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Let $V$ be a $d$-dimensional $\mathbb C$-vector space and the Grassmann algebra $$\mathcal G (V):=\bigoplus_{n=1}^d V^{\wedge n}$$ where $\wedge$ denotes the antisymmetric tensor product.

I was asking myself the following question:

If we take a vector $x\in\mathcal G (V)$, is there a smallest vector subspace $V_x$ of $V$ such that $x\in\mathcal G (V_x)$?

Or with another formulation, given $V_1,V_2$ vector subspaces of $V$ such that, for $j=1,2$, $x\in\mathcal G(V_j)$, is it necessarily true that $x\in\mathcal G(V_1\cap V_2)$?

If this is not true, does it become true if we had the hypothesis $x\in V^{\wedge n}$ for some $n\in\mathbb N$?

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Yes. Let $W$ be the dual vector space to $V$. For $w \in W$ and $\alpha \in \mathcal{G}(V)$, define $w(\alpha)$ by $$w(v_1 \wedge v_2 \wedge \cdots \wedge v_k) = \sum_{i=1}^k (-1)^{i-1} w(v_i) \ {\Large (}v_1 \wedge v_2 \cdots \wedge v_{i-1} \wedge v_{i+1} \wedge \cdots \wedge v_k {\Large )}$$ and extending linearly to $\mathcal{G}(V)$. (Exercise: This is well defined.)

Now, fix $\beta \in \mathcal{G}(V)$. Define $W' \subseteq W$ by $$W' = \{ w \in W : w(\beta) =0 \}.$$ Define $$V' = \{ v \in V : w(v)=0 \ \forall w \in W' \}.$$

I claim that $\beta$ is in $\mathcal{G}(V'')$ if and only if $V''$ contains $V'$.

The easy direction: Suppose that $\beta \in \mathcal{G}(V'')$. Let $W'' = \{ w \in W : w(v)=0 \ \forall v \in V'' \}$. Then we immediately have $w(\beta) =0$ for $w \in W''$, so $W'' \subseteq W'$. Duality reverses containments, so $V'' \supseteq V'$.

The hard direction: Let $\beta$ and $V'$ be as above. We must show that $\beta \in \mathcal{G}(V')$. Choose a basis $e_1$, $e_2$, ..., $e_n$ for $V$ such that $e_{k+1}$, $e_{k+2}$, ..., $e_n$ is a basis for $V'$. For $I = \{ i_1, i_2, \ldots, i_r \} \subseteq \{ 1,2,\ldots, n \}$, define $e_I = e_{i_1} \wedge e_{i_2} \wedge \cdots \wedge e_{i_r}$, with $i_1 < i_2 < \cdots < i_r$. Write $\beta$ in this basis as $$\beta = \sum_{I \subseteq \{ 1,2,\ldots, n\}} \beta_I e_I.$$

Let $f_1$, $f_2$, ..., $f_n$ be the basis of $W$ dual to $V$. So $f_1$, $f_2$, ..., $f_k$ is a basis for $W'$. We have $$f_i(\beta) = \sum_{i \in I} \pm \beta_I e_{I \setminus i}$$ where the rule for which sign to use is unimportant.

For $i \leq k$, we have $f_i(\beta) =0$ since $f_i \in W'$. So $\beta_I=0$ whenever $I$ contains such an $i$. In otherwords, the only nonzero terms in $\beta$ are of the form $\beta_I e_I$ when $I \subseteq \{ k+1, k+2, ..., n \}$. In other words, $\beta \in \mathcal{G}(V')$. QED


Making $W$ act on $\mathcal{G}(V)$ in this way is a standard tool. It is the skew commutative analogue of $W$ acting by derivations on the ring $\mathrm{Sym}^{\bullet}(V)$.

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Nice, thank you very much David. –  Sebastien B Dec 14 '12 at 18:07
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