Let $M$ be a smooth $n$-manifold with $p\in M$. Consider the cotangent space $$\operatorname{Alt}^{1}T_pM = \operatorname{Hom}(T_pM,\mathbb{R}) = T^*_pM$$
Now, choosing a chart $(U,\varphi)$ around $p$ on $M$ determines coordinates $x^1,…,x^n$.
I know that having $\partial_i = \frac{\partial}{\partial x^i}$ as the standard basis vectors of $T_pM$, the differentials ($1$-forms) $\mathrm dx^i \in T_p^*M$ are the dual basis vectors of the cotangent space.
Therefore any element $\alpha$ of the contangent space $T^*_pM$ can be expressed with respect to the standard basis vectors $\mathrm dx^i$ as follows
$$\alpha = \sum_{i=1}^n \alpha_i \mathrm dx^i$$
where $a_i$ are the components of $\alpha$ with respect to the basis $\mathrm dx^1,…,\mathrm dx^n$.
Now my question is the following. In one of our solutions to an exercise, the tutor phrased the above expression as
$$\alpha_q = \sum_{i=1}^n \alpha_q\left(\frac{\partial}{\partial x^i}\right) \mathrm dx^i$$
And this is also something that got brushed over in the lecture, but i don't see how this holds since assuming $\alpha_q \in T_p^*M$ to be a $1$-form, it can certainly be evaluated at basis vectors via $ \alpha_q\left(\frac{\partial}{\partial x^i}\right)$ but what it returns is a real number, so (if i am not mistaken)
$$ \alpha_q\left(\frac{\partial}{\partial x^i}\right) \in \mathbb{R}$$
how does that make sense? The components with respect to the basis vectors $\mathrm dx^i$ are supposed to be elements of $T_p^*M$ and not real numbers, aren't they?
Best Answer
As pointed out in the comments, the cotangent space $T_p^*M$ is a vector space over $\mathbb{R}$. Thus any linear combination of elements in $T_p^*M$ has real valued coefficients.