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I am in the midst of trying to solve problem 13 from chapter 5 of Hamilton's Mathematical Gauge theory text, and have come across some terms I am not sure what to do with. I will elaborate below.

Let $P\rightarrow M$ be a principal $G$ bundle, and $E=P\times_\rho V\rightarrow M$ an associated vector bundle. Fix a connection $A$ on a $P$, the induces the following covariant derivative on $E$, letting $\Phi$ be a local section of $E$:

$$\nabla^A_X\Phi=[s,d\phi(X_x)+\rho_*(A_s(X_x))\phi]$$ where $s$ is a local section of $P$, $\phi$ is a map $U\subset M\rightarrow V$, $X\in \mathfrak{X}(M)$, and where $A_s=s^*A$, can be thought of as a Lie algebra valued one form on the base $M$.

In a principal bundle we define curvature as:

$$F=dA+\frac{1}{2}[A,A]$$

The curvature of a covariant derivative in an arbitrary vector bundle is:

$$F^\nabla(X,Y)\Phi=\nabla_X\nabla_Y\Phi-\nabla_Y\nabla_X\Phi-\nabla_{[X,Y]}\Phi$$

I am trying to show that in an associated vector bundle:

$$F^\nabla(X,Y)\Phi=[s,\rho_*(F_s(X,Y))\phi]$$

where $F_s=s^*F$. So I calculated the following:

$$\nabla^A_X\nabla^A_Y\Phi=[s,d\left(\rho_*(A_s(X_x)\right)(Y_x)\phi+\rho_*(A_s(X_x))d\phi(Y_x)+\rho_*(A_s(X_x))d\phi(Y_x)+\rho_*(A_s(X_x)A_s(Y_x))]$$ $$\nabla^A_X\nabla^A_Y\Phi=[s,d\left(\rho_*(A_s(Y_x)\right)(X_x)\phi+\rho_*(A_s(Y_x))d\phi(X_x)+\rho_*(A_s(Y_x))d\phi(X_x)+\rho_*(A_s(Y_x)A_s(X_x))]$$ $$\nabla^A_{[X,Y]}\Phi=[s,d\phi([X,Y]_x)+\rho_*(A_s([X,Y]_x)\phi]$$

Putting it all together I obtain that:

$$F^\nabla(X,Y)\Phi=[s,d\left(\rho_*(A_s(Y_x)\right)(X_x)\phi-d\left(\rho_*(A_s(X_x)\right)(Y_x)\phi+\rho_*(A_s(X_x)A_s(Y_x)-A_s(Y_x)A_s(X_x))\phi-d\phi([X,Y]_x)-\rho_*(A_s([X,Y]_x)\phi]$$

I feel like I see all the pieces, but am unsure of how to move them around. Like the third term I'm pretty sure simplifies to $\frac{1}{2}[A_s,A_s](X_x,Y_x)$, but I don't know what to do with the exterior derivative of $\rho_*$, or what to do with the Lie bracket terms, especially since it seems like I have all the pieces without those terms, so I feel like they should cancel out some how but that doesn't make a ton of sense to me. Any help would be greatly appreciated.

Edit: I suspect I'm messing up something up with the exterior derivative, perhaps $d(d\phi(X_x)(Y_x)$ isn't zero? Because you technically contract then apply the exterior derivative again? I am unsure...

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  • $\begingroup$ I find this all easier to understand if I start by going backwards. Namely, by starting with a connection on a vector bundle and figuring out how it induces a connection on the appropriate frame bundle. Once you figure that out, you can define what a connection on a frame bundle is and show how it induces a connection on the vector bundle. Finally, that allows you to figure out what a connection on a principal bundle is and how it induces a connection on the vector bundle. $\endgroup$
    – Deane
    Jul 14, 2022 at 12:43
  • $\begingroup$ @Deane I feel like I’m not having issue with the connection, just the actual calculation of curvature using the definition above. I think I may have figured it out using a lie derivative property I had forgotten, will update soon. $\endgroup$
    – Chris
    Jul 14, 2022 at 13:15

1 Answer 1

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My suspicion that I was calculating the exterior derivative incorrectly were correct. Here is the full solution:

