# Tag Info

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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