# Tag Info

6

A rank 0 tensor is a scalar. A rank 1 tensor is a row or column vector. A rank 2 tensor is a matrix, often square. A rank 3 tensor? Think 3D matrix. Instead of a rectangle with data entries for each column and row, think of a cube. Rank 4... go 4D!

6

Coordinate-wise, one could say that a matrix is a "square" of numbers, while a tensor is a $n$-block of numbers. But this is horrible, not insightful and even a bit wrong, since those coordinates must "change in appropriate ways" (this is part of why this is horrible). It may be best to think as follows: given a vector space $V$, a matrix can be seen in ...

4

Yes, your idea is okay. Let $(M,g)$ and $(N,h)$ be arbitrary Riemannian manifolds, then the Ricci curvature of the product manifold $(M\times N, g\oplus h)$ takes the form $$\mathrm{Ric}_{g\oplus h} = \begin{pmatrix} \mathrm{Ric}_g & 0 \\ 0 & \mathrm{Ric}_h \end{pmatrix}$$ So in particular if you arrange for $M$ to have constant scalar ...

3

They're the same. The fact you want to use to show this is that if $f$ is a bilinear map from $V \times V$ to $R$, then it extends to a map on your group $G$, and vanishes on all of the elements of $H$, so it induces a linear map on the tensor product $V \otimes V$. Knowing this, we want the linear maps $L(V \otimes V, R)$ to correspond to the bilinear maps ...

3

Let $D$ be a tensor derivation on a manifold $M$ I assume it means $D$ is $\mathbb{R}$-linear and obeys Leibniz rules with respect to tensor multiplication and contraction See axioms 2 and 3 here. Actually here we only need 2 things: (1) For every vector field $X$ and 1-form $\omega$ we have $$D(\omega(X))=D(\omega)(X)+\omega(D(X)).$$ (2) For function ...

3

Are you sure about the formula from the book? Because it involves twice the derivative of $R^\alpha_\beta$, which shouldn't be the case, while your result seems correct.

3

Matrices are a special type of tensor, rank 2. Scalars, vectors, matrices, are all tensors. Honestly tensors are so general the vast majority of things you deal with in your class are tensors.

2

I was also having trouble with this for a long time. The explanation which finally worked for me was the following: For the purposes of parallel transport along a particular circle of latitude, the sphere can be replaced by the cone which is tangent to the sphere along that circle, since a “flatlander” living on the surface and travelling along the circle ...

2

Let me explain here an actually useful characterisation of a "tensor" that is not as oldfashioned as the one of how a "tensor" transforms under change of coordinates. In the process I hope I can clarify why the difference between two connexions is a "tensor". I shall assume smoothness everywhere. A connexion, the way it is defined by the Original Poster, ...

2

$R_{ip}R_{iq}= \delta_{pq}$ (notice that the indices on the Kronecker delta should be $pq$) can be shown in just a few steps: \begin{align}R_{ip}R_{iq} &= R_{pi}^TR_{iq} &\text{(by definition of transpose)} \\ &= [R^TR]_{pq} &\text{(by definition of matrix products)} \\ &= [I]_{pq} &\text{(by definition of orthogonal matrices)} \\ ... 1 These definitions coincide if V is a finite dimensional vector space. If we denote by V\otimes V the tensor product via the quotient construction, then you can construct an isomorphism \Phi\colon V\otimes V\rightarrow L(V^*,V^*,\mathbb R) as follows: The natural map V\times V\rightarrow L(V^*,V^*,\mathbb R),\; (v,w) \mapsto [(\varphi,\psi)\mapsto ... 1 maybe you tought of this, let S be subspace of Hilbert space V and g_1,...,g_n be a basis of S but not orthogonal basis and let them have unit norm, and lets say you want to find projection of f\in V on S, if g_i , 1\le i \le n were orthogonal you would have P_S f = \sum_{i=1}^n \langle f,g_i\rangle g_i but since they are not find dual basis ... 1 The simplest explanation is the following: Given a basis ({\bf e}_i)_{1\leq i\leq n} of some vector space V over a field F each vector {\bf x}\in V gets coordinates x_i\in F \>(1\leq i\leq n) with respect to that basis:{\bf x}=\sum_{i=1}^n x_i{\bf e}_i\ .$$In fact, for each i, the i^{\rm th} coordinate of {\bf x} depends linearly on ... 1 The Shape Operator S: T_pM \rightarrow T_pM this is true. In particular, this is calculated in terms of dot-products of the coordinate velocities and the normal vector field to the surface; S(\alpha') \cdot \alpha' = \alpha'' \cdot U where U is the normal vector field to M. Some usual notation: (ala O'neill's Elementary Differential Geometry)$$ ...

1

The red and blue vector fields in your picture are not parallel along the pink curve. One way to see this is to note that you can compute the covariant derivative of a vector field along a curve in the sphere by computing its ordinary derivative in $\mathbb R^3$, and then orthogonally projecting that onto the tangent plane. At any point on the pink circle, ...

1

One way to see that the blue vector field is not parallel is by noting that it is the vector field corresponding to velocity. If it were parallel, the red circle would be a geodesic. But geodesics in the sphere are the great circles. (Actually, this is the same thing that Jack Lee is saying, but phrased differently.)

1

Compute first the integral $$\int_S x_k dS$$ which is obviously zero for any $k$ then it will be of course zero after construction with any other tensor.

1

You seem to have figured it all out so here is just a quick answer to your final question. The particular names we use for the summation variable(s) are just labels so we are free to change $(i,j)\to(j,i)$. Performing such a change gives us that $$\partial_i u_j\partial_i u_j = \partial_j u_i\partial_j u_i$$ which is what you need to get to the final ...

1

A connection on $E$ is a map $\nabla: \Gamma(E) \to \Gamma(E \otimes T^*M)$ satisfying certain conditions. By having it act as the Levi-Civita connection on $T^*M$ you inductively also have connections $\nabla: \Gamma(E \otimes (T^*M)^{\otimes_k}) \to \Gamma(E \otimes (T^*M)^{\otimes_{k+1}})$. $\nabla^k$ means a composition of a bunch of these to get a map ...

1

As $T$ and $S$ are inverse of each other, it is $T^i_{\phantom{k}k}S^k_{\phantom{h}h}=\delta^i_{\phantom{h}h}$. Let's write the right-hand side of (1) such we clearly distinguish the bounded indices ($h,i,j$)from the free ones in both, (1), ($k,m,p$), and (2), ($i,j,p$). $$S^k_a\Gamma^a_{bp} A^b_cT^c_m\,+\,S^k_aA^a_{b,p}T^b_m\,+\,S^k_aA^a_b\Gamma^b_{cp} ... 1$$\delta_{ij}v_j = \delta_{ji} v_j = v_i$$because$$\delta_{ij} = \delta_{ji} = \begin{cases} 0, & i\ne j \\ 1, & i=j\end{cases}$$so the sum \sum_j \delta_{ij}v_j is ONLY NONZERO when i=j. That is$$\sum_j \delta_{ij}v_j = \delta_{i1}v_1 + \delta_{i2}v_2 + \cdots + \delta_{ii}v_i + \cdots + \delta_{in}v_n = 0v_1 + 0v_2 + \cdots + 1v_i + ...

1


Only top voted, non community-wiki answers of a minimum length are eligible