Defining a Riemannian metric

Question

Let $M$ be a smooth manifold. I have seen a Riemannian metric be defined in many ways:

• A smooth choice of an inner product $g_p:T_pM\times T_pM\to\mathbb{R}$ which is symmetric and positive-definite, at each point $p\in M$.

• An element of $\Gamma(T^\ast M\otimes T^\ast M)$ (which is positive-definite, symmetric)

• An element of $\Gamma(\mathrm{Hom}(TM\otimes TM,\mathbb{R}))$ (positive-definite, symmetric)

Are these all equivalent? The reason why I wonder is that if $TM$ is not parallelisable, then $\Gamma(TM)$ is not a free module, so we don't have the isomorphism $\Gamma(T^\ast M)\otimes\Gamma(T^\ast M))\cong\mathrm{Hom}(\Gamma(TM)\otimes\Gamma(TM),C^\infty(M))$.

EDIT

Can we equivalently define a Riemannian metric as an element of $\Gamma(T^\ast M)\otimes\Gamma(T^\ast M)$, or is this just utterly wrong?

Answer 1

A section of the bundles involved in the second and the third definition can be zero, and not define a metric.

After the change in your question, you can analyze the situation in a chart, (since metric are defined on charts then glued with partition of the unity) that is in an open subset $U$ of $R^n$, since the tangent bundle of an open subset of $R^n$ is trivial, consider the trivialization $(e_1,...,e_n)$

A metric $\langle,\rangle$ defined on $U$, induce a bilinear map on $TU$ of $b(e_i(x),e_j(x))=\langle e_i(x),e_j(x)\rangle$. The dual of $(e_1,...,e_n)$ defines a trivialization of $TM^*$ and to the metric $\langle,\rangle$ defined on $U$, you can associate the elements of $TM^*\otimes TM^*$ defined by $\sum c_{ij}(x)e_i^*(x)\otimes e_j^*(x)$ where $c_{ij}(x)=\langle e_i(x),e_j(x)\rangle$.

Answer 2

A section of a bundle $E$ (with various properties) is a fancier way of referring to a "smoothly varying" choice of $s_p\in E_p$ (with the same properties) as $p$ varies over $M$. So (1) and (2) are identical. With regard to (2) and (3), we're just using the isomorphism (truly a definition) $\text{Hom}(E,\Bbb R) = E^*$ (where here $E=TM\otimes TM$).

Notice that we're never trying to say that $\Gamma(E\otimes F) \cong \Gamma(E)\otimes\Gamma(F)$.