# Tag Info

2

We say T is a $(a,b)$ tensor $T:T^*_p \times...\times T^*_p \times T_p \times...\times T_p \to \mathbb{R}$ where $a$ is the number of $T^*_p$s and $b$ is the number of $T_p$s. A dual vector is an element of $T_p^*$, which means that it gives a map $T_p\rightarrow \mathbb R$, and so it's a $(0,1)$ tensor. A vector is an element of $T_p$, which means that it ...

1

I don't think there is any standard tensor notation convention for expressions like yours. Both your "tensor-like" proposal and the version suggested by weux082690 have their merits. In my opinion, which version to prefer depends on how often you are going to use expressions of this kind in your paper (or other document), and so how much effort you are ...

0

You don't seem to have set up the approach properly. You should start with the tensor product of the $H_i$'s, i.e. $H=H_1\otimes H_2\otimes \cdots \otimes H_n$. This is a quotient vector space. An expression of the form $\phi_1\otimes \phi_2\otimes \cdots \otimes \phi_n$ represents an equivalence class in $H$. In order for the inner product to be ...

3

Given a vector $v$ and a dual vector $f$, you can produce a scalar $f(v)$. This can be viewed as a map $v \mapsto f(v)$ or $f \mapsto f(v)$. So a vector determines a (linear) map from the space of dual vectors to scalars (i.e. a $(1,0)$-tensor since we have 1 dual vector input and no vector inputs). Likewise a dual vector determines a (linear) map from the ...

3

In the first case you can consider the sets $\{i,j\}$ and $\{k,l\}$ as your indeces. Hence, for each pair you have 6 combinations $11,22,33, 12,13,23$ and therefore total of $36$ independent components. In the second case, the third condition implies that $6$ by $6$ matrix from the case 1 is symmetric which gives you $21$ independent components.

1

For example taking two linear maps $f,g:\Bbb R^n\to\Bbb R$ one can construct a bilinear map $f\otimes g:\Bbb R^n\times\Bbb R^n\to\Bbb R$ via $$f\otimes g(v,w)=f(v)g(w).$$ Attached to $f\otimes g$ there is a matrix associated by evaluating $$f\otimes g(b_i,b_j),$$ that is $$\left(\begin{array}{cccc} f(b_1)g(b_1)&f(b_1)g(b_2)&...&f(b_1)g(b_n)\\ ... 1 Just thinking in terms of real functions of one variable, using the chain rule, we can see that a differential dx gets multiplied by \phi' under a change of variables x=\phi(x'). dx=\phi'\,dx'. Similarly we see that the differential operator d \over dx gets divided by \phi': \frac{d}{dx}=\frac{1}{\phi'}\frac{d}{dx'}, which makes total sense ... 2 You can use the construction of the complex conjugate vector space and identify a sesquilinear form with an element of V^{*} \otimes \overline{V}^{*}. An elementary tensor \varphi \otimes \psi \in V^{*} \otimes \overline{V}^{*} gives rise to a sesquilinear form g_{\varphi \otimes \psi} \colon V \times V \rightarrow \mathbb{C}  defined by$$ g(v, w) := ...

2

This isn't true. For example let $$\mathbf{T}=\begin{pmatrix}2&2&2\\ 0&2&2\\ 0&0&2 \end{pmatrix}$$ everywhere. Then $\mathbf{a}\cdot\mathbf{T}\mathbf{a}=2a_1^2+2a_2^2+2a_3^2+2a_1a_2+2a_2a_3+2a_3a_1$ $=(a_1+a_2)^2+(a_2+a_3)^2+(a_3+a_1)^2>0$ Let $p$ be any scalar field with a constant gradient, for example $p(x,y,z)=x$. Then ...

2

In general $$\sum_{m=1}^N\sum_{n=1}^N A_{ijmn}B_{mnkl} = \delta_{ik}\delta_{jl}$$ for $i,j,k,l = 1,\ldots,N$. Now suppose the tensor $\mathbf{A}$ is unfolded as the $N^2\times N^2$ matrix given by $$A = \left[ \begin{array}{ccccccccc} A_{1111} & A_{1112} & \cdots & A_{111N} & \cdots & A_{11N1} & A_{11N2} & \cdots & ... 1 No. In general, let G be a group (say a Lie group) and let V be a representation of it. One thing we might mean by "invariants" of v \in V are polynomial invariants: that is, elements of the algebra S(V^{\ast})^G of G-invariant polynomial functions on V. This algebra sometimes consists of only the constant functions. In particular, this is true ... 1 Ok, we have the covariant derivative in the k direction for a rank one tensors, contravariantly$$u^i{}_{;k}=u^i{,k}+\Gamma^i{}_{sk}u^s,$$and covariantly$$u_{i;k}=u_{i,k}-\Gamma^s{}_{ik}u_s,$$which indicate a the tensor varies in direction k. Then for rank two tensor we have the possibilities$$A_{ij}\ ,\ A^i{}_j\ ,\ A^{ij}\ ,\ A_iB_j\ ,\ A^iB_j\ ,\ ...

