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

garrett

A few friends have expressed an interest in exploring the geometry of symmetric spaces and Lie groups as they appear in several approaches to describing our universe. Rather than do this over email, I've decided to bring the discussion to PF, where we may draw from the combined wisdom of its denizens.

Whenever possible, I will be adding and referring to material on my related personal research wiki:

http://deferentialgeometry.org

The meta-idea is to have a linear discussion and development (kind of a mini-course) here on PF, while building up the wiki as a reference and personal research tool for this and other physics inquiries. This will provide background material and (hopefully) enforce a consistent mathematical notation for the discussion. I'm hoping this dual resource use will provide the best of both mediums.

The subjects I/we would like to cover in this thread include:

Lie algebra generators, T_A, (using su(2) as simplest nontrivial example)
(matrix representation, Clifford bivectors, or other Clifford elements)
structure coefficients, (maybe Lie algebra roots, weights, and classification)
exponentiation, g = exp(x^A T_A), giving Lie group elements (SU(2) example)
coordinate change, double covers (SU(2) vs SO(3))
symmetry: Killing vector fields related to generator action
local Lie group manifold geometry -- frame, connection, and curvature
symmetric spaces
Kaluza-Klein theory
appearance and incorporation of Higgs scalars
Peter-Weyl theorem and its use for calculating harmonics

And wherever else the discussion takes us. I'd like things to be (perhaps painfully) specific and pedantic -- relying on explicit examples. I'd like to mostly play with SU(2) and SU(3) as the simplest non-trivial and practical examples. What I'm after is to fully describe these groups as manifolds in terms of their local geometry, symmetries, global geometry, harmonics, etc. And show how they can be incorporated into Kaluza-Klein theory.

I'll usually ask questions at the end of posts. Sometimes I'll know the answer, and sometimes I won't. These will either serve as "homework" (I'll wait 'till someone (PF'ers welcome) answers correctly before proceeding) or as open questions hopefully leading to me learning stuff. (If you want to play, it will help if you have Mathematica or Maple available to use -- or it may be possible to do things the hard way.) I'll also happily answer questions (or meta-questions) related to the posts, probably with references to the wiki.

I'm not sure exactly where this will go or how it will evolve as a discussion, but I thought it would be fun to try here on PF. Now I need to add the first post to this zero-eth one...

garrett

Look here for related background material:

http://deferentialgeometry.org/#%5B%5BLie%20group%5D%5D

A Lie group, in contrast to any old group, is also a manifold. This manifold can be given a metric, and hence a geometry, such that the flows induced by the Lie algebra generators corresponds to Killing vector fields. It will be good to work this out explicitly for a specific example.

The three Lie algebra generators for su(2) may be represented by 2x2 traceless anti-Hermitian matrices related to the Pauli matrices,
[tex]
\begin{array}{ccc}
T_1 = i \sigma_{1}^{P} = \left[\begin{array}{cc}
0 & i\\
i & 0\end{array}\right] &
T_2 = i \sigma_{2}^{P}=\left[\begin{array}{cc}
0 & 1\\
-1 & 0\end{array}\right] &
T_3 = i \sigma_{3}^{P}=\left[\begin{array}{cc}
i & 0\\
0 & -i \end{array}\right]\end{array}
[/tex]
From the resulting multiplicative relation,
[tex]
T_A \times T_B = \frac{1}{2} \left( T_A T_B - T_B T_A \right) = - \epsilon_{ABC} T_C
[/tex]
the structure coefficients for this Lie algebra are equal to minus the permutation symbol, [itex]C_{AB}{}^C= -\epsilon_{ABC}[/itex]. Also, the trace of two multiplied su(2) generators provides a useful orthogonality relation,
[tex]
\left< T_A T_B \right> = \frac{1}{2} Tr(T_A T_B) = - \delta_{AB}
[/tex]

Near the identity, elements of a Lie group can be approximately represented using coordinates multiplying the corresponding Lie algebra generators,
[tex]
g \simeq 1 + x^i T_i = 1 + T
[/tex]
In which
[tex]
T = x^i T_i = \left[\begin{array}{cc}
i x^3 & i x^1 + x^2\\
i x^1 - x^2 & -i x^3\end{array}\right]
[/tex]
is a general Lie algebra element labeled by coordinates, [itex]x^i[/itex]. In fact, for SU(2), all group elements can be exactly represented by exponentiating Lie algebra elements,
[tex]
g = e^T = 1 + T + \frac{1}{2!} T^2 + \frac{1}{3!} T^3 + ...
[/tex]
This gives all [itex]g[/itex] as 2x2 coordinatized unitary matrices with unit determinant.

