# Tag Info

7

I think you are asking if every closed manifold is the boundary of a compact manifold. The answer is negative. For instance, real projective plane is not the boundary of any compact 3d manifold (since it has odd Euler characteristic and the Euler characteristic of an odd-dimensional manifold is half of the Euler characteristic of its boundary: This follows ...

5

Hint. The relation $$ad-cb=1,$$ requires that $(a,c)\ne (0,0)$, and says that the point $(d,b)$ lies on the line $$ax-cy=1,$$ which is described parametrically by $$(x,y)=t(c,a)+(a^2+c^2)^{-1/2}(a,-c).$$ Thus $$\left(\begin{matrix}a &b \\ c& d\end{matrix}\right) =\left(\begin{matrix}a &at-c(a^2+c^2)^{-1/2} \\ c& ... 4 The curve \gamma(t) = \operatorname{Exp}_{p}(tv + v_{0}) is not generally a geodesic. Let (M, g) be the round unit sphere in \mathbf{R}^{3}, and let p = (0, 0, 1). In polar coordinates (r, \theta) in the tangent plane T_{p} M, the exponential map is given by$$ \operatorname{Exp}_{p}(r, \theta) = (\sin r \cos\theta, \sin r \sin\theta, \cos r). ...

4

A trivial answer would be that given any $n$-manifold $M$, the $(n+1)$-manifold-with-boundary $M\times [0,1)$ has $M$ as its boundary. But I don't know if there's intended to be any further restrictions on the situation that make a manifold occurring as a boundary of something else more difficult.

3

Every neighbourhood of $\omega_1$ contains uncountably many ordinals, and hence $$(\alpha,\omega_1)\times [0,1)\tag{1}$$ is not homeomorphic to a subset of $\mathbb{R}$, since it is not second countable (there are uncountably many disjoint open subsets $\{\beta\}\times \left(\frac14,\frac34\right)$, $\alpha < \beta \leqslant \omega_1$, so the space is ...

3

If we consider $\nabla_\Gamma$ to be the covariant derivative operator associated with the metric $g_\Gamma$ on the hypersurface $\Gamma$ induced from the metric $g$ in the ambient manifold, then for any two tangent vector fields $X$,$Y$ on $\Gamma$, $(\nabla_\Gamma)_X Y$ is also a vector field on $\Gamma$, and thus we may in addition view the map ...

3

Note that $T_{f(A)} \mathbb S^1$ is one dimensional. To show that $df_A$ is onto, one only needs to find a tangent vector $X$ such that $df_A(X)\neq 0$. (Then the image would be at least one dimensional and thus is onto). Also, it suffices to check that when $A=I$. (Let $B(t)$ be a curve in $U(n)$ such that $B(0)= I$ and $\dot B(0)=X$. So represent a vector ...

3

EDITED Hint: Start over. What are the equations that dictate symmetry and $\det=1$? Show that this gives you a submersion. This manifold has two components: The symmetry condition dictates that the eigenvalues must be real, so having their product be $1$ forces them both to have the same sign. Final comment: In fact, you can write this subset of $\Bbb R^4$ ...

2

1) When composed with the given charts, the identity map gives $(\psi \circ \operatorname{id}_\Bbb R \circ\ \phi^{-1})(x) = x^{1/3}$. This map is not smooth at $0$. Hence the identity map is not a diffeomorphism. 2) Think of a bijection that "cancels out" with the cubic root in $\psi$.

2

If $\phi:M \to \Bbb R^n$ is in fact an isometric embedding, then the image $\text{Im} \phi$ of $M$ under $\phi$ is compact, since $M$ is compact. If the $x_i$, $1 \le i \le n$, are a set of Cartesian coordinates on $\Bbb R^n$ (where by "Cartesian" I mean compatible with the vector space structure on $\Bbb R^n$, so they are related by a nonsingular linear ...

2

$L(\gamma)=L(\sigma)$ is obvious. The difficult thing is to prove that these suprema can be written as an integral. Nevertheless, here is why one has $L(\gamma)=L(\sigma)$: Both $L(\gamma)$ and $L(\sigma)$ are the sup of the same set, namely the set of all sums of the form $$\sum_{k=1}^N |\gamma(t_k)-\gamma(t_{k-1})|$$ with ...

2

The usual definition of compactness of $X$ assumes $X$ is $T_2$ or hausdorff. In case you do assume that, then each compact subset $C$ of a Hausdorff Space $Y$ is closed. To show this assume that $y\in Y\backslash C$. For each $c\in C$ you will find open neighbourhoods $U_c$ of $c$ and $V_c$ of $y$ which are disjoint, since $Y$ is hausdorff. A finite number ...

2

Here's a sequence of propositions which will prove the theorem. Proposition 1 $Gr(n,k)$ is diffeomorphic to the homogeneous space $G/H = SO(n)/S(O(k)\times O(n-k))$, where $S(O(k)\times O(n-k))$ means all $n\times n$ matrices whose top left $k\times k$ and bottom right $n-k\times n-k$ are orthogonal, all other entries are $0$, and the overall determinant ...

2

This is a baby version of the coarea formula, which is a curvilinear version of the Fubini's theorem. By breaking $\Gamma(0)$ into small pieces we assume that we calculate in a small local coordinate with variable $x \in B$ and metric $g_{ij}$. Then locally $(x, t)$ is a chart of $S$ given by $(x, t)\mapsto (\Phi_x(t), t)$, where $\Phi_x(t)$ solves the ODE ...

