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I have recently been introduced to the Tensor bundle and to Riemannian metrics. The tensor bundle is : $$\mathcal{T}^{k,\ell}M=\coprod_{p\in M}(T_pM)^{\otimes k}\otimes(T_p^\ast M)^{\otimes \ell}=\bigcup_{p\in M}\{p\}\times(T_pM)^{\otimes k}\otimes(T_p^\ast M)^{\otimes \ell}.$$

The special case we are interested in is $\mathcal{T}^{0,2}M$, and particularly its sections $g\in\Gamma(\mathcal{T}^{0,2}M)$. Now, to define a Riemannian metric, we need additional hypothesis on $g$. This is due to the following two observations.

One the one hand, from universal property of the tenor product, a bilinear map $B:E\times F\to\mathbb{R}$ corresponds uniquely to a linear map $\tilde{B}:E\otimes F\to\mathbb{R}$ :

That is, $\mathcal{B}(E,F;\mathbb{R})\cong(E\otimes F)^\vee$.

On the other hand, the following linear map (defined on pure tensors and extended by linearity, using this principle twice here) is an isomorphism : $$\begin{matrix}\Phi&:&E^\vee\otimes F^\vee&\to&(E\otimes F)^\vee\\&&\varphi\otimes\psi&\mapsto&[u\otimes v\mapsto\varphi(u)\psi(v)]\end{matrix}$$

Therefore, we have finally, in our case : $$\mathcal{B}(T_pM,T_pM;\mathbb{R})\cong T_p^{0,2}M.$$

Thus, a section $g\in\Gamma(\mathcal{T}^{0,2}M)$ is a collection $g:M\to\mathcal{B}(T_\bullet M,T_\bullet M;\mathbb{R})$ of bilinear forms on the tangent spaces, such that this collections varies regularly in the parameter (since we have constructed a topology on $\mathcal{T}^{0,2}M$). A Riemannian metric is therefore a collection of scalar products that fits the previous requirements. My question is the following :

How to interpret positiveness and definiteness of a bilinear form $B:E\times E\to\mathbb{R}$ in terms of its associated element in $E^\vee\otimes E^\vee$ ? That is, in our case, how to define a Riemannian metric only in terms of the tensor bundle $\mathcal{T}^{0,2}M$ ?

P.S. : I know most of the "tensor bundle" thing is unnecessary to answer the main question, that is about interpreting positiveness and definiteness of a bilinear form in terms of its representation in the tensor product of duals. However, I still wanted to include this discussion as a motivation for my question and as an entry point to some more references I couldn't find by searching.

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The construction $g_p:T_pM\times T_pM\to\mathbb R$ is defined as $$g_p(X,Y)={g_p}_{\mu\nu}X^{\mu}Y^{\nu},$$ where $X=X^{\mu}\partial_{\mu}$, $Y=Y^{\nu}\partial_{\nu}$ and $\partial_i$ are the coordinate basis for $T_pM$. The metric tensor $g_p$ has components ${g_p}_{\mu\nu}$ which are functions of the coordinates.

At a given point, ${g_p}_{\mu\nu}X^{\mu}Y^{\nu}$ is a scalar quadratic form for which on checks positive definiteness as usually one does is our linear algebra courses.

I would stress that $g$ is a section of the bundle $T^{(0,2)}M$ which assign to each $p$ in $M$ a tensor $g_p$ an element of $T_pM\otimes T_pM$, which is a bilinear map $T_pM\times T_pM\to\mathbb R$.

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  • $\begingroup$ I didn't think of writing it in coordinates... Thanks ! $\endgroup$
    – Anthony
    Sep 21, 2020 at 22:19
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    $\begingroup$ in the end, after the concepts, if you are going to do explicit calculations, you use coordinates, right? Thanks for the rewards! $\endgroup$
    – janmarqz
    Sep 21, 2020 at 23:39
  • $\begingroup$ To pull back the metrics for instance ? I guess so then, indeed, so that makes me feel kind of stupid... $\endgroup$
    – Anthony
    Sep 21, 2020 at 23:40
  • $\begingroup$ Maybe you mean $T^{(2,0)}M$ instead of $T^{(0,2)}M$? $\endgroup$
    – kid111
    Mar 29, 2022 at 20:43

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