description/proof of that for finite-dimensional vectors space, components function of interior multiplication of antisymmetric tensor by vector with respect to standard bases is this
Topics
About: vectors space
The table of contents of this article
Starting Context
- The reader knows a definition of interior multiplication of antisymmetric-tensor by vector.
- The reader admits the proposition that the interior multiplication of the antisymmetrization of the tensor product of any \(1\)-covectors or the wedge product of any \(1\)-covectors by any vector is the sum of each covector operated on the vector with the rest antisymmetrization or wedge product with a sign.
Target Context
- The reader will have a description and a proof of the proposition that for any finite-dimensional vectors space, the components function of interior multiplication of antisymmetric tensor by vector with respect to any standard bases 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:
\(F\): \(\in \{\text{ the fields }\}\)
\(V\): \(\in \{\text{ the } d \text{ -dimensional } F \text{ vectors spaces }\}\)
\(V^*\): \(= \text{ the covectors space of } V\)
\(B\): \(\in \{\text{ the bases for } V\}\), \(= \{b_1, ..., b_d\}\)
\(B^*\): \(= \text{ the dual basis of } B \text{ for } V^*\), \(= \{b^1, ..., b^d\}\)
\(B_q\): \(= \text{ the standard basis for } \Lambda_q (V) \text{ with respect to } B\), \(= \{b^{j_1} \wedge ... \wedge b^{j_q} \vert j_1 \lt ... \lt j_q\}\)
\(B_{q - 1}\): \(= \text{ the standard basis for } \Lambda_{q - 1} (V) \text{ with respect to } B\), \(= \{b^{j_1} \wedge ... \wedge b^{j_{q - 1}} \vert j_1 \lt ... \lt j_{q - 1}\}\)
\(v\): \(\in V\), \(= v^j b_j\)
\(i_v\): \(: \Lambda_q (V) \to \Lambda_{q - 1} (V)\), \(= \text{ the interior multiplication }\)
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Statements:
\(i_v: \sum_{(j_1, ..., j_q)} t_{j_1, ..., j_q} b^{j_1} \wedge ... \wedge b^{j_q} \mapsto \sum_{(j_1, ..., j_{q - 1})} t'_{j_1, ..., j_{q - 1}} b^{j_1} \wedge ... \wedge b^{j_{q - 1}}\), where \(t'_{j_1, ..., j_{q - 1}}\) is determined like this:
Add each \(j \notin \{j_1, ..., j_{q - 1}\}\) to \((j_1, ..., j_{q - 1})\) to get \((l_1, ..., l_q)\) such that \((j_1, ..., j_{q - 1}) = (l_1, ..., \hat{l_m} = j, ..., l_q)\): \(j\) uniquely determines \((l_1, ..., l_q)\) and \(m\), so, \(m\) is expressed as \(m (j)\) as the function of \(j\)
\(t'_{j_1, ..., j_{q - 1}} = \sum_{j \notin \{j_1, ..., j_{q - 1}\}} t_{l_1, ..., l_q} (-1)^{m (j) - 1} v^j\)
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2: Note
The formula for \(t'_{j_1, ..., j_{q - 1}}\) is not so simple, but that is the best we have managed to get so far.
When \(q = d\), the formula is simpler: \(\sum_{(j_1, ..., j_q)} t_{j_1, ..., j_q} b^{j_1} \wedge ... \wedge b^{j_q}\) consists of only 1 term, \(t_{1, ..., d} b^1 \wedge ... \wedge b^d\), and for each \((j_1, ..., j_{q - 1})\), \((l_1, ..., l_q)\) is nothing but \((1, ..., d)\), and \(t'_{j_1, ..., j_{q - 1}} = t_{1, ..., d} (-1)^{m - 1} v^{l_m}\).
