Commutativity of tangent projection map and inner product

Question

I stumbled into this simple property, that I can't find a proof of, although I verified it holds in a number of cases.

Let $\mathbb{M}$ be a smooth manifold embedded into an ambient space $\mathbb{A}$ and let $\Pi_x:\mathbb{A}\to T_x\mathbb{M}$ denote an orthogonal projection to the tangent space $T_x\mathbb{M}$, namely $\Pi_x(a)\in T_x\mathbb{M}$ for every $a\in\mathbb{A}$. Furthermore, let $\langle\cdot.\cdot\rangle^\mathbb{A}$ denote an inner product in $\mathbb{A}$.

Now, take a curve $\gamma_{x,v}:[-\epsilon,\ \epsilon]\to\mathbb{M}$ such that $\gamma_{x,v}(0)=x\in\mathbb{M}$ and $\dot{\gamma}_{x,v}(0)=v\in T_x\mathbb{M}$ and consider the function $\Pi_{\gamma_{x,v}(t)}(a)$ for a given $a\in\mathbb{A}$. Define the following map:

\begin{equation} \Pi_x^\bullet(a,v):=\left.\frac{\mathrm{d}}{\mathrm{d}t}\Pi_{\gamma_{x,v}(t)}(a)\right|_{t=0}. \end{equation}

The property I was referring to is the following: Given $x\in\mathbb{M}$ and $a\in\mathbb{A}$, for every pair $v,w\in T_x\mathbb{M}$ it holds that

\begin{equation} \langle\Pi_x^\bullet(a,v),w\rangle^\mathbb{A} = \langle\Pi_x^\bullet(a,w),v\rangle^\mathbb{A}. \end{equation}

Does anyone know why is that so ? Thanks in advance!

Answer 1

This is not an answer to your original question. I find the map you define, $\Pi_x$, to be awkward to work with. For me the easiest way to prove what you found is to use an orthonormal moving frame.

One of your comments indicates that you are trying to use $\Pi_x$ to define the Hessian of a function on the submanifold $M$. You're right that something like that should work. I show below how to define the Hessian of a function first on a submanifold of Euclidean space and then on a Riemannian manifold in a simple way without any mention of the Levi-Civita connection and Christoffel symbols.

$\newcommand\R{\mathbb{R}}$For convenience, let $\mathbb{A} = \R^n$ and $\langle \cdot,\cdot\rangle_{\mathbb{A}}$ be the standard dot product on $\R^n$.

Consider an $m$-dimensional submanifold, a smooth function $f: M \rightarrow \R$. Suppose we want to define the Hessian of $f$ at $x$. Since the Hessian is geometrically invariant, we can move $M$ so that $x$ is the origin and $T_xM$ is the $m$-plane $$ T = \{x^{m+1} = \cdots = x^n = 0 \}$$ Near the origin, $M$ is the graph of a function $\phi: T \rightarrow T^\perp$, where $\phi(0) = 0$ and $d\phi(0) = 0$. In other words, the map $\Phi: T \rightarrow \R^n$ given by $$ \Phi(x) = (x,\phi(x)) $$ is the inclusion map of a neighborhood of $0 \in M$. The Hessian of $f$ at $x$ is $$ \nabla^2_{vw}f(0) = v^jw^k\partial^2_{jk}(f\circ\Phi)(0), $$ where $v = (v^1, \dots, v^m)$ and $w=(w^1, \dots, w^m)$. This definition makes it obvious that the Hessian is symmetric. It is easy to show that it is a symmetric bilinear tensor. It's also worth pointing out that the second fundamental form of $M$ at $0$ is simply the Hessian of $\phi$ at $0$.

The Hessian of a function on a Riemannian manifold is just as easy. Given a point $x_0 \in M$, it is an elementary calculation to show that in a neighborhood of $x_0 \in M$, there are local coordinates such that the metric $g$ satisfies $$ g_{ij}(x_0) = \delta_{ij}\text{ and }\partial_kg_{ij}(x_0) = 0. $$ In every exposition I've ever seen, this is proved using exponential coordinates, but this is a huge overkill. It is a simple exercise in calculus, and a proof can be found here. The Hessian of $f$ at $x_0$ is now given by $$ \nabla^2_{vw}f(x_0) = v^jw^k\partial^2_{jk}f(x_0), $$ where the partial derivatives are with respect to the local coordinates obtained above. It is easily proved that this is a well-defined symmetric bilinear tensor and equal to the standard definition. Again, symmetry is automatic.