Geodesic tangent space is a vector space?

Question

I have very little knowledge of Differential Geometry and I'm stuck while reading about General Relativity.

Consider defining something called a null geodesic tangent space, in analogy with the tangent space $𝑇_𝑃(𝑀)$ at $𝑃\in 𝑀$, where $M$ is a (pseudo-)Riemannian manifold. Here, $𝐺(𝑀)$ is the space of null geodesics defined on $M$ with respect to the Levi-Civita connection $\Gamma$. $$G(M):=\left\{\gamma: \mathbb{I}\rightarrow M \left.\right|~ \mathbb{I}\subset\mathbb{R} \text{ and } \gamma \text{ is a null geodesic}\right\}$$

A linear map $D:C^\infty(M)\rightarrow\mathbb{R}$ is called a derivation at $P$ if it satisfies $$D(\gamma\delta) = \gamma(P) D(\delta) + \delta(P) D(\gamma) ~~\forall \gamma,\delta\in C^\infty(M)$$ The set of all derivations of $G(M)$ at $P$, denoted by $\Lambda_P(M)$ is called the null geodesic tangent space at $P$. An element of $\Lambda_P(M)$ is called a tangent vector of a null geodesic at $P$.

My question is: Will $\Lambda_P(M)$ be a vector space?

My attempt: It is clear that $\Lambda_P(M)\subset T_P(M)$. Now, to prove it is a subspace (i.e., a vector space), we need to show the existence of an additive identity, closure under addition, and closure under scalar multiplication. This is where I'm getting stuck!

I've searched these discussions but have failed to extend those proofs in my case. Any help or guidance would be greatly appreciated.

1. https://physics.stackexchange.com/questions/297618/proof-that-tangent-space-is-a-vector-space
2. Prove that tangent spaces, modeled as equivalence classes of curves, are vector spaces
3. https://math.stackexchange.com/questions/3977724/what-is-nedeed-to-prove-that-the-tangent-space-on-a-manifold-is-a-vector-space
4. https://proofwiki.org/wiki/Tangent_Space_is_Vector_Space
5. https://bose.res.in/~amitabha/diffgeom/chap4.pdf
6. https://www.physicsforums.com/threads/prove-t_-p-m-is-a-vector-space-with-the-axioms.1046947/#google_vignette

Answer 1

First of all, the correct terminology is that each (smooth) curve $c: [a,b]\to M$ has the velocity vector $c'(t)\in T_{c(t)}M$ for every $t\in [a,b]$. One can define tangent vectors in terms of derivations but you do not have to. The thing that you learn in a differential geometry class or reading a textbook is that if $M$ is an open subset of ${\mathbb R}^n$, then tangent vectors (defined as derivations) can be identified with elements of ${\mathbb R}^n$ and $\gamma'(t)$ is the same as the derivative of $\gamma$ at $t$ in terms of vector calculus. (The derivation corresponding to a vector ${\mathbf v}\in T_pM$ is just the directional derivative $D_{{\mathbf v}}$ of a function in the direction of the vector ${\mathbf v}$.)

With this in mind, consider the following example: Take $M={\mathbb R}^2$ and equip it with the standard constant Lorentzian metric $g=dx^2 - dy^2$. Complete geodesics $c: \mathbb R\to M$ with respect to this metric and satisfying $c(0)={\mathbf 0}$ are given by parametric equations $$ c(t)=(at, bt), $$
for various vectors $(a,b)\in {\mathbb R^2}$. The velocity vector of such a geodesic is just $(a,b)$. (The important part is that $c$ passes through the origin, the normalization $c(0)={\mathbf 0}$ does not affect the velocity vector.) The null-geodesics among these, are the ones satisfying the equation $a^2-b^2=0$, i.e. $a^2=b^2$, which means $b=\pm a$. The set of such vectors $(a,b)$ is the union of two lines: $b=a$ and $b=-a$. It is the "null-cone" of the metric $g$ at the origin. Now you will figure out why this set is not a linear subspace of ${\mathbb R}^2$.