# Second fundamental form of an immersion

I cannot quite replicate the following calculation from Aubin's Some Nonlinear Problems in Riemannian Geometry (page 349).

The setting is as follows.

Let $$(M^n,g)$$ and $$(\tilde{M}^m,\tilde{g})$$ be two $$C^\infty$$ Riemannian manifolds and let $$f:M\rightarrow\tilde{M}$$ be a smooth immersion. Let $$\nabla$$ be the Levi-Civita connection of $$(M,g)$$ and let $$\tilde{\nabla}$$ be the Levi-Civita connection of $$(\tilde{M},\tilde{g})$$.

Now let $$\{x^i\}_{i=1}^n$$ be local coordinates in a neighbourhood of $$P\in M$$ and let $$\{y^\alpha\}_{\alpha=1}^m$$ be local coordinates in a neighbourhood of $$f(P)$$ in $$\tilde{M}$$.

Say $$f$$ is injective on a neighbourhood $$\Omega$$ of $$P$$ in $$M$$. Let $$Y$$ be a vector field on $$\Omega$$ and extend $$\tilde{Y}=f_*Y$$ to a neighbourhood of $$f(P)$$. For $$X$$ in $$T_x\Omega$$, let $$\tilde{X}=f_*X$$. Finally, verify that $$\tilde{\nabla}_{\tilde{X}}\tilde{Y}$$ is well defined and define the second fundamental form $$\alpha_x$$ of $$f$$ at $$x\in \Omega$$ by $$\alpha_x(X,Y)=\tilde{\nabla}_{\tilde{X}}\tilde{Y}-f_*(\nabla_X Y)$$

The book now claims that in coordinates we have $$\alpha_x(X,Y)=\left[\partial^2_{ij}f^\gamma(x)-\Gamma_{ij}^k\partial_k f^\gamma+\tilde{\Gamma}_{\alpha\beta}^\gamma(f(x))\partial_if^\alpha(x)\partial_jf^\beta(x)\right]X^iY^j\frac{\partial}{\partial y^\gamma}$$ where $$\nabla_{\frac{\partial}{\partial x^i}}\frac{\partial}{\partial x^j}=\Gamma_{ij}^k\frac{\partial}{\partial x^k}$$ and $$\tilde{\nabla}_{\frac{\partial}{\partial y^\alpha}}\frac{\partial}{\partial y^\beta}=\tilde{\Gamma}_{\alpha\beta}^\gamma\frac{\partial}{\partial y^\gamma}$$.

I can easily calculate the coordinate expression of $$f_*(\nabla_X Y)$$. Indeed I get $$f_*(\nabla_X Y)=(X^i\partial_i Y^j\partial_jf^\gamma+X^iY^j\Gamma_{ij}^k\partial_kf^\gamma)\frac{\partial}{\partial y^\gamma}$$

However, when I try to calculate $$\tilde{\nabla}_{\tilde{X}}\tilde{Y}$$ I run into trouble, since at some point I have to calculate $$\tilde{\nabla}_{\partial_if^\alpha\frac{\partial}{\partial y^\alpha}}\left(Y^j\partial_j f^\beta\frac{\partial}{\partial y^\beta}\right)=\color{red}{\left(\partial_if^\alpha\frac{\partial}{\partial y^\alpha}\right)\left(Y^j\partial_j f^\beta\right)\frac{\partial}{\partial y^\beta}}+Y^j\partial_if^\alpha\partial_j f^\beta\tilde{\nabla}_{\frac{\partial}{\partial y^\alpha}}\frac{\partial}{\partial y^\beta}$$ and the red summand seems syntactically wrong.

Formally, $$\left(\partial_if^\alpha\frac{\partial}{\partial y^\alpha}\right)\left(Y^j\partial_j f^\beta\right)=\frac{\partial}{\partial x^i}\left(Y^j\partial_j f^\beta\right)=\partial_iY^j\partial_j f^\beta+Y^j\partial_{ij}^2f^\beta$$ which gives the correct answer, but I do not quite understand what is going on in this formal calculation. What am I missing?

Moreover, what does the second partial $$\partial_{ij}^2f^\beta$$ of a map between manifolds mean exactly?

• Good question relating to $\partial_{ij}^2 f^{\beta}$ as even in the case $f:M\to \mathbb R$ you cannot sensibly define $D^2_u f : T_u M \times T_u M \to \mathbb R$ unless $u$ is a critical point of $f$.
– user284001
Jun 9, 2021 at 14:35
• @fundamentalform, the Hessian of a scalar function $f$ is well-defined, if there is a Riemannian metric. In local coordinates, it's given by $$\nabla^2_{XY}f = (\partial^2_{ij}f - \Gamma^k_{ij}\partial_kf)X^iY^j$$ Jun 9, 2021 at 17:08

What Aubin calls the second fundamental form (which I don't like, since the second fundamental form is usually for a submanifold with the induced metric) is perhaps better called simply the Hessian of $$f$$. It's best to just use the original formula: $$\nabla^2_{XY}f = X^iY^j(\partial^2_{ij}f^\gamma - \Gamma_{ij}^k\partial_kf^\gamma + (\widetilde{\Gamma}^\gamma_{\alpha\beta}\circ f)\partial_if^\alpha\partial_jf^\beta)\frac{\partial}{\partial y^\gamma}$$ and observe that it is both well-defined and is invariant under changes of coordinate on both $$M$$ and $$\widetilde{M}$$. Aubin's definition is for me less intuitive and works only if $$f$$ is an immersion. The formula above does not require this.
• Thank you Deane for your answer. What is the relationship between the Hessian of the immersion $f$ and the second fundamental form of the submanifold $f(M)\subset\tilde{M}$? Jun 10, 2021 at 13:39
• If the Riemannian metric on $M$ is the pullback of the metric on $\widetilde{M}$, i.e., $g = f^*\tilde{g}$ (such a map is called an isometric immersion), then the second fundamental form of $M$ is in fact equal to the Hessian of $f$. You can check that, in this case, $\nabla^2_{XY}f(p)$ is normal to $f_*T_pM$. This generalizes the standard definition of the second fundamental form of a submanifold in Euclidean space, and has a natural geometric interpretation. Jun 10, 2021 at 16:29