definition of wedge product of multicovectors
Topics
About: vectors space
The table of contents of this article
Starting Context
Target Context
- The reader will have a definition of wedge product of multicovectors.
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:
\( F\): \(\in \{\text{ the fields }\}\)
\( V\): \(\in \{\text{ the } F \text{ vectors spaces }\}\)
\(*\wedge\): \(: \Lambda_{k_1} (V: F) \times \Lambda_{k_2} (V: F) \to \Lambda_{k_1 + k_2} (V: F), (t_1, t_2) \mapsto (k_1 + k_2)! / ({k_1}! {k_2}!) Asym (t_1 \otimes t_2)\)
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Conditions:
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\(\wedge (t_1, t_2)\) is usually denoted as \(t_1 \wedge t_2\).
2: Note
Another definition defines that \(t_1 \wedge t_2 = Asym (t_1 \otimes t_2)\).
The difference in the coefficients results in the differences in the coefficients for some properties of wedge product.
Indeed, \(t_1 \wedge t_2 \in \Lambda_{k_1 + k_2} (V: F)\), because \(t_1 \otimes t_2 \in L (V, ..., V: F)\), \(Asym (t_1 \otimes t_2) \in \Lambda_{k_1 + k_2} (V: F)\), and \((k_1 + k_2)! / ({k_1}! {k_2}!) Asym (t_1 \otimes t_2) \in \Lambda_{k_1 + k_2} (V: F)\).
Let us see some properties of wedge product.
Let us see that wedge product is associative.
\((t_1 \wedge t_2) \wedge t_3 = (k_1 + k_2)! / ({k_1}! {k_2}!) Asym (t_1 \otimes t_2) \wedge t_3 = (k_1 + k_2 + k_3)! / ((k_1 + k_2)! {k_3}!) Asym ((k_1 + k_2)! / ({k_1}! {k_2}!) Asym (t_1 \otimes t_2) \otimes t_3) = (k_1 + k_2 + k_3)! / ((k_1 + k_2)! {k_3}!) (k_1 + k_2)! / ({k_1}! {k_2}!) Asym (Asym (t_1 \otimes t_2) \otimes t_3) = (k_1 + k_2 + k_3)! / ({k_1}! {k_2}! {k_3}!) Asym (t_1 \otimes t_2 \otimes t_3)\), by the proposition that the antisymmetrization of the tensor product of any tensors is the antisymmetrizations applied sequentially.
\(t_1 \wedge (t_2 \wedge t_3) = t_1 \wedge (k_2 + k_3)! / ({k_2}! {k_3}!) Asym (t_2 \otimes t_3) = (k_1 + k_2 + k_3)! / ({k_1}! (k_2 + k_3)!) Asym (t_1 \otimes (k_2 + k_3)! / ({k_2}! {k_3}!) Asym (t_2 \otimes t_3)) = (k_1 + k_2 + k_3)! / ({k_1}! (k_2 + k_3)!) (k_2 + k_3)! / ({k_2}! {k_3}!) Asym (t_1 \otimes Asym (t_2 \otimes t_3)) = (k_1 + k_2 + k_3)! / ({k_1}! {k_2}! {k_3}!) Asym (t_1 \otimes t_2 \otimes t_3)\), by the proposition that the antisymmetrization of the tensor product of any tensors is the antisymmetrizations applied sequentially.
So, \((t_1 \wedge t_2) \wedge t_3 = t_1 \wedge (t_2 \wedge t_3)\).
So, while \(t_1 \wedge ... \wedge t_n := (...((t_1 \wedge t_2) \wedge t_3) ... \wedge t_{n - 1}) \wedge t_n\), it can be associated in any way.
\(t_1 \wedge ... \wedge t_n = (k_1 + ... + k_n)! / ({k_1}! ... {k_n}!) Asym (t_1 \otimes ... \otimes t_n)\): to prove it inductively, it holds when \(n = 2\); supposing that it holds for \(n = 2, ..., n' - 1\), \(t_1 \wedge ... \wedge t_{n'} = (t_1 \wedge ... \wedge t_{n' - 1}) \wedge t_{n'} = (k_1 + ... + k_{n' - 1})! / ({k_1}! ... {k_{n' - 1}}!) Asym (t_1 \otimes ... \otimes t_{n' - 1}) \wedge t_{n'} = (k_1 + ... + k_{n'})! / ((k_1 + ... + k_{n' - 1})! {k_{n'}}!) Asym ((k_1 + ... + k_{n' - 1})! / ({k_1}! ... {k_{n' - 1}}!) Asym (t_1 \otimes ... \otimes t_{n' - 1}) \otimes t_{n'}) = (k_1 + ... + k_{n'})! / ((k_1 + ... + k_{n' - 1})! {k_{n'}}!) (k_1 + ... + k_{n' - 1})! / ({k_1}! ... {k_{n' - 1}}!) Asym (Asym (t_1 \otimes ... \otimes t_{n' - 1}) \otimes t_{n'}) = (k_1 + ... + k_{n'})! / ({k_1}! ... {k_{n'}}!) Asym (t_1 \otimes ... \otimes t_{n'})\), by the proposition that the antisymmetrization of the tensor product of any tensors is the antisymmetrizations applied sequentially.
