A question about $\gamma'$(0)=$d\gamma$($\frac{d}{dt}\mid_{t=0}$)

Question

I saw this equation on a manifold note: $\gamma'$(0)=$d\gamma$($\frac{d}{dt}\mid_{t=0}$), I feel very confused about $d\gamma$($\frac{d}{dt}\mid_{t=0}$), does it mean $\frac{d\gamma}{dt}\mid_{t=0}$? Does it have some special meaning when people use $d\gamma$($\frac{d}{dt}\mid_{t=0}$)?

Answer 1

Here, we're regarding the curve $\gamma$ as a map from $\Bbb R$ (or perhaps just some smaller interval containing $0$) to our manifold $M$. Then,

• $d\gamma$ is the pushforward (tangent map) $$d\gamma: T_0 \Bbb R \to T_{\gamma(0)} M,$$
• $\left.\frac{d}{dt}\right|_{t = 0}$ is the standard coordinate vector in $T_0 \Bbb R$, and so
• $d\gamma\left(\left.\frac{d}{dt}\right|_{t = 0}\right)$ just denotes the image of this vector in $T_{\gamma(0)} M$.

This does indeed mean the same thing as $\left.\frac{d\gamma}{dt}\right|_{t = 0}$, which we can think of the value of the vector field $\frac{d\gamma}{dt}$ along the parameterized curve $\gamma$ at $t = 0$.