Tetrads and differential forms

21.1 Converting coordinates to laboratory components#

Coordinate bases are versatile, but they need not look like a laboratory’s orthogonal ruler-and-clock axes. In spherical coordinates, a change of one radian is not a change of one meter. For a local experiment we often want an orthonormal basis instead.

For example, in flat cylindrical coordinates,

the one-forms $c\,dt$, $dr$, $r\,d\phi$, and $dz$ directly measure components along orthonormal clock-and-ruler directions. The factor $r$ converts an angular coordinate increment into a local length. Generalize this construction by introducing four one-forms

such that

In this chapter only, $a,b,c,d=0,1,2,3$ label internal orthonormal-frame directions, not the spatial indices $i,j,k$ used elsewhere. Greek indices still label spacetime coordinate components.

The $e^a{}_\mu$ are called a tetrad, vierbein, or coframe. Its inverse $e_a{}^\mu$ defines basis vectors $e_a=e_a{}^\mu\partial_\mu$. Their duality says $e^a(e_b)=\delta^a_b$. A vector’s laboratory components are

The metric is a kind of matrix square built from $e$, but not an ordinary unique positive square root. Six choices remain free: at every point we may rotate the three laboratory axes and boost the laboratory’s time axis.

More precisely, if $L^a{}_b(x)$ obeys $L^T\eta L=\eta$, then

produces the same metric. Sixteen tetrad components minus six local Lorentz freedoms leave the metric’s ten components. Coordinate freedom remains as well; this count describes the additional frame representation, not the physical degrees of freedom counted in Chapter 20.

An orthonormal frame makes the metric’s frame components equal to $\eta_{ab}$ at every point where that frame is defined. Curvature can still be nonzero: the frame’s comparison law changes across the region. The polar-frame calculation below will distinguish this comparison law from curvature itself.

21.2 Oriented measurements and differential forms#

A one-form takes one vector and returns a number. A two-form takes two vectors and returns an antisymmetric number, naturally measuring oriented area. A $p$-form generalizes this to $p$ directions.

The wedge product antisymmetrizes:

For one-forms $\alpha$ and $\beta$,

This is the determinant of a two-by-two array. It measures the signed parallelogram area seen by the two measuring devices. Exchange the sides and orientation reverses; use the same side twice and the area vanishes.

For a $p$-form and a $q$-form,

Thus two two-forms commute under the wedge product, while two one-forms anticommute. The sign follows from moving $p$ directions past $q$ directions: $pq$ exchanges.

The exterior derivative differentiates the coefficient functions and adds the derivative’s coordinate one-form on the left. For

its rule is

The coefficient array is antisymmetric; $1/p!$ compensates for summing its $p!$ signed permutations. The derivative raises the form degree by one. For a scalar,

For a one-form $A=A_\mu dx^\mu$,

The factor $1/2$ prevents counting each antisymmetric pair twice. For example, if $A=x^2dy$, then $dA=2x\,dx\wedge dy$.

Two rules do most of the work:

Why does $d^2f=0$? The second partial derivatives are symmetric under index exchange, while the wedge basis is antisymmetric. Their contraction cancels. This is the common engine behind “the curl of a gradient vanishes” and “the divergence of a curl vanishes.”

The exterior derivative needs no metric or connection. It does not tell you the full directional variation of an arbitrary tensor. It extracts a particular antisymmetric derivative of differential forms. It therefore complements $\nabla$; it does not replace it.

Forms also make Stokes’ theorem compact:

The boundary integral and interior derivative are two descriptions of the same accumulated oriented change. Appropriate orientations and smoothness are part of the statement.

WORKED EXAMPLE

Calculate both sides of Stokes’ theorem

Why can a measurement around an edge equal an accumulation over its interior?

Builds onIntegrals add local measurements Covectors and differential measurements

See the idea

Tile a region with little squares and walk around each square counterclockwise. Every shared edge is traversed twice in opposite directions, so its contributions cancel. Only the outer boundary remains. Differential forms organize the signed measurements needed to turn that picture into a calculation.

