tangent space of the manifold $\operatorname{Imm}(S^1,\mathbb{R}^d)$

Question

The space of smooth, periodic, immersed curves $$\operatorname{Imm}(S^1,\mathbb{R}^d)=\{c \in C^\infty(S^1,\mathbb{R}^d) : c'(\theta) \neq 0, \forall \theta \in S^1\}$$ forms an infinite-dimensional manifold. Supposedly the tangent space is the modelling space itself, i.e. $T_c\operatorname{Imm}(S^1,\mathbb{R}^d)=C^\infty(S^1,\mathbb{R}^d).$ I understand why $T_c\operatorname{Imm}(S^1,\mathbb{R}^d) \subset C^\infty(S^1,\mathbb{R}^d)$, but I don't see why we get the reverse inclusion because surely the trivial curve $\gamma(x)=0$ is in $C^\infty(S^1,\mathbb{R}^d),$ but it cant be in $T_c\operatorname{Imm}(S^1,\mathbb{R}^d)$ by definition of $\operatorname{Imm}(S^1,\mathbb{R}^d)$. What am I missing?

Answer 1

I think the non-technical answer to your question is that the subset $Imm(S^1,\mathbb R^d)\subset C^{\infty}(S^1,\mathbb R^d)$ is open. Similarly to the finite dimensional situation the tangent spaces to an open subset is the modelling vector space. Otherwise put, for $c\in Imm(S^1,\mathbb R^d)$, $f\in C^{\infty}(S^1,\mathbb R^d)$ and $t\in\mathbb R$, you get $c+tf\in Imm(S^1,\mathbb R^d)$ is $t$ is suffieciently small. Thus you get a curve emanating from $c$ in direction $f$, and hence $f$ should be in the tangent space at $c$.