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From this note, On section $4.1.1$ we have,

If $\{e_\alpha\}$ be a local frame of a vector bundle $\pi:E\rightarrow X$ then there exist a section $\Theta\in C^\infty(M,\Omega^2_N\otimes E)$, called curvature form, such that, $$(d^\nabla)^2s=\Theta\wedge s\tag1$$ where $\Theta$ can be written as, $$\Theta=\Theta^\beta_{\alpha}e^\alpha\otimes e_\beta\tag2$$

From this M.SE post we have (which seems align with the formula from the book I follow, Differential Geometry by Loring W.Tu),

Let $\Omega_\mu^\nu\in\Omega^2(U)$ be such that, $$\nabla\nabla e_\mu=\Omega^\nu_\mu\otimes e_\nu\tag3$$

So the main confusion was formula $(1)$ use wedge product on the other hand formula $(3)$ use tensor product. I didn't understand why two different product used here? Because if I consider local frame as sections then $(1)$ become $(d^\nabla)^2e_\mu=\Theta\wedge e_\mu$ which seem not same as $(3)$. Where I made wrong?

I don't understand why $e^\alpha\otimes e_\beta$ pop up on the formula in $(1)$, like what's the role of the dual frame $e^\alpha$ here? I know it sound odd, but I think I didn't understand the role of tensor/wedge product used in these formulas completely.

WhyMeasureTheory
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    In the notes you sent, it says $\Theta \in C^\infty (M, \Omega ^2_M \otimes \text{End } E)$, so it is a $2$-form on $M$ with values in $\text{End } E$ (an element in $C^\infty(M, \text{End }E)$ may be expressed as $L^{\alpha}\beta e^\beta \otimes e\alpha$, so that the pairing is given by $(L^\alpha_\beta e^\beta \otimes e_\alpha , c^\gamma e_\gamma) = L^\alpha_\beta c^\beta e_\alpha $). – Pengin Apr 05 '24 at 18:02
  • But isn't a $2$-form should look like $\omega=\omega_{ij}e^i\wedge e^j$ and the action on the vectors $X=X^ie_i,Y=Y^je_j$ was $\omega(X,Y)=\omega_{ij}X^iY^j$? Then why here it was written $L_\beta^\alpha e^\beta\otimes e_\alpha$? @Pengin – WhyMeasureTheory Apr 05 '24 at 19:37
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    The $\Theta^\beta_\alpha$ are real-valued 2-forms in the way you described. $\Theta$ itself is a vector-valued 2-form acting on two vector fields $X,Y$ via $\Theta(X,Y) = \Theta^\beta_\alpha(X,Y) e^\alpha \otimes e_\beta \in C^\infty(M,\text{Hom } E)$. Here $e^\alpha\otimes e_\beta$ are the basis elements of $E^\ast \otimes E \cong \text{Hom } E$, and for each of the basis elements, we have a real valued 2 form $\Theta ^\beta_\alpha$. – Pengin Apr 05 '24 at 22:59
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    Okay, now it makes sense. Can you explicitly show how you get $(L^\alpha_\beta e^\beta\otimes e_\alpha, c^\gamma e_\gamma)=L^\alpha_\beta c^\beta e_\alpha$? I guess that will resolve my all doubt. And thanks @Pengin for your response. – WhyMeasureTheory Apr 06 '24 at 08:12
  • An element $L^\alpha_\beta e^\beta \otimes e_\alpha$ acts on $c^\gamma e_\gamma$ by $(L^\alpha_\beta e^\beta \otimes e_\alpha, c^\gamma e_\gamma ) =L^\alpha_\beta e^\beta (c^\gamma e_\gamma) e_\alpha = L^\alpha_\beta \delta ^\beta_\gamma c^\gamma e_\alpha = L^\alpha_\beta c^\beta e_\alpha$, this is how $\text{End }E$ is identified with $E^* \otimes E$. See also this post @WhyMeasureTheory – Pengin Apr 06 '24 at 10:22
  • If you can say something on the wedge product and tensor product used in $(1)$ and $(3)$ resp. and post everything as an answer I will happy to accept that. @Pengin And thanks for everything. – WhyMeasureTheory Apr 07 '24 at 10:01
  • Perhaps you have this completely sorted out, but note that the author is defining the curvature of the bundle of $E$-valued $k$-forms (rather than just of $E$-valued functions). That's why there's the wedge product rather than tensor product. When you write an $E$-valued $k$-form locally as $\phi\otimes \sigma$ for a $k$-form $\phi$ and section $\sigma$ of $E$, then you have to wedge $\nabla^2\sigma$ and $\phi$ to get an $E$-valued $(k+2)$-form. – Ted Shifrin Apr 07 '24 at 20:29
  • Thank you for your informative comment., I see the notation play a big role like $\Omega^k\otimes E$ vs $\Gamma(E)$. Before jumping into more advance topics I should have a strong understanding of these basics. Can you recommend a book that thoroughly explains these preliminaries? @TedShifrin – WhyMeasureTheory Apr 07 '24 at 20:43

