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4

Consider $X = \mathbb R$ in the discrete metric. Then a set is compact if and only if it is finite. Now, take for example $x_n$ an enumeration of $\mathbb Q$. Then $x_n$ leaves eventually every finite subset of $\mathbb R$. Also note that $d(x,x_n) \to 1$ for all $x \in \mathbb R$.


4

It's actually true that $M$ is complete if and only if its universal cover $\widetilde{M}$ is complete. Let $p: \widetilde{M} \to M$ be the universal covering map, $q \in M$ and $\tilde{q} \in p^{-1}(q)$. As has already been stated, by Hopf-Rinow, all we need to if we want to conclude that $\widetilde{M}$ implies $M$ is complete is prove the corresponding ...


3

Note that $(\nabla_X Y)(p)$ should be thought of as the derivative of $Y$ at $p$ in the direction of $X(p)$ and not the other way around! (This is manifested in the definition of an affine connection by requiring that $\nabla_X Y$ is $C^{\infty}(M)$-linear in $X$ while only $\mathbb{R}$-linear in $Y$). Regarding your questions: Up to replacing the role of $...


3

Here's a way to get a representation as series. It works in a more general situation. Let $M$ and $N$ be smooth closed Riemannian manifolds. Denote by $\{\varphi_i\}_{i=1}^\infty$, $\{\psi_j\}_{i=j}^\infty$ the eigenfunctions with eigenvalues $\lambda_i$ and $\mu_j$ for the Laplace operators on $M$ and $N$ respectively. Let $L_2$ norms of eigenfunctions be ...


1

You can use that $T_{(t,0)}\Theta$ is linear. Compute seperately $T_{(t,0)}\Theta(s,0)=\frac{d}{du}|_0 \Theta(t+us, 0)=\frac{d}{du}|_0c(t+us)=s\dot c(t)$ and $T_{(t,0)}\Theta(0,Y)=\frac{d}{du}|_0\Theta(t, uY)=\frac{d}{du}|_0\gamma_{(c(t), Pt(c,t)Y)}(u)= Pt(c,t)Y$.


1

Firstly, $U$ has to be open to be an $n$-dimensional submanifold of $R^n$. If $g$ is a differentiable metric, then for every $p\in U$, $g_p$ is a scalar product defined on $T_pU=R^n$. So the answer is yes.


1

If $M$ is compact,Hopf Rinow implies that $M$ is complete, let $\hat M$ be the universal cover of $M$, and $p:\hat M\rightarrow M$.Lift the metric defined on $M$ with $p$ and $p:\hat M\rightarrow M$ preserves the metric.Suppose $\hat M$ is incomplete. Let $c:I=(a,b)\rightarrow \hat M$ an incomplete geodesic maximal that you can't extend. $p(c)$ can be ...


1

Unforunately it is not true that $u^{-1}TN$ is a subbundle of $TM$. By definition, let $f : X \to Y$ and $\pi: E \to Y$ be a vector bundle over $Y$. Then the pullback bundle $f^{-1}E$ on $X$ is given by $$f^{-1}E = \{ (x, v) \in X \times E : f(x) = \pi(v)\}.$$ Roughly speaking, it is a vector bundle over $X$, such that for each $x\in X$ the fiber is $E_{...


1

Let $E$ be an Euclidian vector space and $S_1,S_2$ two subspaces of same dimension $k$. Assume that $E = S_1 \oplus S_2$. Then there exists $f$ an orthogonal symetry such that $f(S_1)=S_2$ and $f(S_2)=S_1$. Proof: Let $P$ be the matrix of the scalar product on $E$, and choose a basis of $S_1$ and a basis of $S_2$ such that $P$ has the form : $$ P = \...



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