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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?

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  • $\begingroup$ 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 $f$. $\endgroup$
    – user284001
    Jun 9, 2021 at 14:35
  • $\begingroup$ @fundamentalform, the Hessian of a scalar function $f$ is well-defined, if there is a Riemannian metric. In local coordinates, it's given by $$\nabla^2_{XY}f = (\partial^2_{ij}f - \Gamma^k_{ij}\partial_kf)X^iY^j$$ $\endgroup$
    – Deane
    Jun 9, 2021 at 17:08

1 Answer 1

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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.

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  • $\begingroup$ 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 $f(M)\subset\tilde{M}$? $\endgroup$ Jun 10, 2021 at 13:39
  • $\begingroup$ 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. $\endgroup$
    – Deane
    Jun 10, 2021 at 16:29

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