I have been trying to learn the proof of dimension of exterior power from this text : http://www.thehcmr.org/issue1_2/poincare_lemma.pdf.( Page 16)
I am not able to understand the part of linear independence ( how its been assured) . It would be nice if you could explain . thank you .
Linear independence regarding Exterior Power .
1 Answers
To say that vectors $v_1, \ldots, v_n$ (lets focus on $n \geq 2$) are linearly dependent is equivalent to saying that some fixed $v_i$ is a linear combination of $v_1, \ldots, v_{i-1}, v_{i+1}, \ldots, v_n$. If you know beforehand that $v_i \neq 0$ for every $i$, then evidently the coefficients of this linear combination can't all be zero.
In your particular case, it's not hard to see that $e_{i_1} \wedge \dots \wedge e_{i_s} \neq 0$ by using, for instance, the universal property. Now suppose that for some fixed set of indices $\{i_1, \ldots, i_s\}$ you had $e_{i_1} \wedge \dots \wedge e_{i_s} = \sum_{\{i_1, \ldots, i_s\} \neq \{j_1, \ldots, j_s\}} a_{j_1, \ldots, j_s} e_{j_1} \wedge \dots \wedge e_{j_s}.$ In this case, there should be some multi-index $\{j_1, \ldots, j_s\}$ such that $a_{j_1, \ldots, j_s} \neq 0$, because the left-handed side is not zero. Applying the corresponding $B_{j_1, \ldots, j_s}$ (which is a linear map) to both sides gives you $B_{j_1, \ldots, j_s}(e_{i_1} \wedge \dots \wedge e_{i_s}) = a_{j_1, \ldots, j_s}.$ But since $\{i_1, \ldots, i_s\} \neq \{j_1, \ldots, j_s\}$, the left-handed side must be zero, by construction, and the right-handed side was supposed to be $\neq 0$. Hence the contradiction.
-
0@Ananda $B$ is not defined explicitly. On page $2$, the space $\bigwedge^n(V)$ is defined as having the property that given any multilinear $f : V \times \dots \times V \rightarrow W$, there exists a unique *linear* $f^{\prime} : \bigwedge^n(V) \rightarrow W$ such that $f^{\prime} \circ \wedge = f$. This property is being used to define $B$. Take $f$ to be the (multilinear) map $\Sigma_{i_1, \ldots, i_s}$ on page $4$. Now $B$ is precisely the map $f^{\prime}$ associated to this $f$, which is linear. – 2012-05-29