The first "homework" question is:

What is this g matrix, explicitly, in terms of these coordinates?

Some hints:

Define a new quantity,
[itex]r = \sqrt{(x^1)^2+(x^2)^2+(x^3)^2}[/itex]
What do you get for arbitrary powers of T?
Use the series expansions for sin and cos of r.
Write the answer in terms of sin and cos of r, and T.

Taoy

Ok, I'm going to start to digest this, piece by piece :). (This could get messy if we don't have sub-threads ;o) ).

Flows and Killing vector fields.

I've not seen this explicitly. When I first came across group manifolds and constructed metrics on them it was in terms of the left-invariant one-forms,
[tex]\lambda_a T_a = g^{-1} dg[/tex];
I guess that formally these are refered to as Maurer-Cartan forms. For an N dimensional group there are N of these which form a basis for the manifold,
[tex]ds^2 = \sum_i (\lambda_i)^2 [/tex].
There should also be some vectors dual to these one-forms; how do these relate to the killing vectors fields?

In general there could be up to N(N+1)/2 killing vectors (or is that killing fields?), which come from the infinitessimal isometries of the metric
[tex]\xi_{a;b} + \xi_{b;a} = 0[/tex]
whereas there are only ever going to be N one-forms. :/

garrett

Hi Joe, welcome to Physics Forums.

I was going to get into Killing vectors -- just as soon as someone writes down exactly what g is...

The key expression for calculating the Killing vector fields is going to be:
[tex]
\vec{\xi_A} \underrightarrow{d} g = T_A g
[/tex]
This expresses the fact that the left action of the Lie algebra generator, [itex]T_A[/itex], on group elements is equal to the flow induced on the group manifold by the corresponding Killing vector field, [itex]\vec{\xi_A}[/itex]. Once we know g in terms of coordinates, we can calculate its derivatives and inverse and find [itex]\vec{\xi_A}[/itex] explicitly. I'll go ahead and do that as soon as someone writes down what g is, which should be easy if you play with it for a few minutes.

(If you'd rather have me write out the calculations, instead of tackling illustrative "homework" problems, let me know and I'll just do that.)

We will talk about symmetries of our group manifold. Typically, a group manifold is of higher dimension than it needs to be to have the symmetries corresponding to its Lie algebra. You can reduce this "waste" by "dividing" by proper subgroups to get a "symmetric space." We'll do all this. :)

Taoy

Excellent news. I'm working on the form of g right now; I believe I've just seen the trick - even powers of T appear to have a nice form :).

I'm happy to work through the examples for the time being; there's nothing like doing it to learn it.

In the mean time could you perhaps clarify your use of the upper and lower arrows, I can guess their meaning, but it doesn't hurt to be explicit.

garrett

Great!

The related wiki page is here:

http://deferentialgeometry.org/#%5B%...0algebra%5D%5D

Explicitly, every tangent vector gets an arrow over it,
[tex]
\vec{v}=v^i \vec{\partial_i}[/tex]
and every 1-form gets an arrow under it,
[tex]\underrightarrow{f} = f_i \underrightarrow{dx^i}[/tex]
These vectors and forms all anti-commute with one another. And the coordinate vector and form basis elements contract:
[tex]
\vec{\partial_i} \underrightarrow{dx^j} = \delta_i^j
[/tex]
so
[tex]
\vec{v} \underrightarrow{f} = v^i f_i
[/tex]
And, in the expression I wrote in the post above,
[tex]
\vec{\xi_A} \underrightarrow{d} g = \xi_A{}^i \partial_i g
[/tex]
in which [itex]\partial_i[/itex] is the partial derivative with respect to the [itex]x^i[/itex] coordinate.

The notation is slightly nonstandard, but looks good and works very well, even when extended to vectors and forms of higher order.

Taoy

Ok, here's the answer to the "homework".

We are computing the explicit form of the group element for SU(2) in terms of the generators [itex]T_a[/itex] in the given representation.

We use the power series expansions the for sine and cosine of r,

[tex]
\begin{align*}
\cos(r) & = 1 - \frac{1}{2!}r^2 + \frac{1}{4!}r^4 - \frac{1}{6!}r^6 + \dots \\
\sin(r) & = r - \frac{1}{3!}r^3 + \frac{1}{5!}r^5 - \frac{1}{7!}r^7 + \dots
\end{align*}
[/tex]

and for the exponential of the matrix T:

[tex]
\begin{align*}
e^T & = I + T + \frac{1}{2!}T^2 + \frac{1}{3!}T^3 + \frac{1}{4!}T^4 + \frac{1}{5!}T^5 + \dots \\
& = \left(I + \frac{1}{2!}T^2 + \frac{1}{4!}T^4 + \dots\right) +
\left(T + \frac{1}{3!}T^3 + \frac{1}{5!}T^5 + \dots \right)
\end{align*}
[/tex]

where [itex]I[/itex] is the identity matrix.

We observe that [itex]T.T = -r^2 I[/itex], and that therefore even powers of T take the form

[tex]
T^{2n} = (-1)^n r^{2n} I.
[/tex]

We can substitute this back into the expansion for the exponential to obtain:

[tex]
\begin{align*}
e^T
& = I \left(1 - \frac{1}{2!}r^2 + \frac{1}{4!}r^4 - \frac{1}{6!}r^6 + \dots\right) +
T \left(1 - \frac{1}{3!}r^2 + \frac{1}{5!}r^4 - \frac{1}{7!}r^6 + \dots \right) \\
& = I \cos(r) + \frac{1}{r} T \sin(r)
\end{align*}
[/tex]

garrett

(2) Killing vector fields

Exactly right. So, the matrix expression for a SU(2) element as a function of SU(2) manifold coordinates is
[tex]
g(x) = e^{x^i T_i} = \cos(r) + x^i T_i \frac{\sin(r)}{r}
[/tex]
Since it's a SU(2) element, it has unit determinant and its inverse is its Hermitian conjugate:
[tex]
g^- = \cos(r) - x^i T_i \frac{\sin(r)}{r}
[/tex]

The next thing is to understand the symmetries of the manifold. We can associate a symmetry, or Killing vector field, [itex]\vec{\xi_A}[/itex], with the flow induced by each Lie algebra generator acting from the left:
[tex]
\xi_A{}^i \partial_i g = T_A g
[/tex]
There are also Killing vector fields associated with generators acting from the right:
[tex]
\xi'_A{}^i \partial_i g = g T_A
[/tex]
Notice that the Lie algebra is necessarily the same as the left Killing vector / symmetry algebra under the Lie derivative:
[tex]
(T_A T_B - T_B T_A) g = ( \xi_A{}^i \partial_i \xi_B{}^j \partial_j - \xi_B{}^i \partial_i \xi_A{}^j \partial_j ) g
[/tex]
[tex]
= C_{ABC} T_C g = ( L_{\vec{\xi_A}} \vec{\xi_B} ) \underrightarrow{d} g = C_{ABC} \vec{\xi_C} \underrightarrow{d} g
[/tex]
The sign of the structure coefficients swaps for the "right acting" Killing vector field algebra.

Now I'll go ahead and calculate the set of three "left acting" Killing vector fields over the group manifold. Multiplying the symmetry equation by the inverse group element gives:
[tex]
\xi_A{}^i ( \partial_i g ) g^- = T_A
[/tex]
The three Killing vector fields each have three components, so [itex]\xi_A{}^i[/itex] is a square matrix that can be inverted and multiplied in to give
[tex]
( \partial_i g ) g^- = \xi^-_i{}^A T_A
[/tex]
(Note: If we consider this as an equation relating Lie algebra valued 1-forms, it's
[tex]
( \underrightarrow{d} g ) g^- = \underrightarrow{\xi^-}^A T_A
[/tex]
which we'll use later.) We'll next mutliply both sides by [itex]T_B[/itex] and use the orthogonality of our Pauli matrix generators under the matrix trace to get the inverse Killing vector matrix all by itself on one side:
[tex]
\xi^-_i{}^B = - \xi^-_i{}^A < T_A T_B> = - < ( \partial_i g ) g^- T_B >
[/tex]
So now we can just calculate that out explicitly, which is made easy by the nice form of g you found:
[tex]
\xi^-_i{}^B = - < \left( (T_i - x^i) \frac{\sin(r)}{r} + x^i x^j T_j ( \frac{\cos(r)}{r^2} - \frac{\sin(r)}{r^3}) \right) \left( \cos(r) - x^k T_k \frac{\sin(r)}{r} \right) T_B >
[/tex]
The Pauli matrices are traceless, so only a few terms will survive the trace, with the generator orthogonality under the trace used again to give
[tex]
\xi^-_i{}^B = \delta_{iB} \frac{\sin(r)\cos(r)}{r} + x^i x^B ( \frac{1}{r^2} - \frac{\sin(r)\cos(r)}{r^3} )
[/tex]
Inverting this matrix (ow, my head! Mathematica helped.) gives the matrix of Killing vector fields over the SU(2) manifold:
[tex]
\xi_A{}^i = \delta_{iA} \frac{r}{\sin(r)\cos(r)} + x^A x^i ( \frac{1}{r^2} - \frac{1}{r \sin(r) \cos(r)} )
[/tex]
These are the components of the three Killing vector fields over the group manifold associated with the left action of Lie algebra generators on group elements.

Something interesting to note: In this whole post, we never had to use the matrix representation of the generators -- all we needed were the commutation relations and orthogonality under the trace. In fact, if the generators, [itex]T_A[/itex] are thought of as Clifford algebra bivectors, everything we've done works out exactly the same way, without ever looking at a Pauli matrix explicitly. The trace operator, [itex]<>[/itex], is the same (up to a multiplicative factor equal to the matrix dimension) as the Clifford algebra "scalar part" operator. In the next post I can talk about this Clifford algebra stuff (and rotations and double covers) or go on to talk about the frame, metric, and connection on the group manifold. I'll get to the Clifford algebra stuff soon anyway, but it's your choice what we do next. Clifford and rotations -- or metric, frame, and connection?

So... the next "homework" is...
1) Make sure I didn't mess this calculation up anywhere. ;)
2) What, explicitly, are the other three Killing vector fields, [itex]\vec{\xi'_A}[/itex] associated with the right action of the generators?
3) What would you like to see next: Clifford algebra and rotations, or the group manifold metric and geometry?

(This post represents a bit of work (my whole evening, in fact) so feel free to ask questions about it for a bit. And if we can't get the "right corresponding" Killing fields tomorrow, I'll try to do it so we can move on.)

Taoy

Where does the expression [itex] \xi_A{}^i \partial_i g = T_A g [/tex] come from, i.e. how do we derive it? Also, there appears to be an implicit assumption that there are as many vector fields as there are generators, that is [itex]i[/itex] and [itex]A[/itex] run over the same indexes. Why is this obvious? In general there could be up to N(N+1)/2 killing vector fields, whereas here we have exactly N.

Also, it looks to me, in Geometric Algebra language, that the left hand side operator is something like: [itex] \xi_A \cdot \nabla [/itex] which is a scalar, however we know that the [itex]T_A[/itex] is going to be a bi-vector... what's going on here?

Other thoughts that come to mind are:

• Group element parameters, vs coordinates.
  The [itex]x^i[/itex]s parametrise the elements of the group, and we can obviously define these killing fields in terms of them, and so they can also be considered coordinates on the manifold.
• r is obviously the length of a vector, where the [itex]x^i[/itex] are coordinates in an orthonormal frame. However the appearance of the sine and cosine of this length is a mystery to me, raising the question "When is the length of a vector the same as an angle?".
• I'm looking forward to seeing how this is manifestly the surface of a 3-sphere.

Taoy

Hmm. I'm not sure about this step. Once you have multiplied to the right by T_B on each side, if you want to take the trace you have to trace the whole thing, i.e. the LHS is:

[tex]
\begin{align*}
( \partial_i g ) g^- & = \xi^-_i{}^A T_A \\
( \partial_i g ) g^- T_B & = \xi^-_i{}^A T_A T_B \\
< ( \partial_i g ) g^- T_B> & = <\xi^-_i{}^A T_A T_B>
\end{align*}
[/tex]

What step to do use to remove the [itex]\xi^-_i{}^A[/itex] matrix from the trace on the right hand side so that you can use the orthogonality condition?

Oh, actually I see it. [itex]\xi^-_i{}^A[/itex] is not a matrix in this expression, it's just a scalar, and so it can be pulled out the front.

garrett

Good questions.

