Alternating covariant tensors measure oriented k-dimensional input
Objective: Why is ω(v,w)=0 not enough to conclude that v and w are linearly dependent in R³?
A one-form eats one vector and returns a scalar. A k-form at one point eats k tangent vectors. Linearity in each slot makes the measurement compatible with vector addition and scaling; alternation forces the result to vanish when the inputs lose k-dimensional independence. Before manipulating symbols, name the manifold, its dimension, the degree of every form, the tangent or cotangent space where each input lives, and every orientation, metric, compactness, boundary, or support hypothesis. The exact conclusion is: If dim V=n, then dim Λ^k(V^*)=binomial(n,k) for 0≤k≤n and Λ^k(V^*)={0} for k>n. An alternating k-form vanishes on every linearly dependent k-tuple.
The new object means: For a finite-dimensional vector space V, an alternating covariant k-tensor is a multilinear map ω:V^k→R satisfying ω(v_{σ(1)},…,v_{σ(k)})=sign(σ)ω(v_1,…,v_k) for every permutation σ. The space of these maps is Λ^k(V^*). Set Λ^0(V^*)=R. A differential form is not a decorative string of dx symbols: it is an alternating multilinear measurement with a declared point, degree, and transformation law.
Rebuild the worked problem “On V=R³, let ω((a_1,a_2,a_3),(b_1,b_2,b_3))=a_1b_2−a_2b_1. Evaluate ω on v=(1,2,3), w=(4,5,6), verify alternation, and identify its kernel behavior.” from the definition, including every sign and degree. Keep this failure boundary visible: Alternating does not mean merely antisymmetric in one selected pair; it controls every permutation. A k-form is covariant: vectors are inputs and scalars are outputs. When k>dim V the only alternating form is zero, so a formal dx_1∧⋯∧dx_k cannot be nonzero in too small a space.
If dim V=n, then dim Λ^k(V^*)=binomial(n,k) for 0≤k≤n and Λ^k(V^*)={0} for k>n. An alternating k-form vanishes on every linearly dependent k-tuple.
Choose a basis e_1,…,e_n with dual basis e^1,…,e^n. Alternation says a value is determined by increasing index lists i_1<⋯<i_k.
There are binomial(n,k) such index lists, and the wedge basis e^{i_1}∧⋯∧e^{i_k} is linearly independent, proving the dimension formula.
If v_1,…,v_k are dependent, write one as a combination of the others. Multilinearity expands the value into terms with repeated inputs, and alternation makes each repeated-input term zero.
On V=R³, let ω((a_1,a_2,a_3),(b_1,b_2,b_3))=a_1b_2−a_2b_1. Evaluate ω on v=(1,2,3), w=(4,5,6), verify alternation, and identify its kernel behavior.
- Substitute directly: ω(v,w)=1·5−2·4=−3.
- Swapping inputs gives ω(w,v)=4·2−5·1=3=−ω(v,w), and ω(v,v)=1·2−2·1=0.
- The third coordinates never enter. Any pair whose projections to the first two coordinates are dependent gives zero; dependence in all of R³ also forces zero.
Result: ω(v,w)=−3, ω(w,v)=3, and repeated or projected-dependent inputs give zero; in dual notation ω=e^1∧e^2.