# Hessian form of a real valued function on a submanifold of $\mathbb{R}^{n+m}$

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!

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 $$\text{proj}\nabla z^\text{tan}$$.