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1

The oriented volume spanned by three vectors is a multilinear alternating map of the vectors themselves. That's the starting point. See, e.g., the book by S.Winitzki.


1

The isotropic part is the one that you can represent with a scalar. The condition is that the rest is traceless. Isotropic in physics means independent on the direction, which is exactly the condition that it acts on every vector equally. So, $$A_{ij}v_j=av_i$$ for every $v_j$ simply means $A_{ij}=a\delta_{ij}$. You actually already wrote down the solution ...


0

Denote $B\cdot P$ the action of an invertible matrix $B$ on $P$. Hint. The group $GL_n(\mathbb C)$ is generated by exponentials, so to show that a map $P$ is invariant it is enough to show that it is invariant by matrices of the form $\exp(B)$. Now consider the map $t\mapsto \exp(tB)\cdot P$. It satisfies a certain first order differential equation, and ...


5

The symmetrizer $S: \bigotimes^k V \to \bigotimes^k V$ is idempotent. Hence, $\ker(S) = \mathrm{im}(\mathrm{id}-S)$. This is generated by elements of the form $\alpha-{}^\sigma \alpha$, where $\sigma$ is some permutation.


2

It cannot be that $f'(s\otimes t) = (f(s),f(t))$ because on one hand $f$ isn't necessarily defined on $T$ (as a polynomial function), while on the other hand $(f(s),f(t))$ isn't an element of $S \otimes T$. Now, note that there is a natural map $S \to S \otimes T$, namely $s \mapsto s \otimes 1$. Thus if $f \in S[X]$ is $$ f(X) = a_n X^n + a_{n-1}X^{n-1} + ...


2

Your problem is $\delta_{bd}\ \delta^{bd}=\delta^b_b=\mathrm{Tr}(\mathrm{Id_3})=3$, not one.


0

For a vector space $V$ over a scalar field ${\Bbb F}$, you can see it as a bilinear transformation $$B:V\times V\longrightarrow{\Bbb F}$$ which allow you to impose a metric in $V$. If we consider that $V$ is spawned by a basis $\{b_i\}$, then it is necessary that the matrix $$[B]=[B(b_i,b_j)]$$ complies certain conditions.


2

I am going to suppose that by affine transformation you mean a mapping from $T : V \to V$ from a vector space to itself s.t. $$T(x) = Ax + b$$ where $A : V \to V$ is a linear map, and $b$ is an element of the vector space. If that is the case, then no $T$ is not a tensor. A tensor needs to be a linear map on the tangent space of each point of the space on ...


0

I found a paper that derives what appears to be the correct solution for $I$ in the case $n=2$. The answer turns out to be significantly more complicated already for $n=2$ than the formula given in the question and it is given in terms of hypergeometric functions of the $SL(n)$ invariants of the form $Q$ (the hyperdeterminant is only one of several ...


0

Both are correct, if you define "divide by the metric" correctly. Going back to linear algebra, if you want to solve, for $A: \mathcal{V}\to\mathcal{W}$ a linear mapping the equation $$ A v = w $$ if $A$ is invertible one can multiply on both sides $A^{-1}$ $$ v = (A^{-1} A)v = A^{-1} w $$ In your case the metric $g_{bc}$ is a mapping from $TM \to ...


1

Multiply by $g^{\sigma\tau}$ and use that $$ g^{ab}g_{bc} = \delta^a_c, $$ so the right-hand side becomes $ d^{\sigma} $, which is what you want. I'll leave the left-hand side to you, since you don't say if you mind raising the index on $A_{\mu\nu\tau}$.


2

First note that every vector bundle $E$ gives you $C^{\infty}(M)$-Module structure on global sections $\Gamma(E).$ Additionaly for vector bundles $E,F$ on maniflod $M$ there are $C^{\infty}(M)$-Module isomorphism $$\Gamma(E)\otimes_{C^{\infty}(M)}\Gamma(F)\simeq\Gamma(E\otimes_{\mathbb{R}} ...


2

You'd best think of the cross product by first defining the following: Let $\varepsilon:\mathbb{R}^3\times\mathbb{R}^3\times\mathbb{R}^3\rightarrow\mathbb{R}$ be a (0,3)-type tensor that is totally antisymmetric. Such tensors form a one-dimensional subspace within the space of (0,3)-type tensors, therefore any such tensor (except the zero tensor) differ ...


3

Well, you're correct, in some sense, the cross-product of two vectors is not really a vector of the same type. The language used to distinguish the two types is: polar vectors: flip sign of component under coordinate inversion axial vectors: do not flip sign of component under inversion (also known as pseudovector see this wikipedia article) All of this ...


2

If the Riemannian curvature tensor takes this form, then at any given point all the sectional curvatures are equal. Thus if $n>2$ then Schur's Lemma tells us that the sectional curvature is in fact constant; i.e. $k$ must be constant. If $n=2$ then this is the problem of prescribed Gauss curvature, which you should be able to find many papers about.


2

SECTION A : The linearly independent elements of a Totally Symmetric Tensor $\;T_{i_{1}i_{2}\cdots i_{p-1}i_{p}}\;$(important for the interpretation of Quark Theory of Baryons in Particle Physics) \begin{equation*} \bbox[#FFFF88,8px] {\boldsymbol{3}\boldsymbol{\otimes}\boldsymbol{3}\boldsymbol{\otimes}\boldsymbol{3}= ...


1

Here, $i$ is, if you like, the index corresponding to the direction of the covariant derivative. More precisely, for, say, a $1$-form $\eta$, $\nabla_i \eta_j$ is the abstract index notation for denotes the $2$-tensor $\nabla \eta$ defined by $$(\nabla \eta)(X, Y) := (\nabla_X \eta)(Y).$$ Indeed, in abstract index notation this quantity is $\nabla_i \eta_j ...


0

$\mathtt{Definition:}$ $T$ is an isotropic tensor of type $(0,n)$ if $\;T_{i_1i_2...i_n}=R_{i_1j_1}R_{i_2j_2}...R_{i_nj_n}T_{j_1j_2...j_n}$ whenever $R$ is an orthogonal matrix i.e $R^TR=RR^T=I$. $\mathtt{n=2:}$See my answer here. $\mathtt{n=3:}$For tensors of type $(0,3)$ we can mimick the proof for $n=2$ to deduce skew-symmetricness. Suppose $T_{pqr}$ is ...


0

$\mathtt{Step\,1:Realize\;T\;is\;diagonal}$ $$R_{kl}=\begin{cases} -1 & \text{if }k=l=j\\ \delta_{kl} & \text{otherwise}\\ \end{cases}\\ \\\therefore\text{R is the reflection w.r.t to the hyperplane perpendicular to the j-th vector }\\\text{in the standard ordered basis} \\R=R^T\land ...



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