What's the difference between $S^1, \mathbb{P}^1$? I know that topologically, $S^1, \mathbb{P}^1$ is isomorphic. 
In doing differential geometry can I assume they are diffeomorphic as a smooth manifold?
What is the difference between them?
 A: Let's be precise.  The two spaces $\mathbb{S}^1$ and $\mathbb{RP}^1$ are a priori different objects with different definitions:
$$\mathbb{S}^1 = \{v\in\mathbb{R}^2\ |\ \|v\|^2 = 1\}$$
and
$$\mathbb{RP}^1 = \{\mbox{ codimension $1$ linear subspaces of $\mathbb{R}^2$ }\}.$$
Next we need to specify topologies (and smooth atlases).  I'll leave this to you as an exercise.
Only after we have specified topologies and smooth structures can we precisely explore the question of whether $\mathbb{S}^1$ and $\mathbb{RP}^1$ are the "same."  They're not even the same set: $\mathbb{S}^1$ is a set of two-vectors and $\mathbb{RP}^1$ is a set of lines.  However, they are homeomorphic and diffeomorphic as topological spaces and smooth manifolds, respectively.  
This is a good illustration of the distinction between "isomorphic" and "same".  Often, when two objects are isomorphic but not the same, there are many different isomorphisms between them with no single canonical identification and it is fruitful to study all the different ways of identifying them.
It's worth noting that the isomorphism between $\mathbb{S}^1$ and $\mathbb{RP}^1$ is an accident of low dimension.  The natural relationship between the sphere and projective space is a double-covering, where a point in $\mathbb{S}^n$ is sent to its span.  Since antipodal points span the same subspace in $\mathbb{R}^n$, we can interpret this as $\mathbb{RP}^n \cong \mathbb{S}^n/(\mbox{antipodal identification})$.  In dimension $1$ only, the quotient of the sphere by antipodal identification is again the sphere.
A: It is true that $S^1$ and $\mathbb{R}P^1$ are diffeomorphic as smooth manifolds, so in that sense they are the same. Nonetheless, I think different things when I see $S^1$ written than I do when I see $\mathbb{R}P^1$ written. To me, $S^1$ is the set of unit vectors in $\mathbb{R}^2$, whereas $\mathbb{R}P^1$ is the set of lines through the origin in $\mathbb{R}^2$. It is a fact (that must be proven) that $S^1$ and $\mathbb{R}P^1$ are diffeomorphic.
As an example of this distinction, consider the following statement: 

There is a canonical degree $2$ covering map $\pi\colon S^1\to \mathbb{R}P^1$.

If I was thinking of $\mathbb{R}P^1$ simply as being $S^1$, then the statement wouldn't be true, since there is not a canonical degree $2$ cover $S^1\to S^1$. However, if I'm thinking of $S^1$ as unit vectors in $\mathbb{R}^2$ and $\mathbb{R}P^1$ as lines through the origin in $\mathbb{R}^2$, then the map $\pi$ is canonical: $\pi(x)$ is defined to be the unique line through the origin passing through $x$. Note that you can similarly define a degree $2$ cover $\pi\colon S^n\to \mathbb{R}P^n$ in higher dimensions, when $S^n$ and $\mathbb{R}P^n$ are not diffeomorphic. 
A: $\newcommand{\R}{\mathbb{R}}$
There is a "natural" way of interpreting $\R P^1$ as a circle with half the radius of $S^1$. As froggie said, the pullback of the volume form on $\R P^1$ to $S^1$ along the canonical projection map is precisely the volume form on $S^1$, and since $S^1$ is a double cover of $\R P^1$, we have that $$\int_{S^1} \text{vol}_{S^1}=2\int_{\mathbb{R}P^1} \text{vol}_{\mathbb{R}P^1}$$
Such that if $S^1=S(1)$ is a circle of radius one, then $\R P^1$ is isomorphic (diffeomorphic and isometric) to $S(1/2)$.
