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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 $R(x,y)$ ...


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 $T^{\tau(a)}{T'}_{\lambda(a)}=T^{a}{T'}_{\tau(\lambda(a))}=0$.


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 ...


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 ...


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 ...


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 ...


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 ...


2

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

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 ...


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 ...


1

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


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 = ...


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 ...


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 ...



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