Let $F^A$ be a curvature two form on a principal bundle $P$ corresponding to the connection one form $A$. Let $f:P\rightarrow P$ be a global bundle automorphism; there then exists a map $\sigma_f:P\rightarrow G$ such that:

$$f(p)=p\cdot \sigma_f(p)$$

Finally let $\mu_G$ be the maurer cartan form, given by:


where $X_g$ is a vector field on some Lie group $G$. Under the pullback of $A$ by $f$ we have:"

$$f^*A=\text{Ad}_{\sigma_{f^{-1}} } \circ A+\sigma_f^*\mu_G$$

I am trying to show that: $$F^{f^*A}=\text{Ad}_{\sigma_{f^{-1}}}\circ F^A$$ where: $$F^A=dA+\frac{1}{2}[A,A]$$ and for two Lie algebra valued one forms $\eta,\omega$: $$[\eta,\omega](X,Y)=[\eta(X),\omega(Y)]-[\eta(Y),\omega(X)]$$ which comes from the definition of the wedge product for Lie algebra valued $k$ and $l$ forms.

When replacing $A$ with $f^*A$, we see that:

$$F^{f^*A}=d(\text{Ad}_{\sigma_{f^{-1}}}\circ A)+\sigma_f^*d\mu_G+\frac{1}{2}\text{Ad}_{\sigma_{f^{-1}}}\circ [A,A]+\frac{1}{2}\sigma_f^*[\mu_G,\mu_G]+\frac{1}{2}[\text{Ad}_{\sigma_{f^{-1}}}\circ A,\sigma_f^*\mu_G]+\frac{1}{2}[\sigma_f^*\mu_G,\text{Ad}_{\sigma_{f^{-1}}}\circ A]$$

By the Maurer-Cartan equation we have that: $$\sigma_f^*d\mu_G=-\frac{1}{2}\sigma^*_f[\mu_G,\mu_G]$$ hence: $$F^{f^*A}=d(\text{Ad}_{\sigma_{f^{-1}}}\circ A)+\frac{1}{2}\text{Ad}_{\sigma_{f^{-1}}}\circ [A,A]+\frac{1}{2}[\text{Ad}_{\sigma_{f^{-1}}}\circ A,\sigma_f^*\mu_G]+\frac{1}{2}[\sigma_f^*\mu_G,\text{Ad}_{\sigma_{f^{-1}}}\circ A]$$

My problem is now that I am unsure of how to calculate the exterior derivative of $\text{Ad}_{\sigma_{f^{-1}}}\circ A$. By our previous formula for two Lie algebra valued one forms we have that for $\eta=\text{Ad}_{\sigma_{f^{-1}}}\circ A$, and $\omega=\sigma_f^*\mu_G$:

$$\frac{1}{2}\left([\eta,\omega]+[\omega,\eta]\right)(X,Y)=\frac{1}{2}\left([\eta(X),\omega(Y)]-[\eta(Y),\omega(X)]+[\omega(X),\eta(Y)]-[\omega(Y),\eta(X)]\right)$$ $$=[\eta(X),\omega(Y)]-[\eta(Y),\omega(X]$$ $$=[\eta,\omega](X,Y)$$

So I feel that:

$$d(\text{Ad}_{\sigma_{f^{-1}}}\circ A)=\text{Ad}_{\sigma_{f^{-1}}}\circ dA-[\eta,\omega]$$

but I don't see why this should be true. In fact, since $\text{Ad}$ is a representation of $G$ on $\mathfrak{g}$ I would think that: $$d(\text{Ad}_{\sigma_{f^{-1}}}\circ A)=\text{Ad}_{\sigma_{f^{-1}}}\circ dA$$ but then I have this extra factor of $[\eta,\omega]$, so I am at a loss. Any hints on how to calculate this exterior derivative, or on how that wedge product may actually be zero would be greatly beneficial.


