I am trying to relate the informal and rigorous definitions of tangent vectors on a manifold and am getting stuck on some points.

Let $M$ be a differentiable manifold. Informally, we can say that $x^i(\lambda)$ are the coordinates of some curve $\gamma(\lambda)$ on $M$. Let $\gamma(\lambda_p) = x^i(\lambda_p)$ be the point $p\in M$.

Now it is informally stated that tangent vector components for some $X\in T_pM$ are given by \begin{align} X^i = \left.\frac{dx^i}{d\lambda}\right|_{\lambda = \lambda_p}\end{align}.

Now I will try and do this rigorously.

Let $(U, x)$ be some chart on $M$.

Define a curve \begin{align} \gamma: \mathbb{R} \supseteq I &\longrightarrow M \\ \lambda & \longmapsto \gamma(\lambda) \end{align}

With respect to the chart, the components of $\gamma$ are $(x \circ \gamma)(\lambda) = (x^1(\gamma(\lambda)), ..., x^n(\gamma(\lambda))$. To write this as $x^i(\lambda)$, do I need to redefine $x^i\circ \gamma \rightarrow x^i$? This is essentially my question, as the rest would follow from this.

Define a tangent vector $X_{\gamma, p}$ at $p\in M$ by, for $f\in C^{\infty}(M)$, \begin{align} X_{\gamma, p}(f) = (f \circ\gamma)'(\lambda_p) \end{align}.

Inserting the coordinate identity: \begin{align} (f \circ x^{-1} \circ x \circ \gamma)'(\lambda_P) = (\partial_i(f\circ x^{-1}))(x(p))\cdot (x^i \circ \gamma)'(\lambda_p) \end{align}

The $(\partial_i(f\circ x^{-1}))(x(p))$ form the coordinate basis of the tangent space, whilst the components are $(x^i \circ \gamma)'(\lambda_p)$, which is written in different notation as \begin{align} \left.\frac{d(x^i\circ\gamma)}{d\lambda}\right|_{\lambda = \lambda_p} \end{align}

If the redefinition I asked about above is indeed correct, then these would seem to agree. However, if not, I would appreciate an explanation of how they are equivalent.

  • 1
    $\begingroup$ Yes it's a standard (but annoying) abuse of notation that people use the same symbol $x^i$ to mean both the ith component of the chart map, i.e, $\text{pr}_{\Bbb{R}^n}^i\circ x$ and also ith component of the chart representative of the curve, i.e $\text{pr}_{\Bbb{R}^n}^i\circ x \circ\gamma$. Here I'm using $\text{pr}_{\Bbb{R}^n}^i:\Bbb{R}^n\to \Bbb{R}$ to mean $\text{pr}_{\Bbb{R}^n}^i(a)=\text{pr}_{\Bbb{R}^n}^i(a^1,...,a^n)=a^i$ $\endgroup$
    – peek-a-boo
    Commented Oct 16, 2021 at 16:38
  • $\begingroup$ Thanks, it seemed like that was the case, appreciate the clarification. $\endgroup$
    – Bedge
    Commented Oct 18, 2021 at 16:27


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