0
votes
1answer
19 views

Boundedness of Riemannian curvature gradient

I'm reading a paper by Wan-Xiong Shi "Deforming the metric on complete Riemannian manifolds". And there is a statement without proof. It can be summarized as follows: Let $B(x_{0},\gamma)$ be a ...
1
vote
0answers
21 views

How to change the parametric equations of a hypersurface in $V_N$ to another form…

This exercise was given in the first pages of Synge & Schild Tensor Calculus. The parametric equations of a hypersurface in $V_N$ are $x^1=a\cos{u}$, $x^2 = a\sin{u^1}\cos{u^2}$, $x^3 = ...
0
votes
1answer
34 views

Time evolution of Laplacian

While reading monograph on the Ricci flow, I came accross a fact (at least I think it is a fact), which is not proved explicitly in that book. Assume a smooth 1-parameter family of Riemannian metrics ...
2
votes
0answers
91 views

Prove the Curvature Tensor is a Tensor

For an affine connection $\nabla$, prove the curvature R $R(X,Y,Z,\alpha)=\alpha(\nabla_X \nabla_Y Z - \nabla_Y \nabla_X Z -\nabla_{[X,Y]}Z)$ with $X,Y,Z$ vector fields and $\alpha$ a co-vector, is ...
0
votes
0answers
53 views

The trace of a wedge product of matrices

I'm trying understand a computation on Besse's book (p. 371). I already know the curvature operator $R:\bigwedge^2\to\bigwedge^2$ may be written in block diagonal form relative to the direct sum ...
1
vote
0answers
50 views

Curvature tensor on the sphere

While reading R. Hamilton paper "Three-manifolds with positive Ricci curvature" I came across the sentence: For the sphere we have $$R(u,v,u,v)=R_{ijkl}u^{i}v^{j}u^{k}v^{l}>0,$$ which is the ...
4
votes
1answer
63 views

Symmetry of the Riemannian curvature tensor

The Riemannian curvature tensor, in local coordinates, $R_{ijkl}$, has the following symmetries: $$R_{ijkl}+R_{jikl}=0;$$ $$R_{ijkl}+R_{ijlk}=0;$$ $$R_{ijkl}=R_{klij};$$ ...
3
votes
2answers
77 views

What is the value of the 2-form $X(u(Y))-Y(u(X))-u([X,Y])$, is it $2du(X,Y)$?

Let $X,Y$ be two vector fields on a Riemannian manifold $(M,g,\nabla)$ where $\nabla $ is the symmetric metric connection on $M$ and let $u$ be a 1-form. I want to find the value or a simple form or ...
2
votes
3answers
241 views

Resources for properly developing a modern understanding tensors

I am currently learning about tensors as they come up in the mathematics behind continuum mechanics. I was fairly disappointed with my initial foray into tensors, as presented in the book Classical ...
1
vote
0answers
47 views

Metric on an open subset of $\mathbb{R}^d$ and Christoffel symbol of the second kind under diffeomorphisms

This is a follow-up question to this question I proved the following identity. $\Omega \subset \mathbb{R}^d$ is open and $g$ is a metric field on $\Omega$. Further $\Gamma_{\,kl}^j$ denotes the ...
3
votes
0answers
78 views

Intuition about the Riemann and Ricci tensors

I have started an attempt to self-study Riemannian Geometry. I well understand all the algebraic properties of the Riemann tensor (With a symmetric connection), and how it gradually becomes less and ...
1
vote
1answer
56 views

Metric tensors. Have I got the correct understanding?

My course is covering metric tensors in a slap-dash way, so I want to ensure I have understood correctly how they are described. I hope you can confirm this! So, I believe that a metric tensor is a ...
1
vote
1answer
260 views

Riemann tensor in terms of the metric tensor

The Riemann tensor is a function of the metric tensor when the Levi-Civita connection is used. Most tensor notation based texts give the Riemann tensor in terms of the Christoffel symbols, which are ...
5
votes
1answer
179 views

Metric on an open subset of $\mathbb{R}^d$ and Christoffel symbol of the second kind

I need to prove the following identity. $\Omega \subset \mathbb{R}^d$ is open and $g$ is a metric field on $\Omega$. Further $\Gamma_{\,kl}^j$ denotes the Christoffel symbol of second kind. $g^{ij}$ ...
1
vote
1answer
63 views

Trouble understanding Tensor product in context of Torsion Tensor

I know that there are quite a few threads dealing with this question already. I have pored through them for quite some time and they have been informative. However there are still some clarifications ...
3
votes
1answer
133 views

A linear connection induces a covariant derivative of tensor fields.