Employing Professor Shifrin's notation, I write that:


I have already calculated the exterior derivative of the second term, albeit in a different notation, so I move to the first term:

$$d(\sigma_{f^{-1}}A\sigma_f)=d(\sigma_{f^{-1}})\wedge A\sigma_f+ \text{Ad}_{\sigma_f^{-1}}\circ dA+\sigma_{f^{-1}}A\wedge d(\sigma_f) $$

Note that: $$\sigma_{f^{-1}}\sigma_f=e$$ hence: $$d(\sigma_{f^{-1}})\sigma_f=-\sigma_{f^{-1}}d\sigma_f\Rightarrow d\sigma_{f^{-1}}=-\sigma_f^{-1}d\sigma_f\sigma_{f^{-1}}$$ Therefore we obtain:

$$d(\sigma_{f^{-1}}A\sigma_f)=-\sigma_{f^{-1}}d\sigma_f\wedge \text{Ad}_{\sigma_f^{-1}}A+ \text{Ad}_{\sigma_f^{-1}}\circ dA-\text{Ad}_{\sigma_{f^{-1}}}\circ A \wedge d(\sigma_{f^{-1}})\sigma_f$$

The final term in this sum can be rewritten as:

$$\text{Ad}_{\sigma_{f^{-1}}}\circ A \wedge d(\sigma_{f^{-1}})\sigma_f=-\text{Ad}_{\sigma_{f^{-1}}}\circ A \wedge \sigma_f^{-1}d\sigma_{f}$$


$$d(\sigma_{f^{-1}}A\sigma_f)=-\sigma_{f^{-1}}d\sigma_f\wedge \text{Ad}_{\sigma_f^{-1}}A+ \text{Ad}_{\sigma_f^{-1}}\circ dA+\text{Ad}_{\sigma_{f^{-1}}}\circ A \wedge \sigma_f^{-1}d\sigma_{f}$$

From here, I can kind of see how the pieces should move, but I am unsure why. Reconciling the two notations is proving troublesome to me. To be clearer, it seems that from the definition for the wedge product of twisted forms, this product is actually symmetric for one forms, so I feel like the first term and the last term would cancel, which doesn't make sense. I am then left to conclude that there is something different about the wedge product I have used in this edit, and the wedge product I used in the original post, but I am unsure what that difference actually is, or how to make it precise.

Edit 2:

The definition of wedge product for Lie algebra valued one forms is as follows:

Given a $\omega\in \Omega^k(P,\mathfrak{g})$ and $\eta\in \Omega^k(P,\mathfrak{g})$, then their wedge product is given by:

$$[\omega,\eta](X_1,\dots,X_{k+l})=\frac{1}{k!l!}\sum_{\sigma\in S}\text{sign}(\sigma)\left[\omega(X_{\sigma(1)},\dots,X_{\sigma(k)}), \eta(X_{\sigma(k+1)},\dots,X_{\sigma(k+l)} ) \right]$$

When choosing a basis $\{T_i\}$ for the Lie algebra $\mathfrak{g}$, we have that $\eta$ and $\omega$ can be written as: $$\eta=\sum_{i=1}^n \eta^i\otimes T_i$$ $$\omega=\sum_{i=1}^n \omega^i\otimes T_i$$ where each $\omega^i$ and $\eta^i$ are $k$ and $l$ forms on $P$ respectively. Then the wedge product can be written as $$\omega \wedge \eta =\sum_{i,j=1}^n \omega^i\wedge\eta^j\otimes[T_i, T_J] $$

Hence: $$[\eta,\omega] =\sum_{i,j=1}^n \eta^j\wedge\omega^i\otimes[T_j,T_i] $$ $$=(-1)^{lk}\sum_{i,j=1}^n\eta^i \wedge\omega^j\otimes[T_j,T_i]$$ $$=(-1)^{lk+1}\sum_{i,j=1}^n\eta^i \wedge\omega^j\otimes[T_i,T_j]$$ $$=(-1)^{lk+1}[\omega,\eta]$$ So for two one forms, $\eta$ and $\omega$: $$[\eta,\omega]=(-1)^2[\omega,\eta]=[\omega,\eta]$$ Hence the wedge product of Lie algebra valued forms is symmetric for one forms.

