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I recently came across a result in a text in the field of differential geometry and I'm wondering why it is true. Let $M\subset\mathbb{R}^{n+m}$ be an $n$-dimensional submanifold and $w:M\longrightarrow\mathbb{R},\,w(x)=u(x)-\langle x,z\rangle$, where $u:M\longrightarrow\mathbb{R}$, $z\in\mathbb{R}^{n+m}$ and $\langle\cdot,\cdot\rangle$ denotes the canonical inner product on $\mathbb{R}^{n+m}$. The author states that we have $D_{M}^{2}w(x)=D_{M}^{2}u(x)-\langle II_{x}(\cdot,\cdot),z\rangle$, where $II_{x}$ is the second fundamental form at the point $x$ and $x$ being a critical point of $w$. How is the exact computation to get this result?

What I tried is to first compute the gradient w.r.t $M$ of $w$: $\nabla^{M}w(x)=\nabla^{M}u(x)-z^{tan}$, where $z^{tan}$ is the tangential component of $z$. If this is correct (is it?), I don't know how to get the Hessian of $w$ at $x$. Thanks in advance!

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The derivative operator on an embedded submanifold is just the projection of the ambient space's derivative operator, so: $$ \nabla^M w(x) = \nabla^M (u - \langle{}\cdot, z\rangle{})(x) = \nabla^M u(x) - \text{proj}_{T_xM} z = \nabla^M u - z^{\text{tan}}, $$ like you wrote. This is because the derivative of $x \in \mathbb{R}^d \mapsto \langle{}x, z\rangle{}$ is $z$. The Hessian of a function $u : M \to \mathbb{R}$ is defined as the linear map $H(u)_x : T_x M \to T_x M$ defined by $$ H(u)_x(v) = \nabla_v^M(\nabla^M u)|_x, \ \ \ \ \forall v \in T_x M. $$ The $D^2_M u$ that you write above, considered as a bilinear form, is obtained by just setting $D^2_M u(v, y) = \langle{}H(u)_x(v), y\rangle{}$. We'll take this dot product at the end.

Well, we computed $\nabla^M w$ above, and so let's compute $\nabla^M_v \nabla^M w$: $$ \nabla^M_v \nabla^M w|_x = \text{proj}_{T_x M}(\nabla_v^{\mathbb{R}^d} \nabla^M w) = \text{proj}_{T_x M} \nabla^{\mathbb{R}^d}_v (\nabla^M u - z^\text{tan}) = H(u)_x(v) - \text{proj}_{T_x M}\nabla^{\mathbb{R}^d}_v z^\text{tan}. $$ Now, letting $\nu$ be a unit normal vector field with respect to which we're defining the second fundamental form, $z^\text{tan}$ is just $$ z^\text{tan} = z - \langle{}\nu, z\rangle{}\nu $$ and thus $$ \nabla^{\mathbb{R}^d}_v z^\text{tan} = -\langle{} \nabla^{\mathbb{R}^d}_v \nu, z\rangle{}\nu - \langle{}\nu, z\rangle{}\nabla^{\mathbb{R}^d}_v \nu. $$ Projecting this to $T_x M$ gives $\nabla^M_v$ (and note the first term vanishes): $$ \nabla^M_v z^\text{tan} = - \langle{}\nu, z\rangle{}\text{proj}_{T_x M}(\nabla^{\mathbb{R}^d}_v \nu). $$ Taking an inner product with an arbitrary vector $y \in T_x M$ gives $$ H(w)_x(v)\cdot y = -\langle{}\nu, z\rangle{}\langle{}\nabla^{\mathbb{R}^d}_v \nu, y\rangle{} = \langle{}II_x(v,y) , \nu\rangle{} = \langle{}II_x(v, y), z\rangle. $$ The second equality follows from Weingarten's formula (relating the second fundamental form to the shape operator) and the last from the fact that $II_x(v, y)$ is perpendicular to $M$, thus in dot producting against $z$ we just pick up a perpendicular component, i.e. $\langle{}\nu, z\rangle{}\nu$.

Putting it all together gives that $D^2_M w(x) = D^2_M u(x) - \langle{}II_x(\cdot, \cdot), z\rangle{}$. The minus here comes from the fact that way above, remember we're subtracting off $\text{proj}\nabla z^\text{tan}$.

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  • $\begingroup$ Nice answer, thank you! $\endgroup$ Commented Jul 29, 2020 at 6:31

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