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
    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 $G$. – Ted Shifrin Sep 14 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.

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

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