# Lie group geometry, Clifford algebra, symmetric spaces, Kaluza-Klein, and all that

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Hey Joe,

Wow, MG11 looks cool -- have fun. No hurry on the SU(2) metric, get to it when you feel like it.

The derivation of the version of Killing's equation using the frame or vielbein instead of the metric is pretty straightforward. Start with the expression for the Lie derivative of the metric components, then plug in the expression for the metric in terms of the frame components.

Here's what may be causing confusion though: there is no pretty (index free) way, using the notation I've described, to deal with the metric. So I choose to think of Killing's equation in terms of the Lie derivative of the vielbein as fundamental. The two expressions are equivalent, so it's a matter of taste.

(And, oops, I had an index wrong in a previous expression for Killing's equation. It should be:
$${{\cal L}_\vec{\xi}} \, \vec{e_C} = B_C{}^D \vec{e_D}$$
with
$$B_C{}^D = - B^D{}_C$$
)

If you can't prove the equivalence of the two expressions for Killing's equation to your satisfaction, let me know and I'll go through it.

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Hey Garrett, finally found some wireless connectivity here at Freie University, Berlin; and for my sins I stayed up last night until I'd done my homework.... yes; I'm coked up on coffee to make up for it :).

Ok, here's what I think.

garrett said:
OK, once you believe all this, which may take a while, I'll have three "homework" questions for you:
1) Are the set of three $\vec{\xi'_B}$ also Killing, even though we've chosen them as our orthonormal basis vector fields? (Why?)
I think that the answer is yes, because ${\cal L}_{\vec{\xi'_A}} \vec{\xi'_B} = 2 \epsilon_{ABC} \vec{\xi'_C}$, and for a fixed B, the right hand side is antisymmetric in A and C, and so like rotation - it therefore fulfils the requirement for a killing vector.

However, the $\vec{\xi_A}$ fields are non-killing vectors with respect to this metric; BUT we can also form a metric out of these other fields, and they are killing vectors with respect to that metric. We appear then to have two independant metrics that this manifold can support.

2) What is the metric, $g_{ij}$ corresponding to this choice of orhonormal basis vectors?
I believe it is,

$$g'_{ij} = \frac{r^2}{\sin^2 r} \delta_{ij} + x_i x_j \left(\frac{1}{r^2} - \frac{1}{\sin^2 r} \right)$$

3) Would the metric have been different if we had chosen to use $\vec{\xi_B}$ as the orthonormal basis vectors?
Apparently not. The difference in sign of the $\epsilon$ term doesn't contribute to a change in the form of $g_{ij}$ under the replacement of $\vec{\xi'_A}$ with $\vec{\xi_A}$.

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garrett said:
Hey Joe,
Wow, MG11 looks cool -- have fun.
It should be. I'm primarily here because tomorrow David Hestenes is hosting a parallel session on Geometric Algebra and Gravity; Doran and Lasenby are here too apparently. I'm hoping to find some people who are into the conformal projective framework... I'll definitely let you know how it goes :).

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Taoy said:
It should be. I'm primarily here because tomorrow David Hestenes is hosting a parallel session on Geometric Algebra and Gravity; Doran and Lasenby are here too apparently. I'm hoping to find some people who are into the conformal projective framework... I'll definitely let you know how it goes :).
Well, you've certainly found the best bunch.

Make sure to talk with Chris Doran about black holes -- he's written a couple of especially clear papers on them in the past couple of years.

