I just started studying Riemannian geometry, so excuse me if I don't have all the necessary tools yet.
Let $M$ be a finite dimensional smooth Riemannian manifold and let us denote as $\langle \cdot,\cdot\rangle_x:T_xM\times T_xM\to\mathbb{R}$ the Riemannian metric. Suppose that $\langle \cdot,\cdot\rangle _x':T_xM\times T_xM\to\mathbb{R}$ is another Riemmannian metric for the same manifold. We know that for each $x\in M$ there exists a symmetric positive definite operator $P_x:T_xM\to T_xM$ such that $\langle u, v\rangle _x' = \langle u, P_x v\rangle _x$.
Can I choose such a $P_x$ smoothly, that is, as a smooth map $x\mapsto P_x$ from $M$ to some other manifold? What would the codomain of this map be? I have the feeling that the codomain of this map must be the vector bundle of the endomorphisms of the tangent (?), but I could not even find a definition of this object.
If I can choose $P_x$ smoothly, is the map $x\mapsto||P_x||_x$ smooth or at least continuous? Here, by $||P_x||_x$ I mean the operator norm induced by $\langle \cdot,\cdot\rangle_x$.
Best Answer
I'll use $(\cdot, \cdot)$ to denote the dot product on $\mathbb{R}^n$. In coordinates, for $u, v \in \mathbb{R}^n$, $n = \dim(M)$, $$\langle u, v \rangle_{x} = (u, G(x)v)$$ and $$\langle u, v \rangle'_{x} = (u, G'(x)v).$$ Here $G_{ij}(x) = \langle e_i, e_j \rangle_x$ is the matrix representation of the metric tensor. Thus $\langle u, v \rangle'_{x} = \langle u, P_x v \rangle_x$ for all $u,v$ requires $$(u, G'(x)v) = (u, G(x)P(x)v).$$ Thus $P(x) = G(x)^{-1}G'(x)$. Since both $G$ and $G'$ are smooth, $P$ is smooth.
For each $x \in M$, $P_x \in L(T_xM)$, and $P_x$ varies smoothly with $x$. Hence $P$ is a smooth map from $M$ to $\bigsqcup_{x \in M}L(T_xM) = \bigcup_{x \in M}(\{x\} \times L(T_xM))$
You ask about the smoothness of the map $f(x) = \|P_x\|_{x}$. In coordinates, if $A$ is a constant $n \times n$ matrix, \begin{align} \|A\|_{x}^2 &= \sup_{(u, G(x)u) = 1}\|A u\|_{x}^2 \\ &= \sup_{(u, G(x)u) = 1}(Au,G(x)Au). \end{align} Now let $v = G(x)^{1/2}u$ to get \begin{align} &= \sup_{(u, G(x)u) = 1}(Au,G(x)Au) \\ &= \sup_{(v, v) = 1}(AG(x)^{-1/2}v, G(x)AG(x)^{-1/2}v) \\ &= \sup_{(v, v) = 1}(G(x)^{1/2}AG(x)^{-1/2}v, G(x)^{1/2}AG(x)^{-1/2}v) \\ &= \|G(x)^{1/2}AG(x)^{-1/2}\|^2. \end{align} Hence $$f(x) = \|G(x)^{1/2}P(x)G(x)^{-1/2}\|.$$ Now this brings us to analysis of smoothness of the usual operator norm on $L(\mathbb{R}^n)$. Using $A(t) = \text{diag}(1 + t, 1 - t)$, we have $\|A(t)\| = \max(1 + t, 1 - t) = 1 + |t|$, which shows that the operator norm is only continuous, even when restricted to positive definite matrices. Thus our function $f$ is continuous, but generally not differentiable.