# What does “assigning a Riemannian metric in a differentiable fashion” mean?

In the book of Frankel, The Geometry of Physics, at page 45, he states that

A Riemannian metric on a manifold $$M^n$$ assigns, in a differentiable fashion, a positive definite inner product $$⟨, ⟩$$ in each tangent space $$M_p^n$$ .

However, what does he mean by "in a differentiable fashion" ? I know what is differentiable manifold, but I don't understand how can the assignment of a metric tensor be differentiable ?

• Around each $p$, you have coordinates, so locally a metric tensor can be thought of as assigning matrices. Then you want the matrix entries to be differentiable functions (usually we want the metric to be more than just differentiable, e.g., $C^k$). – user10354138 May 21 at 10:20
• @user10354138 Thanks for your comment; then how do we formulate that condition mathematically in a more appropriate manner ? – onurcanbektas May 21 at 10:21
• Do you know the concept of vector bundles over manifolds? – Paul Frost May 21 at 10:39
• @PaulFrost Unfortunately, but I do know tangent bundles over manifolds. – onurcanbektas May 21 at 10:42
• The coordinate-free way is to define $\langle,\rangle_p\colon T_pM\times T_pM\to\mathbb{R}$ for each $p$ (actually it is a section of $T^*M\otimes T^*M$, if you know that) such that $p\mapsto\langle X(p), Y(p)\rangle_p$ is a differentiable function on $M$ for all differentiable vector fields $X,Y$ (similarly $C^k$, smooth, analytic, etc.). You also need to impose the positive definite and symmetric that you see in inner product. – user10354138 May 21 at 10:52