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I know that if $E$ and $F$ are two vector bundle with connection $\nabla^E$ and $\nabla^F$, then it is natural to define tensor connection $\nabla = \nabla^E\otimes 1 + 1 \otimes \nabla^F$ on $E \otimes F$.

By I have a stupid question: is is $\nabla^E \otimes 1$ also a connection?

It seems to satisfy the conditions: $\nabla(f\sigma_E\otimes \sigma_F) = \nabla^E(f\sigma_E)\otimes\sigma_F = (\nabla f) \sigma_E \otimes \sigma_F +f \nabla^E\sigma_E\otimes \sigma_F = \nabla f \sigma_E \otimes \sigma_F + f\nabla(\sigma_E \otimes \sigma_F)$.

I guess the reason is the above condition is true only if I put $f$ with $\sigma_E$; and this ambiguity is not desired. But I'm not sure.

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I think $0$ is a connection. So it looks like you are correct. –  Stephen Montgomery-Smith Mar 31 at 3:37
    
@Stephen: No, $0$ is not a connection. The Leibniz rule has to hold for multiplication by scalar functions. –  Ted Shifrin Mar 31 at 3:58
    
@TedShifrin Yes, I missed that. So in that case, the OP is probably wrong. –  Stephen Montgomery-Smith Mar 31 at 4:06

2 Answers 2

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It can't be, because, on the one hand, for any vector field $X \in \Gamma(TX)$, you would have $$ (\nabla \otimes I)_X(f(\sigma_E \otimes \sigma_F)) = (\nabla \otimes I)_X((f \sigma_E) \otimes \sigma_F) = \nabla_X(f\sigma_E) \otimes \sigma_F = (X(f)\sigma_E + f\nabla_X\sigma_E) \otimes \sigma_F\\ = X(f)(\sigma_E \otimes \sigma_F) + f(\nabla_X\sigma_E \otimes \sigma_F), $$ whilst on the other, $$ (\nabla \otimes I)_X(f(\sigma_E \otimes \sigma_F)) = (\nabla \otimes I)_X(\sigma_E \otimes (f\sigma_F)) = \nabla_x\sigma_E \otimes(f \sigma_F) = f(\nabla_X \sigma_E \otimes \sigma_F). $$ You really do need an honest connection on $F$ to get a connection on $E \otimes F$. Indeed, even if $F$ is trivial, you need to take at least the flat connection $d$.

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Only if $F$ is a trivial bundle. Note that $f(\sigma_E\otimes\sigma_F)=\sigma_E\otimes(f\sigma_F)$ as well!

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