0

Resolution: Using $\sigma^{i}\sigma^{i}=Id$ Then $$x^{i}(\vec{n} \times \vec{\sigma})^{i}=x^{i}\sigma^{i}\sigma^{i}(\vec{n} \times \vec{\sigma})^{i}=[x] \cdot \vec{\sigma} \cdot \vec{(\vec{n} \times \vec{\sigma})}$$

3

The part of the expression you are referring to is $\Gamma^{\ell}_{ij}\Gamma^m_{\ell m} - \Gamma^m_{i\ell}\Gamma^{\ell}_{jm}$. Suppose now that the dimension is two so that each index takes the value $1$ or $2$. Fixing $i$ and $j$, we have \begin{align*} \Gamma^{\ell}_{ij}\Gamma^m_{\ell m} - \Gamma^m_{i\ell}\Gamma^{\ell}_{jm} &= ...

1

Actually you are done. You should not expect $A^{f, \theta}_{\theta k}= A^{e,\theta}_{\theta k}$, the same way that you do not expect $A^{f, i}_{jk} = A^{e, i}_{jk}$. Indeed, saying that contraction is invariant mean $$A^{f, \theta}_{\theta k}= a^j_kA^{e,\theta}_{\theta j}.$$ Indeed, writing $i, j, k, l, m, n,\cdots$ for both the indices of $e,f$ might ...

1

Note that \begin{align*} T'^{a} &= de'^a + \omega'^a_b\wedge e'^b\\ &= d((\Lambda^{-1})^a_b e^b) + ((\Lambda^{-1})^a_cd\Lambda^c_b + (\Lambda^{-1})^a_c\omega^c_d\Lambda^d_b)\wedge e'^b\\ &= (\Lambda^{-1})^a_bde^b + d(\Lambda^{-1})^a_b\wedge e^b + (\Lambda^{-1})^a_cd\Lambda^c_b\wedge e'^b + (\Lambda^{-1})^a_c\omega^c_d\Lambda^d_b\wedge e'^b\\ ...

1

Note that $T_{abc} = T_{bac} = -T_{bca} = -T_{cba} = T_{cab} = T_{acb} = -T_{abc}$, so $T_{abc} = 0$.

1

I am not 100% sure it would match your expectations but have a look at Peter Olver's Applications of Lie Groups to Differential Equations. Specifically for the mathematics surrounding the Noether theorem you also may wish to look at the book by Yvette Kosmann-Schwarzbach, The Noether theorems. Invariance and conservation laws in the twentieth century, ...

1

In the end for a couple of tensors $A_{ijk}$ and $v^s$ (described by components) their juxtaposition $$A_{ijk}v^s,$$ give you the components of a new tensor, whose rank is the sum of their ranks, this juxtaposition is only a plain multiplication of numbers. One can get the components of a rank two if we "contract" some indexes as in $A_{ijs}v^s$.

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I have found the answer in Kobayashi & Nomizu, volume 1, page 124. If $K$ is a tensor of type $(r,s)$, then one may construct a new tensor $\nabla K$ of type $(r, s+1)$, defined by $$(\nabla K) (X_1, \dots, X_s, Y) = (\nabla _Y K) (X_1, \dots, X_s)$$ and thus define inductively $\nabla ^k K$ as $\nabla (\nabla ^{k-1} K)$. Choosing now $K$ to be $f$, a ...

2

The "type" of a tensor $T$—i.e., of a tensor-valued function, or "tensor field"—determines how $T$ transforms under change of coordinates. If you view a scalar as a scalar-valued function it behaves like a $0$-tensor. If you view a scalar as the operation of scalar multiplication, it behaves like a $2$-tensor. Strictly speaking those are ...

2

Recall for an $R$-module $M$ and $\mathfrak{a}$ ideal of $R$ the canonical isomorphism $$M \otimes_R R/\mathfrak{a} \cong M/\mathfrak{a}M$$ In particular $$I \otimes R/L \cong I/IL$$ since $IL+J \subset I$, you have the inclusion $$\frac{IL+J}{IL} \to \frac{I}{IL} \cong I \otimes \frac{R}{L}$$

0

What you seem to be calling rank is what is usually called degree or order. The term rank is usually reserved for another quantity related to a tensor. If $E$ is an order five tensor on $\mathbb{R}^5$, then for each choice of $i_1, i_2, i_3, i_4, i_5 \in \{1, 2, 3, 4, 5\}$, we have a corresponding component $E_{i_1i_2i_3i_4i_5}$ which is a real number. ...

2

In $n$ dimensions let $\varepsilon_{i_1\dots i_n}$ be the Levi-Civita Symbol defined by $$\varepsilon_{i_1\dots i_n}=\begin{cases} 1&\text{ if }i_1,\dots ,i_n\text{ is an even permutaion of }1,\dots ,n\\ -1&\text{ if }i_1,\dots, i_n\text{ is an odd permutaion of }1,\dots ,n\\ 0&\text{ otherwise.}\\ \end{cases}$$ Then the parallelotope given by ...

1

If the metric is given in a conformal chart, $g= e ^{2 u} (dx^2+dy^2)$, the curvature is $(-\Delta u)e^{-2u}$, where $\Delta$ is the Laplace operator. In your case, $u={x^2+y^2\over 2}$, $K=-e^{x^2+y^2}$

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