Why is first fundamental form considered intrinsic I am reading Kuhnel's differential geometry book, and in chapter 4, it says that "intrinsic geometry of a surface" can be considered to be things that can be determined solely from the first fundamental form. But I am unsure on why the first fundamental form itself can be considered to be something intrinsic to the surface. Kuhnel defines the first fundamental form to be the inner product induced from that of $\mathbb{R}^3$ restricted to $T_pM$. So isn't the larger $\mathbb{R}^3$ used?
So my question is: Supposed I have a smooth 2-dimensional manifold (e.g. in the sense defined by Lee's Introduction to Smooth Manifolds), but I didn't put it in any ambient space. Is there a way for me to define the first fundamental form? $T_pM$ is intrinsically defined (as the space of derivations in Lee's book), and so what is the inner product that I should put on $T_pM$?
 A: Like you said, a property is intrinsic if it can be computed with knowledge of the first fundamental form (metric). I take it as meaning a property that a 'being' living on the surface could calculate. For example, Gaussian curvature is intrinsic. A being on a sphere, with only their metric, could see their world was 'curved' by finding a triangle whose interior angles sum up to greater than 180. 
A: What you're looking for is known as a Riemannian metric. One way of phrasing it is as follows. (Lee talks about these later in a slightly different way of phrasing later in his book.)
A Riemannian metric on a smooth manifold $M$ is a choice of inner product $g_p$ on each tangent space $T_pM$ that varies smoothly, in the sense that given any two smooth vector fields $X$ and $Y$ on $M$, the function $g(X,Y) = p \mapsto g_p(X_p,Y_p)$ is smooth. 
Now let's start with an embedding $f: M \hookrightarrow \Bbb R^n$. The first fundamental form gives us a Riemannian metric on $M$ by setting $g_p(x,y) = Df(x) \cdot Df(y)$. To see that this is smoothly varying as defined above, consider the maps $TM \to \Bbb R^n \times \Bbb R^n \to \Bbb R^n, (p,v) \mapsto (f(p), Df(v)) \mapsto Df(v)$. Write a smooth vector field $X$ as $p \mapsto (p,X(p))$. Then the map $M \to \Bbb R^n, p \mapsto Df(X(p))$ is smooth, and so is the inner product map $\Bbb R^n \times \Bbb R^n \to \Bbb R, (x,y) \mapsto x\cdot y$. It is readily seen that $g(X,Y) = Df(X(p)) \cdot Df(Y(p))$ and that this latter map is smooth, as desired.
I should also say that a smooth manifold does not automatically come with a Riemannian metric, which is much more structure.
