Change of Riemannian metric as a smooth map

Question

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 $\langle \cdot,\cdot\rangle_x$.

Answer 1

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.

Answer 2

The right language to define these objects is provided by vector bundles. A vector bundle over a smooth manifold $M$ is a smooth manifold $E$ with a surjective map $$ \pi : E \to M $$ which is locally trivial at any point. You can check the definition in any differential geometry book. The picture to keep in mind is the trivial bundle $$ M \times \mathbb{R}^n \to M $$ where the map is the projection on the first component. Any vector bundle over $M$ is locally of this form. The fibre over a point is $$ E_x = \pi^{-1}(x). $$ Maps of vector bundles are smooth maps which restrict to linear maps of the corresponding fibres.

Going back to your question, what you need is to define the tangent bundle of $M$ as the vector bundle which assigns to any $x \in M$ the tangent space $T_xM$. On vector bundles you can naturally define linear constructions as the dual bundle and the tensor product construction, thus you can easily define the bundle $$ End(TM) $$ as the bundle whose fibre over $x \in X$ is the space of endomorphism of $T_xM$. So it's not difficult to see that your $P$ must be a section of the bundle, in particular it is a section of the sub-bundle of symmetric endomorphisms. So a smooth map $$ M \to Sym(TM) $$ which assigns to any $x \in X$ a symmetric endomorphism of $T_xM$.