Second fundamental form and metrics

Question

Suppose $M$ and $N$ are orientable manifolds, $f: M \to N$ is a smooth embedding and $g$ is a Riemannian metric on $N$. When $M$ has codimension $1$ and $\vec{n}$ is a prefered unit normal section of $f$, the second fundamental form of $f$ with respect to $\vec{n}$ can be regarded as a section $II: M \to T^*M \otimes T^*M$ defined by $II_p(X,Y) = g_p(\nabla_X Y, \vec{n}_p)$. It is a quadratic form on each tangent space, and, when it is additionally positive definte, it defines a Riemannian metric on $M$. I wonder if there is a nice geometric illustration of the resulting metric on $M$ in this case. In particular, in how far does it differ from the pullback-metric $f^*(g)$ ?

Answer 1

There is no nice geometric illustration of the "resulting metric". One sees an immediate problem with homogeneity. Let $N$ be in fact $\mathbb{R}^n$. And compare the embedding $f_1, f_2$ of $M$ such that $f_2$ is obtained by scaling $f_1$ up by a factor of 2. The pullback metrics satisfy

$$ 4 (f_1)^*(g) = (f_2)^*(g) $$

as we expect, since the image under $f_2$ is twice as big and $g$ measures the infinitesimal square distance.

The second fundamental form for the $f_1$ embedding however is only twice that of the second embedding, instead of 1/4. So the scaling of the "metric" is counter to our geometric intuition and you cannot really then draw a good geometric illustration for it.

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For your second question, the only reasonable answer is "very far". Try for yourself a few test cases. Observe that the hyperplane in Euclidean space is totally geodesic (has vanishing second fundamental form). This represents in some sense the limit of how bad things can be. So now take $\mathbb{S}^n\subseteq \mathbb{R}^{n+1}$, and scale it up and down as discussed in the previous section. You easily see that the difference between the two can be made as big as possible.