For example, when \(d = 4\), \(t_{1, ..., d} b^1 \wedge b^2 \wedge b^3 \wedge b^4 \mapsto t_{1, ..., d} ((-1)^{1 - 1} v^1 b^2 \wedge b^3 \wedge b^4 + (-1)^{2 - 1} v^2 b^1 \wedge b^3 \wedge b^4 + (-1)^{3 - 1} v^3 b^1 \wedge b^2 \wedge b^4 + (-1)^{4 - 1} v^4 b^1 \wedge b^2 \wedge b^3)\), which can be denoted as \((t_{1, ..., d}) \mapsto (t_{1, ..., d} (-1)^{1 - 1} v^1, t_{1, ..., d} (-1)^{2 - 1} v^2, t_{1, ..., d} (-1)^{3 - 1} v^3, t_{1, ..., d} (-1)^{4 - 1} v^4)\).
For an example for \(q \lt d\), let \(d = 4\) and \(q = 3\); \(t_{2, 3, 4} b^2 \wedge b^3 \wedge b^4 + t_{1, 3, 4} b^1 \wedge b^3 \wedge b^4 + t_{1, 2, 4} b^1 \wedge b^2 \wedge b^4 + t_{1, 2, 3} b^1 \wedge b^2 \wedge b^3 \mapsto t_{2, 3, 4} i_v (b^2 \wedge b^3 \wedge b^4) + t_{1, 3, 4} i_v (b^1 \wedge b^3 \wedge b^4) + t_{1, 2, 4} i_v (b^1 \wedge b^2 \wedge b^4) + t_{1, 2, 3} i_v (b^1 \wedge b^2 \wedge b^3) = t_{2, 3, 4} (b^2 (v) b^3 \wedge b^4 - b^3 (v) b^2 \wedge b^4 + b^4 (v) b^2 \wedge b^3) + t_{1, 3, 4} (b^1 (v) b^3 \wedge b^4 - b^3 (v) b^1 \wedge b^4 + b^4 (v) b^1 \wedge b^3) + t_{1, 2, 4} (b^1 (v) b^2 \wedge b^4 - b^2 (v) b^1 \wedge b^4 + b^4 (v) b^1 \wedge b^2) + t_{1, 2, 3} (b^1 (v) b^2 \wedge b^3 - b^2 (v) b^1 \wedge b^3 + b^3 (v) b^1 \wedge b^2) = (t_{1, 2, 3} v^3 + t_{1, 2, 4} v^4) b^1 \wedge b^2 + (- t_{1, 2, 3} v^2 + t_{1, 3, 4} v^4) b^1 \wedge b^3 + (- t_{1, 2, 4} v^2 - t_{1, 3, 4} v^3) b^1 \wedge b^4 + (t_{1, 2, 3} v^1 + t_{2, 3, 4} v^4) b^2 \wedge b^3 + (t_{1, 2, 4} v^1 - t_{2, 3, 4} v^3) b^2 \wedge b^4 + (t_{1, 3, 4} v^1 + t_{2, 3, 4} v^2) b^3 \wedge b^4\), which conforms to this proposition: for example, \((t_{1, 2, 4} v^1 - t_{2, 3, 4} v^3) b^2 \wedge b^4\) is \(j \notin \{2, 4\}\) are \(\{1, 3\}\) and \({2, 4}\) is augmented to \((1, 2, 4)\) and \((2, 3, 4)\), and the plus of \(t_{1, 2, 4} v^1\) is because \(1\) is the 1st item of \((1, 2, 4)\) and the minus of \(- t_{2, 3, 4} v^3\) is because \(3\) is the 2nd item of \((2, 3, 4)\).
3: Proof
Whole Strategy: Step 1: calculate \(i_v (\sum_{(j_1, ..., j_q)} t_{j_1, ..., j_q} b^{j_1} \wedge ... \wedge b^{j_q})\) using the proposition that the interior multiplication of the antisymmetrization of the tensor product of any \(1\)-covectors or the wedge product of any \(1\)-covectors by any vector is the sum of each covector operated on the vector with the rest antisymmetrization or wedge product with a sign; Step 2: get the coefficient of each \(b^{j_1} \wedge ... \wedge b^{j_{q - 1}}\).