And, err, don't expect them to talk about differential forms. ;)

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Taoy said:
Hey Garrett, finally found some wireless connectivity here at Freie University, Berlin; and for my sins I stayed up last night until I'd done my homework.... yes;
Great.
I'm coked up on coffee to make up for it :).
OK, we'll see how that works out... ;)
For our set of three orthonormal vector fields, we choose
$$\vec{e_A}=\vec{\xi'_A}$$
The first question was whether these vector fields are Killing with respect to the metric associated with this orthnormal basis.
I think that the answer is yes, because ${\cal L}_{\vec{\xi'_A}} \vec{\xi'_B} = 2 \epsilon_{ABC} \vec{\xi'_C}$, and for a fixed B, the right hand side is antisymmetric in A and C, and so like rotation - it therefore fulfils the requirement for a killing vector.
Yes. Except your wording is funny. The relevant equation is:
$${\cal L}_{\vec{\xi'_A}} \vec{e_B} = 2 \epsilon_{ABC} \vec{e_C}$$
Each vector field, $\vec{\xi'_A}$, is Killing because it generates a rotation of the orthonormal basis -- $2 \epsilon_{ABC}$ is antisymmetric in B and C.
However, the $\vec{\xi_A}$ fields are non-killing vectors with respect to this metric;
Nope, actually they are also Killing.
$${\cal L}_{\vec{\xi_A}} \vec{e_B} = 0$$
The Lie derivative of the orthonormal basis vectors with respect to these vectors vanishes. If you wish to be pedantic, note that 0 is technically antisymmetric in its indices, since 0=-0.
BUT we can also form a metric out of these other fields, and they are killing vectors with respect to that metric. We appear then to have two independant metrics that this manifold can support.
You answer this below -- both choices of orthonormal basis vectors produce the same metric.
I believe it is,
$$g'_{ij} = \frac{r^2}{\sin^2 r} \delta_{ij} + x_i x_j \left(\frac{1}{r^2} - \frac{1}{\sin^2 r} \right)$$
Yes, this is what I got for the "inverse metric":
$$g^{ij} = \frac{r^2}{\sin^2 r} \delta^{ij} + x^i x^j \left(\frac{1}{r^2} - \frac{1}{\sin^2 r} \right)$$
and for the metric I got
$$g_{ij} = \frac{\sin^2 }{r^2} \delta_{ij} + x^i x^j \frac{1}{r^2} \left( 1 - \frac{\sin^2 r}{r^2} \right)$$
The difference in sign of the $\epsilon$ term doesn't contribute to a change in the form of $g_{ij}$ under the replacement of $\vec{\xi'_A}$ with $\vec{\xi_A}$.
Correct! So, choosing either set of Killing vector fields to be the orthonormal basis vectors gives the same metric for SU(2). I like to choose $\vec{e_A}=\vec{\xi'_A}$ because they have the nice property that ${\cal L}_{\vec{\xi_A}} \vec{e_B} = 0$, and the $\vec{\xi_A}$ Killing vector fields have the same commutation relations as the su(2) generators, $T_A$.

One more quick question, when you get a chance:
What's the expression for the frame (vielbein) 1-form -- corresponding to the chosen orthonormal basis vectors?
$$\underrightarrow{e}=?$$

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Gold Member
Let me answer that last question, since I was just being lazy. Also, I want to put delta functions in to raise and lower the indices properly -- which has been bothering me.

$$\underrightarrow{e}=\underrightarrow{dx^i} \left( \delta_i^B \frac{\sin(r)\cos(r)}{r} + \delta_{ij} x^j x^B ( \frac{1}{r^2} - \frac{\sin(r)\cos(r)}{r^3} ) - \delta^{BC} \epsilon_{ikC} x^k \frac{\sin^2(r)}{r^2} \right) \sigma_B$$

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garrett said:
Let me answer that last question, since I was just being lazy. Also, I want to put delta functions in to raise and lower the indices properly -- which has been bothering me.
Yay :) That was bothering me too.

p.s. have had great conversations with Hestenes and Lasenby (more the former); Doran had to cancel so no low-down on BHs.

garrett said:
Correct! So, choosing either set of Killing vector fields to be the orthonormal basis vectors gives the same metric for SU(2). I like to choose $\vec{e_A}=\vec{\xi'_A}$ because they have the nice property that ${\cal L}_{\vec{\xi_A}} \vec{e_B} = 0$, and the $\vec{\xi_A}$ Killing vector fields have the same commutation relations as the su(2) generators, $T_A$.
Hmm, so from the perspective of the metric the different between left and right invariant fields gets hidden; they are separate and distinct at the level of the vierbein though. What's the geometric meaning of this?