1 Answers1

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$e^\alpha \otimes e_\beta$ pops up because $\Theta\in C^\infty(M, \Omega^2_M\otimes \text{End } E)$ is a 2-form which takes values in $\text{End } E \cong E^* \otimes E$ . $\Theta = \Theta^ \beta_{\alpha} e^\alpha\otimes e_\beta$ acts on a section $c^\gamma e_\gamma \in C^\infty(M, E)$ via $$(\Theta^\beta _\alpha e^\alpha \otimes e_\beta )(c^\gamma e_\gamma) = \Theta ^\beta_\alpha e^\alpha (c^\gamma e_\gamma) e_\beta = c^\alpha \Theta ^\beta_\alpha e_\beta \in C^\infty(M,\Omega^2_M\otimes E).$$ Similarly, for $s = s^\alpha e_\alpha \in C^\infty (M, \Omega^k_M \otimes E)$, we have $$\Theta \wedge s = (\Theta^\beta_\alpha \wedge s^\gamma) e^\alpha(e_\gamma) e_\beta = \Theta^\beta_\alpha \wedge s^\alpha e_\beta . $$ If $s=e_\mu\in C^\infty(M, E)$, this is the same as $\nabla \nabla e_\mu=\Omega ^\nu_\mu \otimes e_\nu$ (which, following the previous notation, should be written as $\Omega ^\nu_\mu e_\nu$), as $\Omega^\nu_\mu e_\nu = (\mathrm{d}{\omega}^{\nu}_{\mu} + {\omega}^{\nu}_{\rho} \wedge {\omega}^{\rho}_{\mu}) e_\nu$ and $$\Theta \wedge s = \Theta^\nu_\mu e_\nu= (\mathrm{d}{\omega}^{\nu}_{\mu} - {\omega}^{\rho}_{\mu} \wedge {\omega}^{\nu}_{\rho}) e_\nu= (\mathrm{d}{\omega}^{\nu}_{\mu} + {\omega}^{\nu}_{\rho} \wedge {\omega}^{\rho}_{\mu}) e_\nu= \nabla \nabla e_\mu.$$

Pengin
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  • Why you use two different notation, one for $\Omega_\mu^\nu e_\nu=\left(\mathrm{d} \omega_\mu^\nu+\omega_\rho^\nu \wedge \omega_\mu^\rho\right) e_\nu$ and another for $\Theta_\mu^\nu e_\nu=\left(\mathrm{d} \omega_\mu^\nu-\omega_\mu^\gamma \wedge \omega_\gamma^\nu\right) e_\nu$ ? @Pengin – WhyMeasureTheory Apr 07 '24 at 18:44
  • As per my book I know, $(\nabla)^2\left(e_\alpha\right)=\nabla\left(\omega_\alpha^\beta e_\beta\right)=\mathrm{d} \omega_\alpha^\beta e_\beta-\omega_\alpha^\beta \wedge \nabla e_\beta=\mathrm{d} \omega_\alpha^\beta e_\beta-\omega_\alpha^\beta \wedge \omega_\beta^\gamma e_\gamma=\left(\mathrm{d} \omega_\alpha^\beta-\omega_\alpha^\gamma \wedge \omega_\gamma^\beta\right) e_\beta$ – WhyMeasureTheory Apr 07 '24 at 18:45
  • And if I understand correctly, then in your last equation you want $\Theta\wedge e_\mu$ instead of $\Theta\wedge s$, right? Then I compute it as, $$\Theta\wedge e_\mu=(\Theta^\beta_\alpha e^\alpha\otimes e_\beta)\wedge e_\mu\stackrel{?}{=}\Theta^\beta_\alpha e_\beta\stackrel{?}{=}(\mathrm{d}{\omega}^{\beta}{\alpha} - {\omega}^{\gamma}{\alpha} \wedge {\omega}^{\beta}{\gamma}) e\beta$$ – WhyMeasureTheory Apr 07 '24 at 18:45
  • Regarding your first comment, I followed the notation from your notes and from the linked M.SE. post. Regarding your second question, I slightly edited the answer, and yes, we do this for $s=e_\mu$. However, your last equation is wrong, we have $s=e_\mu = 1\otimes e_\mu \in C^\infty(M,E) = C^\infty(M, \Omega^0_M \otimes E)$ and so $\Theta \wedge (1 \otimes e_\mu) = \Theta^\nu_\rho \wedge 1 (e^\rho \otimes e_\nu)(e_\mu) = \Theta^\nu_\mu e_\nu$. – Pengin Apr 07 '24 at 20:08
  • Aha, I forgot to cross check them, Thanks. I guess now I can relate every formula. Do you have any insight about the difference curvature form (page 25 on the note) to curvature tensor (page 26 on the note) here? I just want to know your perspective. I was reading the difference between tensoring and wedging. That why asked. And yes, I accept your answer. @Pengin – WhyMeasureTheory Apr 07 '24 at 20:28
  • I would say that both $R$ and $\Omega$ are basically the same thing, as they may both be regarded as sections over $T^M \otimes T^M \otimes \text{End }E$, on which they agree (this is also the discussion on page 27). – Pengin Apr 07 '24 at 21:48