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6

Hodge theory is a vast subject, especially since Deligne and Griffiths revolutionized it and algebraized it around 1970 with the introduction of variations of Hodge structures and mixed Hodge structures, axiomatic approaches which found amazing applications in algebraic geometry. At the most elementary level I would say that Hodge theory is a refinement ...


4

We can use the result proved in this question to build one. There it is shown that $f : (-\frac{\pi}{2}, \frac{\pi}{2}) \to \mathbb{R}$ defined by $$f(x) = \tan(x)$$ is a diffeomorphism. To obtain the function you are looking for, use four copies of that one: $f : (-\frac{\pi}{2}, \frac{\pi}{2})^4 \to \mathbb{R}^4$ $$f(x_1, x_2, x_3, x_4) = ...


4

The factor $a$ makes the length of each circle centered at the origin equal to $2\pi a r$, where $r$ is the radius. Case $a<1$: not enough circumference To realize this case in practice, cut off a part of the circle, specifically $(1-a)$ part of it (in angle terms, $2\pi (1-a)$). This is well explained in wikiHow: Note that this surface is ...


3

No, $\operatorname{Aut}(E)$ is not a principal bundle. In a local (orthonormal) trivialization of $E$ over $U\subseteq M$, we get a local trivialization of $\operatorname{Aut}(E)$ of the form $\pi_{\operatorname{Aut}(E)}^{-1}(U) \leftrightarrow U\times \operatorname{O}(n)$; however, when two such local trivializations overlap, the transition functions act on ...


3

You are correct that $S$ is not a smooth surface in $\mathbb{R}^3$. Here is a rough sketch of a proof. Lemma 1: $S$ is a surface iff $T = \{(x,y,z)\in \mathbb{R}^3: x^3 + y^3 + z^3 =0 \}$ is a surface. Proof idea: Consider $f:\mathbb{R}^3\rightarrow \mathbb{R}^3$ with $f(x,y,z) = (x, \sqrt[3]{2}y, \sqrt[3]{3}z)$. Then $f$ is a diffeomorphism, and ...


3

I think it's important to know first how deeply you want to study differential geometry/differentiable manifolds. I agree completely with Mike Miller's comment above, but would like to add a few thoughts. There is a big distinction between just studying differentiable manifolds and differential geometry. Geometry relies entirely on the definition of ...


3

It follows from a much more general fact: Whenever a Lie group $G$ acts smoothly on a smooth manifold $M$, its orbits are immersed smooth manifolds. This can be derived from the following observations: For each $p\in M$, the isotropy group $G_p = \{g\in G: g\centerdot p=p\}$ is a closed subgroup of $G$. The quotient space $G/G_p$ has a unique smooth ...


3

Always, since $p\not\in C_p(M).$


3

I agree with Mike that you should ask a faculty member about this, since we have no concrete idea about how deep your lecture program goes, nor what is of interest to your professor. That being said, there is an answer lying perfectly in the intersection of your two areas, and that is Klein Geometry. Klein started the so called Erlangen Program in the ...


2

I think the question is probably too broad to have a very satisfying general answer. But (assuming you're looking for a smooth Lie group action) here are a couple of necessary, but certainly not sufficient, conditions: Each level set of $f$ must be a smooth submanifold, because every orbit of a smooth Lie group action is an (immersed) submanifold. Each ...


2

As you pointed out, Stein manifolds are Kähler, so one way to find non-compact complex manifolds which are not Stein is to look for non-compact complex manifolds which are not Kähler. Let $S$ be a non-Kähler complex surface (such as a Hopf surface), then for any $k > 0$, $S\times\mathbb{C}^k$ is a non-compact complex manifold which is not Kähler ...


2

In the general non-commuting case, the flow $\phi^t_{V+W}$ equals to first order both $\phi^t_V \circ \phi^t_W$ and $\phi^t_W \circ \phi^t_V$. Morally, the second order approximation should be 'halfway between' the two aforementioned flows. Since $\phi^{t}_V \circ \phi^{t}_W \circ \phi^{-t}_V \circ \phi^{-t}_W$ is approximated by $\phi^{t^2}_{[V,W]}$, we ...


2

I gave an answar here as well. Theorem. Every open star-shaped set $\Omega$ in $\mathbb{R}^n$ is $C^\infty$-diffeomorphic to $\mathbb{R}^n.$ Proof. For convenience assume that $\Omega$ is star-shaped at $0.$ Let $F=\mathbb{R}^n\setminus\Omega$ and $\phi:\mathbb{R}^n\rightarrow\mathbb{R}_+$ (here $\mathbb{R}_+=[0,\infty)$) be a $C^\infty$-function ...


2

In order to fix the ideas, let the Lorentzian metric on the Minkowski space $\mathbb{L}^3$ have signature $-++$. The Lorentz group $O(1,2)$ of $\mathbb{L}^3$, on the one hand, acts transitively on the subset $\mathbb{S}^2_1 \subset \mathbb{L}^3$ and, on the other hand, is a subgroup of the linear group $\mathrm{Gl}(3, \mathbb{R})$. The first fact implies ...


