I'm reading Loring Tu's "Introduction to Manifolds", and came across a question (3.9), titled: Vanishing of a covector of top degree:
Let $V$ be a vector space of dimension $n$. Prove that if an n-covector $\omega$ vanishes on a basis $e_1, …e_n$ for $V$, then $\omega$ is the zero n-covector on V
There is one thing I do not understand, and one I'm curious about:
- What does it mean that $\omega$ vanishes on a basis of $V$? Isn't $\omega$ is defined on n-tuples of $V$'s elements, and not on single elements?
- Why is this question titled like so? What is a covector of top degree?
Many thanks!
Best Answer
(1) It means that $\omega(e_1,...,e_n)=0$. They are asking that if $\omega(e_1,...,e_n)=0$ for some basis $e_1,...,e_n$, then $\omega=0$, meaning that it gives zero for all inputs of $n$-tuples.
(2) A covector is a multilinear functional satisfying the alternating property. A multilinear functional is, like you said, something that takes in $k$ vectors as inputs, while being linear on each entry, and outputting a number. The alternating property requires that if we permute the input vectors by some permutation $\sigma:\{1,...,n\}\to\{1,...,n\}$, then $\omega(v_{\sigma(1)},...,v_{\sigma(n)})=|\mathrm{sgn}(\sigma)|\omega(v_1,...,v_n)$, where $\mathrm{sgn}$ is the sign of the permutation. Since the vector space $V$ has dimension $n$, it turns out that $n$-covectors are the highest numbers of inputs we can take without being trivial. That means that if we have an $k$-covector for $k>\dim V=n$, then $\omega=0$ automatically. This is because that given $k$ many vectors as inputs, we always have linear dependence $v_k=\sum_{i=1}^{k-1} a_i v_i$. So $$\omega(v_1,...,v_k)=\sum_{i=1}^{k-1}a_i\omega(v_1,...,v_{k-1}, v_i)$$
But each of $\omega(v_1,...,v_{k-1},v_i)$ must be $0$ because if you can take the permutation $\sigma$ that swaps the $i^{\text{th}}$ entry with the last entry, this is an odd permutation and so $|\mathrm{sgn}|=-1$. Therefore $\omega(v_1,...,v_{k-1},v_i)=-\omega(v_1,...,v_{k-1},v_i)$. So $\omega(v_1,...,v_k)=0$ for any inputs, as long as $k>n$. This justifies why an $n$-covector is called a covector of top degree, since there are no nontrivial higher degree covectors.