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Let $k$ be a field, and $V$ a finite-dimensional vector space over $n$. We may define some quotient spaces of $V^{\otimes n}$, by stating their universal property in terms of factoring multilinear maps:

  1. The symmetric quotient $S^n(V)$, defined by the universal property that whenever $f: V^{\times n} \to W$ is symmetric in its arguments (swapping two arguments does not change the value), $f$ factors through $S^n(V)$.
  2. The antisymmetric quotient $A^n(V)$, defined by the universal property that whenever $f: V^{\times n} \to W$ is antisymmetric in its arguments (swapping two arguments negates the value), $f$ factors through $A^n(V)$.
  3. The alternating quotient $\Lambda^k(V)$, defined by the universal property that whenever $f: V^{\times n} \to W$ is alternating in its arguments (if two arguments are equal, the value is zero), $f$ factors through $\Lambda^n(V)$.

I'd like to remark that while $S^n(V)$ and $A^n(V)$ are neatly described as a quotient of $V^{\otimes n}$ by some subspace associated to the $S_n$ group action, for example $A^n(V)$ is the quotient of $V^{\otimes n}$ by the subspace $$\{\sigma \cdot \omega - (-1)^{\sigma} \omega \mid \omega \in V^{\otimes n}, \sigma \in S_n\}, $$ the quotient of $\Lambda^n(V)$ is "quadratic", the denominator being the linear span of $$\{ v_1 \otimes \cdots \otimes v_n \mid v_i = v_j \text{ for some } i \neq j\}.$$ By "quadratic", I mean that I don't think this ideal is the image of any collection of natural linear maps, rather it seems to be the linear span of the image of some quadratic maps. (In the case $n = 2$, we would be taking the span of the image of the quadratic map $v \mapsto v \otimes v$).

Now, let's dualise everything to define some subspaces of $V^{\otimes n}$. We can define:

  1. The symmetric subspace $\mathrm{Sym}^n(V)$, defined by the universal property that whenever $f: W \to V^{\otimes n}$ is symmetric in its outputs, $f$ factors through $\mathrm{Sym}^n(V)$.
  2. The antisymmetric subspace $\mathrm{Anti}^n(V)$, defined by the universal property that whenever $f: W \to V^{\otimes n}$ is antisymmetric in its outputs, $f$ factors through $\mathrm{Anti}^n(V)$.
  3. The exterior subspace $\mathrm{Ext}^n(V)$, defined by the universal property ......

I've trailed off because I don't quite know how to define $\mathrm{Ext}^n(V)$ --- I can define the other two in terms of group actions (they are the fixed subspace, and the sign-isotypic subspace of the $S_n$-action).

Question: What should the definition of $\mathrm{Ext}^k(V)$ be? (Note this question is only really interesting in characteristic 2)

Here is some food for thought: suppose that $V$ and $W$ are dual under some nondegenerate pairing $\langle -, - \rangle: V \times W \to k$. Then the natural nondegenerate pairing $V^{\otimes n} \times W^{\otimes n} \to k$ defined by $$ \langle v_1 \otimes \cdots \otimes v_n, w_1 \otimes \cdots \otimes w_n \rangle = \langle v_1, w_1 \rangle \cdots \langle v_n, w_n \rangle$$ naturally restricts to give nondegenerate pairings $$ \mathrm{Sym}^n(V) \times S^n(W) \to k $$ and $$ \mathrm{Alt}^n(V) \times A^n(W) \to k $$ (To see the naturality, first restrict the pairing to $\mathrm{Sym}^n(V) \times W^{\otimes n}$, and notice that once the argument on the left is fixed, the map is multilinear and symmetric, and hence factors through $S^n(W)$. Then use a pair of dual bases to argue nondegeneracy).

So potentially a definition for $\mathrm{Ext}^n(V)$ would be the subspace orthogonal to $$ \{ w_1 \otimes \cdots \otimes w_n \mid w_i = w_j \text{ for some } i \neq j\} \subseteq W^{\otimes n},$$ which would then give a natural nondegenerate pairing between $\mathrm{Ext}^n(V)$ and $\Lambda^n(V)$. However, this seems quite roundabout, and I'm also having trouble describing what this subspace is in simpler terms.

Joppy
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