ is not conjugate to $p$, by \entryref{d8300a62544d} there are two minimizing
 geodesics $\mu,\sigma$ from $p$ to $q$ with $\mu'(\ell)=-\sigma'(\ell)$,
 $\ell=d(p,q)$. Since $q\in C_m(p)$ we have $p\in C_m(q)$ and $p$ realizes the
 distance from $q$ to $C_m(q)$, so $\mu'(0)=-\sigma'(0)$; thus $\mu$ and $\sigma$ form
-a closed geodesic $\gamma$ satisfying (b). $\square$
+a closed geodesic $\gamma$ satisfying (b). $\square$

projects/riemannian-geometry/.astrolabe/atoms/29c2442300b5.md

@@ -29,9 +29,9 @@ with $\varepsilon_j'\leq\varepsilon_j$ gives
 $\exp_p(t_0+\varepsilon_j)\gamma'(0)=\exp_p(t_0+\varepsilon_j')\sigma_j'(0)$, hence
 $\gamma'(0)=\sigma_j'(0)$, contradicting the definition of $t_0$. Conversely, if
 (a) holds, since a geodesic does not minimize after the first conjugate point
-(\entryref{31ea0ff2d398} to the Index Theorem, Chap. 11), the cut point occurs at
+(\entryref{be5e1aca2bd6} to the Index Theorem, Chap. 11), the cut point occurs at
 $\gamma(\tilde t)$, $\tilde t\leq t_0$. If (b) holds, joining $\sigma(t_0-\varepsilon)$
 to $\gamma(t_0+\varepsilon)$ by the unique minimizing geodesic $\tau$ inside a
 totally normal neighborhood of $\gamma(t_0)$ yields a curve of length strictly less
 than $t_0+\varepsilon$, so again the cut point occurs at $\gamma(\tilde t)$,
-$\tilde t\leq t_0$. $\square$
+$\tilde t\leq t_0$. $\square$

projects/riemannian-geometry/.astrolabe/atoms/2b94970df630.md

@@ -8,9 +8,9 @@ source: tex
 src: docarmo
 title: Corollary 2.5
 ---
-Let $\gamma:[0,a]\to M$ be a geodesic. Then a Jacobi field $J$ along $\gamma$ with
-$J(0)=0$ is given by
+Let $\gamma:[0,a]\to M$ be a geodesic and put $p=\gamma(0)$. Then a Jacobi field
+$J$ along $\gamma$ with $J(0)=0$ is given by
 
 $$
 J(t)=(d\exp_p)_{t\gamma'(0)}(tJ'(0)),\quad t\in[0,a].
-$$
+$$

projects/riemannian-geometry/.astrolabe/atoms/317503003357.md

@@ -13,7 +13,10 @@ $q\in M$ to the geodesic sphere $S_r(p)$ of radius $r<c$ stays outside the
 geodesic ball $B_r(p)$ in some neighborhood of $q$.
 *Proof.* Let $W$ be a totally normal neighborhood of $p$; by the lemma of
 homogeneity restrict to unit-speed geodesics, so to the unit bundle
-$T_1W=\{(q,v);\ q\in W,\ v\in T_qM,\ |v|=1\}$. Let
+$T_1W=\{(q,v);\ q\in W,\ v\in T_qM,\ |v|=1\}$. After shrinking $W$, assume that
+$\overline{W}$ is compactly contained in a normal neighborhood $U$ of $p$; since
+$T_1\overline{W}$ is compact, decrease $I$ so that
+$\gamma(I\times T_1W)\subset U$. Let
 $\gamma:I\times T_1W\to M$, $I=(-\varepsilon,\varepsilon)$, be the differentiable
 map with $t\mapsto\gamma(t,q,v)$ the unit-speed geodesic through $q$ with velocity
 $v$ at $t=0$. Set $u(t,q,v)=\exp_p^{-1}(\gamma(t,q,v))$ and
@@ -33,4 +36,4 @@ $\frac{\partial^2 F}{\partial t^2}(0,p,v)=2|v|^2=2>0$. Hence there is a
 neighborhood $V\subset W$ of $p$ with $\frac{\partial^2 F}{\partial t^2}(0,q,v)>0$
 for all $q\in V$, $|v|=1$. Take $c>0$ with $\exp_p B_c(0)\subset V$: any geodesic
 in $B_c(p)$ tangent to $S_r(p)$, $r<c$, has a strict local minimum of $F$ at
-$(0,q,v)$, so in a neighborhood of $q$ its points stay outside $B_r(p)$. $\square$
+$(0,q,v)$, so in a neighborhood of $q$ its points stay outside $B_r(p)$. $\square$