  • 3
    $\begingroup$ Your last comment cannot be right. Obviously, $\sigma_f$ needs to be differentiated, as well. Have you tried working this out in the less abstract matrix notation, knowing that $R_g^*A = g^{-1}Ag + g^{-1}\,dg$? $\endgroup$ Oct 2, 2022 at 22:02
  • $\begingroup$ @TedShifrin I was under the impression that I can only write $A$ like that in a local trivialization where I can view $A$ as a Lie algebra valued one form on the base manifold $M$ plus the Maurer Cartan form. Since $f$ is a global bundle automorphism, I felt as though I couldn't work it out this way. Locally I can show something similar, but globally I am struggling with that exterior derivative term. $\endgroup$
    – Chris
    Oct 2, 2022 at 22:18
  • $\begingroup$ No, this works quite generally. The Maurer-Cartan form in my expression is coming from the change of gauge. My $g$ will be your $\sigma_f$. Even having worked 40+ years in differential geometry, I avoided the notation of your post whenever I could. $\endgroup$ Oct 2, 2022 at 22:28
  • $\begingroup$ @TedShifrin I think I see, in what you have written, is your $g$ constant, or is it also a function $P\rightarrow G$? Because I feel like for a constant $g$ it should still be just be $g^{-1}Ag$ $\endgroup$
    – Chris
    Oct 2, 2022 at 22:50
  • 1
    $\begingroup$ For the record, if $\omega$ and $\eta$ are matrix-valued $1$-forms, then $(\omega\wedge\eta)(X,Y) = \omega(X)\eta(Y) - \omega(Y)\eta(X)$, "as usual." This, in general, is different from $[\omega,\eta](X,Y) = [\omega(X),\eta(Y)]-[\omega(Y),\eta(X)]$. You have a sign error in your differentiation: Passing $d$ across $A$ produces a negative, as usual, since $A$ is a (matrix-valued) $1$-form. If $\tilde A = g^{-1}Ag$ and $\omega = g^{-1}dg$, I get $d\tilde A = -\omega\wedge\tilde A + g^{-1}dA\,g - \tilde A\wedge \omega$. $\endgroup$ Oct 5, 2022 at 5:21

1 Answer 1


Here is an easier proof: (Following your notation, it seems that you are using Hamilton's book on mathematical gauge theory, right? I try to stick to this notation too).

First of all, recall that the curvature $F^{A}\in\Omega^{2}(P,\mathfrak{g})$ is "of type $\mathrm{Ad}$'', what essentially means that $$r_{g}^{\ast}F^{A}=\mathrm{Ad}_{g^{-1}}\circ F^{A},$$ where $r_{g}:P\to G$ is the right group action $r_{g}(p):=p\cdot g$. In other words, the curvature is an element of the subset $\Omega^{2}_{\mathrm{hor}}(P,\mathfrak{g})^{\mathrm{Ad}}$.

The secons thing to realize is that


which follows from the fact that $\ast$ commutes with both $\mathrm{d}$ and $[\cdot,\cdot]$. As a next step, note that $f$ can be written as


Hence, combining everything from above, we have that

$$F^{f^{\ast}A}_{p}=(f^{\ast}F^{A})_{p}=(r^{\ast}_{\sigma_{f}(p)}F^{A})_{p}=\mathrm{Ad}_{\sigma_{f}^{-1}(p)}\circ F^{A}_{p}.$$

Last but not least, note that $\sigma_{f}^{-1}=\sigma_{f^{-1}}$ and hence, we get your claimed equality

$$F^{f^{\ast}A}=\mathrm{Ad}_{\sigma_{f^{-1}}}\circ F^{A}.$$

  • $\begingroup$ This is a nice proof, and if I can't figure out the more brute force approach, or if no one else does I will select this answer as the correct one, but it would be nice to see how the $d(\text{Ad}_{\sigma_{f^{-1}}}\circ A)$ can be calculated out. $\endgroup$
    – Chris
    Oct 6, 2022 at 21:11
  • $\begingroup$ Well, it is true that $\mathrm{d}(\mathrm{Ad}_{g}\circ A)=\mathrm{Ad}_{g}\circ\mathrm{d}A$. Right now, it I don't see how your additional term cancels. I will have a look lateron, if I have time. $\endgroup$ Oct 7, 2022 at 8:43
  • $\begingroup$ that's true for a constant $g$, but $\sigma_f$ depends on the point $p$ so the exterior derivative should do something to it. Thank you again though, and don't worry if you don't have the time. $\endgroup$
    – Chris
    Oct 7, 2022 at 13:09

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .