smooth connection on exterior power Suppose $E$ is a smooth vector bundle with a smooth connection $\nabla$. Then it induces a smooth connection on tensor powers of $E$. Does it also induce smooth connection on say exterior powers of $E$.
thanks.
 A: If two vector bundles $V_1$ and $V_2$ have connections $\nabla_1$ and $\nabla_2$, then the vector bundle $V=V_1 \otimes V_2$ has a unique connection $\nabla$ for which $\nabla(s \otimes t)=(\nabla_1 s)\otimes t + s \otimes \nabla_2 t$. In particular, taking $V_1=V_2$, we get a connection on both the symmetric and the exterior squares, and this generalizes to all tensor powers and exterior and symmetric powers.
A: Yes, connections can be transported along any functor $F$ from vector spaces to vector spaces.  Exterior powers are standard examples of such functors.
Say $E$ is a vector bundle on $M$.  The connection $\nabla$ is equivalent to an isomorphism $p_1^*E \to p_2^*E$ (where $p_1, p_2: M^{(2)} \to M$ are the projections from the first infinitesimal neighborhood of the diagonal) that restricts to the identity on the diagonal.  The functor $F$ induces a corresponding functor on vector bundles over $M^{(2)}$, and hence a corresponding connection.
The equivalence is explained in the beginning of section 2 (page 5) of Osserman's notes for schemes, but it holds for anything smooth (e.g., manifolds as ringed spaces).