projects/riemannian-geometry/.astrolabe/atoms/3403ca7591a6.md

@@ -14,6 +14,6 @@ orthogonal complement of $\gamma'(0)$. We say that $\gamma$ is *focal point free
 at $(0,a]$ if there exists some $\varepsilon>0$ such that $\gamma$ has no focal
 points relative to the submanifold $\Sigma_\varepsilon=\exp_{\gamma(0)}(B_\varepsilon(0))$.
 
-Observe that since $\Sigma_\varepsilon$ is geodesic at $p$, any Jacobi field $J$
+Observe that since $\Sigma_\varepsilon$ is geodesic at $\gamma(0)$, any Jacobi field $J$
 along $\gamma$, with $J(0)\neq 0$ and $J'(0)=0$, automatically satisfies
-$S_{\gamma'(0)}(J(0))=0$.
+$S_{\gamma'(0)}(J(0))=0$.

projects/riemannian-geometry/.astrolabe/atoms/413b1274c2c2.md

@@ -64,11 +64,12 @@ $$
 $$
 
 which we write as $1/\lambda=\rho=a_1|p-p_0|^2+k_1$ with $a_1=\sigma/2$,
-$k_1=\text{const.}$, $p_0\in\mathbb{R}^n$. Taking the inversion
+$k_1=\text{const.}$, $p_0\in\mathbb{R}^n$. The proof in this case will be complete
+once we show $k_1=0$: if $k_1=0$, taking the inversion
 $g(p)=\frac{p-p_0}{|p-p_0|^2}+p_0$ and $h=g\circ f^{-1}$, $h$ is conformal with
 coefficient $a_1$, hence an isometry followed by a dilatation; thus
 $f=h^{-1}\circ g$ is an inversion followed by a dilatation, followed by an
-isometry. To show $k_1=0$: applying the argument to $f^{-1}$,
+isometry. To show $k_1=0$, apply the argument to $f^{-1}$:
 $\lambda=a_2|f(p)-q_0|^2+k_2$, hence
 
 $$
@@ -84,7 +85,8 @@ $$
 $$
 
 If $k_1\neq 0$, the left side is a transcendental function of $|p(s_0)-p_0|$, while
-$(11)$ implies it is algebraic. This contradiction shows $k_1=0$.
+$(11)$ implies it is algebraic. This contradiction shows $k_1=0$, completing the
+case $\sigma\neq0$.
 
 *Case $\sigma=0$.* Here $\rho=1/\lambda=\sum a_i x_i+c_1=A_1(x)+c_1$. Applying the
 initial argument to $f^{-1}$,
@@ -103,4 +105,4 @@ $$
 
 which contradicts $(11')$ unless $A_1(p(s))\equiv 0$. We conclude that if
 $\sigma=0$, $\lambda=\text{const.}$; then lengths of tangent vectors are multiplied
-by a constant $\lambda$ and $f$ is an isometry followed by a dilatation. $\square$
+by a constant $\lambda$ and $f$ is an isometry followed by a dilatation. $\square$

projects/riemannian-geometry/.astrolabe/atoms/45bcded293f4.md

@@ -10,7 +10,8 @@ title: the Lobatchevski plane
 ---
 Let $G=\{(x,y)\in\mathbb{R}^2;\ y>0\}$ with metric $g_{11}=g_{22}=\frac{1}{y^2}$,
 $g_{12}=g_{21}=0$. The $y$-axis segment $\gamma(t)=(0,t)$, $a\le t\le b$ ($a>0$),
-is a geodesic: for any arc $c(t)=(x(t),y(t))$ with $c(a)=(0,a)$, $c(b)=(0,b)$,
+is the image of a geodesic: for any arc $c(t)=(x(t),y(t))$ with $c(a)=(0,a)$,
+$c(b)=(0,b)$,
 
 $$
 \ell(c)=\int_a^b\sqrt{(x')^2+(y')^2}\,\frac{dt}{y}\ge\int_a^b\frac{|y'|}{y}\,dt\ge\int_a^b\frac{dy}{y}=\ell(\gamma),
@@ -21,4 +22,4 @@ isometries $z\mapsto\frac{az+b}{cz+d}$, $z=x+iy$, $ad-bc=1$ (cf. Exercise 4 of
 Chap. 1) carry the $y$-axis into upper semicircles or rays $x=x_0$, $y>0$, which
 are therefore geodesics; these are all the geodesics, since through each $p\in G$
 and each direction there passes such a circle centered on the $x$-axis (a normal
-direction giving a vertical ray).
+direction giving a vertical ray).

