# Geometric interpretation of connection forms, torsion forms, curvature forms, etc

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.

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I'm treating part of this in my bachelors thesis, focussing on the geometrical intuition. Have a look at: enigmage.de/bachelor.pdf –  Turion Jul 5 '11 at 9:30
Relevant: mathoverflow.net/questions/61878/… –  Jesse Madnick Sep 19 '11 at 21:11

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"
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(Re: there isn't one for manifolds) Well, in fact, there is (for Riemannian manifolds, I mean) –  Grigory M Jun 23 '11 at 9:48
Thank you for your answer, though I realize that I might not have been clear. I've edited the question accordingly. –  Jesse Madnick Jun 23 '11 at 9:49

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 $M$: $$d\omega^2_1 = -K\,\phi^1 \wedge \phi^2.$$

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