A question on the definition of $0$-tensor

Question

I'm working on the following question in Munkres.

Let $\omega$ be a $k$-form defined in an open set $A \subset \mathbb{R}^n$. We say that $\omega$ vanishes at $x$ if $\omega(x)$ is the $0$-tensor. Show that if $\omega$ vanishes at each $x$ in a neighborhood of $x_0$, then $d\omega$ vanishes at $x_0$.

I have a question on the meaning of $0$-tensor. Consider a generic $k$-form: $$\omega = \sum_{[I]} f_I \,dx_I$$ This means all $f_I$ are zero right?

In particular, can I choose $f_I\, dx_I$ in the sum so that they all non-zero, but their sum evaluates to zero?

Answer 1

Q1: yes.

Q2: no.

When evaluating at a point, $\omega(x)$ belongs a to vector space of dimension $\binom{n}{k}$ and the $dx_I$ are a linear basis of this space. In more advanced language, you start from a vector space $V$ of dimension $n$, then you build the $k$-th exterior power $\wedge^kV$. If $e_1,\ldots,e_n$ is a basis of $V$, then $e_{i_1}\wedge\cdots \wedge e_{i_k}$ where $I=\{i_1<\cdots<i_k\}$ where $I$ ranges over the size $k$ subsets of $\{1,\ldots,n\}$, forms a basis of $\wedge^k V$. In your situation the elements of this basis are denoted $dx_I$.

Answer 2

I guess it depends on what do you mean by the $dx_I$'s or from which set exactly you consider your multiindices $I$.

We have that $dx_1\wedge dx_2+dx_2\wedge dx_1=0$. So, in this sense, the answer is yes, the sum can be zero.

But if you consider just $dx_I=dx_{i_1}\wedge\cdots\wedge dx_{i_n}$, where $i_1<\dots<i_n$, then those are linearly independent. So, the answer is no.