I'm reading this paper, and I'm confused about something.

Let $A$ be a connection on a principal $G$-bundle $P$ over $\mathbb{R}^4$, and $F(A)$ its curvature. Let $\mathrm{ad}(P)=P\times_G\mathfrak{g}$ be the fiber bundle associated to the adjoint representation. We have that the curvature $F(A)$ is a $2$-form with values in $\mathrm{ad}(P)$.

In terms of a trivialization of $P$ over $\mathbb{R}^4$, and the basic coordinates $(x_1,x_2, x_3, x_4)$, $F(A)$ may be written as a Lie algebra-valued $2$-form $$ F(A)=\sum_{i<j}F_{ij}dx_i\wedge dx_j\quad(\star) $$

In the article he states the following:

With respect to this trivialization, the connection is described by a Lie algebra-valued $1$-form $$ A=A_1 dx_1+A_2 dx_2+A_3 dx_3+A_1 dx_4\quad(\star \star) $$

I don't understand why it is possible to write the connection $1$-form like in $(\star \star)$. To explain why it sounds too strange to me, let's see why we can write $F(A)$ like in $(\star)$. Let $\psi:U\times G\rightarrow \pi^{-1}(U)$ be a trivialization, where $U\subset \mathbb{R}^4$ and $\pi:P\rightarrow M$ is the projection. Then, taking coordinates on $U\times G$, namely $(x_1, x_2, x_3, x_4, g_1, \dots, g_n)$, we have: $$ F(A)=\sum_{i<j}F_{ij}dx_i\wedge dx_j+\sum_{ij}H_{ij}dg_i\wedge dx_j+\sum_{i<j}R_{ij}dg_i\wedge dg_j $$

Now we have that $H_{ij}$ and $R_{ij}$ must be zero because $F(A)$ is horizontal and $d\psi_u^{-1}(V_u)=0\times T_{\psi^{-1}(u)}G\subset \mathbb{R}^4\times T_{\psi^{-1}(u)}G$, where $V_u\subset T_uP$ is the subspace of vertical vectors. So the equation $(\star)$ holds.

Now we can see that this only works because $F(A)$ is horizontal. And the connection $1$-form is the complete opposite, it is zero on $X$ iff $X$ is horizontal. So, by this argument we would have that in this coordinates: $$ A=\sum_{i=1}^n A_i dg_i $$

What is wrong with my argument? And how can I see that $(\star \star)$ really holds?

  • 1
    $\begingroup$ Your statement in the last paragraph is definitely wrong. If the connection form took nonzero values only on vertical vectors, then every lift of a curve in $M$ would be horizontal. That's obviously nonsense. (In the case of the Levi-Civita connection, you'd be saying that every curve in $M$ is a geodesic.) What is correct is that the vertical part of $A$ is the Maurer-Cartan form on $G$. $\endgroup$ – Ted Shifrin Sep 14 '18 at 20:38

If I understand well, then the object $A$ in $(**)$ is not the connection $1$-form. (The text only says "... can be described by a Lie algebra valued $1$-form ..." and not that this is the connection $1$-form.) In principal bundle language you would consider the pullback of the connection form along the local section of the principal bundle defined by a local trivialization. This does encode the full connection $1$-form since given the section $\sigma$, and point in the domain of the trivialization can be written uniquely as $\sigma(x)g$ for $x\in M$ and $g\in G$. Equivariancy of $A$ then tells you that it suffices to know the connection $1$-form in the point $\sigma(x)$. But there any tangent vector can be written as the sum of an element of the form $T_x\sigma\cdot\xi$ and a vertical vector, on which the behavior of the connection form is prescribed. Thus you can recover the connection form $A$ from $\sigma^*A$. I think the explicit relation is that $A$ is simply the sum of $\sigma^*A$ and the Maurer-Cartan, but I am not entirely sure about that.

  • $\begingroup$ That's a good suggestion. Thank you! I have just one question: what do you mean by $T_x\sigma\cdot\xi$? $\endgroup$ – Leonardo Schultz Sep 17 '18 at 18:57
  • 1
    $\begingroup$ It's just the tangent map (derivative) of $\sigma$ applied to the tangent vector $\xi$, so you may prefer to write it as $\sigma_*\xi$. $\endgroup$ – Andreas Cap Sep 18 '18 at 8:29

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.