9

Every orientable 3-manifold is parallelizable, i.e. its tangent bundle is trivial. This implies that any normal bundle is stably trivial. So $N\Sigma$ is a stably trivial bundle of rank 4 over a 3-dimensional manifold. Milnor proved that if a vector bundle has rank higher than the dimension of your CW complex, then the vector bundle is trivial, if and only ...


5

Associated to $T : \mathfrak{X}(M)\times\mathfrak{X}(M) \to \mathfrak{X}(M)$ is a map $T' : \Omega^1(M)\times\mathfrak{X}(M)\times\mathfrak{X}(M) \to C^{\infty}(M)$ given by $$T'(\alpha, X, Y) = \alpha(T(X, Y)).$$ When people say that $T$ is a $(1, 2)$-tensor field, they really mean the associated map $T'$ is a $(1, 2)$-tensor field. To check that $T'$ is a ...


4

By the choice of F, dF/dx = f(x). In the vocabulary of differential forms, this is saying that f(x) dx is the exterior derivative of the 0-form, namely the function, F. In other words, that dF = f dx. The general Stokes theorem applies to higher differential forms ω instead of just 0-forms such as F. A closed interval [a, b] is a simple example of a one-...


4

A single point is $0$-dimensional, so the tangent space should also be a zero dimensional vector space. Similarly, you'll only have $\Omega^0(M)$ being nonzero ($\Omega^i(M)=0$ for $i>0$). These are just the (constant) functions as all functions on a single point are constant.


3

Terminology aside, the idea is that we can define a contravariant functor $$\Omega^*:(\text{Smooth Manifolds})\to (\text{Chain Complexes of Vector Spaces})$$ by assigning to a manifold $M$ its de Rham Complex $$ \Omega^0(M)\xrightarrow{d}\Omega^1(M)\xrightarrow{d}\Omega^2(M)\xrightarrow{d}\cdots\xrightarrow{d} \Omega^n(M). $$ You can find the definition of a ...


3

You might not be familiar with the concept of vector bundles but this is related to it. So when you have the cotagent bundle or the tangent bundle we have a canonical smooth map $\pi : T^*M\rightarrow M$ and a smooth section will be a smooth map $s:M\rightarrow T^*M$ such that $\pi\circ s=id$, which is exactly saying that at each point we get an element in $...


3

If you know the product rule -- derivative of $fg$ at $A$ is $H\mapsto f'(A)(H)\cdot g(A) + f(A)\cdot g'(A)(H)$ -- then you can apply it to the function $G=\mathrm{id}\cdot F$, i.e. $G(A)=AA^{-1}=I_n$ to get \begin{align*} 0=G'(A)&=\mathrm{id}'(A)(H)\cdot F(A)+\mathrm{id}(A)\cdot F'(A)(H)\\ &= H\cdot F(A)+A\cdot F'(A)(H), \end{align*} which ...


3

It's isomorphic to the subgroup of $\operatorname{GL}(2,\mathbb C)$ consisting of matrices of the form $$\begin{pmatrix}1&0\\z&1\end{pmatrix}$$ via the isomorphism $$\varphi:z\mapsto\begin{pmatrix}1&0\\z&1\end{pmatrix},$$ since $\varphi(0)=E_2$ and $$\varphi(a)\varphi(b)=\begin{pmatrix}1&0\\a&1\end{pmatrix}\begin{pmatrix}1&0\\b&...


3

I'll answer in reverse order. First, recall the definition of curvature: $\Omega = d\omega-\omega\wedge\omega$, so that $$d\omega = \Omega + \omega\wedge\omega.$$ If you differentiate $$\Omega - d\omega + \omega\wedge\omega = 0,$$ you get $$0=d\Omega + (\Omega+\omega\wedge\omega)\wedge\omega - \omega\wedge(\Omega+\omega\wedge\omega) = d\Omega +\Omega\wedge\...


3

You've made two mistakes. The first mistake is that $\int_M S_g\Omega_g \neq S_g\operatorname{vol}_g$; the left hand side is a number, while the right hand side is a function unless $S_g$ is constant, in which case you do get an equality. Secondly, $\Omega_{cg} = c^{\frac{n}{2}}\Omega_g$ not $c^n$ (think about what happens when you replace $g$ by $cg$ in ...


