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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

Let $$f(x,y) = x_1y_2\qquad \text{and}\qquad g(x,y)=y_1x_2\qquad \forall x=(x_1,x_2),y=(y_1,y_2)\in\Bbb R^2$$ Then $f(x,x)=g(x,x)$ for all $x\in\Bbb R^2$ but $f\neq g$.

2

Hint: First, you want to show that the $0$ map is bounded and multilinear. Then, you need to demonstrate that if $f,g$ are bounded multilinear maps, and $\alpha$ is an element of your original field, then $\alpha f + g$ is also a bounded multilinear map.

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

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 think that in some sense there can't be any "worthwhile" canonical form: If $V$ is $n$-dimensional then the space of such tensors is $n^3$-dimensional, and $GL(V)$ is only $n^2$-dimensional. So any canonical form would have $n^3-n^2$ degrees of freedom, which is almost as many as if you didn't use the canonical form.

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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

$A(\lambda x_1,\ldots,\lambda x_m) = \lambda^{m-j}\overline{\lambda}^jA(x_1,\ldots,x_m)= \lambda^{m-k}\overline{\lambda}^k A(x_1,\ldots,x_m)$ From the above, since $\lambda \neq 0$, we have: $$\left[ \left(\frac{\lambda}{\overline{\lambda}}\right)^{k-j} - 1\right] A(x_1,\ldots,x_m) = 0$$ Thus: \$\left[ \left(\frac{\lambda}{|\lambda|}\right)^{2(k-j)} - ...

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