I have the following question:

The one parameter family of diffeomorphism $\phi_{t}$ of $\mathbb{R}^2$ to itself for $t\in (\pi,\pi)$ is defined in polar coordinates $(r,\theta)$ by $$\phi_t(r,\theta)=(rcos(\theta+t),rsin(\theta+t))$$ and $\phi_{t}(0)=0$ for all $t$, i.e. origin is fixed.

Compute the derivative $X=\frac{d\phi_t}{dt}_{t=0}$.

Let $M_{s}:\mathbb{R}^2\rightarrow \mathbb{R}^2$ be the derivative of the map $\phi_s$ for a fixed $s$. What is the relation between $M_{s}(X)$ and $\frac{d\phi_t}{dt}_{t=s}$.

My efforts:

First I changed everything into Cartesian coordinates by putting $x=rcos(\theta)$ and $y=rsin(\theta)$.

So $\phi_t(x,y)=(xcos(t)-ysin(t), xsin(t)+ycost(t))$. So $$\phi_t=\begin{pmatrix} cost(t) & -sin(t) \\ sin(t) & cos(t) \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}$$.

Observe that this matrix is rotation matrix.

Now $$\frac{d\phi_t}{dt}=\begin{pmatrix} -sin(t) & -cos(t) \\ cos(t) & -sin(t) \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}$$

Putting $t=0$ we get $X(x,y)=(-y,x)$

Now I find $$M_s=\begin{pmatrix} cos(s) & -sin(s) \\ sin(s) & cos(s) \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}$$

I hope my calculations are correct.

I am not able to solve the last part. Relation between $M_{s}(X)$ and $\frac{d\phi_t}{dt}_{t=s}$.

Second thing, I don't understand what is happening here geometrically.

Is this question related to concept of flow, vector field, integral curves? (Just a wild guess!!)

I have done the calculations (as it was easy) but I don't understand the theory. What's going on here in background ? I guess it is related to vector field ...

  • $\begingroup$ Instead of changing everything to cartesian coordinates, try changing everything to polar coordinates. $\endgroup$ – Tyrone Aug 17 '17 at 12:20
  • $\begingroup$ Does the below help? $\endgroup$ – Faraad Armwood Aug 17 '17 at 19:10
  • $\begingroup$ @FaraadArmwood Yes surely it helped a lot. I have upvoted your answer. Can you please look into my calculations and tell me whether they are correct or not. $\endgroup$ – Tensor_Product Aug 17 '17 at 20:49
  • $\begingroup$ I'm glad it helped. I can, but that'll require some time. I'm working on some notes right now. I'll try to look back at it later. I'm sorry. $\endgroup$ – Faraad Armwood Aug 17 '17 at 21:02

$\bullet$ Just use the chain rule for smooth functions $f: \mathbb{R}^n \to \mathbb{R}^1$ i.e,

\begin{align*} \frac{d \phi_t}{d t}\Bigr|_{t = 0} &= \nabla \phi_0 \cdot \gamma'(0) \\ \\ &= \begin{pmatrix} \frac{\partial \phi_t}{\partial r}\Bigr|_{t=0} & \frac{\partial \phi_t}{\partial \theta}\Bigr|_{t = 0} \end{pmatrix} \begin{pmatrix} \frac{\partial x}{\partial t}\Bigr|_{t=0} \\ \frac{\partial y}{\partial t}\Bigr|_{t=0} \end{pmatrix}\end{align*}


\begin{cases} \gamma(t) = (r \cos ( \theta + t), r \sin (\theta + t))\\ \\ \phi_t(r, \theta) = (x(r, \theta,t), y(r, \theta,t)) = (x,y)\end{cases}

$\bullet$ For the relationship between the derivatives , first recall that $\phi_{m+n} = \phi_{m} \circ \phi_n = \phi_n \circ \phi_m$. Next we define the vector field,

$$\textbf{v}(x_0)=\frac{d}{dt}\Bigr|_{t = 0} \phi_t(x_0), x_0 \in \mathbb{R}^2$$

i.e $\textbf{v}$ gives the initial rate of change of $x_0$ under the $1$-parameter subgroup. Now we just manipulate the leibniz notation,

$$ \frac{d \phi_t}{dt} \Bigr|_{t = s} = \frac{d (\phi_{\epsilon + s})}{d\epsilon} \Bigr|_{\epsilon = 0} = \frac{d}{d \epsilon} \Bigr|_{\epsilon = 0} (\phi_{\epsilon} \circ \phi_s) = \textbf{v}(\phi_s)$$

Hence, if you want to know how $\phi_s$ moves $p \in\mathbb{R}^2$, you just take the velocity field $\textbf{v}$, and translate it to the point $p$.

$\bullet$ Geometrically what is going on is that you are just taking a particle on a circle of radius $r$ centered at the origin and moving it along this circle by $t$ adding to the angle.


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