Work it out

1. 01
   Make a one-form into a measurable path integral

   Use dimensionless coordinates on a rectangle $0\le x\le a$, $0\le y\le b$. Let $\alpha=x^2dy$. Along a path $s\mapsto(x(s),y(s))$, this means integrate $x(s)^2y'(s)\,ds$. Reversing the path reverses the sign. Horizontal segments contribute zero because $dy=0$.

   $$\int_\gamma\alpha=\int_{s_0}^{s_1}x(s)^2\frac{dy}{ds}\,ds.$$

   Why this step works A one-form acts on the path’s tangent at each point; an ordinary single-variable integral adds those local readings.

2. 02
   Walk the four edges

   Choose counterclockwise orientation. The bottom edge has $dy=0$. The right edge has $x=a$ and $y:0\to b$, contributing $a^2b$. The top is horizontal. The left has $x=0$, so it contributes zero even though it is traversed downward.

   $$\oint_{\partial\Omega}\alpha=0+\int_0^b a^2\,dy+0+\int_b^0 0^2\,dy=a^2b.$$

   Why this step works The boundary’s direction matters. Clockwise traversal would negate the answer.

3. 03
   Measure oriented area with a determinant

   For two arrows $V=(V^x,V^y)$ and $W=(W^x,W^y)$, define $(dx\wedge dy)(V,W)=V^xW^y-V^yW^x$. It is their signed area. Swapping the arrows flips the sign, so $dy\wedge dx=-dx\wedge dy$. Here $d\alpha=d(x^2)\wedge dy=2x\,dx\wedge dy$.

   $$\int_\Omega d\alpha=\int_0^a\!\int_0^b2x\,dy\,dx=a^2b.$$

   Why this step works We choose dx wedge dy as the positive orientation. The compatible boundary traversal is counterclockwise.

4. 04
   Recover the general planar rule

   For $\alpha=P\,dx+Q\,dy$, differentiate the coefficients. Terms with $dx\wedge dx$ or $dy\wedge dy$ vanish; the reversed pair contributes a minus sign. This gives the signed interior density whose integral equals the counterclockwise boundary integral.

   $$d\alpha=(\partial_xQ-\partial_yP)\,dx\wedge dy,\qquad \int_\Omega d\alpha=\oint_{\partial\Omega}\alpha.$$

   Why this step works Oppositely traversed internal edges cancel. The difference of neighboring edge readings becomes a derivative as the tiles shrink.

Go deeper

The general theorem needs an oriented manifold with boundary, a sufficiently smooth form of the appropriate degree, and the boundary orientation induced by the interior. On a compact oriented surface, $\alpha$ is a one-form and $d\alpha$ a two-form. In $n$ dimensions, one integrates an $(n-1)$-form on the boundary and its exterior derivative on the interior. Singularities and omitted points matter: a form undefined at a puncture cannot be passed through that puncture by invoking Stokes’ theorem. The remaining text develops the algebra that makes the same calculation independent of the chosen coordinates.

Further readingMIT · Measures on surfaces and Stokes’ theorem

Test the idea

FIRST, PREDICT

If we reverse both the chosen surface orientation and its induced boundary direction, what happens?

Only the boundary integral changes sign.Neither changes sign because geometric area is positive.Both integrals change sign, and they remain equal.

Compare the reasoning

Only the boundary integral changes sign.

An integral of a differential form over an oriented surface also changes sign when the surface orientation is reversed.

Neither changes sign because geometric area is positive.

A differential form measures oriented area. It is not an unsigned area density.

Both integrals change sign, and they remain equal.

Yes. Orientation records which order of directions counts as positive. Stokes’ theorem uses compatible orientations on both sides.

A hint

The wedge product changes sign when its two input directions are exchanged.

NOW CHANGE THE EXAMPLE

For $\alpha=xy\,dy$, integrate counterclockwise around the dimensionless rectangle $0\le x\le2$, $0\le y\le3$.

A hint

Either integrate along the right edge, or use $d\alpha=y\,dx\wedge dy$.

