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Let $V$ be a k-dimensional vector space and $m,n \in \mathbb{Z}$ are such that $m+n \le k$. We consider the wedge product: $\wedge: \Lambda^nV \times \Lambda^mV \to \Lambda^{m+n}V$ defined by $(v,w) \to \pi(v \otimes w)$ where $\pi$ is anti-symmetrisation projector. Is it true that if for fixed v we have $v\wedge w = 0$ for all $w \in \Lambda^mV$ then $v$ must be zero?

I'm trying to show this by taking $v_{i_1} \wedge ... \wedge v_{i_n}$ as a basis for $\Lambda^nV$ so that we can write $v = v^{i_1, ..., i_n}v_{i_1} \wedge ... \wedge v_{i_n}$. Now I want to be able to pick out particular elements from $\Lambda^mV$ to show that each $v^{i_1, ..., i_n}$ is zero. I'm struggling to do this.

Wooster
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1 Answers1

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Actually you dont have to pick particular elements of $\bigwedge^m V$, just pick any non-zero $v_{j_1} \wedge \dotsc \wedge v_{j_m}, ~ n+1 \leq j_1 < \dotsb j_m \leq k$ (This where you need $m+n \leq k$). Then the elements

$$v_{i_1} \wedge \dotsc \wedge v_{i_n} \wedge v_{j_1} \wedge \dotsc \wedge v_{j_m}, ~ 1 \leq i_1 < i_2 < \dotsc i_n \leq n$$

are linear independent in $\bigwedge^{n+m} V$, i.e. all coefficient vanish if the linear combination vanishes.

MooS
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  • Thank you for this. I'm struggling to see that as $V$ is $k$-dimensional why should it be that the basis expansion of $v$ has $1 \le i_1 < i_2 < .. < i_n \le n$? Surely the indices should be allowed to go all the way to $k$? – Wooster Feb 23 '16 at 12:18
  • I skipped the step where I changed the ordering. I meant the following: Take any summand $f v_{i_1} \wedge \dotsc \wedge v_{i_n}$ of your $v$. Then pick $j_1, \dotsc, j_m$, such that ${i_1, \dotsc, i_n, j_1, \dotsc, j_m}$ are all different. Then apply the wedge-product and you will get $f=0$. – MooS Feb 23 '16 at 12:23
  • Okay I understand, great! This is also using the fact that the wedge product of linearly independent vectors is zero? – Wooster Feb 23 '16 at 12:28
  • What do you mean by that? $e_1 \wedge e_2 \neq 0$ for $e_1,e_2 \in \mathbb R$ the standard basis. – MooS Feb 23 '16 at 12:30
  • I think I misunderstood the argument. So if we choose $j_1,...,j_m$ such that ${i_1,...,i_n,j_1,...,j_m}$ are all different and let $w = v_{j_1} \wedge ... \wedge v_{j_m}$ then the only term surviving from $v \wedge w$ is $f v_{i_1} \wedge ... \wedge v_{i_n} \wedge v_{j_1} \wedge ... \wedge v_{j_m}$? – Wooster Feb 23 '16 at 12:36
  • No, not the only surviving term. But one of the surviving terms. And by linear independece, you get that all coefficients of all surviving terms are zero. – MooS Feb 23 '16 at 12:45
  • Ah okay yes. If $m+n = k$ it would be the only surviving term? – Wooster Feb 23 '16 at 12:45
  • Yes, certainly, since we have $\bigwedge^{k}V$ is one-dimensional, there can be only one surviving term. – MooS Feb 23 '16 at 12:47