\(t_1 \wedge ... \wedge (r t_j + r' t'_j) \wedge ... \wedge t_n = r t_1 \wedge ... \wedge t_j \wedge ... \wedge t_n + r' t_1 \wedge ... \wedge t'_j \wedge ... \wedge t_n\), because \(t_1 \wedge ... \wedge (r t_j + r' t'_j) \wedge ... \wedge t_n = (k_1 + ... + k_n)! / ({k_1}! ... {k_n}!) Asym (t_1 \otimes ... \otimes (r t_j + r' t'_j) \otimes ... \otimes t_n) = (k_1 + ... + k_n)! / ({k_1}! ... {k_n}!) Asym (r t_1 \otimes ... \otimes t_j \otimes ... \otimes t_n + r' t_1 \otimes ... \otimes t'_j \otimes ... \otimes t_n)\), by the property of tensor product of tensors, \(= (k_1 + ... + k_n)! / ({k_1}! ... {k_n}!) (r Asym (t_1 \otimes ... \otimes t_j \otimes ... \otimes t_n) + r' Asym (t_1 \otimes ... \otimes t'_j \otimes ... \otimes t_n))\), by the proposition that any antisymmetrization-of-tensor map is linear, \(= r (k_1 + ... + k_n)! / ({k_1}! ... {k_n}!) Asym (t_1 \otimes ... \otimes t_j \otimes ... \otimes t_n) + r' (k_1 + ... + k_n)! / ({k_1}! ... {k_n}!) Asym (t_1 \otimes ... \otimes t'_j \otimes ... \otimes t_n) = r t_1 \wedge ... \wedge t_j \wedge ... \wedge t_n) + r' t_1 \wedge ... \wedge t'_j \wedge ... \wedge t_n\).
When \((t^1, ..., t^k)\) is any combination of elements of \(V^*\) and \(\sigma \in S_k\) is any, for each \(v_1, ..., v_k \in V\), \(t^{\sigma_1} \wedge ... \wedge t^{\sigma_k} (v_{\sigma_1}, ..., v_{\sigma_1}) = t^1 \wedge ... \wedge t^k (v_1, ..., v_k)\), by the proposition that for any vectors space and its any covectors combination, the antisymmetrization of the tensor product of any permutation of the covectors combination operated on the same permutation of any vectors combination is the antisymmetrization of the tensor product of the covectors combination operated on the vectors combination, because \(t^{\sigma_1} \wedge ... \wedge t^{\sigma_k} = k! Asym (t^{\sigma_1} \otimes ... \otimes t^{\sigma_k})\) and \(t^1 \wedge ... \wedge t^k = k! Asym (t^1 \otimes ... \otimes t^k)\).
When \((t^1, ..., t^k)\) is any combination of elements of \(V^*\) and \(\sigma \in S_k\) is any, \(t^{\sigma_1} \wedge ... \wedge t^{\sigma_k} = sgn \sigma t^1 \wedge ... \wedge t^k\), because \(t^{\sigma_1} \wedge ... \wedge t^{\sigma_k} (v_1, ..., v_k) = t^{\sigma_1} \wedge ... \wedge t^{\sigma_k} (v_{(\sigma \circ \sigma^{-1})_1}, ..., v_{(\sigma \circ \sigma^{-1})_k}) = t^1 \wedge ... \wedge t^k (v_{\sigma^{-1}_1}, ..., v_{\sigma^{-1}_k})\), by the above result, \(= sgn \sigma^{-1} t^1 \wedge ... \wedge t^k (v_1, ..., v_k)\), because \(t^1 \wedge ... \wedge t^k\) is antisymmetric, but \(sgn \sigma^{-1} = sgn \sigma\).
