I have started an Analysis on Manifolds course and I am struggling to understand a lot about differentials. In particular I have been trying to understand the following question:
Consider the map $\eta:\chi(\mathbb{R}^3)\rightarrow C^\infty(\mathbb{R}^3)$ which maps a vector field $$ X = a^1\frac{\partial}{\partial x^1} + a^2\frac{\partial }{\partial x^2} + a^3\frac{\partial }{\partial x^3} $$ on $\mathbb{R}^3$ to the function $\eta(X)$ on $\mathbb{R}^3$ given by $$ \eta(X) = x^1a^2+x^2a^3+x^3a^1. $$ Prove that $\eta$ is a differential 1-form.
What is wrong with my following understanding?
A differnetial 1 form should assign to each point p in our open set (in this case $\mathbb{R}^3$) a covector $\omega_p\in T_p^*(\mathbb{R}^3)$. But in this scenario, it seems to me that if we evaluate eta at a point p, we get $$\eta(X)(p)=\eta(x^1,x^2,x^3)=x^1a^2+x^2a^3+x^3a^1=c,$$ and this equals a constant, but we should be getting a vector that is in fact a covector. Clearly I have misunderstood things.
Our function eta takes in a vector field, but according to my definition (and I assume other definitons are analogous) the input should be a point in the space. How can eta then be a differential 1-form?
We can also show that eta is $C^\infty(\mathbb{R}^3)$-linear map to prove the question. Why is this an acceptable solution?
You have evaluated $\eta$ for both a vector field and a point. For fixed $p$ the mapping $X \mapsto \eta(X)(p)$ is a covector. For fixed $X$ the mapping $p \mapsto \eta(X)(p)$ is a scalar field, an element in $C^\infty(\mathbb{R}^3)$.
Also note that a mapping $f$ that takes an element in $X$ and returns a mapping $Y \to Z$ also can be seen as a mapping that takes an element in $Y$ and returns a mapping $X \to Z$, or as a mapping that takes a pair from $X \times Y$ and returns an element in $Z$.