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I got motivation from twisted Cohomology where we twisted the derivative $d_\psi=d+\psi\wedge$ and find cohomology class $H_{\psi}^k(M)$ where $\psi$ is closed one form.

I try to define twisted connection similarly, which lead me to define $\nabla_{X,\psi}e_j=\omega^i_j(X)e_i+\psi(X)e_j.\tag1$ which satisfy:

  • $\nabla_{X,\psi}(fs)=X_{\psi}(f)s+f\nabla_{X,\psi}(s)$ where we define $X_\psi(f)=X(f)+\psi(X)\cdot f\tag2$

Here if $\psi=0$ then this exactly matches with our classical notion of connection satisfying the ordinary Leibniz rule. Does this really define a connection? Is there any existing work related with this? It will be a great help if anyone provide some information or guidance.

I tried to compute associate curvature form using $(1)$ and $(2)$, $\nabla_{X,\psi}\nabla_{Y,\psi}e_j=[X(\omega^i_j(Y))+\psi(X)\cdot\omega^i_j(Y)]e_i+\omega^k_j(Y)\omega^i_k(X)e_i+[X(\psi(Y))+\psi(X)\cdot\psi(Y)]e_j+\psi(Y)[\omega^k_j(X)e_k+\psi(X)e_j]$ and Interchanging $X$ and $Y$ will give $\nabla_{Y,\psi}\nabla_{X,\psi}e_j$. And $$\nabla_{[X,Y],\psi}e_j\stackrel{1}{=}\omega^i_j([X,Y])e_i+\psi([X,Y])e_j.$$ Combining all we get, \begin{align} &\nabla_{X,\psi}\nabla_{Y,\psi}e_j-\nabla_{Y,\psi}\nabla_{X,\psi}e_j-\nabla_{[X,Y],\psi}e_j\\ &=d\omega^i_j(X,Y)e_i+\omega^k_j\wedge\omega^i_k(X,Y)e_i\\&+[X(\psi(Y))-Y(\psi(X))-\psi([X,Y])]e_j+\psi\wedge\omega^k_j(Y,X)e_k\\ &=d\omega^i_j(X,Y)e_i+\psi\wedge\omega^i_j(Y,X)e_i+\omega^k_j\wedge\omega^i_k(X,Y)e_i\\&+[X(\psi(Y))-Y(\psi(X))-\psi([X,Y])]e_j \end{align} I see some inconsistency, Like I was guessing to have $$d_\psi\omega^i_j(X,Y)e_j=d\omega^i_j(X,Y)e_i+\psi\wedge\omega^{i}_j(X,Y)e_i$$ but from the calculation I got something like $$\cdots=[\psi(Y)\omega^k_j(X)-\psi(X)\omega^k_j(Y)]e_k=\psi\wedge\omega^k_j(Y,X)e_k=\psi\wedge\omega^i_j(Y,X)e_i.$$ Another problem is, I couldn't interpret what $[X(\psi(Y))-Y(\psi(X))-\psi([X,Y])]e_j$ represent here?
N00BMaster
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    The formula makes no sense as it stands. Do you mean $\psi(X)e_j$? And then your formula for $X_\psi$ doesn’t make sense to me. Please give explicitly the domain and codomain of $\nabla_{X,\psi}$. – Ted Shifrin Apr 28 '24 at 20:45
  • I fixed the typos, thanks @TedShifrin. We hypothesized this $\nabla_{X,\psi}$ as connection which take a smooth section and give a section as usual connection do. – N00BMaster Apr 29 '24 at 07:55
  • Why use the cumbersome $\nabla$ definition of curvature rather than the differential form definition? – Ted Shifrin Apr 29 '24 at 23:15

1 Answers1

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If $\nabla$ is a connection on the vector bundle $E$, $\nabla_\psi$ appears just to be the natural connection on $E\otimes\underline{\Bbb R}$, where $\underline{\Bbb R}$ is the trivial bundle with connection $d_\psi$.

Let $\tilde\omega$ be the connection matrix for $\nabla_\psi$. We have $\tilde\omega = \omega + \psi\otimes I$, i.e., $\tilde\omega_j^i = \omega_j^i + \delta_j^i\psi$. Since $\psi$ is closed, we have $d\tilde\omega = d\omega + d\psi\otimes I = d\omega$. Note that $\omega\wedge\psi\otimes I + \psi\otimes I\wedge\omega = 0$ since $\omega_j^i\wedge\psi + \psi\wedge\omega_j^i=0$ for all $i,j$. Therefore, depending on your sign conventions, $$\tilde\Omega = d\tilde\omega \pm \tilde\omega\wedge\tilde\omega = d\omega \pm \omega\wedge\omega = \Omega.$$

By the way, this post gives the general form of the curvature of the tensor product of two bundles.

Ted Shifrin
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  • Thanks for your answer. I did some more computation and tried to come up with curvature like thing using the defined connection. But the end result wasn't look like something I can related with. Can you say something on it? @TedShifrin – N00BMaster Apr 29 '24 at 20:40
  • What does twisted derivative do? I never see such operator but it seems like do some algebraic manipulation rather capture any geometric information. Can you shed some light on its geometric intuition? @TedShifrin – falamiw Jul 12 '24 at 18:00
  • @falamiw Connections on vector bundles (and their curvatures) are certainly geometric. – Ted Shifrin Jul 12 '24 at 18:47
  • Yes, and I know a little bit about how connections and curvatures relate to geometry (from my 3rd year manifold course). But I was wondering what additional geometric things are happening when we add a twist (which is 1-form here) @TedShifrin I want to explore it. Any resource where geometric insight was discussed would be a great suggestion for me. – falamiw Aug 01 '24 at 21:11