When furthermore \((t^1, ..., t^k)\) has any duplication, \(t^1 \wedge ... \wedge t^k = 0\), because supposing \(t^m = t^n\), taking \(\sigma\) as the permutation that switches \(m\) and \(n\) (\(sgn \sigma = -1\)), \(t^1 \wedge ... t^m \wedge ... \wedge t^n ... \wedge t^k = - t^1 \wedge ... t^n \wedge ... \wedge t^m ... \wedge t^k = - t^1 \wedge ... t^m \wedge ... \wedge t^n ... \wedge t^k\).
When \(t_1 \in \Lambda_{k_1} (V: F)\), a \(k_1\)-covector, and \(t_2 \in \Lambda_{k_2} (V: F)\), a \(k_2\)-covector, \(t_2 \wedge t_1 = (-1)^{k_1 k_2} t_1 \wedge t_2\), because 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, \(t_1 = {t_1}_{j_1, ..., j_{k_1}} b^{j_1} \wedge ... \wedge b^{j_{k_1}}\) and \(t_2 = {t_2}_{l_1, ..., l_{k_2}} b^{l_1} \wedge ... \wedge b^{l_{k_2}}\), and \(t_2 \wedge t_1 = ({t_2}_{l_1, ..., l_{k_2}} b^{l_1} \wedge ... \wedge b^{l_{k_2}}) \wedge ({t_1}_{j_1, ..., j_{k_1}} b^{j_1} \wedge ... \wedge b^{j_{k_1}}) = {t_1}_{j_1, ..., j_{k_1}} {t_2}_{l_1, ..., l_{k_2}} b^{l_1} \wedge ... \wedge b^{l_{k_2}} \wedge b^{j_1} \wedge ... \wedge b^{j_{k_1}}\), but as each \(b^m\) is an element of \(V^*\), 1st, \(b^{j_1}\) can be moved to in front of \(b^{l_1}\) by the \(k_2\) transpositions each of which gives the \(-1\) factor with the total \((-1)^{k_2}\) factor, then, \(b^{j_2}\) can be moved to in front of \(b^{l_1}\) with the \((-1)^{k_2}\) factor, ..., after all, \(b^{l_1} \wedge ... \wedge b^{l_{k_2}} \wedge b^{j_1} \wedge ... \wedge b^{j_{k_1}}\) can be changed to \(b^{j_1} \wedge ... \wedge b^{j_{k_1}} \wedge b^{l_1} \wedge ... \wedge b^{l_{k_2}}\) with the \({{-1}^{k_2}}^{k_1} = {-1}^{k_1 k_2}\) factor, and \(= {-1}^{k_1 k_2} {t_1}_{j_1, ..., j_{k_1}} {t_2}_{l_1, ..., l_{k_2}} b^{j_1} \wedge ... \wedge b^{j_{k_1}} \wedge b^{l_1} \wedge ... \wedge b^{l_{k_2}} = {-1}^{k_1 k_2} t_1 \wedge t_2\).
When furthermore \(k_1 = k_2 = k\) and \(t_1 = t_2 = t\), when \(k\) is odd, \(t \wedge t = 0\), because \(t \wedge t = (-1)^{k^2} t \wedge t = - t \wedge t\); when \(k\) is even, \(t \wedge t\) is not necessarily \(0\): \(t \wedge t = (-1)^{k^2} t \wedge t = t \wedge t\) does not imply that \(t \wedge t = 0\): for example, \(k = 2\) and \(t = t_{1, 2} b^1 \wedge b^2 + t_{3, 4} b^3 \wedge b^4\), then, \(t \wedge t = (t_{1, 2} b^1 \wedge b^2 + t_{3, 4} b^3 \wedge b^4) \wedge (t_{1, 2} b^1 \wedge b^2 + t_{3, 4} b^3 \wedge b^4) = t_{1, 2} t_{3, 4} b^1 \wedge b^2 \wedge b^3 \wedge b^4 + t_{1, 2} t_{3, 4} b^3 \wedge b^4 \wedge b^1 \wedge b^2 = t_{1, 2} t_{3, 4} b^1 \wedge b^2 \wedge b^3 \wedge b^4 + t_{1, 2} t_{3, 4} b^1 \wedge b^2 \wedge b^3 \wedge b^4 = 2 t_{1, 2} t_{3, 4} b^1 \wedge b^2 \wedge b^3 \wedge b^4 \neq 0\); if \(t = t_{1, 2} b^1 \wedge b^2\), \(t \wedge t = 0\), so, sometimes \(t \wedge t = 0\) for a \(t \neq 0\).
