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Suppose I choose $\{e_{i_1}\wedge e_{i_2}\wedge\cdots\wedge e_{i_k}:i_1 < \cdots <i_k \}$ to be my basis for the exterior product space $\bigwedge^k(V)$ where $\{e_1, \cdots, e_n\}$ is the standard basis for the vector space $V$. I would like to prove that this is a basis. Linear independence is easy, but spanning seems like a massive undertaking. Suppose I take any arbitrary vector $v_1 \wedge v_2 \wedge \cdots \wedge v_k$ in my exterior product space, and I wish to express this vector in terms of the basis elements above. It seems like I have a massive chain of sums and products to deal with here. Is there any way to simplify the computations? How would you do this?

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  • $\begingroup$ If you already know what the basis is for the $k$-fold tensor product, you can use the image of that basis under the quotient map into the exterior algebra as a spanning set of $\bigwedge^k(V)$. Then it should be pretty easy to see the candidate basis you have above is indeed a basis of the exterior product. $\endgroup$ – yousuf soliman Jun 20 '17 at 18:28
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    $\begingroup$ The element $v_1 \wedge v_2 \wedge \cdots \wedge v_k$ isn't an arbitrary $k$-vector. An arbitrary $k$-vector would be a sum of $k$-vectors of this form (what you have is called a $k$-blade). $\endgroup$ – Aweygan Jun 20 '17 at 18:31
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As a comment above said, not every $v \in \bigwedge^kV$ can be written in the form $v_1 \wedge v_2 \wedge \cdots \wedge v_k$. However, every vector can be written as a linear combination of vectors of this form. It therefore suffices to show that every $v$ of that form (i.e. every $k$-blade) can be expressed as a linear combination of your basis.

With that, suppose that $v_j = \sum_{i=1}^n \alpha_{ij} e_i$. We can then write $$ v_1 \wedge \cdots \wedge v_k = \left(\sum_{i_1=1}^n \alpha_{i_11}e_i\right) \wedge \cdots \wedge \left(\sum_{i_k=1}^n \alpha_{i_kk}e_i\right) $$ By multilinearity, expand this into $$ \sum_{1 \leq i_1,\dots,i_k \leq n} \alpha_{i_11}\cdots\alpha_{i_kk}(e_{i_1} \wedge \cdots \wedge e_{i_k}) $$ Now, using the alternating property of the wedge product, why is this ultimately a linear combination of your basis vectors?

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