5
$\begingroup$

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?

$\endgroup$
1
  • 1
    $\begingroup$ $D\varphi$ 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 $\gamma\circ \varphi_t$ $\endgroup$
    – J.V.Gaiter
    Commented Jan 19, 2022 at 17:48

2 Answers 2

2
$\begingroup$

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 $t_0\in(a,b)$).

$\endgroup$
2
$\begingroup$

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 $\mathbf{S}^n$.

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .