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Let $M$ be a smooth manifold. Choosing a metric tensor $g$ on $TM$ one gets a vector bundle isomorphism $g^\flat\colon TM\to T^*M$. Changing $g$ we get different maps $g^\flat$.

I would like to understand the statement and its consequences for algebraic topology:

Every two metric choices are homotopic.

My interpretation would be that the metric tensor is a (positive-definite) section $g\colon M\to \mathrm{Sym}^2(T^*M)$ and every two of these are homotopic (through positive-definite sections) by $$H_t(x) = t g_1(x) + (1-t)g_2(x)$$

Similarily if I treat $g^\flat$ as a section of the bundle $\mathrm{Hom}(TM, T^*M)$ which is an isomorphism on each fiber, the above homotopy yields a homotopy (through isomorphisms as well) between any $g_1^\flat$ and $g_2^\flat$.

This would mean that from the point of view of algebraic topology, the equivalence between $TM$ and $T^*M$ is "canonical" (induced isomorphisms on (co)homology and similar concepts do not depend on the metric chosen).

Have I got this right?

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  • $\begingroup$ "Canonical" would mean there is a preferred choice of metric but there is not, what we say instead is that we have a "contractible choice" of metric. There always exists a metric (over paracompact spaces) and the space of all metrics is contractible, but it's still important to bear in mind that we've made a choice. This distinction is especially important when studying subspaces of metrics with particular properties which are not homotopy-invariant (e.g. positive scalar curvature metrics). $\endgroup$ – William Apr 20 '20 at 12:59
  • $\begingroup$ In some contexts like classifying spaces there is not a canonical model of $BG$ but there is a canonical homotopy type and since all the constructions we do starting with $BG$ are homotopical in nature that's "good enough" and we can pretend like there is "the" $BG$, but in geometry the constructions that arise from a metric aren't always homotopy invariant, like curvature for example. (Also some topologists might still be pedantic enough to demand you say "a $BG$" and only talk about the uniqueness of its homotopy type.) $\endgroup$ – William Apr 20 '20 at 13:09
  • $\begingroup$ (Sorry for all the comments, I think I forgot to address your actual question.) As for wether the isomorphism $TM\cong T^*M$ is considered "canonical" I would personally say no. As you said, it depends on a choice of metric, which I don't consider canonical. For me it's analogous to vector space bases: you can always choose a basis for a vector space $V^n$ (even though the choice isn't contractible) and this induces an isomorphism $V \cong \mathbb{R}^n$, but I would not say "$V$ is canonically isomorphic to $\mathbb{R}^n$" because everything is relative to the basis. $\endgroup$ – William Apr 20 '20 at 13:18
  • $\begingroup$ @William Thanks for your comments. An isomorphism $f\colon TX\to T^*X$ will induce a map $f^*$ between cohomology and if $g$ is homotopic is to $f$, then $g^*=f^*$. Choosing any isomorphism $f$ induced from some metric seem to generate the same $f^*$? I understand that for a differential geometer these identifications are totally different – but if someone is interested only in functors that factor through $\textbf{hTop}$, the choice does not matter, does it? $\endgroup$ – Paweł Czyż Apr 20 '20 at 13:24
  • $\begingroup$ Which cohomology theory do you mean? I'm used to maps between cohomology being induced by maps between spaces $f\colon X \to Y$, rather than an isomorphism $TX \to T^*X$. But I think you are right, if you are only concerned about functors that factor through hTop, then choosing a different metric will give you the same functor, unless I'm overlooking something. In that case I would call the functor canonical rather than the choices you had to make to produce it (like how there is a canonical homotopy type for $BG$). $\endgroup$ – William Apr 20 '20 at 13:36
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Being continuously isomorphic is stronger than being homotopy equivalent. The metric, as you say, provides a continuous isomorphism. In this sense, homotopy invariants are 'canonically' the same.

Notice that this is not the same as saying that the map induced on a homotopy invariant by two continuous isomorphisms are the same, but if they differ by a choice of the metric then, as you write, they are indeed the same.

However, this is not the only isomorphism possible between the tangent and the cotangent bundle of the manifold. Suppose your manifold admits a non degenerate closed two form $\omega:M\to \Omega^{2}(M)$, i.e. it is symplectic. Then there is an isomorphism between the tangent and the cotangent bundle induced via $v\mapsto \omega(v,-).$

This is not even a section of the same bundle and as you can check, two choices of the symplectic form are homotopic. Thus they give the same isomorphism on homotopy invariants, but in general different from the one induced by the metric.

Edit: Let me also mention a fact, which escaped me when I originally wrote the answer, which might also be helpful.

In general vector bundles over a manifold (more generally CW complexes) have the homotopy type of the base space. There are many ways to see this, the quickest being that if $p: E\to M$ is a vector bundle with fibre $V$, then the fibration

$$V \to E\xrightarrow{p} M $$

has contractible fiber and so $p_*:\pi_{i}(E)\to \pi_i(M)$ is an isomorphism for all $i>0$ which then implies that $E\simeq M$ as both $E$ and $M$ have the homotopy type of a CW complex.

So the homotopy type of a vector bundle is not that interesting of a question. There are more interesting questions one can ask, such as classifying vector bundles up to isomorphisms which takes us to the realm of topological $K$-theory, which is an interesting topic in and of itself.

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    $\begingroup$ Thank you for the edit! Indeed, this question was mainly motivated by the "canonical" identification of $K^0(TX)\simeq K^0(T^*X)$ (in the context of studying pseudo-differential operators on a manifold). $\endgroup$ – Paweł Czyż 20 hours ago
  • $\begingroup$ Yes, indeed, $K^0(-)$ is a topological invariant $\endgroup$ – S.S. 19 hours ago

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