Tag Info

New answers tagged

0

The answer is that the second contracted pair of deltas amounts to a single delta contracted which gives $4$ whereas the first amounts to the product of two contracted deltas hence $(4)(4)=16$. Thus, $-2(16-4)=-24$. I'm not sure about the source of the minus just yet. I'll add some detail as tensor arithmetic seems unclear to some users. First, the ...


0

The two notations are only slightly different. The metric tensor in the Klein model is $$ ds^2 \;=\; \frac{{dx_1}^2 + \cdots + {dx_n}^2 }{1 - {x_1}^2-\cdots - {x_n}^2}+\frac{(x_1\,dx_1 + \cdots + x_n\, dx_n)^2}{\bigl(1-{x_1}^2-\cdots-{x_n}^2\bigr)^2} $$ and the metric tensor in the Poincaré model is $$ ds^2 \;=\; 4\frac{{dx_1}^2+ \cdots + ...


1

Metric tensor would be the matrix $(g_{ij})$ given by the products $\langle f_i,f_j \rangle$. Covariant means (probably) that matrix (the inverse matrix transforms contravariantly).


1

$$\dfrac{d(g_{ab})}{dt}=\dfrac{\partial(g_{ab})}{\partial x^c}\cdot\dfrac{\partial x^c}{\partial t}=\partial_cg_{ab}\cdot\dot{x}^c$$ $$\dfrac{d(g_{ab}\dot{x}^b)}{dt}=\dfrac{d(g_{ab})}{dt}\cdot\dot{x}^b+g_{ab}\cdot\dfrac{d(\dot{x}^b)}{dt}=\partial_cg_{ab}\dot{x}^c\dot{x}^b+g_{ab}\ddot{x}^b$$


0

Since this question didn't get much response (4 days later), I'll post my own answer. $$\bf{U}\bf{V} = \bf{V}\bf{D}$$ where $\bf{V}$ is the tensor with the eigenvectors, and $\bf{D}$ is the diagonal tensor with the eigenvalues on the diagonal. One has $$\bf{U} = \bf{V}\bf{D}\bf{V}^{-1}$$ and since $$\bf{U} = \bf{U}^{T}$$ it follows that ...


1

Let $V(M)$ denote the space of vector fields on $M$. There are two equivalent requirements for a function $V(M)\times\ldots\times V(M)\to C^\infty(M)$ in order for it to be a tensor. One can require the function to be linear over $C^\infty(M)$ or alternatively require the function to induce a multilinear form at every point, depending only on the value of ...


1

It is a quadratic form straight from the definition: A quadratic form is a homogeneous polynomial of degree 2 in $n$ variables (over a field $\mathbb{F}$, but take $\mathbb{F} = \mathbb{R}$ for simplicity). To see this, let me write $\mathrm{d} s^2$ as $g : T_p M \times T_p M \rightarrow \mathbb{R}$, where $T_p M$ is the tangent space at a point $p$ to the ...


3

In fact $a\otimes b\mapsto a\otimes b$ is an isomorphism $A\otimes_DB\leftrightarrow A\otimes_FB$ as spaces or $D$-algebras, for any spaces/algebras $A$ and $B$ over a field $F$ which is the fraction field of a domain $D$. The reason is that $$\begin{array}{ll} \displaystyle \color{Blue}{\frac{1}{y}}a\otimes b & \displaystyle =\frac{1}{y}a\otimes ...


0

I figured out the following proof. Considering the following primitive idempotent $$\tilde{e}_\lambda \equiv a_\lambda s_\lambda,$$ one can show that it is equivalent to $e_\lambda$, since $\tilde{e}_\lambda e_\lambda = a_\lambda s_\lambda s_\lambda a_\lambda = \eta a_\lambda e_\lambda \ne 0$. Therefore the left ideal generated by $e_\tau$ can be ...


3

If $\sigma$ is a permutation, then $T^{\sigma(a)}{T'}_{a}=T^{a}{T'}_{\sigma^{-1}(a)}$. If a permutation is part of a symmetrizer for a Young tableaux, then so is its inverse, so $T^{\tau(a)}{T'}_{a}=T^{a}{T'}_{\tau(a)}$. So $T^{\tau(a)}{T'}_{\lambda(a)}=T^{a}{T'}_{\tau(\lambda(a))}=0$.


0

If you consider the definition of tensor product by universal property, which is unavoidably equal to one you have in mind, then you have two natural isomorphisms $V\otimes W\cong W\otimes V$ and $(V_1\otimes V_2)\otimes V_3\cong V_1\otimes (V_2\otimes V_3).$ This implies that you can simply say that the position on which we write particular factor of tensor ...


1

$(|1\rangle \otimes |0\rangle \otimes |1\rangle) + (|0\rangle \otimes |0\rangle \otimes |1\rangle)$ is clearly a product state. It's tricky to find good notation to denote this and I don't think there's an established standard way. Here's one option: Let $\sigma_{ij}$ be an automorphism that flips the $i$th and $j$th position of a tensor product ...


3

A different reason why it can't be true is that there just aren't enough $(\varepsilon,v)$ pairs to represent all the functionals when $F$ is finite. Suppose for example $F=\mathbb F_p$ and $n=3$. $\mathcal L(V)$ is a 9-dimensional vector space and therefore there are $p^9$ different linear functionals on $\mathcal L(V)$. On the other hand there are only ...


