description/proof of that for \(q\)-covectors space, transition of components of covector w.r.t. 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 admits 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.
Target Context
- The reader will have a description and a proof of the proposition that for any \(q\)-covectors space, the transition of the components of any covector with respect to 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:
\(F\): \(\in \{\text{ the fields }\}\)
\(V\): \(\in \{\text{ the finite-dimensional } F \text{ vectors spaces }\}\)
\(\Lambda_q (V: F)\): \(= \text{ the } q \text{ -covectors space }\)
\(B\): \(\in \{\text{ the bases for } V\} = \{b_l \vert 1 \le l \le dim V\}\)
\(B'\): \(\in \{\text{ the bases for } V\} = \{b'_l \vert 1 \le l \le dim V\}\)
\(B^*\): \(= \text{ the dual basis of } B = \{b^l \vert 1 \le l \le dim V\}\)
\(B'^*\): \(= \text{ the dual basis of } B' = \{b'^l \vert 1 \le l \le dim V\}\)
\(\widetilde{B^*}\): \(= \text{ the standard basis for } \Lambda_q (V: F) \text{ with respect to } B = \{b^{j_1} \wedge ... \wedge b^{j_k} \vert \forall l \in \{1, ..., k\} (1 \le j_l \le dim V) \land j_1 \lt ... \lt j_k\}\)
\(\widetilde{B'^*}\): \(= \text{ the standard basis for } \Lambda_q (V: F) \text{ with respect to } B' = \{b'^{j_1} \wedge ... \wedge b'^{j_k} \vert \forall l \in \{1, ..., k\} (1 \le j_l \le dim V) \land j_1 \lt ... \lt j_k\}\)
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Statements:
\(b'_l = b_m M^m_l\)
\(\implies\)
\(\forall f = f_{j_1, ..., j_q} b^{j_1} \wedge ... \wedge b^{j_q} = f'_{l_1, ..., l_q} b'^{l_1} \wedge ... \wedge b'^{l_q} \in \Lambda_q (V: F) (f'_{n_1, ..., n_q} = \sum_{(j_1, ..., j_q)} \sum_{\lambda} sgn \lambda f_{j_1, ..., j_q} M^{j_{\lambda_1}}_{n_1} ... M^{j_{\lambda_q}}_{n_q}\)
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2: Proof
Whole Strategy: Step 1: see that \(b'^{l_1} \wedge ... \wedge b'^{l_k} = \sum_{(j_1, ..., j_q)} \sum_\sigma sgn \sigma {M^{-1}}^{l_1}_{j_{\sigma_1}} ... {M^{-1}}^{l_q}_{j_{\sigma_q}} b^{j_1} \wedge ... \wedge b^{j_q}\); Step 2: put the result of Step 1 into \(f_{j_1, ..., j_q} b^{j_1} \wedge ... \wedge b^{j_q} = f'_{l_1, ..., l_q} b'^{l_1} \wedge ... \wedge b'^{l_q}\), and express \(f_{j_1, ..., j_q}\) with \(f'_{l_1, ..., l_q}\) s; Step 3: express \(f'_{n_1, ..., n_q}\) with \(f_{j_1, ..., j_q}\) s.
Step 1:
\(b'^{l_1} \wedge ... \wedge b'^{l_q} = \sum_{(j_1, ..., j_q)} \sum_\sigma sgn \sigma {M^{-1}}^{l_1}_{j_{\sigma_1}} ... {M^{-1}}^{l_q}_{j_{\sigma_q}} b^{j_1} \wedge ... \wedge b^{j_q}\), by 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.
Step 2:
Putting the result of Step 1 into \(f_{j_1, ..., j_q} b^{j_1} \wedge ... \wedge b^{j_q} = f'_{l_1, ..., l_q} b'^{l_1} \wedge ... \wedge b'^{l_q}\), \(f_{j_1, ..., j_q} b^{j_1} \wedge ... \wedge b^{j_q} = f'_{l_1, ..., l_q} \sum_{(j_1, .... j_q)} \sum_\sigma sgn \sigma {M^{-1}}^{l_1}_{j_{\sigma_1}} ... {M^{-1}}^{l_q}_{j_{\sigma_q}} b^{j_1} \wedge ... \wedge b^{j_q}\).
