I have just begun learning about the connection 1-forms, torsion 2-forms, and curvature 2-forms in the context of Riemannian manifolds. However, I am finding it hard to relate these notions to any sort of geometric intuition.

How can one interpret these differential forms geometrically? Or at least, what is the motivation for considering such objects? What sort of information do they provide?

Edit: I should clarify that I think I have somewhat of an intuition for connections, the curvature tensor, and (to a lesser extent) the torsion tensor. What I am asking about are the connection forms, torsion forms, and curvature forms.


3 Answers 3


Almost seventeen months later, I think I found the answers I was looking for in Barrett O'Neill's Elementary Differential Geometry.

Since O'Neill explains things better than I can, I'm going to paraphrase excerpts of his text -- nearly quoting verbatim -- modifying some notation and terminology to fit my own (such is differential geometry!). In short, I make zero claims of originality.

Note: Since the treatment is "elementary," O'Neill works in $\mathbb{R}^3$.

Connection $1$-forms

The power of the Frenet-Serret formulas stems from the fact that they express the derivatives $T', N', B'$ in terms of $T, N, B$ -- and thereby define curvature and torsion. The connection $1$-forms do the same thing, expressing the covariant derivatives of a local frame in terms of the vector fields themselves.

Let $\{E_i\}$ be a local orthonormal frame. The definition $$\omega^j_i(X) = \langle \nabla_X E_i, E_j \rangle$$ shows that $\omega^j_i(X)$ is the initial rate at which $E_i$ rotates toward $E_j$ as a point moves in the $X$ direction.

The connection equations $\nabla_{X}E_i = \omega^j_i(X)E_j$ give information about the "rate of rotation" of the local orthonormal frame $\{E_i\}$.

The Cartan Structure Equations in $\mathbb{R}^3$

In $\mathbb{R}^3$, Cartan's first structure equation reads $$d\phi^j = \phi^i \wedge \omega^j_i,$$ where $\{\phi^i\}$ is the dual coframe to $\{E_i\}$. This equation may be recognized as the dual of the connection equations $\nabla_{X}E_i = \omega^j_i(X)E_j$.

In $\mathbb{R}^3$, Cartan's second structure equation reads $$d\omega^j_i = \omega^k_i \wedge \omega^j_k.$$ These equations show that $\mathbb{R}^3$ is flat.

Surfaces in $\mathbb{R}^3$

Let $M \subset \mathbb{R}^3$ be an isometrically immersed surface. We let $\{E_1, E_2, N\}$ denote a local orthonormal frame for $\mathbb{R}^3$ that has $N$ normal to $M$. Thus, if $\{\phi^1, \phi^2, \phi^3\}$ denotes the dual coframe, then $\phi^3|_M = 0$.

Taking $E_i = N$ in the connection equations $\nabla_X E_i = \omega^j_i(X)E_j$ shows that $$-\nabla_XN = \omega^3_1(X)E_1 + \omega^3_2(X)E_2.$$ By definition, the term on the left-hand side is the shape operator $S(X) = -\nabla_XN$. Thus, we interpret the connection $1$-forms as follows:

  • $\omega^2_1$ -- Gives the rate of rotation of $\{E_1, E_2\}$
  • $\omega^3_1$ and $\omega^3_2$ -- Describes the shape operator

Cartan's second structure equation in $\mathbb{R}^3$ then implies the Gauss-Codazzi equations: $$\begin{align*} d\omega^2_1 & = \omega^3_2 \wedge \omega^3_1 \ \text{ (Gauss equation)} \\ d\omega^3_1 & = \omega^2_1 \wedge \omega^3_2 \ \text{ (Codazzi)} \\ d\omega^3_2 & = \omega^1_2 \wedge \omega^3_1 \ \text{ (Codazzi)} \end{align*}$$

By analogy with the formula $S(X) \times S(Y) = K\,(X \times Y)$, where $K$ is the Gaussian curvature, one can prove that $$\omega^3_1 \wedge \omega^3_2 = K\,\phi^1 \wedge \phi^2.$$ Together with the Gauss equation, this implies Cartan's second structure equation for $M$: $$d\omega^2_1 = -K\,\phi^1 \wedge \phi^2.$$


For starters, a connection on a manifold is all about parallel transport of tangent vectors. Think about a surface embedded in three dimensional space and a curve on that surface. Fix a point and a vector at this point that is tangent to the surface.

In $\mathbb{R}^3$ there is an intuitive sense of parallel transport of this vector along the curve (which I won't explain, unless you want me to). If you parallel transport a vector along a curve on the surface from a point $p_1$ to a point $p_2$ in this sense, then the vector at point $p_2$ won't necessarily be tangent to the surface anymore.

This means that while there is a canonical way of parallel transport for tangent vectors in $\mathbb{R}^n$, there isn't one for manifolds. This is where the connection comes in: Specifying a connection on a surface (or more generally on a manifold) is the same as saying how to project the vector at the point $p_2$ to the tangent space of the point $p_2$. Or, to rephrase it: How to transport a tangent vector along a curve in a way that the transported vector is still a tangent vector. This kind of transport is then the "parallel transport with respect to the specified connection".

The curvature measures how much the parallel transport on the surface deviates from the parallel transport in the ambient $\mathbb{R}^3$.

Torsion is a little bit more involved: It measures how "tangent vectors rotate" when they are parallel transported.