2

For putting a manifold structure on the quotient, it's irrelevant whether $G_x$ is normal. The quotient of a Lie group modulo a closed subgroup always has a unique smooth manifold structure such that the group action is smooth -- this is the Quotient Manifold Theorem (see Theorem 21.10 in my Introduction to Smooth Manifolds, 2nd ed.). Normality of the ...


2

Hint The elements of the Lie algebra $\mathfrak{g} \cong T_{\Bbb I} G$ are the tangent vectors at the identity element $\Bbb I \in G$ to curves in $G$ through that point.


2

It seems natural to require that the operators $d$, $\nabla$, $\text{div}$, $\Delta$, etc., are complex linear and to work with a Hermitian metric $g$ on vector fields that's conjugate linear, say, in its second entry, as you describe. Then we have Hermitian $L^2$ inner products on the spaces of complex-valued functions: $$ \langle f, h \rangle_{L^2, ...


2

A global section would indeed trivialize the bundle - if it were a principal bundle! I'm going to work more generally here and talk about the gauge group of a principal $G$-bundle $\pi: P \to M$. The gauge group is defined to be the group of $G$-equivariant diffeomorphisms $\varphi: P \to P$ that cover the identity (take fibers to the same fiber). That is, ...


2

One interesting generalization follows Fourier's original analysis for Ordinary Differential Equations on $\mathbb{R}$. It's easiest to break first into ODEs on a half line $[0,\infty)$ for Sturm-Liouville problems $$ Lf=-\frac{d^{2}f}{dx^{2}}+q(x)f(x) = g(x),\\ \cos\alpha f(0)+\sin\alpha f'(0) = 0. $$ where ...


1

We have $$ \begin{aligned}{d\over dt}\int_c \Phi_t^* \omega &= \int_{\Phi_t\circ c} \left.{\partial \over \partial t}\right|_{t=0}\Phi_t^* \omega \\ &= \int_{\Phi_t\circ c} \mathcal{L}_\mathbb{X}\omega \\ &= \int_{\Phi_t\circ c} \iota_\mathbb{X} \underbrace{d \omega}_{=0} + d \iota_\mathbb{X} \omega~\text{(by Cartan's Formula)} \\ &= ...


1

"Why would one define $B(X,Y)=tr(ad(X)ad(Y))$ ?" You are right, for matrix algebras I would define a bilinear form more simply, namely just by $C(X,Y)=tr(X)tr(Y)$. This is very natural, because a trace form for linear operators is the easiest thing you can imagine. If we do not have linear operators $X,Y$, then we can enforce this, by using the adjoint ...


1

This is, in my opinion, one of the most interesting question one may ask about symplectic topology. Not only is it related to the classification problem of symplectic structures on manifolds, but the results obtained in this direction show all the complexity and the subtlety of the subject. Let $M^{2n}$ be a manifold. Write $S(M)$ for the space of ...


1

Let $\{e_a\}$ be a local orthonormal frame for the tangent bundle, so $\delta_{ab} = \langle e_a, e_b\rangle$, and let $\nabla$ be a Riemannian connection. We have $\nabla e_a = e_c\otimes {\omega^c}_a$ where ${\omega^c}_a$ are the connection one-forms. Taking the covariant derivative of both sides of the identity $\delta_{ab} = \langle e_a, e_b\rangle$ we ...


1

To check that $D(g^k s)(x)$ is well-defined, we need to show that it's independent of the choices of the function $g$ and the section $s$, i.e., that $D(g^k s)(x)$ only depends on the value of $dg_x = v \in T_x^\ast M$ and $s(x) = e \in E_x$. (I will ignore the constant $\frac{i^k}{k!}$. It's there to normalize things.) There may be a cleaner way to do ...


1

I would say that the basic intuition is that a Riemannian metric defines an inner product on the space of differential forms, which can be used to define an adjoint operator to the exterior derivative. Nowconsider the adjoint $(d_{k-1})^*:\Omega^k\to\Omega^{k-1}$ of $d_{k-1}:\Omega^{k-1}\to\Omega^k$. If the spaces $\Omega^\ell$ were all finite dimensional, ...


1

I have no idea what you mean "without using coordinates". The best you can hope to do without a lot of work is to determine the conjugacy class of the stabiliser. For that, all you need to know is the spectrum. Clearly, if $b \in \mathfrak{su}(n)$, then $ib$ is traceless and hermitian. It can therefore be diagonalised via (special) unitary ...


1

Yes. For each $p\in M$, the set $U_p = M\smallsetminus C_p$ is a neighborhood of $p$, and these neighborhoods cover $M$. Every open cover of a manifold has a countable subcover. [Note that the question in your title is different from the one you asked in the text, and has a different answer -- it's certainly possible to find a countable collection of cut ...


1

The usual adapted chart is formed by using $TU$ which is the disjoint union of $\{ x \} \times T_xM$ for $x \in U$. I assume $U$ is the domain of a chart on $M$ then the formula for $\Psi: TU \rightarrow \mathbb{R}^{2n}$ is simply formed by stringing together the coordinates of the point and the vector which form a bundle-point. That is: assuming $\phi: U ...


1

There is a homology theory that looks like this called (co)bordism, and it is very interesting. It comes in many flavors depending on what kind of extra structure you ask for on the manifolds. The basic problem with your proposal is functoriality: the image of a submanifold need not be a submanifold. The correct definition of bordism fixes this by allowing ...



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