Work through the solution

Only the right edge contributes: $\int_0^3 2y\,dy=9$. The interior gives $\int_0^2\int_0^3 y\,dy\,dx=9$ as well.

Orientation makes neighboring boundary contributions cancel; Stokes’ theorem records the resulting equality.

21.3 Why an orthonormal frame needs a spin connection#

Two nearby laboratories may choose differently rotated or boosted axes. Differentiating their component lists naively confuses physical change with a change of reference frame. That is precisely the problem that connections solve.

Introduce the matrix of one-forms

and define, for internal vector components,

This is the spin connection, despite its usefulness even before spinors appear. Metric compatibility in the orthonormal frame implies

Antisymmetry applies after lowering the first frame index. A boost component can therefore satisfy $\omega^0{}_1=\omega^1{}_0$ with both indices in the displayed mixed positions. Lowering the first index of the time component multiplies it by $\eta_{00}=-1$, which accounts for the apparent difference between these two statements.

The relationship to Christoffel symbols follows by demanding that conversion between coordinate and frame components commute with differentiation:

This is the tetrad postulate. It is a compatibility relation among two representations of the same connection and the map between their bases. Once the tetrad and Levi-Civita connection are specified, it determines the associated spin connection.

Under $V'=LV$, require $DV$ to transform as $D'V'=L(DV)$. Expanding $d(LV)$ produces an unwanted $(dL)V$ term. The connection cancels it precisely if

This is why a connection transforms inhomogeneously. The derivative of the position-dependent frame change produces the extra term.

21.4 Torsion and curvature in a moving frame#

The torsion two-form is

For the Levi-Civita connection of ordinary GR, $\mathcal T^a=0$, so

This is the torsion-free first Cartan structure equation. To connect it to Chapter 7, antisymmetrize the tetrad postulate in $\mu,\nu$. The derivative and spin-connection terms give

Thus the two-form expresses the same torsion tensor in frame components. Zero torsion together with metric compatibility determines $\omega$ from the tetrad.

The curvature two-form is

This is the second structure equation. The matrix product includes a sum over $c$ and a wedge product of its one-form entries. Curvature has components

With our Riemann convention these agree with the curvature defined by $[\nabla_\mu,\nabla_\nu]$. For a vector-valued form, $D=d+\omega\wedge$ combines exterior differentiation with the frame correction. Apply it twice to the component functions $V$:

The derivatives of $V$ cancel. The remaining matrix is the same curvature that controls infinitesimal loop transport.

The connection’s inhomogeneous transformation disappears from curvature:

This does not mean its component matrix is invariant. It means curvature transforms tensorially, so observers can compare the same geometric object in their different frames. For a detailed frame-based development, see David Tong’s author-written chapter on connections and Cartan geometry.

21.5 Worked example: a rotating frame can have connection without curvature#

Consider the Euclidean plane, away from the origin, in polar coordinates:

This two-dimensional example uses only spatial frame indices. Compute

Choose

The second first-structure equation reads

The first equation vanishes too, because $d\phi\wedge d\phi=0$. The nonzero connection records how the radial and angular unit vectors rotate as $\phi$ changes.

But

on a regular angular coordinate patch; the matrix wedge contribution also vanishes here. The plane is flat. The polar frame is undefined at the origin, but a bad frame does not create a physical curvature singularity there.

Now replace the plane by a sphere of radius $a$:

Then $de^2=a\cos\theta\,d\theta\wedge d\phi$. The same cancellation method gives

Now the connection’s coefficient varies with latitude:

Its Gaussian curvature is $1/a^2$, and its scalar curvature is $2/a^2$. The same two-line machinery has distinguished a rotating choice of axes from actual curved geometry.

21.6 Frame symmetry and other gauge theories#

Electromagnetism uses a potential one-form $A$ and field strength $F=dA$. In a non-Abelian gauge theory the internal transformations need not commute. Its connection is matrix-valued, so products of different connection matrices need not cancel. The curvature has the form $F=dA+A\wedge A$. The spin connection obeys the same geometric pattern.