Step 1:
Let us calculate \(i_v (\sum_{(j_1, ..., j_q)} t_{j_1, ..., j_q} b^{j_1} \wedge ... \wedge b^{j_q})\).
\(= v^j i_{b_j} (\sum_{(j_1, ..., j_q)} t_{j_1, ..., j_q} b^{j_1} \wedge ... \wedge b^{j_q})\), because \(i_v (\sum_{(j_1, ..., j_q)} t_{j_1, ..., j_q} b^{j_1} \wedge ... \wedge b^{j_q}) = (\sum_{(j_1, ..., j_q)} t_{j_1, ..., j_q} b^{j_1} \wedge ... \wedge b^{j_q}) (v, \bullet) = (\sum_{(j_1, ..., j_q)} t_{j_1, ..., j_q} b^{j_1} \wedge ... \wedge b^{j_q}) (v^j b_j, \bullet) = v^j (\sum_{(j_1, ..., j_q)} t_{j_1, ..., j_q} b^{j_1} \wedge ... \wedge b^{j_q}) (b_j, \bullet) = v^j i_{b_j} (\sum_{(j_1, ..., j_q)} t_{j_1, ..., j_q} b^{j_1} \wedge ... \wedge b^{j_q})\): \(\sum_{(j_1, ..., j_q)} t_{j_1, ..., j_q} b^{j_1} \wedge ... \wedge b^{j_q} \in \Lambda_q (V)\) is multilinear.
\(= v^j \sum_{(j_1, ..., j_q)} t_{j_1, ..., j_q} i_{b_j} (b^{j_1} \wedge ... \wedge b^{j_q})\), because interior multiplication is \(F\)-linear.
By the proposition that the interior multiplication of the antisymmetrization of the tensor product of any \(1\)-covectors or the wedge product of any \(1\)-covectors by any vector is the sum of each covector operated on the vector with the rest antisymmetrization or wedge product with a sign, \(= v^j \sum_{(j_1, ..., j_q)} t_{j_1, ..., j_q} \sum_{m \in \{1, ..., q\}} (-1)^{m - 1} b^{j_m} (b_j) b^{j_1} \wedge ... \wedge \hat{b^{j_m}} \wedge ... \wedge b^{j_q} = v^j \sum_{(j_1, ..., j_q)} t_{j_1, ..., j_q} \sum_{m \in \{1, ..., q\}} (-1)^{m - 1} \delta_{j_m, j} b^{j_1} \wedge ... \wedge \hat{b^{j_m}} \wedge ... \wedge b^{j_q} = v^j \sum_{(j_1, ..., j_q)} t_{j_1, ..., j_q} (-1)^{m (j) - 1} b^{j_1} \wedge ... \wedge \hat{b^j} \wedge ... \wedge b^{j_q}\), where \(m (j)\) is determined by \(j_m = j\): when \(j \notin \{j_1, ..., j_q\}\), the term of \((j_1, ..., j_q)\) vanishes, so, it would be rather expressed as \(= v^j \sum_{(l_1, ..., l_q) \text{ such that } j \in \{l_1, ..., l_q\}} t_{l_1, ..., l_q} (-1)^{m (j) - 1} b^{l_1} \wedge ... \wedge \hat{b^j} \wedge ... \wedge b^{l_q}\).
That is the expression with respect to each \(j\) and each \((l_1, ..., l_q)\) such that \(j \in \{l_1, ..., l_q\}\); we want the expression with respect to each \(b^{j_1} \wedge ... \wedge b^{j_{q - 1}}\).
For each fixed \(b^{j_1} \wedge ... \wedge b^{j_{q - 1}}\), the terms that contribute to it are the ones that \(j \notin \{j_1, ..., j_{q - 1}\}\) and \((l_1, ..., l_q)\) with \(j\) added to \((j_1, ..., j_{q - 1})\).
So, \(t'_{j_1, ..., j_{q - 1}} = \sum_{j \notin \{j_1, ..., j_{q - 1}\}} t_{l_1, ..., l_q} (-1)^{m (j) - 1} v^j\).