Let $M$ be a smooth manifold. notation: $\mathcal T(M)^{(k,l)}$ is the $C^{\infty}(M)$-module of all tensor fields of type $(k,l)$ on $M$ ($k$ indicates the covariant part). $\mathcal ...
1
vote
1answer
116 views

Decomposition of the Curvature operator and Matrix representation

I'm trying do this question from Peter Petersen's Book and I can't do some parts. I know that $$R=\frac{scal}{2n(n-1)}g\circ g+\left(Ric-\frac{scal}{n}g\right)\circ g+W$$ Where, $R$ is the ...
2
votes
1answer
64 views

Self-dual and anti-self-dual decomposition

Please take a look at the following: Let $(M,g)$ be a four-dimensional oriented Riemannian manifold. The Hodge star operator $*$ obeys $**=Id$ acting on 2-forms. This allow us to decompose the ...
4
votes
1answer
230 views

The divergence of the Weyl tensor

First, the Weyl tensor is given by $$W_{ijkl}=R_{ijkl}-\frac{1}{n-2}(g_{ik}A_{jl}-g_{il}A_{jk}-g_{jk}A_{il}+g_{jl}A_{ik})$$ where, $A_{ij}$ is the Schouten tensor, given by ...
2
votes
0answers
75 views

Tensors: summing over indices

Would anybody mind teaching me how to work these indices? Definitions: Throughout the following, repeated indices are to be summed over. Hodge dual of a p-form $X$: $$(*X)_{a_1...a_{n-p}}\equiv ...
2
votes
1answer
225 views

Contraction of the second Bianchi identity

The second Bianchi identity is $${R^a}_{b[cd;e]}=0$$ And contracting it with respect to $a$ and $e$ we get $${R^a}_{b[cd;a]}=0 \Leftrightarrow $$ $${R^a}_{bcd;a}+R_{bc;d}-R_{bd;c}=0$$ What I don't ...
2
votes
1answer
391 views

Quotient theorem for tensors

Can somebody please explain to me how the following statement is true? The Riemann curvature tensor $R^c_{dab}$ is given by the Ricci identity $$(\nabla_a\nabla_b-\nabla_b\nabla_a)V^c\equiv ...
1
vote
0answers
180 views

Riemannian curvature and its application on covariant derivative of tensors

This identity can be generalized to get the commutators for two covariant derivatives of arbitrary tensors as follows $ \begin{align} T^{\alpha_1 \cdots \alpha_r}{}_{\beta_1 \cdots \beta_s ; ...
1
vote
0answers
310 views

How to derive covariant derivative and Lie derivative of tensors

1) As title says, how does one derive the following equation for covariant derivate of tensor: $A^{\alpha}{}_{;\beta} = A^{\alpha}{}_{,\beta} + \Gamma^{\alpha} {}_{\gamma\beta}A^\gamma$ where ...
4
votes
2answers
150 views

Tensor Components

I would like to ask something On the Barrett Oneill's Semi-Riemann Geometry there is a definition of tensor component: Let $\xi=(x^1,\dots ,x^n)$ be a coordinate system on $\upsilon\subset M$. If $A ...
4
votes
0answers
164 views

What are spinor fields?

For example, a tensor field $T=T_{a,b}^{\ \ \ \ c}$ on a manifold $M$ should be thought as $$ T_{a,b}^{\ \ \ \ c}dx^{a}\otimes dx^{b}\otimes \frac{\partial}{\partial x^{c}} $$ with local coordinate ...
2
votes
1answer
56 views

Question on tensor calculation in Reimannian geometry

Given a Riemannian manfiold $M$ with metric $g=(g_{i,j})$. Let $T=T_{A,B,C,\dots}^{a,b,c,\dots}$ be a tensor on $M$. I would like to compute for example $T_{A,B,C,\dots}^{a,n,c,\dots}g_{n,m}$. We ...
4
votes
2answers
273 views

Is there such a thing as discrete Riemannian geometry?

General relativity expresses gravity as a curvature in space time, created by the stress energy tensor. $$T_{\mu\nu} \approx R_{\mu\nu} - \frac{R}{2} g_{\mu\nu}$$ Given I put the fact that energy is ...
2
votes
0answers
162 views

Extending Tensor Fields defined on Manifolds to Ambient Space

I am currently reading about tensor fields on manifolds, and I came across two comments that sound contradictory to me. The first comment is made in the book by James Munkres "Analysis on Manifolds", ...
2
votes
1answer
164 views

Simple problem with the normal curvature tensor

If $M$ is a s-R(semi-Riemannian) submanifold of a s-R manifold $\overline{M}$ the function $R^{\perp}:\mathfrak{X}(M)\times\mathfrak{X}(M)\times\mathfrak{X}(M)^\perp\rightarrow\mathfrak{X}(M)^\perp$ ...