3

I don't think what you say is true. $\mathcal{L}(V) = \text{Hom}(V, V) = V \otimes V^*$. Your map $\Theta_v: V \otimes V^* \rightarrow V$ is given by tensoring the evaluation map $ev_v: V^* \rightarrow k$ with $V$. A functional on $\text{Hom}(V, V)$ is the same as an element of $\text{Hom}(V, V)$, the relationship being taking the dual of an endomorphism ...


0

$\nabla^2 (\phi \vec A)=\partial_i^2(\phi A_j)=\phi \partial_i^2(A_j)+2\partial_i (\phi) \partial_i(A_j)+A_j\partial_i^2(\phi)=\phi \nabla^2 \vec A+2(\nabla \phi \cdot \nabla)\vec A+\vec A \nabla^2 \phi$ And the rest is trivial.


1

Your question indicates you should read a book on tensors or multilinear algebra, but here is a brief reply. Tensors of type (m,n) on a vector space can be viewed as objects belonging to the space $V \otimes V \ldots \text{(m times)} V^* \otimes V^* \otimes \ldots \text{(n times)}$, where $V^*$ is the dual space to the vector space $V$ and $\otimes$ is the ...


2

Rather than convert the bivector $R(x \wedge y)$ to a 2-form, one can use clifford algebra instead. Define a product operation on arbitrary vectors $a, b, c$ such that $$aa = g(a,a), \quad (ab)c = a(bc)$$ I use $g$ here for the Riemannian metric instead of $\langle,\rangle$, for reasons that will become clear shortly. This "geometric product" produces a ...


4

Well, Singer and Thorpe do explain the notation (see p.356 in the original text) by saying that Using the usual isomorphisms defined by the inner product, a curvature tensor on $V$ may be regarded as a $2$-form on $V$ with values in the vector space of skew symmetric endomorphisms of $V$. So the meaning of $R(x,y)z$ is an endomorphism $R(x,y)$ ...


0

For a more mathematically flavored introduction, consider studying: Vector Calculus, Linear Algebra, and Differential Forms, by Hubbard and Hubbard Semi-Riemannian Geometry with an Introduction to Relativity, by O'Neill A good advanced linear algebra book. This should be one that focuses on abstract vector spaces and multilinear algebra, not Gaussian ...


0

Have a look at: http://www.amazon.com/Einsteins-Theory-Introduction-Mathematically-Untrained/dp/1461407052/ref=sr_1_fkmr0_1?s=books&ie=UTF8&qid=1431089294&sr=1-1-fkmr0&keywords=relativity+theory+for+the+mathematically+unsohisticated which is Øyvind Grøn, Arne Næss: Einstein's Theory: A Rigorous Introduction for the Mathematically ...


3

What is a tensor? In short, a tensor is a generalization of a vector which is needed to express physical quantities which have more data than we can fit into a single vector field. However, it's more than that. We also need tensors of different transformation type. Ultimately, in physics, we wish to write equations which are independent of the choice of ...


1

In the special case of two dimensions the shear operator is known as the Cauchy-Riemann operator (or $\bar \partial$ operator, or one of two Wirtinger derivatives), and is denoted $\dfrac{\partial}{\partial \bar z}$. It is certainly useful in complex analysis. The $n$-dimensional case comes up in the theory of quasiconformal maps where the operator ...


2

The object $\tfrac{\partial}{\partial x}$ is understood here as a vector field (the coordinate vector field corresponding to the function $x \colon U \to \mathbb{R}$ for some open set $U \subseteq M$, where $M$ is the manifold in the question, but one can think of $\mathbb{R}^3$ for simplicity). Hence, $\tfrac{\partial}{\partial x}$ is a particular example ...


0

Well, the partial derivative operator in your question is indeed (locally) a tensor, being a section of the tangent bundle. The dual one-form is $dx,$ i.e. it is the one-form defined by the following relationships: it evaluates to one on the vector field $\frac{\partial}{\partial x},$ and it evaluates to zero on the other basis vector fields ...


1

Clifford algebra can make this manipulation simpler. To use it here, first define the geometric product of vectors. This is traditionally denoted by juxtaposition: i.e. the geometric product of a vector $a$ and $b$ is denoted $ab$. Given vectors $a, b, c$, the geometric product has the following properties: $$aa \equiv \langle a,a\rangle \implies ab = ...


0

Assume that $\{e_i\}$ is a basis for $V$. Note that by linearity we suffice to show that $$ v=e_{i_1},\ (i_v\phi^1\wedge\cdots \wedge \phi^k, e_{i_2} \wedge\cdots \wedge e_{i_{k}} )=( \phi^1\wedge\cdots \wedge \phi^k,v\wedge e_{i_2} \wedge\cdots \wedge e_{i_{k}} )$$ Proof : $$ i_v\phi^1\wedge\cdots \wedge \phi^k ( e_{i_2}, \cdots , e_{i_{k}} )= ...


0

Let $E$ be a matrix with the same size as $F$, and with all elements $E_{jl}=1$. Then there are several different ways to write the expression using the Hadamard ($A\!\circ\!B$), Matrix ($AB$), and Frobenius ($A\!:\!B$) products. $$\eqalign{ S &= (F\circ F):WE + (F\circ F):W^TE - 2\,FF^T:W \cr &= (F\circ F):(W+W^T)E - 2\,FF^T:W \cr ...


2

I don't have a complete answer, but I think I can add some information. In the spirit of you last comment, the way toweards characterizing the cone of simple tensors is looking at maximal linear subspaces contained in it. Each line in $V$ determins a linear subspace of dimension $dim(W)$ contained in the cone and vice versa. These two families are exactly ...



Top 50 recent answers are included