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Sorry to take so long to reply -- I was away for a few days, at my sister's wedding.

What's the geometric meaning of this?
Well, remember that the vielbein and metric are things we just decided to impose on our group manifold. All we really have are the two sets of three vector fields associated with the flows generated by the three generators acting from the left or right. We then chose a metric so that these flows would be symmetries, and chose a vielbein so the Lie derivative of the vielbein with respect to the Killing vectors associated with left acting generators was zero. As it happens, the vector fields associated with right acting generators, equal to the vielbein vectors, are also Killing.

What this implies, relating to Kaluza-Klein theory, is that this SU(2) manifold has more symmetries than just SU(2) -- it has SU(2)xSU(2) as its symmetry group. We can however use group theory to trim a dimension off of our SU(2) manifold to make one that only has SU(2) symmetry -- this is called a symmetric space, or coset space.

But first, we should define what a covariant derivative is, and a connection, and calculate what the connection should be for our SU(2) manifold and choice of frame.

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OK, the set of Clifford algebra vectors, $\gamma_\alpha$, are the basis elements for the flat space that is the rest frame at each manifold point. In differential geometry, these basis elements comprise a local trivialization of the Clifford fiber. The covariant derivative,
$$\underrightarrow{\nabla} = \underrightarrow{dx^i} \nabla_i$$
keeps track of how these basis elements rotate as one moves around on the base manifold:
$$\underrightarrow{\nabla} \gamma_\alpha = \underrightarrow{\omega} \times \gamma_\alpha$$
(That should be somewhat familiar from a month ago in this thread, when we discussed Clifford rotations.) The spin connection is a bivector valued 1-form,
$$\underrightarrow{\omega} = \underrightarrow{dx^i} \frac{1}{2} \omega_i{}^{\alpha \beta} \gamma_{\alpha \beta}$$
So, using this, the covariant derivative of any Clifford valued field is
$$\underrightarrow{\nabla} C = \underrightarrow{\partial} C + \underrightarrow{\omega} \times C$$

Now, since we now have a frame and a covariant derivative, the first nice object we build is the torsion, a Clifford vector valued 2-form,
$$\underrightarrow{\underrightarrow{T}} = \underrightarrow{\nabla} \underrightarrow{e} = \underrightarrow{\partial} \underrightarrow{e} + \underrightarrow{\omega} \times \underrightarrow{e}$$
If we insist that our connection be torsion free, as we often will, then we can solve
$$0 = \underrightarrow{\partial} \underrightarrow{e} + \underrightarrow{\omega} \times \underrightarrow{e}$$
explicitly for the spin connection. It's not so easy to find, but this equation has a closed form solution!

Here's a question:
What's the above equation look like if we write it out in components?
I'll begin:
$$0 = \partial_{\left[ i \right.} \left( e_{\left. j \right]} \right)^\alpha + ?$$

garrett said:
Let me answer that last question, since I was just being lazy. Also, I want to put delta functions in to raise and lower the indices properly -- which has been bothering me.

$$\underrightarrow{e}=\underrightarrow{dx^i} \left( \delta_i^B \frac{\sin(r)\cos(r)}{r} + \delta_{ij} x^j x^B ( \frac{1}{r^2} - \frac{\sin(r)\cos(r)}{r^3} ) - \delta^{BC} \epsilon_{ikC} x^k \frac{\sin^2(r)}{r^2} \right) \sigma_B$$
Wait; how many Clifford bases are we using here? Are you using $\sigma_B \equiv \gamma_B$, or are these bases separate and distinct?

