Exterior algebra of a subspace

Let $E$ and $E^\star$ be two vector spaces in duality according to a (possibly symmetric) non-degenerate bilinear form $\langle\cdot,\cdot\rangle:E^\star\times E\to\mathbb{R}$. Let $F$ be a subspace of $E$. I would like to show that: $$\Lambda^\bullet F=\bigcap_{u^\star\in F^\bot}\mathrm{Ker}(i_{u^\star})$$ where $\Lambda^\bullet F$ is the exterior algebra of $F$, $i_{u^\star}$ is the interior product and: $$F^\bot=\{\phi\in E^\star:\langle\phi,f\rangle=0\ \forall f\in F\}$$ This is the exercise 2 page 140 of the book "Multilinear Algebra" of Werner H. Greub. Basically it means that: $$f\in\Lambda^\bullet F\Leftrightarrow f\in\Lambda^\bullet E\ \text{and}\ \forall u^\star\in F^\bot\ i_{u^\star}f=0$$

From left to right it is quite simple I think: we can suppose we are working with decomposable mulitvectors since these span the whole exterior algebra. Let $f=f_1\wedge\dots\wedge f_k\in\Lambda^k F$. Let $u^\star=:\phi_1\in F^\bot$, we want to show that $i_{u^\star}f=0$ so we show that $\langle i_{u^\star}f,\phi\rangle=0$ for all $\phi\in\Lambda^{k-1}E^\star$. Let $\phi$ be a $(k-1)$-decomposable form $\phi_2\wedge\dots\wedge\phi_k\in\Lambda^{k-1}E^\star$, then we have: \begin{align*} \langle i_{u^\star}v,\phi\rangle&=\langle v,u^\star\wedge\phi\rangle\\ &=\langle f_1\wedge\dots\wedge f_k,\phi_1\wedge\phi_2\dots\wedge\phi_k\rangle\\ &=\mathrm{det}\{\langle f_i,\phi_j\rangle_{1\leq i,j\leq k}\}\\ &=0 \end{align*} since the first column $\{\langle v_i,\phi_1\rangle=\langle v_i,u^\star\rangle\}_{1\leq i\leq k}$ is all zero by orthogonality.

Conversely, let $f\in\bigcap_{u^\star\in F^\bot}\mathrm{Ker}(i_{u^\star})$. Pick some $u^\star\in F^\bot$ then $f\in\mathrm{Ker}(i_{u^\star})$. We have: $$0=\langle i_{u^\star}f,\phi\rangle=\langle f,u^\star\wedge\phi\rangle\quad\forall\phi\in\Lambda^\bullet E^\star$$ So $f\in (I_{F^\bot})^\bot$ where $I_A$ is the ideal generated by $A$ a subspace of $E$ (or $E^\star$). But $\Lambda^\bullet(F^\bot)=(I_F)^\bot$ according to equation (5.76) page 137, so $(I_{F^\bot})^\bot=\Lambda^\bullet F$ and $f\in\Lambda^\bullet F$.

But I am not sure. Could somebody agree or disagree with what I wrote, or provide other ideas?

Thank you!

• It would help to know what you mean by "interior product". There is probably a classic one for the exterior algebra, but I'm more used to working with one of the four or five in Clifford algebra. Does $i_{u^*}$ mean the map $x\mapsto \langle u^*,x\rangle$ extended somehow on $\Lambda^\bullet E$? – rschwieb Oct 9 '12 at 12:44
• Nevermind, wikipedia explains :) – rschwieb Oct 9 '12 at 12:49
• @rschwieb: The interior product in an operator that inserts a vector (or a 1-form) into the first slot of a $k$-form (or a $k$-vector). For instance, in case the of $k$-forms: $i_v:\Lambda^k E^\star\to\Lambda^{k-1}E^\star$, $\omega\mapsto\omega(v,\dots)$ (insertion in the first slot, the $k-1$ other slots remaining free). – Benjamin Oct 9 '12 at 12:57