# Tag Info

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.

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

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} &= ...

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 $$\frac{IL+J}{IL} \to \frac{I}{IL} \cong I \otimes \frac{R}{L}$$

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

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 ...

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 $$g(v, w) := ... 2 In general$$ \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 & ...

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 ...

1

I asked this question also in the forum MathOverflow even though this question is so basic. Indeed somone solved my problem almost immediately in the comments. I write the answer here because maybe can be useful to someone in a future. The answer (as I espected) was obvious and I had this problem just because tensors are a new thing for me. The object ...

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 $$A_{ij}\ ,\ A^i{}_j\ ,\ A^{ij}\ ,\ A_iB_j\ ,\ A^iB_j\ ,\ ... 1 Just thinking in terms of real functions of one variable, using the chain rule, we can see that a differential dx gets multiplied by \phi' under a change of variables x=\phi(x'). dx=\phi'\,dx'. Similarly we see that the differential operator d \over dx gets divided by \phi': \frac{d}{dx}=\frac{1}{\phi'}\frac{d}{dx'}, which makes total sense ... 1 For example taking two linear maps f,g:\Bbb R^n\to\Bbb R one can construct a bilinear map f\otimes g:\Bbb R^n\times\Bbb R^n\to\Bbb R via$$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

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

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 ...

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

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

1

They do cancel, if you keep track of the signs correctly. For example, the second term produces $(Yf) \omega(Z,X)$, while the fourth produces $-\omega( -(Yf)X,Z) = (Yf)\omega(X,Z)$. These are negatives of each other because of the antisymmetry of $\omega$.

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

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 ...

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 $A_{ijs}v^s$.

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

That's because the metric itself is nabla-flat, so by Leibniz rule you get what you want.

1

I think your formula switched $f$ and $g$, I found: $(g\circ f)''(x)[u,v]=g''(f(x)[f'(x)[u], f'(x)[v]]+g'(f(x))\circ f''(x)[u,v]$. Write $f=(f^1,...,f^k)$ and plugging in $e_i=u, e_j=v$ gives: \$\frac{\partial (g\circ f)}{\partial x_i\partial x_j}=\frac{\partial f^h}{\partial x_i}\frac{\partial f^\ell}{\partial x_j}\frac{\partial^2 g}{\partial x_h\partial ...

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