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Taoy said:
Wait; how many Clifford bases are we using here? Are you using $\sigma_B \equiv \gamma_B$, or are these bases separate and distinct?
Ah, sorry, good question -- and answer. I use $\gamma_\alpha$ for the general case of Clifford vector elements, and in this specific case of our three dimensional group manifold we do have $\sigma_B = \gamma_B$ -- so feel free to interchange sigma's for gamma's, and Greek indices for capital latin ones.

(This post has been corrected.)

garrett said:
What's the above equation look like if we write it out in components?
I'll begin:
$$0 = \partial_{\left[ i \right.} \left( e_{\left. j \right]} \right)^\alpha + ?$$
Well expanded it looks like,

$$0 = \underrightarrow{dx^i} \underrightarrow{dx^j} \left( \partial_i (e_j)^\alpha \gamma_\alpha + \frac{1}{2} \omega_i{}^{\mu \nu} (e_j)^\alpha \gamma_{\mu \nu} \times \gamma_\alpha \right)$$

So, what is the clifford commutator? I would write:

\begin{align*} \gamma_{\mu \nu} \times \gamma_\alpha & = \frac{1}{2} \left( \gamma_\mu \wedge \gamma_\nu \gamma_\alpha - \gamma_\alpha \gamma_\mu \wedge \gamma_\nu \right) \\ & = \frac{1}{2} \left( \gamma_\mu \wedge \gamma_\nu \cdot \gamma_\alpha + \gamma_\mu \wedge \gamma_\nu \wedge \gamma_\alpha - \gamma_\alpha \cdot \gamma_\mu \wedge \gamma_\nu - \gamma_\alpha \wedge \gamma_\mu \wedge \gamma_\nu \right) \\ & = \frac{1}{2} \left( \gamma_\mu \wedge \gamma_\nu \cdot \gamma_\alpha - \gamma_\alpha \cdot \gamma_\mu \wedge \gamma_\nu \right) \\ & = \frac{1}{2} \left( \gamma_\mu \delta_{\nu \alpha} - \gamma_\nu \delta_{\mu \alpha} - \delta_{\alpha \mu} \gamma_\nu + \delta_{\alpha \nu} \gamma_{\mu} \right) \\ & = \gamma_\mu \delta_{\nu \alpha} - \gamma_\nu \delta_{\mu \alpha} \end{align*}

therefore,

$$0 = \underrightarrow{dx^i} \underrightarrow{dx^j} \left( \partial_i (e_j)^\mu + \frac{1}{2} (\omega_i{}^{\mu \nu} - \omega_i{}^{\nu \mu}) (e_j)^\alpha \delta_{\nu \alpha} \right) \gamma_\mu$$

and so the components are,

$$\partial_{[i} (e_{j]})^\mu + \frac{1}{2} (\omega_{[i}{}^{\mu \nu} - \omega_{[i}{}^{\nu \mu}) (e_{j]})^\alpha \delta_{\nu \alpha}$$

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Gold Member
Sign error, probably because you insist on using that evil wedge. ;)
The cross product is defined as
$$A \times B = \frac{1}{2} ( AB - BA )$$
and the bivector basis elements are
$$\gamma_{\mu \nu} = \gamma_\mu \times \gamma_\nu$$
The derived identity you need is
$$\gamma_{\mu \nu} \times \gamma_\alpha = \gamma_\mu \delta_{\nu \alpha} - \gamma_\nu \delta_{\mu \alpha}$$

garrett said:
Sign error, probably because you insist on using that evil wedge. ;)
Pah! :) In that context the wedge is just as evil as that evil $\times$ :). No, it was just me using the wrong sign :). I was in a hurry to get to a class, so I wasn't as careful as I should have been. I'll fix the original post (so as to avoid unnecessary bad posts :).