Let's say you fix a point $p$ and two tangent vectors $\vec{v}$ and $\vec{u}$ at p. What happens if you parallel transport $\vec{v}$ into the direction of $\vec{u}$ of by an "infinitesimal amount $\epsilon$" and compare this to the result of transporting $\vec{u}$ into the direction of $\vec{v}$ by an "infinitesimal amount $\epsilon$"? If the torsion is zero, then the tips of $\vec{v}$ and $\vec{u}$ will touch up to an error of $\epsilon^3$.

It is not that easy to make these intuitive ideas precise and to relate them to the usual definitions, but I can recommend a nice book where this is explained in more detail:

  • John Baez, Javier Muniain: "Gauge Fields, Knots and Gravity"
  • $\begingroup$ (Re: there isn't one for manifolds) Well, in fact, there is (for Riemannian manifolds, I mean) $\endgroup$
    – Grigory M
    Jun 23, 2011 at 9:48
  • $\begingroup$ Thank you for your answer, though I realize that I might not have been clear. I've edited the question accordingly. $\endgroup$ Jun 23, 2011 at 9:49

Check out Kock's Synthetic Geometry of Manifolds for a nice geometric description of connection forms, curvature forms, and torsion in terms of parallel transport. My mathoverflow answer here gives an explanation of torsion from this point of view.

I'll give a run down for how to intuitively think of the connection form. Let $\nabla$ be a choice of connection (i.e. a parallel transport map) in a bundle $E \to M$. If $x$ and $y$ are infinitesimally close points (to first order, technically) in $M$ and $u$ is in the fibre $E_x$, then we can transport $u$ along the (horizontal lift of) the infinitesimal line segment linking $x$ to $y$ to get an element $\nabla(x,y)u$ in the fibre $E_y$.

In other words, for any pair of infinitesimal neighbours $x, y \in M$ we have a map $$\nabla(x,y): E_x \to E_y.$$ We just require that $\nabla(x,x) = \text{id}_{E_x}$.

If the bundle $E$ has some extra structure, we often want the parallel transport to preserve that structure. For example, for a vector bundle we probably want the maps $\nabla(x,y)$ to be linear, and for a $G$-principal bundle we probably want the maps to be $G$-equivariant. This gives us the notion of a linear connection, and a principal connection, respectively.

Now, lets see where connection forms come in. Say we have a connection $\nabla$ on a (right) $G$-principal bundle $p: P \to M$. We don't need it to be $G$-equivariant, but that would obviously be nice. Anyway, if $u$ and $v$ are infinitesimal neighbours in $P$, then we can project down to $M$ to get a pair of infinitesimal neighbours $p(u)$ and $p(v)$. Let's transport $u$ to the fibre containing $v$ along the corresponding infinitesimal curve to get the element $\nabla(p(u), p(v))u$ which lies in the same fibre as $v$. But the action of $G$ on $P$ is simply transitive, so there is a unique element of $\omega(u,v) \in G$ such that $$v.\omega(u,v) = \nabla(p(u), p(v))u.$$ The $\omega$ here is just the connection form!

But wait, you might ask, why does $\omega$ take a pair of infinitesimally close points in $P$ as arguments rather than a tangent vector to $P$, and why does it have values in $G$ rather than the lie algebra $\frak{g}$? Well, the answer to the first question should be obvious: if two points in $P$ are infinitesimally close (to first order), they lie on a tangent vector to $P$. The answer to the second question is that $\omega(u,v)$ is always an infinitesimal neighbour of the group identity $e \in G$, so it lies on a tangent vector to $G$ at $e$, and hence an element of the lie algebra $\mathfrak{g}$.

If you draw out a picture of this situation you can see that the connection form is the "vertical part" of the parallel transport on a principal bundle.

It is easy to check that $G$-equivariance of a principal connection (transport map) is equivalent to the usual equivariance of the connection form (with respect to the principal and adjoint action) in this picture.

Now, to interpret the curvature, lets just think about any bundle $\pi: E \to M$ with connection $\nabla$. Let's use the notation $x \sim y$ to indicate that two points are infinitesimal neighbours. If we have three points $x,y,z \in M$ such that $x \sim y$, $y\sim z$ and $z\sim x$. These three define an infinitesimal 2-simplex in $M$. Lets consider the transport around (the boundary of) this simplex: $$R(x,y,z) = \nabla(z,x) \circ \nabla(y,z) \circ \nabla(x,y): E_x \to E_x$$ If we transport a point $w \in E_x$ around the simplex, we have no guarantee that we end up back where we started. This is precisely the notion of curvature. The curvature measures the extent to which parallel transport around infinitesimal 2-simplices deviates from the identity.

If $E$ is a vector bundle with a linear connection, then $R(x,y,z): E_x \to E_x$ is linear, as expected. If $E$ is a principal bundle, then we can use the same trick we used for the connection form to define the $\frak{g}$-valued curvature form $\Omega(x,y,z)$ from the the curvature $R$.

  • $\begingroup$ What does it mean (precisely) for two points $x,y \in M$ to be "infinitesimally close" or "close to first order"? $\endgroup$ Jan 28, 2016 at 9:34
  • 1
    $\begingroup$ Basically, synthetic differential geometry takes inspiration from algebraic geometry, and takes place in a topos containing infinitesimal objects; The category of smooth manifolds embeds fully faithfully into this topos, so you can reason rigorously about smooth manifolds using infinitesimals like how physicists do. Two points are infinitesimal neighbours to first order if they lie on the same tangent vector, which is just an infinitesimal curve segment in SDG. $\endgroup$
    – ಠ_ಠ
    Jan 28, 2016 at 9:43

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