description/proof of that for \(q\)-covectors space, transition of standard bases w.r.t. bases for vectors space is this
Topics
About: vectors space
The table of contents of this article
Starting Context
- The reader knows a definition of antisymmetric-tensors-space with respect to field and \(k\) same vectors spaces and vectors space over field.
- The reader knows a definition of wedge product of multicovectors.
- The reader admits the proposition that the \(q\)-covectors space of any vectors space has the basis that consists of the wedge products of the increasing elements of the dual basis of the vectors space.
- The reader admits the proposition that for any finite-dimensional vectors space, the transition of the dual bases for the covectors space with respect to any bases for the vectors space is this.
Target Context
- The reader will have a description and a proof of the proposition that for any \(q\)-covectors space, the transition of the standard bases with respect to any bases for the vectors space is this.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(F\): \(\in \{\text{ the fields }\}\)
\(V\): \(\in \{\text{ the finite-dimensional } F \text{ vectors spaces }\}\)
\(\Lambda_q (V: F)\): \(= \text{ the } q \text{ -covectors space }\)
\(B\): \(\in \{\text{ the bases for } V\} = \{b_l \vert 1 \le l \le dim V\}\)
\(B'\): \(\in \{\text{ the bases for } V\} = \{b'_l \vert 1 \le l \le dim V\}\)
\(B^*\): \(= \text{ the dual basis of } B = \{b^l \vert 1 \le l \le dim V\}\)
\(B'^*\): \(= \text{ the dual basis of } B' = \{b'^l \vert 1 \le l \le dim V\}\)
\(\widetilde{B^*}\): \(= \{b^{j_1} \wedge ... \wedge b^{j_q} \vert \forall l \in \{1, ..., q\} (1 \le j_l \le dim V) \land j_1 \lt ... \lt j_q\}\)
\(\widetilde{B'^*}\): \(= \{b'^{j_1} \wedge ... \wedge b'^{j_q} \vert \forall l \in \{1, ..., q\} (1 \le j_l \le dim V) \land j_1 \lt ... \lt j_q\}\)
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Statements:
\(b'_l = b_m M^m_l\)
\(\implies\)
\(b'^{j_1} \wedge ... \wedge b'^{j_q} = \sum_{(l_1, ..., l_q)} \sum_\sigma sgn \sigma {M^{-1}}^{j_1}_{l_{\sigma_1}} ... {M^{-1}}^{j_q}_{l_{\sigma_q}} b^{l_1} \wedge ... \wedge b^{l_q}\)
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2: Proof
Whole Strategy: Step 1: see that \(\widetilde{B^*}\) and \(\widetilde{B'^*}\) are some bases for \(\Lambda_q (V: F)\); Step 2: see that \(b'^l = {M^{-1}}^l_m b^m\); Step 3: conclude the proposition.
Step 1:
\(\widetilde{B^*}\) and \(\widetilde{B'^*}\) are indeed some bases for \(\Lambda_q (V: F)\), by the proposition that the \(q\)-covectors space of any vectors space has the basis that consists of the wedge products of the increasing elements of the dual basis of the vectors space.
Step 2:
\(b'^l = {M^{-1}}^l_m b^m\), by the proposition that for any finite-dimensional vectors space, the transition of the dual bases for the covectors space with respect to any bases for the vectors space is this.
Step 3:
\(b'^{j_1} \wedge ... \wedge b'^{j_q} = ({M^{-1}}^{j_1}_{m_1} b^{m_1}) \wedge ... \wedge ({M^{-1}}^{j_q}_{m_q} b^{m_q})\).
We note the fact that in general, for each \(f_j, f'_j \in \Lambda_1 (V: F)\) and each \(r, r' \in F\), \(f_1 \wedge ... \wedge (r f_j + r' f'_j) \wedge ... \wedge f_l = r f_1 \wedge ... \wedge f_j \wedge ... \wedge f_l + r' f_1 \wedge ... \wedge f'_j \wedge ... \wedge f_l\): refer to Note for the definition of wedge product of multicovectors.
