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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

Remember: for finite dimensional spaces, two vectors spaces (over the same field) are isomorphic if and only if they have the same dimension. So, for all these, it suffices to simply find a basis, and therefore conclude the dimension. In particular, we have $$ \dim(\mathcal L(V,W)) = \dim(V \otimes W) = \dim(V) \dim(W) $$ When $V$ and $W$ are finite ...


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

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 $A_{\mu\nu\tau}$.


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$0$-tensors are just scalars, so the tensor product in this case is just scalar multiplication.


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Symbolically, it is an $n\times n$ matrix. Don't expand the $\vec{e}_i$ into coordinates. Just take the determinant according to however you normally do so, and whenever multiplication involves the scalars from below, multiply accordingly, and when it involves a scalar times one of these vectors from the top row, multiply the scalar times the vector ...


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I assume that by transoposition you mean dual mapping $T^*$. That is if $T:V\rightarrow V$ then $T^*:V^*\rightarrow V^*$ in following manner $$\left[T^*\alpha\right](v):=\alpha(T(v)).$$ Let $\alpha_1,\dots,\alpha_n\in V^*.$ Fix arbitrary $v_1,\dots,v_n\in V.$ Then $$\left[\bigwedge^n T^*(\alpha_1\wedge\dots\wedge\alpha_n)\right](v_1\wedge\dots\wedge ...



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