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0

I believe there is a error in the book. Note that the book performs the expansion in two steps: $$ a_{ij} x^i x^j = a_{1j} x^1 x^j + a_{2j} x^2 x^j + \cdots + a_{nj} x^n x^j \\ = a_{11} x^1 x^1 + a_{22} x^2 x^2 + \cdots + a_{nn} x^n x^n $$ The first step has the cross products, as you correctly expected, but the second step ...


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As others have pointed out, you have an implicit summation going over the index $i$, called Einstein Summation. I would like to do an explicit example. Consider two basis $\beta = \{ e_1, e_2, e_3 \}$ and $\beta' = \{ e_1', e_2', e_3' \}$ of $\Bbb R^3$ where $$\left\{ \begin{align} e_1 & = (1,1,0), \\ e_2 & = (1,0,1), \\ e_3 & = (0,0,1), ...


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The last sentence means exactly what's written in the equation: If you write $e'_j = S_{ij} e_i = S_{1j} e_1 + S_{2j} e_2 + \cdots + S_{nj} e_n$ over the $\{e_n\}$ basis, then the coefficient of $e_i$ is exactly $S_{ij}$. (Since the $e_i$ are a basis, the $S_{ij}$ are well-defined and unique.) It might be throwing you off that the book is presumably using ...


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Here the formula $$e'_j=S_{ij}e_i$$ really means that $$e'_j=\sum_{i=1}^nS_{ij}e_i.$$ Now it is clear that $S_{ij}$ is the ith component of the vector $e'_j$ with respect to the unprimed basis.


1

Note that for general curvilinear coordinates, the property $$ \delta_{ij} = 0, \: i \neq j \quad \delta_{ij}= 1, \: i=j $$ even if true in one basis, is NOT true in general, because $\delta_{ij}$ transforms as a (0,2) tensor. However, the (mixed) (1,1) Kronecker delta $\delta^i_j$ does have this property.


2

This is simple and you are able to prove it on your own. Write down the definitions on both sides and compare what the result is. Hint: You have to check two different cases.


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There is a certain point of view from which this is obvious. Parallel transport along a curve $C$ lying on a surface $S$ can be achived as follows, let $T$ be a developable surface which is tangent to $S$ along $C$, then $T$ can be mapped onto the plane. Now just move the vector along the surface $T$ in the ordinary planar sense. This gives one immediately ...


1

Let's take an orthogonal basis of $\mathbb{R}^3$: $e_1, e_2, e_3$. We can write vectors $a = \sum_i a_i e_i$ and $b = \sum_i b_i e_i$. The oriented subspace spanned by these vectors is a 2-blade, denoted $a \wedge b$: $$a \wedge b = \left(\sum_i a_i e_i\right) \wedge \left(\sum_j b_j e_j\right) = \sum_{ij} a_i b_j e_i \wedge e_j$$ where in the last line we ...


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Abstractly, tensors are elements of the tensor product of vector spaces. For instance, a vector, which is a (0,1)-tensor, is an element of $V$. These are also called covariant tensors and written (in coordinate form) $a_i$. A co-vector, or contravariant vector, is an element of the dual space $V^*$. Elements of the dual space are linear maps from $V$ to your ...


1

I think you have some misconceptions about what you are talking about. In any given basis any linear operator is represented by a fixed matrix. When we talk about linear operators or tensors we are talking coordinate free, when we talk about matrices then we are not. When you write $a_{ij}b_{kj}$ you already have some coordinates. If $A=a_{ij}$ and $ ...


1

I guess I'd write that as $A B^{t}$. But I don't think that gets at the heart of what you're asking. And the following may help or not...but I offer it up anyhow. To really get a grasp of the difference of the two approaches, pick up a classical differential geometry book -- something like O'Neill's Elementary Differential Geometry, or perhaps do Carmo's ...


2

In Lee's 'Intro to Smooth Manifolds', $\Lambda^k(V)$ refers to the space of alternating $k$-tensors on a vector space $V$, as you mentioned. However, the space $\Omega^k(M)$ is the space of smooth $k$-forms on a smooth manifold $M$. That is, an element of $\omega \in \Omega^k(M)$ is a smooth map $M \to \Lambda^k(T^* M)$ (called a smooth section of ...


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To add to all the answers above, there is a delightful example in the text "Mathematics for Physics" by Stone and Goldbart, (Appendix A.3), to clarify the difference between vectors and co-vectors, which I can't resist quoting here. One way of driving home the distinction between $V$ and $V^*$ is to consider the space $V$ of fruit orders at a grocers. ...


