I am somewhat confused about the connection between the dimension of tangent spaces and the definition of tangent vectors:
Let's say we have a d-dimensional differentiable manifold $M^d$. Further, consider its tangent space $T_p M^d$ at a point $p \in M^d$. According to Königsberger (p. 120) $T_p M^d$ is a vector space with dimension $d$. That is, $T_p M^d$ has the same dimension as $M^d$.
Example: Consider a surface in the $\mathbb{R}^3$, i.e. $M^2$ is an open set embedded in $\mathbb{R}^3$. Following Königsberger, every $T_p M^2$ is 2-dimensional. Hence, $\forall X \in T_p M^2: X \in \mathbb{R}^2$. This makes sense as we know the (affine) tangent space coincides with the tangent plane in this case.
Bredon (p.76f.) defines the tangent vector as a differential operator on the germ of a smooth function $f: U \subset M \rightarrow \mathbb{R}$:
$$ D_{\gamma}(f) = \sum_{i=1}^n \frac{d \gamma_i}{d t} \frac{\partial }{\partial x_i} f\, \Bigg|_{t=0},$$
where $\gamma = (\gamma_i(t), …, \gamma_n(t))$ represents a smooth curve in $M$. He defines the tangent space as
$$T_p M = \{D_{\gamma}\, |\, \gamma(0) = p\}$$
Point of confusion: Because $f \in C^{\infty}(M, \mathbb{R})$ it follows that $D_{\gamma}(f) \in \mathbb{R}$ irrespective of the dimensionality of $M$. This contradicts what Königsberger said (and also the example).
My thought on that: Define $D_{\gamma}$ on the germ of functions $f \in C^{\infty}(M, \mathbb{R}^d)$ such that $D_{\gamma}(f) \in \mathbb{R}^d$. However, as I am fairly new to this topic, I am unsure whether something like this would be possible.
Best Answer
Of course $D_{\gamma}$ evaluated on a function is a number, but the members of tangent space are actually the differential operators on the function. Roughly speaking if the space is n-dimensional, then you can move in n-different direction in the space, that means n- different directions you can take the derivative.
Hence, dimension of manifold=dimension of Tp