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 $f^*(g)$ ?


1 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.

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.

  • $\begingroup$ Thank you. So my guess is that it serves no reasonable purpose to study the Riemannian metric induced by a positive definite 2nd fundamental form, am I right ? $\endgroup$
    – H1ghfiv3
    Oct 27, 2015 at 21:32
  • $\begingroup$ Technically, no. There is one special case where it is studied, but from a different angle. When the second fundamental form is in fact proportional to the induced Riemannian metric (from the embedding), we say that the submanifold is umbilic. Such submanifolds have many special properties and is frequently studied in Riemannian and pseudo-Riemannian geometry. But even in this case the metric being studied is the Riemannian metric induced from the embedding. $\endgroup$ Oct 28, 2015 at 13:02

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