So, \(({M^{-1}}^{j_1}_{m_1} b^{m_1}) \wedge ... \wedge ({M^{-1}}^{j_q}_{m_q} b^{m_q}) = {M^{-1}}^{j_1}_{m_1} ... {M^{-1}}^{j_q}_{m_q} b^{m_1} \wedge ... \wedge b^{m_q}\).
But the issue is that \((m_1, ..., m_q)\) is not necessarily in the increasing order; in fact, it may even contain some duplications.
When \((m_1, ..., m_q)\) contains any duplication, \(b^{m_1} \wedge ... \wedge b^{m_q} = 0\): refer to Note for the definition of wedge product of multicovectors. So, we can ignore such terms and suppose that \((m_1, ..., m_q)\) has no duplication.
Let \(\sigma\) be the permutation of \((m_1, ..., m_q)\) to bring it in the increasing order (\(\sigma\) depends on \((m_1, ..., m_q)\)): \((m_{\sigma_1}, ..., m_{\sigma_q})\) is in the increasing order.
Let \((m_{\sigma_1}, ..., m_{\sigma_q})\) be denoted as \((o_1, ..., o_q)\). For each \(p \in \{1, ..., q\}\), \(m_p = o_r = m_{\sigma_r}\), which means that \(p = \sigma_r = \sigma (r)\), so, \(r = \sigma^{-1} (p)\), so, \(m_p = o_r = o_{\sigma^{-1} (p)}\).
\(b^{m_1} \wedge ... \wedge b^{m_q} = sgn \sigma b^{m_{\sigma_1}} \wedge ... \wedge b^{m_{\sigma_q}}\).
So, without the summation not intended, \({M^{-1}}^{j_1}_{m_1} ... {M^{-1}}^{j_q}_{m_q} b^{m_1} \wedge ... \wedge b^{m_q} = {M^{-1}}^{j_1}_{m_1} ... {M^{-1}}^{j_q}_{m_q} sgn \sigma b^{m_{\sigma_1}} \wedge ... \wedge b^{m_{\sigma_q}} = sgn \sigma {M^{-1}}^{j_1}_{m_1} ... {M^{-1}}^{j_q}_{m_q} b^{m_{\sigma_1}} \wedge ... \wedge b^{m_{\sigma_q}} = sgn \sigma {M^{-1}}^{j_1}_{o_{\sigma^{-1} (1)}} ... {M^{-1}}^{j_q}_{o_{\sigma^{-1} (q)}} b^{o_1} \wedge ... \wedge b^{o_q} = sgn \sigma^{-1} {M^{-1}}^{j_1}_{o_{\sigma^{-1} (1)}} ... {M^{-1}}^{j_q}_{o_{\sigma^{-1} (q)}} b^{o_1} \wedge ... \wedge b^{o_q}\).
Such terms that \((m_1, ..., m_q)\) are some permutations of \((o_1, ...., o_q)\) can be summed up as \(\sum_{\sigma} sgn \sigma^{-1} {M^{-1}}^{j_1}_{o_{\sigma^{-1} (1)}} ... {M^{-1}}^{j_q}_{o_{\sigma^{-1} (q)}} b^{o_1} \wedge ... \wedge b^{o_q} = \sum_{\sigma^{-1}} sgn \sigma^{-1} {M^{-1}}^{j_1}_{o_{\sigma^{-1} (1)}} ... {M^{-1}}^{j_q}_{o_{\sigma^{-1} (q)}} b^{o_1} \wedge ... \wedge b^{o_q} = \sum_{\sigma} sgn \sigma {M^{-1}}^{j_1}_{o_{\sigma_1}} ... {M^{-1}}^{j_q}_{o_{\sigma_q}} b^{o_1} \wedge ... \wedge b^{o_q}\).
So, \(b'^{j_1} \wedge ... \wedge b'^{j_q} = \sum_{(l_1, ..., l_q)} \sum_\sigma sgn \sigma {M^{-1}}^{j_1}_{l_{\sigma_1}} ... {M^{-1}}^{j_q}_{l_{\sigma_q}} b^{l_1} \wedge ... \wedge b^{l_q}\).
