Can every canonical transformation be broken down into a large number of infinitesimal canonical transformations?

Question

Consider a canonical transformation from $(q,p)$ to $(Q,P)$ depending upon a continuous parameter $\alpha$ such that: $$Q_i=Q_i(q,p,t,\alpha), \space P_i=P_i(q,p,t,\alpha)$$ where $q$ and $p$ represent the set of all $q_i$s and $p_i$s respectively.

From what I leant, The generator corresponding to the infinitesimal Canonical transformations has to be represented in the form: $$F(q,P,t)=\sum q_iP_i +\alpha G(q,P,t). \tag{1}\label{1}$$ Because: Since ICTs have to be infinitesimally close to original coordinates, transformation equations have to be of the form: $$Q_i=q_i+\alpha(something), \space P_i=p_i+\alpha(something) \tag{2}\label{2}$$ (where $\alpha$ is small) and this is made sure by eq(1) because $F=\sum q_iP_i$ corresponds to an identity transformation and hence eq(1) gives identity + something, just like ICT demands in eq(2).

Question: If we consider any general CT given by generator function $F$(need not be a function of $q,P$), depending on a continuous parameter $\alpha$; and we shrink $\alpha$ and neglect the second and higher order terms in $\alpha$, are we guaranteed to get it in the form of eq(1)? (see below for an example)Intuitively I wouldn't think so, because $F$ need not be a function of $q,P$; However, does it mean that when we shrink $\alpha$ the resultant GF doesn't correspond to an ICT? How can it be possible that a finite CT cannot be broken down into many ICTs?

PS: example: if we consider a rotation of coordinate systems by an angle $\alpha$, the GF will be given by: $$F=qP\sec(\alpha) -\frac 12 (q^2+P^2)\tan(\alpha).$$ For small angles: $$F=qP -\frac 12 (q^2+P^2)\alpha$$ which is clearly in the form of eq. (1).

Answer 1

1. TL;DR: A finite CT can not always be composed from infinitely many infinitesimal CTs because the space of CTs is not always path connected for topological reasons, cf. this related Phys.SE post.

2. Let us elaborate on OP's example. OP considers a CT in a 2D phase space $M=\mathbb{R}^2$ with a generator $$ F_2(q,p)~=~\frac{\alpha}{2}q^2 + \beta qP + \frac{\gamma}{2}P^2 \tag{i}$$ of type 2, where $\alpha,\beta,\gamma\in\mathbb{R}$ are 3 real parameters.

3. It is straightforward to check that the corresponding CT is $$\begin{align} \begin{bmatrix} Q \cr P \end{bmatrix} ~=~& A\begin{bmatrix} q \cr p \end{bmatrix}, \cr A~:=~& \begin{bmatrix}a & b\cr c & d \end{bmatrix} ~=~\frac{1}{\beta}\begin{bmatrix} \beta^2-\alpha\gamma & \gamma\cr -\alpha & 1 \end{bmatrix}\cr ~\in~&Sp(2,\mathbb{R}) ~=~SL(2,\mathbb{R}), \cr 1~=~&\det A~=~ad-bc.\end{align}\tag{ii}$$

4. Notice that the CT (ii) is only well-defined if $\beta\neq 0$.

5. Also notice that the identity CT $A={\bf 1}_{2\times 2}$ corresponds to $(\alpha,\beta,\gamma)=(0,1,0)$.

6. This suggests that if we are using a single type 2 CT (i) then we can only deform continuously the identity CT $A={\bf 1}_{2\times 2}$ into a CT (ii) with positive $\beta$ value.

7. In particular, OP considers a rotation $$ A~=~ \begin{bmatrix}\cos\theta & \sin\theta\cr -\sin\theta & \cos\theta \end{bmatrix}.\tag{iii} $$ It is easy to see that $\beta >0 \Leftrightarrow \cos\theta >0$. But any rotation (iii) is a CT. This suggests that we in this case can overcome the topological restriction $\beta >0$ by composing several type 2 CTs (i).