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Expanding just a touch on my comment: One simple way to see that vectors and covectors are not as naively identifiable as your friend says is to realize that there are settings where they aren't even isomorphic. They are developed independently on abstract manifolds, via the tangent and cotangent structures, respectively. In particular, the only reason ...


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If $V$ is a finite-dimensional vector space then there exists an isomorphism $V\cong V^*$, under which every vector in $V$ can be interpreted as a covector (linear map $V\to $ the scalar field). However, there are going to be many such maps, and there is in general no canonical such isomorphism, so given a vector $v\in V$, which covector in $V^*$ it ...


0

Sorry, this is a bit too late but here goes. The first thing is to rewrite the cross product using indices and the summation convention $\mathbf{x} \times \mathbf{t}|_r = \epsilon_{rmn} x_m t_n$ where $\epsilon_{rmn}$ is the Levi-Civita symbol. Now rewrite the traction vector $t_n$ as $\sigma_{jn} n_j$ which is how you sneak a normal into the equation. ...


0

Constructing the complete set of isotropic 8th order tensors is messy - there are 105(!) fundamental tensors, of which only 91 are linearly independent. There is a paper "Linearly Independent Sets of Isotropic Cartesian Tensors of Ranks up to Eight" which gives a procedure to construct them (free download from NIST) and also contains a "minimal" list. The ...


0

Your computation is correct. In the book by Chow, Lu and Ni, it is said $$(\nabla ^2 \beta)(X,Y,Z_1,\dots,Z_r) = \nabla_X (\nabla \beta)(Y,Z_1,\dots Z_r) - (\nabla_{\nabla _X Y} \beta)(Z_1,\dots,Z_r)$$ which is different from $$(\nabla ^2 \beta)(X,Y,Z_1,\dots,Z_r) = X[ (\nabla \beta)(Y,Z_1,\dots Z_r)] - (\nabla \beta)(\nabla _X Y,Z_1,\dots,Z_r).$$ Note ...


1

You've forgotten to multiply $T\cdot\Lambda $ by $T^{-1}$ on the right - when we change bases/coordinates we have to make sure that our linear map is expressing a transformation in this new basis. The $T^{-1}$ above ensures this description is complete. The resulting matrix $T\cdot\Lambda \cdot T^{-1}$is then $similar$ to our original matrix $\Lambda$. As ...


2

Actually, when you write $(A \times B)_i$, this is no longer a vector, but just one of its components. The vector itself is $$(A \times B) = ((A \times B)_1, (A \times B)_2, (A \times B)_3)$$ Having this in mind, the expression $$(A \times B)_i = \epsilon_{ijk}A_j B_k$$ means three equations, for $i$ ranges from $1$ to three. The equations are: $$(A \times ...


1

We have (where $(\cdot, \cdot)$ denotes the inner product) \begin{align*} Sv &= \sum_i (Sv, e_i)e_i\\ &= \sum_i \left(\sum_j v_je'_j, e_j\right)e_i\\ &= \sum_i \left(\sum_j v_j(e'_j, e_j)\right)e_i\\ &= \sum_i \left(\sum_j v_j\beta_{ji}\right)e_i \end{align*} So $(Sv)_i = \beta_{ji}v_j$, that is $R= S$.


0

Harold Jeffreys (1973). On isotropic tensors. Mathematical Proceedings of the Cambridge Philosophical Society, 73, pp 173-176. The proof given is a lot more concrete and "hands on" than Weyl's proof linked to by user_of_math.


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I had this same question and I found this: http://www.physics.ucc.ie/apeer/PY4112/Tensors.pdf Page 11, it explains that the Levi-Civita Symbol is a Tensor Density and transforms it using to a tensor the same way that Sharipov did it (and explains the steps).


0

To add to what is written above, an early and well known application of higher order tensors was in the mechanics of deformable bodies (more particularly, in the linearized theory of elasticity), due to Valdemar Voigt (circa 1898) For instance, $\sigma$, a stress tensor and $\epsilon$, a strain tensor, are related by the the tensor equation $$ ...


1

This is somewhat late in the day / year, but I suspect the author is asking about representations of isotropic Cartesian tensors, where "isotropic" means "invariant under the action of proper orthogonal transformations" and "Cartesian" means the underlying space is Euclidean $R^n$ ($R^3$ is a case of great practical interest). The proofs for the two cases ...


3

1), 2): yes. On Manifolds you can introduce a metric, which makes the manifold a Riemmannian manifold (assuming the metric is positive definite). A metric, by definition, is a (0,2) tensor field which defines a scalar product on the tangent bundle, i.e. the metric in a point $p$ is a scalar product on the tangent space at $p$. This, in local coordinates, has ...



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