The shared idea is a freedom to choose a local reference convention, accompanied by a connection that compares neighboring conventions. It is not an assertion that gravity is ordinary electromagnetism with a larger alphabet.

In GR, the tetrad ties the internal Lorentz frame to actual tangent directions: it connects the gauge description to rods, clocks, causal cones, and volume. The Einstein-Hilbert action is linear in curvature, while the usual Yang-Mills action is quadratic in its field strength. The choice of action gives these theories different equations of motion even though their connection and curvature formulas resemble one another.

Further calculation: how spinor frames rotate

Tetrads also let us couple spin-$1/2$ fields to gravity. A spinor has complex components whose rotation and boost rules differ from those of a spacetime vector. For example, a spin-$1/2$ state acquires a minus sign under a full $2\pi$ rotation and returns to itself after $4\pi$; an overall sign alone does not change its measurement probabilities. This is a property of a quantum transformation law, not a small object literally spinning inside the particle. Constructing that representation is new material from quantum theory, not a consequence we have already proved using tensors.

For the four-component Dirac spinor $\psi$, use four $4\times4$ gamma matrices $\gamma^a$, one for each local frame direction. Their entries act on the spinor components; the label $a$ is not a matrix-row index. The anticommutator is $\{A,B\}=AB+BA$, where matrix multiplication need not commute. Choose matrices satisfying

$$\{\gamma^a,\gamma^b\}=2\eta^{ab}I.$$

Here $I$ is the identity matrix. In our signature, the relation says $(\gamma^0)^2=-I$, $(\gamma^i)^2=I$ for each spatial direction, and distinct gamma matrices anticommute. The relation imports the algebra used for spinors; it does not supply the dynamics of a quantum field. With this convention, a compatible spinor derivative is

$$D_\mu\psi=\partial_\mu\psi +\frac14\omega_{ab\mu}\gamma^a\gamma^b\psi.$$

The matrix algebra lets us check this frame law. Define $\Sigma^{ab}=[\gamma^a,\gamma^b]/4$. Using the anticommutator twice gives

$$[\Sigma^{ab},\gamma^c]=\eta^{bc}\gamma^a-\eta^{ac}\gamma^b.$$

The spinor connection $\omega_{ab\mu}\Sigma^{ab}/2=\omega_{ab\mu}\gamma^a\gamma^b/4$ therefore transforms the gamma matrices consistently with a Lorentz vector index. For a rotation in the 1–2 plane, its finite matrix can be written

$$S(\theta)=\exp(\theta\gamma^1\gamma^2/2) =I\cos(\theta/2)+\gamma^1\gamma^2\sin(\theta/2).$$

The equality follows by separating the even and odd powers and using $(\gamma^1\gamma^2)^2=-I$. At $2\pi$ it is $-I$; at $4\pi$ it is $I$. This derives the half-angle rotation rule from the supplied matrix algebra. The direction sign depends on whether we rotate a frame or a state; either convention has the same full-turn behavior. Constructing quantum-field dynamics still requires additional physical input. The connection is still solving the familiar problem of comparing components defined using different local frames. Globally, the spinor transformation rules on overlapping patches must fit together consistently. Such a choice is called a spin structure, and its existence depends on topology. Local gamma matrices alone do not establish it. Tong’s introduction to the spinor representation provides that continuation; translate its metric-sign convention when comparing formulas.

21.7 What changes with an independent connection#

Three different geometric properties deserve three different names:

Standard GR uses a torsion-free, metric-compatible connection, while allowing curvature. More general theories can change these assumptions. Calling a curved spacetime “twisted” in ordinary speech does not imply nonzero mathematical torsion.

In the Palatini approach, vary the metric and connection independently in an Einstein-Hilbert-type action. Under the usual assumptions—four dimensions, a nondegenerate metric, a torsion-free independent connection, and matter action independent of that connection—the connection equation enforces the Levi-Civita connection. Substituting it back recovers metric GR.

Why does this work? The action’s curvature is linear in derivatives of the connection. Integrating by parts transfers those derivatives onto $\sqrt{-g}g^{\mu\nu}$. The resulting equation requires compatibility of the connection with that metric density; in dimensions above two, under these assumptions, it reduces to metric compatibility. Torsion freedom then selects the unique Levi-Civita connection.

