Derivative of local flow of vector field at origin

I'm reading one lemma about local flow of vector field, and there's some point in the proof that I don't understand. Here's the lemma and the proof:

What I don't understand is the line (line 5 counted from the end):

By the definition of local flow, $d \bar{\alpha}$ maps the unit vector of the $t$ axis into $w$ and maps the unit vector of the $y$ axis into itself.

From what I understand, $d\bar{\alpha} = (\frac{\partial{\bar\alpha}}{\partial x}, \frac{\partial{\bar\alpha}}{\partial y}, \frac{\partial{\bar\alpha}}{\partial t} )$. But from the definition of local flow, we only have $\frac{\partial{\bar\alpha}}{\partial t} (q, t) = w(\bar\alpha(q,t))$. So what is the reason for that statement? I know it's hard for everybody to trace when my question is too specific like this, so if anyone needs more details, please feel free to ask. Thanks a lot, I really appreciate.

The statement in the proof is just a reformulation of the definition of the flow. Make yourself clear that $d\alpha_{(p,t)}(\partial_t)$ is just the directional derivative of $\alpha$ in $t$-direction:

$$d\alpha_{(q,t)}(\partial_t)=\frac{d}{ds}\bigg|_{s=0}\alpha(q,s+t)=\frac{\partial{\alpha}}{\partial t}(q,t)$$

In the situation of the proof this means that $$d\alpha_{(p,0)}(\partial_t)=w(p)\neq 0$$ Furthermore we have $\alpha(q,0)=q$ by definition and hence $$d\alpha_{(p,0)}(\partial_y)=\frac{d}{ds}\bigg|_{s=0}\alpha(p+s\,e_y, 0)=\frac{d}{ds}\bigg|_{s=0}(p+s\,e_y)=\partial_y|_p.$$ This essentially means that $\alpha$ locally maps the $t$-axis to the $x$-axis and leaves the $y$-axis invariant. This is the idea of the proof. $\bar\alpha$ is then the restriction of $\alpha$ to the $(y,t)$-submanifold.

I think the author uses a very sloppy language here. Maybe this increases your confusion. The problem is that the domain of $\bar\alpha$ as it is defined is not an open set, so strictly speaking, $d\bar\alpha$ is not well-defined. The domain the author intended to use is the embedding of the specified set into $\mathbb{R}^2$ via $(0,y,t)\rightarrow(y,t)$ and $\bar\alpha$ is defined on this set.

Having this properly set up, the equations above also hold for $\bar\alpha$ and one can use the inverse function theorem the conclude the proof.

• Sorry for asking this, but I'm not very familiar with your symbols. Can you explain what is $d\alpha_{(q,t)}(\partial_t)$ ? – le duc quang Feb 23 '16 at 2:56
• I think the problem is $\partial_t$? In differential geometry it is common to denote the coordinate vector field with $\partial$. Here, $\partial_t$ is the vector field on $V\times I$ which is given everywhere by the unit vector in $t$-direction, explicitly we thus have in $x,y,t$-coordinates: $\partial_t|_{(q,t)}=(0,0,1)$. So my calculation reads in these coordinates $d\alpha_{(q,t)}=\langle (\frac{\partial\alpha}{\partial x}, \frac{\partial\alpha}{\partial y}, \frac{\partial\alpha}{\partial t}),(0,0,1)\rangle = \frac{\partial\alpha}{\partial t}(q,t)$ – Markus Heinrich Feb 23 '16 at 9:00
• I understand. Thanks for your detailed explanation. – le duc quang Feb 24 '16 at 3:01