3

Hint: With some algebraic manipulation you can get that the cartesian equation of the surface is $$x^2-y^2-4z^2=0$$ This is a cone.


2

Subtract the two equations. We get $y=0$ Plug in the first $$x^2-z^2=1$$ A parametrization is $$(x=\cosh t,y=0,z=\sinh t); (x=-\cosh t , y=0, z=-\sinh t)$$ In the image below the two surfaces and their intersection. $$...$$


2

Smooth sections are defined to be maps $\sigma : M \to E$ between the base space of a smooth fiber bundle (which is a smooth manifold) and the total space of the bundle (which is another smooth manifold) which are smooth as maps in the usual sense, and satisfy some particular properties. In other contexts you might want to restrict to bundles and manifolds ...


2

The formula I conjectured is almost true. The correct formula is $ch_{CS}(\overline{\mathcal E})=\overline\partial ch_{BC}$. See the last line of the proof of proposition 3.15 of "Hermitian vector bundles and the equidistribution of the zeroes of holomorphic sections", in which Bott and Chern introduce the Bott-Chern form.


2

Every space curve with zero torsion is a plane curve. If the (time) derivative of the acceleration is zero, then so is torsion.


2

Suppose $D$ is a derivation and $f\in C(X)$. For $x\in X$ look at $(Df)\,(x)$, since $D$ applied to any constant function is necessarily $0$ you have $$(Df)\,(x) = D(f-c )\, (x)$$ with $c$ a constant function. Now for $c=f(x)$ you have that $f-c$ is $0$ at $x$. You may check that there exist $g,h$ continuous with $f-c= g\cdot h$ and $g(x)=h(x)=0$ (this is a ...


2

We may assume $u_0 = I$ wlog, as multiplication is continuous and $$ (u_0 + h)^{-1} - u_0^{-1} + u_0^{-1} h u_0^{-1} = u_0^{-1}\left( (I + hu_0^{-1})^{-1} - I + hu_0^{-1} \right). $$ Using the Neumann series, we have $$ (I+h)^{-1} - I + h = \sum_{k\ge 2} (-h)^k = (-h)^2 \sum_{k\ge 0} (-h)^k = h^2 (I+h)^{-1} = o(\|h\|), $$ as $$ \frac{\| h^2 (I+h)^{-1} \|}{\|...


2

Yes, it is isomorphic to the subgroup of $GL(2,\Bbb C)$ which consists of the matrices of the form$$\begin{bmatrix}1&z\\0&1\end{bmatrix},$$because$$(\forall z_1,z_2\in\Bbb C):\begin{bmatrix}1&z_1\\0&1\end{bmatrix}.\begin{bmatrix}1&z_2\\0&1\end{bmatrix}=\begin{bmatrix}1&z_1+z_2\\0&1\end{bmatrix}.$$


2

If the $C^1$ curve $\gamma\colon I\to\mathbb R^n$ is regular, then by definition, $\gamma'\left(x\right)\ne\mathbf 0$ for all $x\in I$. This means that curve $\gamma$ never "slows down" to stop. For example, constant curve $\gamma_1\left(t\right)=\mathrm{x}_0$ is not regular because $\gamma'_1=\mathbf 0$. On the other hand, straight line $\gamma_2\...


2

Fix a point $x_0\in M$, and then for a given point $x\in M$, define $$f(x):=\int_\gamma \alpha$$ where $\gamma$ is any path from $x_0$ to $x$.


2

Here's the core step of the computation: $$ \nabla_X^\ast \nabla_Y^\ast (\xi)(s) = \nabla_X^\ast (\nabla_Y^\ast (\xi))(s) = L_X (( \nabla_y^\ast \xi)(s)) - (\nabla_Y^\ast)(\nabla_Xs) \\ = L_X(L_Y\xi(s) - \xi(\nabla_Ys))-L_Y\xi(\nabla_X s)+\xi(\nabla_Y\nabla_Xs) \\= L_X L_Y\xi(s) - L_X \xi(\nabla_Ys)-L_Y\xi(\nabla_X s)+\xi(\nabla_Y\nabla_Xs) .$$ Can you take ...