2

Yes, in a rather boring way. Take a point $p\in S$. In a neighborhood of $p$, the surface can be represented (in a suitable coordinate system) by equation $x_{n+1}=g(x_1,\dots,x_n)$ for some smooth function $g$. For a small $r>0$, let $$S_1=\{x: x_{n+1}=g(x_1,\dots,x_n), \ x_1^2+\dots+x_n^2<r^2\}$$ $$\Gamma=\{x: x_{n+1}=g(x_1,\dots,x_n), \ ... 2 This is true, provided you are using correct definition of a smooth function on a closed subset A; dimension of A is irrelevant. Smoothness of f means that there exists a smooth extension h of f to an open neighborhood U of A in M. Let V, W be another pair of open neighborhoods of A, such that$$ \bar{W}\subset V, \bar{V}\subset U. $$... 2 Here is a proof in the smooth setting. One of the (many) equivalent definitions of orientability is the following: Let M be a smooth connected manifold, p:\tilde M\to M is its universal cover and G the group of covering transformations. Note that \tilde M is always orientable. Then M is orientable if and only if G preserves orientation on \tilde ... 2 The cochain complex of sheaves$$0 \to \mathbb{R} \to \Omega^0 \to \Omega^1 \to \cdots$$is exact: this follows from the Poincaré lemma. (Any closed differential (n+1)-form on a sufficiently small open neighbourhood must be the exterior derivative of some differential n-form.) Thus, the cochain complex$$\Omega^0 \to \Omega^1 \to \Omega^2 \to \cdots$$... 2 U \cap V consists of all triples whose first coordinate is nonzero, but whose second and third coordinate are not both zero. A point (x_1 : x_2 : x_3) corresponds to the points \pm s(x_1, x_2, x_3) in S^2, where s = 1/\sqrt{x_1^2 + x_2^2 + x_3^2}. Now look at the portion V_1 = \{(a, b, c) \in S^2 : a > 0 \}. This is in bijective ... 2 The idea is, as always when dealing with local notions on manifolds, to pass to a chart and see what happens. On a chart (i.e. Euclidean space) you have an obvious correspondence between vectors and directional derivatives, in the sense that derivation in direction v is given by Df\cdot v. If you write this down component-wise and lift it up to the ... 1 The theorem is that (\alpha,\beta) has to be a regular value for (x_1+x_4,x_1x_4-x_2x_3). That is, at every (x_1,x_2,x_3,x_4) mapping to (\alpha,\beta) the differential (Jacobian matrix) has to be surjective. If you remember your linear algebra, you know that your differential is surjective except when x_3=x_2=0 and x_1=x_4. Then the equations ... 1 If you wait long enough the polynomial becomes monotone and eventually overcomes all local extrema. For odd degree, it tends to \epsilon\infty when x\to\infty where \epsilon is the sign of the leading coefficient; and to -\epsilon\infty when x\to-\infty. Then the inverse image of a sufficiently large x will contain only one point. Therefore the ... 1 Unless n=2 you are not going to get a curve, but rather a manifold of dimension n-1. Intuitively, you start with a space of dimension n and you impose one constraint, so you end up with something of dimension n-1. Second, unless the polynomial is non-singular, i.e. its gradient is nowhere zero, you are not going to get a smooth manifold but ... 1 Indeed, \partial Q contains S as a subset. This is easiest to see in the special case when \Omega_i(t)\equiv \Omega_i and S(t) \equiv S for all t. Therefore, a function \varphi\in C_c^\infty(Q) must vanish on S. I suspect that the author of the text you are reading actually meant to define Q=\Omega\times (0,T). The latter looks like a ... 1 This is indeed the definition of distributional derivatives. The book "Heat Kernel and Analysis on Manifolds" by Grigoryan (I can even access that part as a free preview on books.google.com) contains formula (7.30): locally integrable function u(t,x) satisfies the heat equation on a manifold N=(0,\infty)\times M in a distributional sense if and only if ... 1 What you wrote does make perfect sense, as integration by parts on a sufficiently smooth manifold S is permissible. Note that you have made a little mistake, and this is the correct version$$ \left\langle \frac{d}{dt}u - \Delta_\Gamma u, \varphi \right\rangle_{\mathscr D^*(S), \mathscr D(S)} = -\left\langle u, \frac{d}{dt}\varphi \right\rangle - \langle ...

1

It is implicitly in $\mathbb R^2$; otherwise, '$x$' and '$y$' don't make sense. As far as a general approach, there are some theorems that can ensure global flows, but they can be hard to check. In this case, I think we are best off working directly with the given vector field. Write down the condition for the integral curves $\gamma(t)$:  \dot\gamma(t) ...

1

There is a kind of "cheating" way to do this. Consider the obvious embeddings $\iota_j:G\to G\times G$ where $\iota_1(g)=(g,1)$ and $\iota_2(g)=(1,g)$. Then, we obtain embeddings $(\iota_j)_\ast:T_e G\to T_{(e,e)}(G\times G)$. Taking the product yields an isomorphism $(\iota_1)_\ast\times(\iota_2)_\ast:T_e G\times T_e G\to T_{(e,e)}(G\times G)$. So, now, ...

1

Here is a concrete perspective on tangent bundles that might help. Take an embedding of your manifold $M$ into $\mathbb{R}^N$ (by Whitney we can always do this). This gives an embedding of $TM$ into $T\mathbb{R}^N = \mathbb{R}^N\times \mathbb{R}^N$ (it's injective and you can write down the differential to check it is an immersion). This already gives you ...

1

This question was essentially answered here, namely, that the cut-locus has measure zero (see the references provided in the link). I assume that your manifold $M$ is smooth, otherwise, I am not sure what notion of measure zero you would be using. I will also assume that $M$ is connected. (If not, apply this argument to each connected component.) Then, put a ...

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