First examine the term $\nabla^A_Y\Phi$:

$$\nabla^A_Y\Phi=[s,\nabla^A_Y\phi]=[s,d\phi(Y)+\rho_*(A_s(Y))\phi]$$

Note that $d\phi$ and $A_s$ are vector ($d\phi$ takes values in $V$, and $A_s$ takes values in $\mathfrak{g}$) valued one forms on $M$, since we are working in a local gauge. This means that the contraction with the vector the vector field $Y$ gives us two vector valued zero forms. Since $d\phi$ is the exterior derivative of a zero form, we can write the following:

$$\nabla^A_Y\Phi=[s,L_Y\phi+\rho_*(A_s(Y))\phi]$$ Where $L_Y$ denote the Lie derivative along $Y$. We then see that: $$\nabla^A_X\nabla^A_Y\Phi=[s,d(\nabla^A_Y\phi)(X)+\rho_*(A_s(X))\nabla^A_Y\phi]$$ Examine the exterior derivative term: $$d(\nabla^A_Y\phi)(X)=d(L_Y\phi)(X)+d\rho_*(A_s(Y))(X)\phi+\rho_*(A_s(Y))d\phi(X)$$ Following the same logic as before we can rewrite this as: $$d(\nabla^A_Y\phi)(X)=L_x(L_Y\phi)+\rho_*(L_XA_s(Y))\phi+\rho_*(A_s(Y))d\phi(X)$$ Note that the exterior derivative 'commutes' with $\rho_*$ as $A_s$ can be written as the tensor product: $$A_s=\sum_{a}^{\text{dim }\mathfrak{g}}\omega^a\otimes T_a$$ where $T_a$ are a basis for $\mathfrak{g}$, and $\omega^a$ are one forms on $M$, thus we see that: $$d\rho_*(A_s)=\sum_{a}^{\text{dim }\mathfrak{g}}d(\omega^a)\otimes \rho_*(T_a)$$ as the $T_a$'s are constant. Thus we can write: $$d\rho_*(A_s(Y))(X)=\rho_*(dA_s(Y))(X)=\sum_{a}^{\text{dim }\mathfrak{g}}d(\omega^a(Y))(X)\otimes \rho_*(T_a)=\sum_{a}^{\text{dim }\mathfrak{g}}L_X(\omega^a(Y))\otimes \rho_*(T_a)=\rho_*(L_X(A_s(Y)))$$ Looking at the second term in $\nabla^A_X\nabla^A_Y\Phi$ we see: $$\rho_*(A_s(X))\nabla^A_y\phi=\rho_*(A_s(X)A_s(Y))\phi+\rho_*(A_s(X))d\phi(Y)$$ By symmetry we easily deduce the form of $\nabla^A_Y\nabla^A_X\Phi$, and see that $\nabla^A_X\nabla^A_Y\Phi-\nabla^A_Y\nabla^A_X\Phi$ is: $$[s,L_XL_Y\phi-L_YL_X\phi+\rho_*(L_XA_s(Y)-L_YA_s(X))+\rho_*([A_s(X),A_s(Y)])]$$ Finally we see that: $$\nabla_{[X,Y]}\Phi=[s,d\phi([X,Y])+\rho_*(A_s([X,Y]))\phi]$$ $$=[s,L_{[X,Y]}\phi+\rho_*(A_s([X,Y]))\phi]$$ With the identity: $$L_XL_Y f-L_YL_X f=L_{[X,Y]}f$$ we then easily see that the terms of $F^\nabla\Phi$ that contain the Lie derivatives of $\phi$ cancel, i.e.: $$L_XL_Y \phi-L_YL_X \phi-L_{[X,Y]}\phi=0$$ Hence we see that $F^\nabla\Phi$ is: $$F^\nabla\Phi=[s,\rho_*(L_XAs_(Y)-L_YA_s(X)-A_s([X,Y])+[A_s(X),A_s(Y)])\phi]$$ Note that: $$[A_s(X),A_s(Y)]=\frac{1}{2}[A_s,A_s](X,Y)$$ and that for any one form $\omega$: $$d\omega(X,Y)=L_X\omega(Y)-L_Y\omega(X)-\omega([X,Y])$$ Therefore we obtain: $$F^\nabla\Phi=[s,\rho_*(dA_s(X,Y)+\frac{1}{2}[A_s,A_s](X,Y))\phi]$$ Using the fact that: $$F_s=dA_s+\frac{1}{2}[A_s,A_s]$$ we finally obtain the desired result:

$$F^\nabla\Phi=[s,\rho_*(F_s(X,Y))\phi]$$

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