There are important exceptions to the slogan “independent connection variation always gives GR.” For a completely general connection, a change $\Gamma^\rho{}_{\mu\nu}\mapsto\Gamma^\rho{}_{\mu\nu}+\delta^\rho{}_\nu A_\mu$ changes Ricci only by $\partial_\mu A_\nu-\partial_\nu A_\mu$. Contracting that antisymmetric change with $g^{\mu\nu}$ gives zero, so the scalar-curvature action alone cannot determine $A_\mu$. This is projective freedom. The added term generally introduces torsion, so it was excluded by the torsion-free assumption in Chapter 14. Connection-dependent matter changes the connection equation; spinor matter can source torsion in Einstein-Cartan formulations. Replacing $R$ by a nonlinear function $f(R)$ generally makes metric and Palatini variation different theories. The careful equivalence statement is analyzed in Dadhich and Pons’s paper on Einstein-Hilbert and Einstein-Palatini formulations.

Using forms, a corresponding first-order gravitational action can be written, with $c=1$ and a consistently chosen orientation,

Here $\varepsilon_{0123}=+1$ is the internal alternating symbol and $\mathcal R^{cd}=\eta^{de}\mathcal R^c{}_e$. For an invertible tetrad, a Lorentz-compatible independent connection, and no torsion-sourcing matter, varying the connection imposes zero torsion; varying the tetrad gives Einstein’s equation. The apparent change of language has exposed a new organization of the same dynamics.

To check the action’s normalization, the alternating-symbol contraction gives $\varepsilon_{abcd}e^a\wedge e^b\wedge\mathcal R^{cd}=2R\,\mathrm{vol}$ and $\varepsilon_{abcd}e^a\wedge e^b\wedge e^c\wedge e^d=24\,\mathrm{vol}$, where $\mathrm{vol}=\sqrt{-g}\,d^4x$ in the chosen orientation. The two terms therefore reproduce $(R-2\Lambda)\mathrm{vol}/(16\pi G_N)$. The frame variables give the same gravitational action when the stated compatibility and torsion conditions hold.

21.8 Your turn: make the sphere calculation work harder#

The plane and sphere calculations in Section 21.5 used the same structure equations. Recover that method on a surface of revolution,

Here $u$ and $f$ have units of length and $\phi$ is dimensionless. Choose $e^1=du$ and $e^2=f(u)d\phi$. Before calculating, predict what distinguishes a rotating coframe from a genuinely curved surface.

Try independently. Find the connection one-forms and $\mathcal R^1{}_2$. Obtain the Gaussian curvature. Then compare $f(u)=u$, $f(u)=a\sin(u/a)$, and $f(u)=a\sinh(u/a)$ on regular patches. Finally, explain what happens for a cylinder with constant $f=a$.

Compare your derivation, including the signs

Differentiation gives $de^2=f'(u)du\wedge d\phi$. The torsion-free equation requires

$$\omega^2{}_1=f'(u)d\phi,\qquad \omega^1{}_2=-f'(u)d\phi.$$

The diagonal connection entries vanish, so the matrix wedge term in this curvature component vanishes. Therefore

$$\mathcal R^1{}_2=-f''(u)du\wedge d\phi =-\frac{f''(u)}{f(u)}e^1\wedge e^2, \qquad K=-\frac{f''}{f},\qquad R=2K.$$

The plane has $K=0$ despite a nonzero connection in this polar coframe. The sphere has $K=1/a^2$; the hyperbolic metric has $K=-1/a^2$. The cylinder has $K=0$: its bending in an ambient three-dimensional picture is extrinsic and does not create intrinsic Gaussian curvature. The condition $f>0$ defines the regular coordinate/frame patch, not a claim that every excluded endpoint is a physical singularity.

Change the problem. If $f(u)=a\exp(u/a)$, determine $K$ and decide whether a nonzero connection alone would have told you its sign. Check the answer by differentiating $f$ twice, without referring to the three cases above.