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This only answers part of your question 0 unfortunatly. The construction is certainly functorial, but the two notions of symmetric/alternating power do not always agree. Let's write $\operatorname{Sym}^n (V)$ for the symmetric tensors, and $\operatorname{Alt} ^n (V)$ for the alternating tensors. I wish this were established notation, but it probably isn't. Let $p$ be the characteristic of the field. Note $V^{\otimes n}$ is a $kGL(V) - \Sigma_n$ bimodule ($\Sigma_n$ is the symmetric group). Then if $r$ is less than $p$, or if $p=0$, $\operatorname{Sym}^r (V) \cong S^r(V)$ and $\operatorname{Alt}^r (V) \cong \Lambda ^r(V)$ as $GL(V)$-modules (this is proved by writing down maps explicitly).

If $r \geq p$ then $S^r$ and $\operatorname{Sym}^r$ are the contravariant (i.e. transpose) duals of one another as $GL$ modules. I imagine the same is true of the alternating power/antisymmetric tensors.

1

Let's write $\operatorname{Sym}^n (V)$ for the symmetric tensors, and $\operatorname{Alt} ^n (V)$ for the alternating tensors. I wish this were established notation, but it probably isn't. Let $p$ be the characteristic of the field. Note $V^{\otimes n}$ is a $kGL(V) - \Sigma_n$ bimodule ($\Sigma_n$ is the symmetric group). Then if $r$ is less than $p$, or if $p=0$, $\operatorname{Sym}^r (V) \cong S^r(V)$ and $\operatorname{Alt}^r (V) \cong \Lambda ^r(V)$ as $GL(V)$-modules (this is proved by writing down maps explicitly).

If $r \geq p$ then $S^r$ and $\operatorname{Sym}^r$ are the contravariant (i.e. transpose) duals of one another as $GL$ modules. I imagine the same is true of the alternating power/antisymmetric tensors.
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