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I came across this following way of defining connection and curvature which is not so obviously equivalent to the definitions as familiar in Riemannian Geometry books like say by Jost.

If $E$ is a complex vector bundle over a manifold $M$ then one defines the space of vector valued $p$-differential forms on them as $\Omega^p(M,E) = \Gamma ( \wedge ^p (T^*M) \otimes E) $

The connection be defined as the map , $\nabla: \Gamma(E) \rightarrow \Omega^1(M,E)$ satisfying $\nabla(fX) = (df)X+f\nabla X$ where $f \in C^{\infty}(M)$ and $X \in \Gamma(E)$

{It might help if someone can make explicit as to what exactly is $(df)X$"? To make sense this has to be an element of $\Omega^1(M,E)$}

Now one defines curvature of the connection as the map, $$R = \nabla \circ \nabla : \Gamma(E) \rightarrow \Omega^2(M,E)$$

Now it is not clear to me as to how this makes $ R \in \Omega^2(M,End(E))$ ?

And how does this match up with the more familiar form as,

$R(X,Y) = \nabla _X \nabla _Y - \nabla _Y \nabla _X - \nabla _{[X,Y]}$

One can extend the definition of connection to be a map $\nabla: \Omega^p(M,E) \rightarrow \Omega^{p+1}(M,E)$ such that for $\omega \in \Omega^p(M)$ and $X \in \Gamma(E)$ one has,

$$\nabla(\omega X) = (d\omega)X + (-1)^{deg \omega}\omega \wedge \nabla X$$

{Here again it would be helpful if someone can can explain what is the meaning of $(d\omega)X$"? To make sense it has to be an element of $\Omega^{p+1}(M,E)$}

This looks very much like the de-Rham derivative of vector valued differential forms but the notation looks uneasy.

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$dfX$ means $df \otimes X$ which is a section of $\wedge^p T^*M \otimes E$. Similarily $d\omega X$ means $d\omega \otimes X$. –  Diego Matessi Jun 16 '10 at 15:03
    
...in my previous comment $p=1$ in the first case and $p=2$ in the second.... –  Diego Matessi Jun 16 '10 at 15:06
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3 Answers

up vote 2 down vote accepted

You need to write $\omega X$ as $\omega \otimes X$.

If you prove that $R: \Gamma(E) \to \Omega^2(M,E)$ is tensorial - i.e. $R(X,Y)(fS) = fR(X,Y)S$ where $S\in\Gamma(E)$, $f\in\mathcal{C}^\infty(M)$ and $X,Y\in TM$, then it follows that this mapping is indeed an element of ${\Omega} ^2(M,End(TM))$, because then the value of $R(X,Y)S$ at a point $m\in M$ depends only on $X_m,Y_m$ and $S_m$ and not on their derivatives (even though we have defined $R$ via differentiation). The desired identification then follows from linear algebra $Hom(\Lambda^2 V \otimes E,E) \simeq Hom(\Lambda^2 V, \otimes E^* \otimes E)$ & $End(E) \simeq E^*\otimes E$.

The classical formula for curvature follows directly from the definition of the action of $\nabla$ on $\Omega^p(M,E)$.

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Probably it is a bit embarrassing, but I did prove the identity about R(X,Y) that you said but even then it wasn't obvious that R is in $\Omega^2(M,End(TM))$. May be you can give some reference or some explanation. –  Anirbit Jun 16 '10 at 17:14
    
I've expanded my answer. It's basically a question of multilinear algebra. –  Vít Tuček Jun 16 '10 at 17:29
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The definition you give is for the curvature of a connection on a vector bundle $E$ and therefore is more general than the Riemann curvature which is for the Levi-Civita connection only. But if you set $E$ equal to the tangent bundle of $M$ and $\nabla$ equal to the Levi-Civita connection of a Riemannian metric, and write out the definition of $R$, as defined above, in local co-ordinates or with respect to a local orthonormal frame, then you will recognize the formula for the Riemann curvature tensor.

As for $(df)X$, note that $\Omega^1(M,E)$ denotes sections of the vector bundle $T^*M\otimes E$ and by $(df)X$ we really mean the section $x \mapsto df(x)\otimes X(x)$. The definition of $(d\omega)X$ is similar.

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Thanks. The definition of a connection is generally made for any vector bundle but writing that in differential forms language for the general case was slightly new to me. Like I suppose definition 3.1.1 of Jost is at the same level of generality as this one but the above restatement of Jost's condition 3.1.4 was not obvious. –  Anirbit Jun 17 '10 at 10:04
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If you consider a frame of your bundle E, say $E_1$...$E_k$ then there will be forms $\omega_{jk}$ such that

$$ \nabla E_j = \omega_{jk} E_k$$

(summation over repeated indices). The notation $\nabla_X E_j$ becomes $\omega_{jk}(X) E_k$. So for instance $$\nabla_X \nabla_Y (E_j) = X(\omega_{jk}(Y))E_k + \omega_{jk}(Y)\omega_{kl}(X) E_l $$ where I have applyied the rules you mentioned. On the other hand $$\nabla \circ \nabla (E_j) = \nabla(\omega_{jk} E_k) =d\omega_{jk}E_k - \omega_{jk}\wedge (\nabla E_k) = (d\omega_{jl} - \omega_{jk} \wedge \omega_{kl}) E_l, $$ (again applying the rules). Now you can apply the above two forms to vectors X and Y. You will need the formula $$ d\omega_{jl}(X,Y) = X(\omega_{jl}(Y)) - Y(\omega_{jl}(X)) - \omega_{jl}([X,Y]), $$ this is how the $\nabla_{[X,Y]}$ part comes out in the curvature. By applying all these ideas in the end you get the "usual" formula for the curvature.

Reference: a good book on this is "From calculus to cohomology" by Madsen and Tornehave.

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I was aware of this above manipulation for connection on tangent bundles. Thats how one defines connection forms there. Thanks for showing how it might generalize. –  Anirbit Jun 17 '10 at 10:05
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