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Yep, that's it. Only $\omega_{i}{}^{\nu \mu}$ is antisymmetric in the last two indices, and we can use the delta to lower an index, so in components the equation
$$0 = \underrightarrow{\partial} \underrightarrow{e} + \underrightarrow{\omega} \times \underrightarrow{e}$$
is equivalent to
$$0 = \partial_{[i} (e_{j]})^\mu + \omega_{[i}{}^\mu{}_\alpha (e_{j]})^\alpha$$
or, if you like doing things half way, and want to write it in terms of the frame 1-forms,
$$0 = \underrightarrow{\partial} \underrightarrow{e}^\mu + \underrightarrow{\omega}^\mu{}_\alpha \underrightarrow{e}^\alpha$$

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This is a famous equation called "Cartan's first structure equation." Remarkably, it is solvable for the spin connection. To find the solution, it is convenient to define this intermediate quantity:
$$\underrightarrow{\underrightarrow{f}} = \underrightarrow{\partial} \underrightarrow{e}$$
the "anholonomy," a Clifford vector valued 2-form. The not so famous solution of Cartan's first structure equation,
$$0=\underrightarrow{\underrightarrow{f}} + \underrightarrow{\omega} \times \underrightarrow{e}$$
is
$$\underrightarrow{\omega} = - \vec{e} \times \underrightarrow{\underrightarrow{f}} + \frac{1}{4} \left( \vec{e} \times \vec{e} \right) ( \underrightarrow{e} \cdot \underrightarrow{\underrightarrow{f}} )$$
How's that for a blaze of notation! Of course, to actually calculate anything we're going to have to slug it out with indices. To make things easier, we introduce the idea of using the frame or orthonormal basis matrices to change an index from latin coordinate indices to Greek frame labels and back again -- similar to the way we use delta to raise and lower Greek indices. This is consistent with the definition of the metric in terms of the frame, and the way it's often used to raise and lower indices. So, for example:
$$\omega_{i j}{}^\mu = \omega_{i \alpha}{}^\mu \left( e_j \right)^\alpha$$
In this way, Cartan's first structure equation is
$$0 = \frac{1}{2} f_{ij}{}^\mu - \omega_{\left[ i j \right]}{}^\mu$$
or, using
$$f_{\alpha \beta}{}^\mu = \left( e_\alpha \right)^i \left( e_\beta \right)^j \left( \partial_i \left( e_j \right)^\mu - \partial_j \left( e_i \right)^\mu \right)$$
it's simply
$$\frac{1}{2} f_{\alpha \beta}{}^\mu = \omega_{\left[ \alpha \beta \right]}{}^\mu$$
And using some index gymnastics gives the explicit solution,
$$\omega_{\alpha \beta \nu} = \frac{1}{2} \left( f_{\alpha \beta \nu} - f_{\beta \nu \alpha} + f_{\nu \alpha \beta} \right)$$
This is the closed form solution to Cartan's first structure equation, giving the spin connection coefficients in terms of the frame and its derivatives. It's use is equivalent in pain and practicality to calculating the Christoffel symbol coefficients from a metric. One could just plug in our frame and calculate... but we're going to be smarter. I think.