A symmetry is a map from the manifold to itself. A continuos map, or flow, can be visualized as moving the manifold coordinates by a little bit:
[tex]
x^i \rightarrow x'^i = x^i + \epsilon^A \xi_A^i
[/tex]
in which [itex]\vec{\xi} = \epsilon^A \vec{\xi_A} [/itex] is a vector field on the manifold, with "small" parameters, [itex]\epsilon_A[/itex]. Under a flow (also known as a diffeomorphism), a function of manifold points, such as the group element g(x), changes as:
[tex]
g(x) \rightarrow g'(x) = g(x + \epsilon^A \xi_A) \simeq g(x) + \epsilon^A \xi_A^i \partial_i g(x)
[/tex]
to first order via Taylor expansion. Now, there is also a map on group elements induced by the Lie algebra generators:
[tex]
g \rightarrow g' \simeq (1 + \epsilon^A T_A) g = g + \epsilon^A T_A g
[/tex]
(and another map for the group element acting from the other side)
The symmetry relation we want comes from equating the maps induced by the Lie algebra generators with the corresponding diffeomorphisms,
[tex]
\xi_A^i \partial_i g = T_A g
[/tex]
Wha-la.

Now, as for the Clifford algebra question: The group element is an exponential of a bivector, [itex]g = e^T[/itex], so it is a mixed, even graded multivector. Taking its derivative "brings down" a bivector, so there is no grade inconsistency. Grade consistency is a good thing to keep an eye on though, and we'll use it later.

Did the rest of the previous post make sense?

garrett

Yes, exactly, [itex]\xi^-_i{}^A[/itex] is a bunch of scalars labeled by indices, which each run from 1 to 3. You can then think of that as a set of three 1-forms, or as a 3x3 "matrix" -- but not a matrix in the same algebra as g.

By the way, the [itex]g^- = g^{-1}[/itex] notation, indicating an inverse element, comes from Donald Knuth -- I also like it so I stole it.

garrett

Answering your other questions:

For now, [itex]r = \sqrt{x1^2 + x2^2 + x3^2}[/itex] is best thought of as just a notational convenience.

We should see the relationship to spheres when we establish the geometry of the group manifold.

Yes, the group parameters are the group manifold coordinates. The A and i indices are, for now, in the same class and are interchangeable. This will be different when we investigate symmetric spaces.

Taoy

Ok, I get this. Another way of getting at it is to study what happens to the components of the metric [itex]g_{ij}(x) = g'_{ij}(x')[/itex] (which we've not come to yet, but I'll mention it anyway), as a function of [itex]x[/itex] is transformed into a different set of coordinates (or basis), via [itex]x -> x' = x + \epsilion \xi[/itex]. If we try and find the condition such that that [itex]g_{ij}(x) = g_{ij}(x')[/itex], i.e. the components don't change, we end up with the condition on [itex]\xi[/itex] that I mentioned in an earlier post, namely, [itex]\xi_{a;b} + \xi_{b;a} = 0[/itex]. These are also killing vector fields, or isometries of the metric.

Hmm, there is an inconsistency. Acting on it with a scalar derivative [itex]\frac{\partial}{\partial x^i}[/itex], which we appear to be doing, doesn't change the grade at all. I would agree with you if we were contracting it with the vector derivative, [itex]\nabla = e^i \partial_i[/itex]. That's not what's happening here though is it?

Yes, it's making sense. I've not expanded the trace out yet, or calculated the fields associated with the right action. I was hoping to do it tonight, but I'm not going to get the chance it seems.

Let's do the clifford stuff as there is an open question about the grade lowering stuff. I'm going to be out of the country over the weekend, and already know most of the clifford stuff - so it will give the others (are there any others? :) a chance to catch up.

p.s. using [itex]g^-[/itex] to indicate the inverse; I like that. It like it more that it was Knuth's :). I don't use enough of his stuff (directly).

Mehdi_

Hi garrett

This PF is very interresting... just reading the answer let you learn a lot...

garrett

OK, we'll talk about Clifford algebra a bit.

First, to answer the grade consistency question: For our three dimensional Clifford algebra (it's actually $2^3=8$ dimensional, but with three basis vectors) our su(2) group element, g, is a scalar plus a bivector. What grades do you get if you multiply this times an arbitrary bivector? You can't get a four-vector, since there is none in the algebra, so you get... a scalar plus a bivector. The grades match on both sides. Happy?

garrett

Hi Mehdi, welcome to Physics Forums.

I'll try to be back later tonight to relate this group and Lie algebra stuff to Clifford algebra, which you should find interesting.