That means that \(f_{j_1, ..., j_q} = f'_{l_1, ..., l_q} \sum_\sigma sgn \sigma {M^{-1}}^{l_1}_{j_{\sigma_1}} ... {M^{-1}}^{l_q}_{j_{\sigma_q}}\).
Step 3:
While \(f_{j_1, ..., j_q}\) is only with \(j_1 \lt ... \lt j_q\), let us expand it for all \(1 \le m_1 \le dim V, ..., 1 \le m_q \le dim V\) to be \(g_{m_1, ..., m_q}\).
When \(\{m_1, ..., m_q\}\) is distinct, it is a permutated \((j_1, ..., j_q)\) by a permutation, \(\lambda\), which means that \((m_1, ..., m_q) = (j_{\lambda_1}, ..., j_{\lambda_q})\), and let \(g_{m_1, ..., m_q} := sgn \lambda f_{j_1, ..., j_q}\).
When \(\{m_1, ..., m_q\}\) is not distinct, let \(g_{m_1, ..., m_q} := 0\).
Let us see that \(g_{m_1, ..., m_q} = f'_{l_1, ..., l_q} \sum_\sigma sgn \sigma {M^{-1}}^{l_1}_{m_{\sigma_1}} ... {M^{-1}}^{l_q}_{m_{\sigma_q}}\).
When \((m_1, ..., m_q) = (j_1, ..., j_q)\), it is nothing but \(g_{m_1, ..., m_q} = f_{j_1, ..., j_q} = f'_{l_1, ..., l_q} \sum_\sigma sgn \sigma {M^{-1}}^{l_1}_{j_{\sigma_1}} ... {M^{-1}}^{l_q}_{j_{\sigma_q}}\), which holds by Step 2.
When \((m_1, ..., m_q)\) is the permutated \((j_1, ..., j_q)\) by \(\lambda\), it is \(g_{m_1, ..., m_q} = sgn \lambda f_{j_1, ..., j_q} = sgn \lambda f'_{l_1, ..., l_q} \sum_\sigma sgn \sigma {M^{-1}}^{l_1}_{j_{\sigma_1}} ... {M^{-1}}^{l_q}_{j_{\sigma_q}} = f'_{l_1, ..., l_q} \sum_\sigma sgn \lambda^{-1} \circ \sigma {M^{-1}}^{l_1}_{j_{(\lambda \circ \lambda^{-1} \circ \sigma)_1}} ... {M^{-1}}^{l_q}_{j_{(\lambda \circ \lambda^{-1} \circ \sigma)_q}}\), because \(sgn \lambda = sgn \lambda^{-1}\) and \(sgn \lambda^{-1} sgn \sigma = sgn \lambda^{-1} \circ \sigma\).
But while \(j_{(\lambda \circ \lambda^{-1} \circ \sigma)_r} = m_s\), \(m_s = j_{\lambda_s}\), so, \(j_{(\lambda \circ \lambda^{-1} \circ \sigma)_r} = j_{\lambda_s}\), which means that \(\lambda_s = (\lambda \circ \lambda^{-1} \circ \sigma)_r\), which means that \(s = (\lambda^{-1} \circ \sigma)_r\).
So, \(g_{m_1, ..., m_q} = f'_{l_1, ..., l_q} \sum_\sigma sgn \lambda^{-1} \circ \sigma {M^{-1}}^{l_1}_{m_{(\lambda^{-1} \circ \sigma)_1}} ... {M^{-1}}^{l_q}_{m_{(\lambda^{-1} \circ \sigma)_q}} = f'_{l_1, ..., l_q} \sum_{\lambda^{-1} \circ \sigma} sgn \lambda^{-1} \circ \sigma {M^{-1}}^{l_1}_{m_{(\lambda^{-1} \circ \sigma)_1}} ... {M^{-1}}^{l_q}_{m_{(\lambda^{-1} \circ \sigma)_q}}\), because when \(\sigma\) goes around all the permutations of \((1, ..., q)\), \(\lambda^{-1} \circ \sigma\) with \(\lambda^{-1}\) fixed goes around all the permutations of \((1, ..., q)\).
That equals \(f'_{l_1, ..., l_q} \sum_\sigma sgn \sigma {M^{-1}}^{l_1}_{m_{\sigma_1}} ... {M^{-1}}^{l_q}_{m_{\sigma_q}}\), because it is just \(\lambda^{-1} \circ \sigma\) replaced by \(\sigma\) while they both go around all the permutations of \((1, ..., q)\).
