Is covariant derivative of the connection one-form defined?

This is in regards to the definition of curvature two-form $$\Omega$$ defined in Nakahara (Sec. 10.3.2, Def. 10.5, Pg. 386) as the covariant derivative of the connection one-form $$\omega$$ $$\Omega \equiv D\omega$$ This is actually in stark contrast to what we have been taught in physics that the covariant derivative of the connection is not defined. Can anyone please explain which understanding is the correct one?

I don't think the above answer is actually correct, or at least does not address the (apparent) source of your confusion, which is as follows, as far as I can tell. In physics classes you've been taught that one cannot take the covariant derivative of the connection, and was more than likely correct in the setting in which it was discussed. One takes derivatives of tensor fields, or more generally, sections of bundles. The covariant derivative from e.g. relativity is not a tensor field.

It can, however, locally, over a trivialisation of the vector bundle $$E$$, say $$E|_U\cong U\times\mathbb{R}^r$$, be represented by the connection $$1$$-form $$A$$, satisfying $$\nabla=d+A$$, with $$A\in\Omega^1(U,\text{End}(E))$$. And of course, then you can take the (covariant) derivative of $$A$$, get some $$2$$-form $$F_A\in\Omega^2(U,\text{End}(E))$$ which does transform like a tensor, and so gives rise to the curvature $$2$$-form $$F_A\in\Omega^2(M,\text{End}(E))$$ at the global level. But again, this is not done by taking the covariant derivative of the connection itself, which is a mapping $$\nabla:\Gamma(TM)\times\Gamma(E)\to\Gamma(E)$$, or a mapping $$\Gamma(E)\to\Gamma(T^*M\otimes E)$$, whatever you prefer. This is why I find the accepted answer incorrect from a pedagogical viewpoint. The fact that the local $$1$$-form can be differentiated and then glued together to the curvature $$2$$-form is not just some obvious fact, it's a deep result in differential geometry.

There is a distinction between connections on vector bundles, also called linear connections, and connections on principal bundles. A connection on a principal bundle $$P$$ really is a genuine Lie algebra valued $$1$$-form on the total space $$P$$, and the curvature really is the covariant derivative of the connection $$1$$-form itself. This is the situation that Nakahara is addressing in the chapter that you're citing. But it (likely) isn't the same situation as in your physics class; that was (more than likely) concerned with the Levi-Civita connection of a (semi-)Riemannnian metric, which is a linear connection on $$E=TM$$, i.e. a mapping $$\nabla:\Gamma(TM)\to\Gamma(T^*M\otimes TM)$$ as described above.

PS. Principal connections and linear connections are indeed related. Every linear connection gives rise to a principal connection on the frame bundle of the given vector bundle. Likewise, their curvatures are related. But they aren't the same thing, and it is import to understand the difference.

• Can you define the curvature two-form by taking a covariant derivative of the spin connection which is a connection on the frame bundle? Is that well defined? Feb 11 at 10:09
• The spin connection is not a connection on the frame bundle of the manifold. It's a connection on the spinor bundle of the manifold. In local coordinates these might seem the same, but they are not. So yes, you can define a curvature $2$-form by taking the covariant derivative of the spin connection, but it's not the same as the curvature $2$-form which, I am guessing, refers to the curvature $2$-form of the metric on $M$. But this is a whole separate topic, so if you would like to ask a question about it, I would suggest creating a separate question for it. Feb 11 at 10:17
• Oh, I see. My bad. Feb 11 at 10:18

Why would the covariant derivative of a connection be not defined? In such case you wouldn't be able to define the curvature tensor, which is the commutator of covariant derivatives. If $$A_{\mu \ i}^{j}dx^{\mu}$$ is your connection 1-form, its (exterior) covariant derivative is $$D_{\nu}(A_{\mu \ j}^{i}dx^{\mu})=\partial_{\nu}A^{i}_{\mu j}dx^{\nu}\wedge dx^{\mu}+A_{\nu j}^{k}A^{i}_{k\mu }dx^{\nu}\wedge dx^{\mu}$$

• So can you write the curvature tensor as the covariant derivative of the Christoffel connection as well? Is that also well-defined? Feb 11 at 8:39
• The curvature tensor is the commutator of covariant derivatives in the Christoffel connection; note the note exterior in my answer; since $dx^{\nu} \wedge dx^{\mu}=-dx^{\mu} \wedge dx^{\nu}$, the index swap results in the sign change. When you just write covariant derivatives, $R^{\alpha}_{\beta \mu\nu}=[\nabla_{\mu}, \nabla_{\nu}]^{\alpha}_{\beta}$ where $R^{\alpha}_{\beta \mu\nu}$ is the Riemann tensor, and $\nabla_{\nu}=\partial_{\nu}A^{\alpha}+\Gamma^{\alpha}_{\beta \nu}A^{\beta}$ Feb 11 at 8:43
• Ok. I just asked to clarify my confusion. I was once gaslighted for believing this by my one of my project supervisors in the past. Feb 11 at 8:51
• *beware the sign convention, specifically with Christoffel connection it is actually $\nabla_{\nu}A^{\alpha}=\partial_{\nu}A^{\alpha}-\Gamma^{\alpha}_{\beta\nu}A^{\beta}$ by tradition. Feb 11 at 8:54
• That minus sign seems incorrect. Feb 11 at 10:21

It makes sense to define curvature for principal connections as the covariant exterior derivative of the connection one-form. The connection one-form $$\omega$$ is a one-form on the total space of a principal bundle with values in a Lie algebra $$\mathfrak g$$. You can take the exterior derivative $$d\omega$$ to obtain a $$\mathfrak g$$-valued two-form on the total space. On the other hand, the connection one-form gives rise to a complementary subbundle to the vertical subbundle in the tangent bundle of your principal bundle. This is called the horizontal subbundle it defines a horizontal projection. Writing this as $$X\mapsto H(X)$$ the covariant exterior derivative of $$\omega$$ is defined as $$(X,Y)\mapsto d\omega(H(X),H(Y))$$. The result is exactly the curvature of the prinicpal connection, viewed as a two form on the total space of the principal bundle (which by construction is horizontal and equivariant).