Lie algebras of vector fields over $\mathbb{R}$ and over $S^1$ An exercise in the book "Moonshine beyond the Monster" has me stumped.  It asks whether the real Lie algebras of smooth vector fields over the reals $V(\mathbb{R})$ and over the circle $V(S^1)$ are isomorphic or not.
The two are isomorphic as real vector spaces assuming the axiom of choice, but they should not be isomorphic as Lie algebras: I know that the Lie algebra determines the topology of a manifold and the two underlying manifolds are not homeomorphic.
Pick coordinates $\theta$ and $t$ on $S^1$ and $\mathbb{R}$ respectively to write typical elements of the corresponding algebras as $\phi(\theta)\frac{d}{d\theta}$ and $f(t)\frac{d}{dt}$ respectively with smooth coefficients.  
My first question is: is the map which takes $\phi(\theta)\frac{d}{d\theta}$ to $\phi(t)\frac{d}{dt}$ (seen as a periodic function on the reals) a well defined map between the algebras?  If so, it seems to me that this embeds the Lie algebra $V(S^1)\hookrightarrow V(\mathbb{R})$.
My second question is, how can we show that the algebras are not isomorphic?  I tried to construct some relation involving brackets that holds in $V(\mathbb{R})$ but not $V(S^1)$ but without success.  I thought I had to force a lack of periodicity for all solutions of some differential equation given solely in terms of the bracket, but I haven't managed it so far: every relation I try is satisfied in $V(S^1)$ as well.  Am I completely off target?  If so, please provide a hint to get me closer.  If not, please suggest a relation that can do the trick.
Some more information: the section in the book that contains the exercise mentions the Witt subalgebra of the complexification of $V(S^1)$, but unless the answer to my first question is 'no', I do not see how one can use the relations there to get a contradiction in $V(\mathbb{R})$.
 A: Intriguing problem. Here's a rather ad hoc proof that they're not isomorphic. There might be an easier way to do this, but this argument has the advantage that it doesn't use any sophisticated theory of infinite-dimensional Lie algebras or Lie groups.
Note that $V(\mathbb R)$ contains an element $X$ such that the map $\operatorname{ad}_X\colon V(\mathbb R)\to V(\mathbb R)$ is surjective (where $\operatorname{ad}_X(Y) = [X,Y]$): For example, $X=d/dt$ is such an element, because given any vector field $Z = w(t)d/dt$, we can let $Y = W(t)d/dt$, where $W$ is an antiderivative of $w$, and then $[X,Y]=Z$.
Now suppose $V(\mathbb R)$ and $V(S^1)$ are isomorphic. Then the isomorphism carries $X$ to some element $\overline X\in V(S^1)$ such that $\operatorname{ad}_{\overline X}\colon V(S^1)\to V(S^1)$ is surjective. We can write $\overline X = f\,d/d\theta$ for some smooth $2\pi$-periodic function $f\colon\mathbb R\to\mathbb R$.  For any vector field $u\,d/d\theta\in V(S^1)$, $[\overline X,u\,d/d\theta] = (fu'-uf')d/d\theta$, so we conclude that for any $2\pi$-periodic smooth function $h$, there is a $2\pi$-periodic smooth function $u$ such that $fu'-uf'=h$.
First take $h\equiv 1$, so there is a $u$ satisfying $fu'-uf'=1$. At any $\theta_0$ where $f(\theta_0)=0$, this equation implies $f'(\theta_0)\ne 0$.  
Next take $h=f$, so there is some $v$ satisfying 
$$fv'-vf'=f.$$
Differentiating with respect to $\theta$ yields
$$fv''-vf''=f',$$
since the $f'v'$ terms cancel. 
If there is any $\theta_0$ where $f(\theta_0)=0$, the first equation implies $v(\theta_0)=0$ also, but then the second equation implies $f'(\theta_0)=0$, contradicting the result in the previous paragraph. Therefore $f$ never vanishes.
Finally, take $h=f^2$, so there is some $w$ satisfying 
$$fw'-wf'=f^2.$$
Because $f$ never vanishes and both $f$ and $w$ are periodic, the fundamental theorem of calculus gives
\begin{align*}
0 &= \left.\frac{w}{f}\right|_0^{2\pi}
= \int_0^{2\pi} \frac{d}{d\theta} \left(\frac{w}{f}\right)\,d\theta
=  \int_0^{2\pi} \frac{f w'- wf'}{f^2}\,d\theta
=\int_0^{2\pi} 1\,d\theta =2\pi,
\end{align*}
which is a contradiction.
A: For any Lie algebra $L$, denote the $q$-th cohomology of $L$ by $H^q(L)$. By De-Rham theorem we know that:
$$H^1(V(\mathbb{S}^1))=H_{De-Rham}^1(\mathbb{S}^1)\cong H_{Singular}^1(\mathbb{S}^1; \mathbb{R})\cong \mathbb{R}$$
$$H^1(V(\mathbb{R}^1))=H_{De-Rham}^1(\mathbb{R}^1)\cong H_{Singular}^1(\mathbb{R}^1; \mathbb{R})\cong 0$$
Thus, $H^1(V(\mathbb{S}^1))\not\cong H^1(V(\mathbb{R}^1))$, and so we get $V(\mathbb{S}^1)\not\cong V(\mathbb{R}^1)$
