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If $X$ is a compact smooth manifold, $K^{0}(X)$ can be defined as the algebraic $K_{0}$-group of $C^{\infty}(X)$. In order to do that we use the following equivalence relation: we say that two idempotents $p$ and $q$ in $M_{\infty}(C^{\infty}(X))$ (the square matrices of arbitrary size over $C^{\infty}(X)$) are equivalent if they are similar, that is, if there exists an invertible $u$ in $M_{\infty}(C^{\infty}(X))$ such that $p=uqu^{-1}$.

My question is very naive: what if we change this equivalence relation by requiring that in addition to $p=uqu^{-1}$ we have $dp=u\cdot dq\cdot u^{-1}$ ? (Here $d$ is the exterior differential.) Is such a definition known? How could this new equivalence relation be described in terms of vector bundles (possibly endowed with some additional structure)? Could it be that what we get is somehow related to differential K-theory?

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I don't even think the formula makes sense. – Fernando Muro Dec 19 '11 at 22:23
Is dp the matrix whose components are d(pij)? In that case it seems to make sense in a certain module over $C^\infty(X)$. – Amin Dec 19 '11 at 22:42
Yes, this is certainly what $dp$ means. The multiplication of matrix valued differential forms is given by the usual combination of exterior multiplication and matrix multiplication. – Svetoslav Zahariev Dec 21 '11 at 21:30
It is true that one can define a connection on $\mathrm{im}(p)$ by $\nabla s=p(ds)p$. But the condition above does not seem to guarantee that this connection is preserved under equivalences. – Sebastian Goette Dec 2 '15 at 8:40

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