# 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 $T_pM$?

• There is another way to attack the intrinsic geometry problem. The equations of Cartan and the frame-based geometry. See Oneill's book where a natural proof is given for Gauss' big theorem on the fact that curvature is intrinsic. See supermath.info/NotesDiffGeometryOneilChapter7a.pdf for a taste. The metric is supplanted by the study of how the coframe varies according to the change in the connection coefficients. All in the wonderful world of differential forms. Jan 2, 2015 at 14:26

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.

• I'm not entirely sure if this was what you were looking for. If not, let me know, and I can try to say something more helpful as appropriate.
– user98602
Jan 2, 2015 at 7:34
• Just to make sure I understand what is happening... if $M$ is 2D-surface embedded in $\mathbb{R}^3$, and the first fundamental form is the induced inner product form $\mathbb{R}^3$, then the Riemannian metric from the first fundamental form is actually the same as the $\mathbb{R}^3$ Euclidean metric? Jan 2, 2015 at 8:04
• Yep, @suncup224.
– user98602
Jan 2, 2015 at 8:09