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For a (smooth) vector bundle $F$ over a manifold $M$, one normally defines a connection to be a linear map $$ \nabla:\Gamma^{\infty}(V) \to \Omega^1(M) \otimes \Gamma^{\infty}(V), $$ satisfying $\nabla(fv) = $d$f \otimes v + f \nabla(v)$. One can then extend this to a map $$ \nabla:\Omega^k(M) \otimes_{C^{\infty}(M)} \Gamma^{\infty}(V) \to \Omega^{k+1}(M) \otimes_{C^{\infty}(M)} \Gamma^{\infty}(V) $$ by defining $$ \nabla(\omega \otimes v) = \text{d}\omega \otimes v + (-1)^k\omega \wedge \nabla(v). $$ The factor of $(-1)^k$ ensures that the definition is well-defined over the tensor product.

Alternatively, one can use the equivalent definition of a connection as a linear map $$ \nabla:\Gamma^{\infty}(V) \to \Gamma^{\infty}(V) \otimes \Omega^1(M), $$ satisfying $\nabla(vf) = v \otimes $d$f + \nabla(v)f$. However, this has the simpler extension to a map $$ \nabla:\Gamma^{\infty}(V) \otimes_{C^{\infty}(M)} \Omega^k(M) \to \Gamma^{\infty}(V) \otimes_{C^{\infty}(M)}\Omega^{k+1}(M) $$ defined by $$ \nabla(v \otimes \omega) = v \otimes \text{d}\omega + \nabla(v) \wedge \omega. $$ I.E. there is no $(-1)^k$ factor. My question is why isn't this simpler formulation the one that is normally used?

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Probably because people often skip parentheses, whereupon $\nabla \nu \wedge \omega$ is ambiguous as opposed to $\omega \wedge \nabla \nu$ which is not. –  Loop Space Feb 7 '13 at 18:18
    
The way I survived, heuristically, so far, is as follows. Notice that the factors in the tensor products are permuted in the two approaches. If from the second approach, you want to get to the first, you have to "pass" the $\omega$ (on the right) above the $\nabla(v)$ (which is on the left in the 2nd approach). This second factor is a bundle valuated one form, ie locally a one form tensor a section of $V$, hence you reobtain the $(-1)^k$. It's a bit sketchy but I've been ok with it. –  Amin Feb 7 '13 at 21:37
    
Oups sorry I just looked at A. Stacey's comment; sorry if this is what it was saying... –  Amin Feb 7 '13 at 21:39
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I am inclined to disagree with your claim that begins with "one normally defines". It is a convention that depends on the author, and varies across the literature. –  S. Carnahan Feb 8 '13 at 0:32

1 Answer 1

The "usual" definition tantamounts to saying that $[\nabla, \omega ]=d\omega$, where $[,]$ is the (graded) commutator of (graded) endomorphisms.

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