Let $P$ be a principal bundle, and $\omega$ a vector-valued 1-form on it. In standard textbooks, one define the exterior covariant derivative of $\omega$ to be

$$D\omega(X_1, X_2) = d\omega(X_1^h, X_2^h),$$

where $X_1, X_2\in TP$, and $^h$ means taking the horizontal part of the vector.

Further, one can show that in Yang-Mills theory, the field strength as the curvature two-form pulled back by a section:


I have zero intuition about why taking the horizontal part in the definition of the exterior covariant derivative. Why would one even think about this in the beginning? Why does this definition make sense?

In the seemly straightforward (from a physics point of view) definition of Yang-Mills field strength

$$\mathcal{F}=dA+A\wedge A$$

how could I tell to formulate this in the language of connections on fibre bundles, one would need that "horizontal part" in the definition?

  • $\begingroup$ Could you provide precise details of your question? $\endgroup$
    – Deane
    Jul 14, 2022 at 15:42
  • $\begingroup$ I re-wrote the question, hopefully, it now conveys my confusion. Thank you! $\endgroup$
    – Alex
    Jul 15, 2022 at 4:51

2 Answers 2


The way I think about horizontality of such a form is as follows. If we have a principal $G$-bundle $P\to M$, we can take the curvature $F_A$ of a connection $A$ on $P$. But now we have information which lives on $P$, namely $F_A\in\Omega^2(P,\mathfrak{g})$, where $\mathfrak{g}$ is the Lie algebra of $G$. We want to have this information on the base manifold $M$. You should be able to find the following theorem, in any text book that discusses gauge theory: there exists an isomorphism between $\Omega^k(M,\text{Ad}(P))$ and $\Omega^k_\text{basic}(P,\mathfrak{g})$, where the space of basic forms are Ad-equivariant horizontal $k$-forms on $P$.

So if we keep in mind that we actually want data on the base manifold $M$, it makes sense to consider horizontal (and equivariant) forms. But there is generally no canonical way of choosing a horizontal sub-bundle of $TP$, and so we are left with this "gauge freedom" of choosing our own horizontal distribution. Note that the Yang-Mills equation is in fact phrased in terms of such forms on the base manifold, because $M$ is where we would have a Riemannian metric - not $P$.

As an example, consider the trivial bundle $S^1\times S^1$. Then the way I think about a horizontal distribution on this total space, i.e. the torus, is as a foliation of $S^1\times S^1$, where the leaves of the foliation are the integral submanifolds of the horizontal distribution. Each of these leaves is diffeomorphic to the base space $S^1$, and the tangent space of each of these leaves is the horizontal distribution restricted to the leaf. If $\omega\in\Omega^1_\text{basic}(P,\mathfrak{g})$, then we can restrict this to a form $\iota^*\omega\in\Omega^1(L,\mathfrak{g})$ where $L$ denotes a leaf of the foliation. The fact that $\omega$ is $\text{Ad}$-equivariant implies that this restriction does not depend on the choice of leaf in the foliation, and the fact that $\omega$ is horizontal means we lose nothing by considering it as a $1$-form on this leaf, because any component of an input vector which is not tangent to the leaf will be killed off.

When I say that $\iota^*\omega$ does not depend on the choice of leaf, I mean the following. Suppose we are at a point $x\in S^1$, i.e. on the base space. If we have two points $p_1,p_2$ in the fibre $\pi^{-1}(x)$, then we can evaluate $\omega$ on the leaf through $p_1$ or the leaf through $p_2$, and these yield the same result. Thus, from the $1$-form on $P$, we have now obtained a $1$-form on $S^1$, which is where we actually wanted it.

As a final note: a word of caution. The above picture helps (me, at least) with understanding how connections and horizontality make sense, but for a general connection there will no integral submanifolds through each point - this is because integrability of the horizontal distribution is equivalent to the connection being flat.


This is not an answer to your question, but one possible explanation of how horizontal tangent vectors appear in the definition of a connection.

Recall that a principal $G-$bundle $P$ over a manifold $M$ has a projection map $\pi: P \rightarrow M$ such that, for each $m \in M$, $\pi^{-1}(m)$ is diffeomorphic to the same fiber $F$. A section of $P$ is a map $s: M \rightarrow P$ such that $\pi(s(m)) = m$, for every $p \in M$.

Suppose we want to differentiate a section $s$. There already is a natural derivative, namely the Jacobian of $s$, \begin{align*} Ds(m): T_mM &\rightarrow T_{s(m)}P\\ v &\mapsto D_vs(m), \end{align*} but this does not depend at all on the principal bundle structure of $M$.

However, we view a section of $P$ as a generalization of the graph of a map $f: M \rightarrow F$. In particular, if $P = M \times F$, then a section of $P$ is given by $$ s(m) = (m,f(m)). $$ We would like to define the analogue of the Jacobian of $f$, $$ Df(m): T_mM \rightarrow T_{f(m)}F, $$ for the section of a principal bundle.

In particular, given a section $s$ of $P$, we'd like to define a bundle map \begin{align*} \nabla s(m): T_mM &\rightarrow T_{s(m)}F_p\\ v &\mapsto \nabla_vs(m), \end{align*} where $F_p = \pi^{-1}(p)$. In other words, the covariant derivative of $s$ should map a tangent vector at a point $p$ on the manifold to a vertical tangent vector, i.e., a tangent vector of the fiber of $P$ at $s(p)$.

There is no canonical way of doing this. Howver, if, for each $p \in P$, we define a projection map $$ \Pi_p: T_pP \rightarrow T_pF_m, $$ where $m = \pi(p)$, then we can define for each $m \in M$ and $v \in T_mM$, $$ \nabla_vs(m) = \Pi_{s(m)}(D_vs(m)). $$ The projection map $\Pi$ must satisfy additional properties, in order for $\nabla$ to be a derivation.

By definition, a subspace $H_p \subset T_pP$ is horizontal if it is transverse to $T_pF_p$. A horizontal subspace $H_p$ uniquely determines a projection map $\Pi_p: T_pP \rightarrow T_pF_p$ and vice versa.

Therefore, a connection is specified by, for each $p \in P$, a horizontal subspace $H_p$ or, equivalently, a projection map $\Pi_p$. The properties that $\Pi_p$ must satisfy in order for $\nabla$ to be a connection can also be formulated as properties that the horizontal subspaces $H_p$ must satisfy. These are the properties you see in most definitions of a connection on a principal bundle.

  • $\begingroup$ Thank you for the explanation. I think I am comfortable with the notion that the horizontal subspace "is" the connection. However, I still have zero intuition of why taking horizontal in the definition of the curvature. Why would one think of taking the horizontal at all? Is there any significance with the quantity $d\omega(X, Y)$ that doesn't take the horizontal part? $\endgroup$
    – Alex
    Oct 6, 2022 at 14:09
  • $\begingroup$ If you move vertically, then it's equivalent to moving around in the group $G$. In particular, the vertical part of $\omega$ is essentially $g^{-1}\,dg$ and satisfies the Maurer-Cartan equation. This effectively means that the curvature in the vertical direction is zero. $\endgroup$
    – Deane
    Oct 6, 2022 at 19:15
  • $\begingroup$ In that case why bother taking the horizontal part in the definition of $d\omega(X^H, Y^H)$? Shouldn't we just say $d\omega(X, Y)$ is curvature? Historically, was there some thinking about $d\omega(X, Y)$ and then we moved to $d\omega(X^H, Y^H)$? $\endgroup$
    – Alex
    Nov 5, 2022 at 2:39

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