2

Your space $\mathcal F$ is usually denoted $L^1(\mathbb R^n,\mu )$ and, if you are denoting by $\mathcal{F}^*$ its dual as a normed space, equipped with the the norm $$ \|f\|_1 = \int_{{\mathbb R}^n}|f(x)|\,d\mu(x), $$ then $L^1(\mathbb R^n,\mu )^*\simeq L^\infty (\mathbb R^n,\mu )$. According to this isomorphism your functional $\iota _U$ corresponds ...


2

Connected abelian Lie groups can be classified in that all of them are $\mathbb{R}^a \times (S^1)^b$ for some numbers $a$ and $b$. (If you want to prove this on your own, it helps to investigate the exponential map from the Lie algebra to the Lie group more closely.) Now, do you know the Lie algebra of this Lie group? Do you know of a compact Lie group with ...


2

The map that you wrote is not an isometry, I'll give you the correct definition. It's a very long calculation but not so difficult. Recall that a smooth map $f:(X,g)\rightarrow (Y,h)$ between riemannian manifold is an isometry if it is a diffeomorphism and $f^*h=g$, where $\forall u,v\in T_pX$ one has $f^*h(u,v):=h(d_pf(u),d_pf(v))$ and $d_pf: T_pX\...


1

If you take the first restriction and plug it into the second, you get $$ x^2-y^2-(x^2+y^2-1) - 1 =0 \Leftrightarrow -2y^2 = 0 \Leftrightarrow y=0 $$ So, any point on the curve satisfies $y=0$ and $x^2-z^2 = 1$ and you can consider two separate parametrizations: $$ (\sqrt{t^2+1}, 0, t), t\in \mathbb{R} $$ $$ (- \sqrt{t^2+1}, 0, t), t\in \mathbb{R} $$


1

In my answer, I will assume that $$ d(A,B)=\inf_{a\in A, b\in B} d(a,b). $$ Here is a simple and classical example. Consider the modular group $\Gamma=PSL(2, {\mathbb Z})$ and its standard fundamental domain $F$. Let $u_n\in \Gamma$ be any sequence corresponding to (strictly) upper-triangular matrices. Then for each $u_n$, $d(u_nF,F)=0$. Or, for a simpler ...


1

For any compact, locally contractible, nonempty, proper subspace of $K$ of $n$ dimensional sphere $\mathbb{S}^n$, Alexander Duality says: $\tilde{H}_i(\mathbb{S}^n-K;\mathbb{Z})\simeq\tilde{H}^{n-i-1}(K;\mathbb{Z})$ for all $i$. Thinking $\mathbb{S}^n$ as one point compactification of $\mathbb{R}^n$ and applying Alexander Duality to the cases when $K$ is ...


1

All you need is the exterior derivative and the Hodge star; you can already anticipate this because the expression you're looking at is $\nabla \times (\mathbf{w}\times \mathbf{u})$, and the curl is really a manifestation of $d$, the metric tensor and the Hodge star, while the cross product is "essentially" the wedge product of two 1-forms once you ...


1

The following definition is from the book you mention but I couldn't find your example there. If $\omega$ is a 1 -form on $M,$ we define its pullback $F^{*} \omega$ to be the 1 -form on $N$ given by $$ \left(F^{*} \omega\right)_{p}\left(X_{p}\right)=\omega_{F(p)}\left(F_{*} X_{p}\right) $$ for any $p \in N$ and $X_{p} \in T_{p} N$ You need to be certain of ...


1

Consider the hypersurface $S=\{\eta_{\mu \nu} x^{\mu}x^{\nu}=0$} and a point $ P \in S$ different than the origin. A (tangent) vector $\mathbf{v}$ in $P$ is a lightlike vector if, by definition, $\eta(\mathbf{v},\mathbf{v})=0$. This is a quadratic equation. On the other hand the condition that defines the tangent space of $S$ in $P$ is linear. For simplicity,...


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