projects/riemannian-geometry/.astrolabe/atoms/48f7c97ce541.md

@@ -42,9 +42,11 @@ $$
 differential $\mathrm{Ad}(a)=dR_{a^{-1}}dL_a:\mathcal{G}\to\mathcal{G}$, so
 $\mathrm{Ad}(a)Y=dR_{a^{-1}}Y$. Let $x_t$ be the flow of $X$. From \entryref{453ecde8947b}, $[Y,X]=\lim_{t\to0}\frac1t(dx_t(Y)-Y)$. Since $X$ is left invariant,
 $L_y\circ x_t=x_t\circ L_y$, giving $x_t(y)=R_{x_t(e)}(y)$, hence $dx_t=dR_{x_t(e)}$
-and $[Y,X]=\lim_{t\to0}\frac1t(\mathrm{Ad}(x_t^{-1}(e))Y-Y)$. With $\langle\ ,\ \rangle$
+and $[Y,X]=\lim_{t\to0}\frac1t(\mathrm{Ad}((x_t(e))^{-1})Y-Y)$. With $\langle\ ,\ \rangle$
 bi-invariant, $\langle U,V\rangle=\langle dR_{x_t(e)}U,dR_{x_t(e)}V\rangle$;
 differentiating in $t$ at $t=0$ gives $0=\langle[U,X],V\rangle+\langle U,[V,X]\rangle$,
-which is (3). The relation (3) characterizes bi-invariant metrics: a positive
-bilinear form on $\mathcal{G}$ satisfying (3) yields, via (2), a bi-invariant
-metric. $\blacksquare$
+which is (3). $\blacksquare$
+
+The relation (3) characterizes bi-invariant metrics: a positive bilinear form on
+$\mathcal{G}$ satisfying (3) yields, via (2), a bi-invariant metric. This last
+fact is not proved here.

projects/riemannian-geometry/.astrolabe/atoms/49a53e470b68.md

@@ -11,10 +11,10 @@ title: convex neighborhoods
 For any $p\in M$ there exists $\beta>0$ such that the geodesic ball $B_\beta(p)$ is
 strongly convex.
 *Proof.* Let $c$ be as in \entryref{317503003357}. Choose $\delta>0$ and $W$ as in \entryref{f29bd1e6ebb3}
-with $\delta<\frac{c}{2}$, and take $\beta<\delta$ with $B_\beta(p)\subset W$. Let
+with $\delta<\frac{c}{2}$, and take $\beta<\delta$ with $\overline{B_\beta(p)}\subset W$. Let
 $q_1,q_2\in\overline{B_\beta(p)}$ and let $\gamma$ be the unique geodesic of length
 $<2\delta<c$ joining them; then $\gamma\subset B_c(p)$. If the interior of
 $\gamma$ is not contained in $B_\beta(p)$, there is an interior point $m$ where the
 maximum distance $r$ from $p$ to $\gamma$ is attained; near $m$ the points of
 $\gamma$ remain in $\overline{B_r(p)}$. Since $m\in B_c(p)$ this contradicts
-\entryref{317503003357}, proving the proposition. $\square$
+\entryref{317503003357}, proving the proposition. $\square$

projects/riemannian-geometry/.astrolabe/atoms/4c21e0cecff0.md

@@ -8,6 +8,6 @@ source: tex
 src: docarmo
 title: Remark 3.8
 ---
-The torus $T^{n+1}$ with the flat metric (Cf. \entryref{5ecedf4a250b}, Chap. 1) contains the
+The torus $T^{n+1}$ with the flat metric (Cf. \entryref{a47005f1c0b9}, Chap. 1) contains the
 torus $T^n$ as a totally geodesic submanifold. This shows that the hypothesis
-that $\overline{M}$ be simply connected in \entryref{d483a90d3a7d} is necessary.
+that $\overline{M}$ be simply connected in \entryref{d483a90d3a7d} is necessary.