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Let $X$ be a compact Hausdorff space.Put $Vec(X)$, the space of all real (or complex) vector bundles over $X$.We put also $Vec_g(X)$, the space of all Riemannian vector bundles over $X$, that is the space of all vector bundles equiped with a continuous fiber wise inner product. By imitation of three equivalent relations defined on $Vec(X)$ in the theory of topological $K$-theory, we define a metric version of the same equivalent relations as follows:

1)Two Riemannian vector bundles $(E,g)$ and $(F,h)$ are equivalent if there is isometric bundle isomorphism between them.

2)Two Riemannian vector bundles $(E,g)$ and $(F,h)$ are equivalent if there is a trivial bundle $\epsilon_k$ with the obvious metric structure such that $E\oplus \epsilon_k$ is equivalent to $F \oplus \epsilon_k$ in the sense of 1) where each direct sum bundle is equipped with direct sum.metric

3)Two Riemannian vector bundles $(E,g)$ and $(F,h)$ are equivalent if there are two trivial bundles $\epsilon_m$ and $\epsilon_n$ with obvious metrics such that $E\oplus \epsilon_m$ is equivalent to $F \oplus \epsilon_n$ in the sense of 1) where each direct sum bundle is equipped with direct sum.metric.

Does this idea produce a new kind of $K$_theory? Does it introduce a functor from the category of compact Hausdorff space to the category of Groups? Can we extend this functor from the category of commutative unital $C^*$ algebras to the category of non commutative unital $C^*$ algebras.

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    $\begingroup$ Two Riemannian vector bundles are (1)-equivalent iff they are isomorphic as vector bundles, so you can get nothing new from this. One way of thinking of this is the fact that a Riemannian vector bundle is classified by a map to $BO(n)$, while a vector bundle is classified by a map to $BGL(n)$. Because the inclusion $O(n) \to GL(n)$ is a homotopy equivalence the induced map on classifying spaces is too. $\endgroup$
    – mme
    Commented Aug 17, 2020 at 20:19
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    $\begingroup$ Said another way, the space of Riemannian metrics on a given vector bundle is convex, hence contractible. So the natural forgetful map from Riemannian vector bundles to vector bundles has contractible fibers. $\endgroup$
    – Qiaochu Yuan
    Commented Aug 17, 2020 at 21:32

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