Natural connection on tautological bundle over real Grassmannian Let me get to the point immediately:

Is there a natural connection on the tautological vector bundle over a Grassmannian (of a real vector space equipped with an inner product)?

In a paper I'm reading there is a smooth 1-parameter family of $k$-dimensional subspaces $W_t$ of a vector space $V$. From this, we need to get a 1-parameter family of injections $\phi_t : W_0 \to V$ such that the image of $\phi_t$ is $W_t$, and such that $\phi_0$ is the inclusion of $W_0$.
In other words, we need to identify all the $W_t$ with the initial subspace $W_0$.
I have no trouble setting up such an identification (in an explicit way using projections and charts on Grassmannians, and small steps of the parameter $t$), but I'm looking for a natural way, mostly because I want to do these things for vector bundles later). I figured one way would be to have a connection on the tautological vector bundle of $k$-planes in $V$ and identify $W_0$ with $W_t$ using parallel transport along the path $\phi_t$.
There is an inner product on $V$.
 A: Let $G=O(V)$ be the group of orthogonal transformations of $V$, and let $E$ denote the tautological bundle on $M=G(k,V)$. The canonical $G$-invariant connection on $E$ is given exactly as one gives the induced connection on a $k$-dimensional submanifold of $V$: We view a section $s$ of $E$ as a $V$-valued function, differentiate it, and, using the inner product structure, project the result back to $E$. That is,
$$(\nabla_\xi s)(p) = \operatorname{proj}_{E_p} ds(p)(\xi) = \operatorname{proj}_{E_p}\xi_p(s).$$
In terms of the moving frame set-up, if $e_1,\dots,e_k,e_{k+1},\dots,e_n$ is an adapted (local) orthonormal frame field, with $e_1,\dots,e_k$ a basis for $E$ and $e_{k+1},\dots,e_n$ a basis for $E^\perp$, we define an $n\times n$ matrix of $1$-forms by $de_i = \sum\limits_{j=1}^n \omega_{ij}e_j$ (once again viewing $e_i$ as a $V$-valued function). The canonical connection on $E$ is given by taking the upper-left block of this matrix: For $1\le\alpha,\beta\le k$, we define $$\nabla e_\alpha = \sum_{\beta=1}^k \omega_{\alpha\beta}e_\beta,$$ and $\big[\omega_{\alpha\beta}\big]$ is the connection matrix. (And we get curvature from the matrix equation $\Omega = d\omega - \omega\wedge\omega$.)