When \(\{m_1, ..., m_q\}\) is not distinct, \(f'_{l_1, ..., l_q} \sum_\sigma sgn \sigma {M^{-1}}^{l_1}_{m_{\sigma_1}} ... {M^{-1}}^{l_q}_{m_{\sigma_q}} = 0\), because for each fixed \(\sigma\), while \(m_{\sigma_r} = m_{\sigma_s}\) for a \(r \neq s\) pair, there is the corresponding \(\sigma'\) that swaps \(\sigma_r\) and \(\sigma_s\) after \(\sigma\), and \(sgn \sigma' = - sgn \sigma\), while \({M^{-1}}^{l_1}_{m_{\sigma_1}} ... {M^{-1}}^{l_q}_{m_{\sigma_q}} = {M^{-1}}^{l_1}_{m_{\sigma'_1}} ... {M^{-1}}^{l_q}_{m_{\sigma'_q}}\). So, \(g_{m_1, ..., m_q} = f'_{l_1, ..., l_q} \sum_\sigma sgn \sigma {M^{-1}}^{l_1}_{m_{\sigma_1}} ... {M^{-1}}^{l_q}_{m_{\sigma_q}}\).
Then, let us take \(g_{m_1, ..., m_q} M^{m_1}_{n_1} ... M^{m_q}_{n_q} = f'_{l_1, ..., l_q} \sum_\sigma sgn \sigma {M^{-1}}^{l_1}_{m_{\sigma_1}} ... {M^{-1}}^{l_q}_{m_{\sigma_q}} M^{m_1}_{n_1} ... M^{m_q}_{n_q}\), where \(n_1 \lt ... \lt n_q\).
For the right hand side, for each fixed \(\sigma\), for each \(\sigma_r\), \(\sigma_r = s\), and \({M^{-1}}^{l_r}_{m_{\sigma_r}} M^{m_s}_{n_s} = {M^{-1}}^{l_r}_{m_{\sigma_r}} M^{m_{\sigma_r}}_{n_{\sigma_r}} = \delta^{l_r}_{n_{\sigma_r}}\), so, the right hand side is \(f'_{l_1, ..., l_q} \sum_\sigma sgn \sigma \delta^{l_1}_{n_{\sigma_1}} ... \delta^{l_q}_{n_{\sigma_q}}\), where only the term with \((l_1, ..., l_q) = (n_{\sigma_1}, ..., n_{\sigma_q})\) is nonzero, but as \(l_1 \lt ... \lt l_q\) and \(n_1 \lt ... \lt n_q\), the term has \(\sigma = id\), and the right hand side is \(f'_{n_1, ..., n_q}\).
So, \(f'_{n_1, ..., n_q} = g_{m_1, ..., m_q} M^{m_1}_{n_1} ... M^{m_q}_{n_q}\).
\(g_{m_1, ..., m_q} M^{m_1}_{n_1} ... M^{m_q}_{n_q} = \sum_{(j_1, .... j_q)} \sum_{\lambda} sgn \lambda f_{j_1, ..., j_q} M^{j_{\lambda_1}}_{n_1} ... M^{j_{\lambda_q}}_{n_q}\).
So, \(f'_{n_1, ..., n_q} = \sum_{(j_1, .... j_q)} \sum_{\lambda} sgn \lambda f_{j_1, ..., j_q} M^{j_{\lambda_1}}_{n_1} ... M^{j_{\lambda_q}}_{n_q}\).
3: Note
Of course, the result can be gotten also in another (probably, simpler) way: in Step 1, instead of expressing \(b'^{l_1} \wedge ... \wedge b'^{l_q}\) with \(b^{j_1} \wedge ... \wedge b^{j_q}\) s, we express \(b^{l_1} \wedge ... \wedge b^{l_q}\) with \(b'^{j_1} \wedge ... \wedge b'^{j_q}\) s (\(M\) appears instead of \(M^{-1}\)); in Step 2, we get \(f'_{j_1, ..., j_q}\) expressed with \(f_{l_1, ..., l_q}\) s; that expression is not bad, but we can transform that to the result of this proposition just by renaming some indexes.
