# Integral curve definition in terms of the tangent vector: why $D \varphi_{t} (d/dt) = X_{\varphi(t)}$?

I have this definition of an integral curve from these notes, page 29, on differentiable manifolds.

With a manifold $$M$$, an integral curve of a vector field $$X$$ is a smooth map $$\varphi: (\alpha,\beta) \subset \mathbb{R} \to M$$ such that $$D \varphi_{t} \left(\frac{d}{dt}\right) = X_{\varphi(t)}$$ An example is given where $$M = \mathbb{R}^{2}$$, and so the derivative of the smooth function $$\varphi(t) = (x(t),y(t))$$ is $$D \varphi\left(\frac{d}{dt}\right) = \frac{dx}{dt}\frac{\partial}{\partial x} + \frac{dy}{dt}\frac{\partial}{\partial y}$$

What is going on here with the $$d/dt$$ in the brackets on the final line? i.e. this part

$$D\varphi\left(\underbrace{\frac{d}{dt}}_{\text{this part here}}\right) = \dots$$ I am assuming there is a function $$f: M \to \mathbb{R}$$ on which $$D\varphi(d/dt)$$ acts, as $$D\varphi(df/dt)$$, but I thought $$\varphi$$ takes in real numbers i.e. the time parameter, since its defined as a map $$(\alpha,\beta) \subset \mathbf{R}$$? Perhaps it's not a function, just a term in brackets to multiply $$D\varphi$$ by?

• $D\varphi$ is the map $T_t\mathbb{R}\cong \mathbb{R}\to T_{\phi(t)}M$ which sends a tangent vector $v$ represented by a curve $\gamma$ to the tangent vector represented by the curve $\gamma\circ \varphi_t$ Commented Jan 19, 2022 at 17:48

Let $$X$$ be a smooth vector field on a manifold $$M$$. Recall that $$X$$ is a section of the tangent bundle $$TM$$, i.e. $$X:M\rightarrow TM$$. So for every point $$p\in M$$, $$X(p)=X_p\in T_pM$$ is a tangent vector.
Now, let $$\varphi:(a,b)\rightarrow M$$ be a curve in $$M$$. Recall that $$(a,b)$$ is a manifold with the global coordinate $$t$$. In particular, at a point $$t_0\in (a,b)$$, the vector $$(d/dt)\vert_{t_0}$$ spans the tangent space $$T_{t_0}(a,b)$$. We say that $$\varphi$$ is an integral curve of $$X$$ if the following holds. For all $$t_0\in (a,b)$$,
$$D(\varphi)_{t_0}\left(\dfrac{d}{dt}\Big\vert_{t_0}\right)=X_{\varphi(t_0)}.$$ What this means is that the map
$$D(\varphi)_{t_0}:T_{t_0}(a,b)\rightarrow T_{\varphi(t_0)}M$$ has the property that $$(d/dt)\vert_{t_0}$$ is mapped to $$X_{\varphi(t_0)}$$ (for all $$t_0\in(a,b)$$).
You are correct in saying that $$\varphi$$ is a map from $$\mathbf{R}$$ to $$M$$. The differential $$D\varphi_t$$, however, is the pushforward; it takes tangent vectors in $$T_{t} (\alpha,\beta) \cong \mathbf{R}$$ over to a tangent vector in $$T_{\varphi(t)} M$$. In our case, the tangent vector $$d/dt$$ is pushed forward by $$\varphi$$ to get $$X_{\varphi(t)} \in T_{\varphi(t)}M$$. It may be useful to compute $$\varphi_t$$ for a few different vector fields $$X$$ on $$\mathbf{R}^2$$ or $$\mathbf{R}^3$$ to get a better idea of what is going on, and then try to carry this over to higher dimensions and other manifolds like $$\mathbf{S}^n$$.