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Gold Member
$$0 = \underrightarrow{\partial} \underrightarrow{e}^\mu + \underrightarrow{\omega}^\mu{}_\alpha \underrightarrow{e}^\alpha$$
and we want to calculate $\omega$, having already chosen a set of orthonormal basis vectors for our group manifold. Where did we get them again? We started with a relation between Killing vector flows and the action of our Lie algebra generators:
$$\vec{\xi'_B} \underrightarrow{\partial} g = g T_B$$
solved this for the Killing vector fields, and chose them to be our orthonormal basis vectors. Since it pops up everywhere, lets go ahead and assign a symbol to this Lie algebra valued 1-form, the Cartan form:
$$\underrightarrow{w} = g^- \underrightarrow{\partial} g = \underrightarrow{\xi'}^B T_B = \underrightarrow{e}^B T_B$$
The generator orthogonality relation then let us calculate the frame 1-forms as
$$\underrightarrow{e}^B = - \left< \underrightarrow{w} T^B \right>$$
We can also use this to calculate the exterior derivative of these 1-forms. The derivative of the Cartan form is
$$\underrightarrow{\partial} \underrightarrow{w} = (\underrightarrow{\partial} g^-)(\underrightarrow{\partial} g) = - g^- (\underrightarrow{\partial} g) g^- (\underrightarrow{\partial} g) = - \underrightarrow{w} \underrightarrow{w}$$
So the exterior derivative of the frame 1-forms is
$$\underrightarrow{\partial} \underrightarrow{e}^B = - \left< \underrightarrow{\partial} \underrightarrow{w} T^B \right> = \left< \underrightarrow{w} \underrightarrow{w} T^B \right> = \underrightarrow{e}^A \underrightarrow{e^C} \left< T_A T_C T^B \right> = ( \underrightarrow{e}^A \epsilon_{AC}{}^B ) \underrightarrow{e^C}$$
and comparing this with Cartan's first structure equation, which we're trying to solve, we see the solution:
$$\underrightarrow{\omega}^B{}_C = - \underrightarrow{e}^A \epsilon_{AC}{}^B$$
This means the spin connection coefficients (with frame label indices) are equal to minus the group structure constants
$$\omega_{ABC} = \epsilon_{ABC}$$
This is really neat, and it gets better. The spin connection bivector is
$$\underrightarrow{\omega} = \underrightarrow{e}^A \frac{1}{2} \epsilon_{A}{}^{CB} \sigma_{CB}$$
and, if you'll recall, the Lie algebra generators could also be expressed as Clifford bivectors,
$$T_A = \frac{1}{2} \epsilon_{A}{}^{CB} \sigma_{CB}$$
$$\underrightarrow{\omega} = \underrightarrow{e}^A T_A = \underrightarrow{w}$$
The spin connection bivector is Cartan's form.

Homework assignment:

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What is then the real use Cartan's first structure equation and the spin connection in what is above ?

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The spin connection describes how the frame rotates as you move around on the manifold. Cartan's first structure equation is the mathematical embodiment of this statement.

How can we write the spin connection (and the Cartan's first structure equation ) if the manifold is a simple torus ?

same question if the manifold now is $$z=5-x^2-y^2$$

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Gold Member
Mehdi_ said:
How can we write the spin connection (and the Cartan's first structure equation ) if the manifold is a simple torus ?
0

same question if the manifold now is $$z=5-x^2-y^2$$
Ah, an embedded surface. That's going to take a couple of pages of algebra to work out the frame and connection. I encourage you to tackle it on a different thread. ;) I'll give you a hint though: treat this manifold as a parameterized surface, embedded in 3D, with coordinates (parameters) x and y. It's going to take some work to get the coefficients of two orthonormal tangent vectors over this surface,
$$\vec{e_1} = a \vec{\frac{\partial}{\partial x}} + b \vec{\frac{\partial}{\partial y}}$$
$$\vec{e_2} = c \vec{\frac{\partial}{\partial x}} + d \vec{\frac{\partial}{\partial y}}$$
, then more work to get the frame, and more to get the connection. Maybe start by choosing c=0, and solve for a,b,c.

I'll look in on the other thread and see how you're doing.

Question 1: To calculate the spin connection of $$z=5-x^2-y^2$$ could we calculate first :
The metric, Ricci Rotation coefficients, christoffel symbols, those orthonomal basis (why not nonholonomic basis?), Riemann tensor, Ricci tensor, Ricci scalar, tetrad method and curvature one forms ??

Question 2: Why it is so important to know the curvature ? Does the spin connection (or maybe Cartan's first structure equation) give information about the curvature ?
Probably Riemann curvature tensor does ?...
What is the relation between the spin connection and Riemann curvature tensor ?

Question 3: Does the curvature give the strength of the field ?

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