A description of how wedge product as an equivalence class of elements of tensor algebra is related with the tensor products construct
Topics
About: exterior algebra
The table of contents of this article
Starting Context
- The reader knows a definition of exterior algebra.
- The reader knows a definition of wedge product.
- The reader admits the proposition that the pair of the k-th exterior algebra and wedge product has the universal mapping property for alternating k-linear maps for any left R-module.
- The reader admits the proposition that k-th tensor algebra is a left R-module.
Target Context
- The reader will have a description of how wedge product as an equivalence class of elements of any tensor algebra is related with the tensor products construct.
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: Description
In a formal definition, wedge product is defined to be an element of an exterior algebra, which means that any wedge product is an equivalence class of elements of a tensor algebra.
For example, for \(\land (V)\), the wedge product, \(v_1 \land v_2\) where \(v_1, v_2 \in V\), is an equivalence class, \([v_1 \otimes v_2]\), which includes \(v_1 \otimes v_2, - v_2 \otimes v_1, 2^{-1} (v_1 \otimes v_2 - v_2 \otimes v_1)\).
On the other hand, in a less formal definition, the wedge product, \(v_1 \land v_2\), is defined as \(v_1 \otimes v_2 - v_2 \otimes v_1\) or \(2^{-1} (v_1 \otimes v_2 - v_2 \otimes v_1)\).
How has the 2nd definition come from the 1st definition?
You know, 'class' is a collection of some elements, and does not correspond to any specific element, like "\(v_1 \otimes v_2 - v_2 \otimes v_1\)". But the class has been somehow surreptitiously identified with the specific tensor products construct . . .
Especially, \(v_1 \otimes v_2 - v_2 \otimes v_1\) is not even include in the class while \(2^{-1} (v_1 \otimes v_2 - v_2 \otimes v_1)\) is included, but even if the latter is chosen as the definition, why should the element be chosen among the various elements of the class?
In fact, the correspondence from the class to an element of \(T (V)\) does not automatically occur logically speaking (how can it, indeed?), and we have to explicitly establish the correspondence by a somewhat arbitrary definition, although there is a set of certain prerequisites the definition has to satisfy.
"prerequisites" there means for example, the definition has to satisfy, for example, \(v_1 \land v_2 = - v_2 \land v_1\). So, it cannot be \(v_1 \land v_2 := v_1 \otimes v_2\).
So, let us make a somewhat arbitrary definition.
\(T^k (V)\) is a left R-module, and by the universal mapping property of the pair of the k-th exterior algebra and wedge product for alternating k-linear maps for any left R-module, for any alternating k-linear map, \(f: V^k \rightarrow T^k (V)\), there is the unique linear map, \(\tilde{f}: \land ^k (V) \rightarrow T^k (V)\). Take f as \(k! A (v_1 \otimes v_2 \otimes . . . \otimes v_k)\) where A is the antisymmetrization operator, which is \(2^{-1} (v_1 \otimes v_2 - v_2 \otimes v_1)\) for \(k = 2\), which is alternating k-linear indeed. Then, there is the unique linear map, \(\tilde{f}\), which offers a \(T^k (V)\) element for each \(\land ^k (V)\) element, which we accept.
That definition is well-defined, although I do not meticulously prove it here, while the well-defined-ness is prevalently accepted, as the definition is nothing but one of the prevalently-accepted definitions. Roughly speaking, it is well-defined because it is antisymmetric.
So, it is not really \(v_1 \land v_2 = v_1 \otimes v_2 - v_2 \otimes v_1\), but \(\tilde{f} (v_1 \land v_2) = v_1 \otimes v_2 - v_2 \otimes v_1\), which is really reasonable, because a class and a specific element cannot be the same object.
While there are 2 prevalently-accepted definitions with the difference of the factors, both are logically OK because any definition essentially includes the leeway in the factor.
