5
$\begingroup$

I know this is a math question; however, physicists are more likely to be familiar with what I'm asking (also, I'm directly trying to utilize it in the context of general relativity).

I may have worded this question initially somewhat backwards. Given some metric, I'm essentially trying to figure out how you can recover the group properties associated with that metric.

I would guess that the tensor locally adheres to the properties of the Lie algebra corresponding to the group that our tensor falls into topologically, However I'm unclear precisely how this would manifest.

My first thought is to lead with a decomposition of the metric into gamma matrices. As is well known, one can decompose a metric $g_{\mu\nu}$ into gamma matrices $\gamma_{\mu}$ such that:

$$Ig_{\mu\nu}=\frac{1}{2}\left(\gamma_{\mu}\gamma_{\nu}+\gamma_{\nu}\gamma_{\mu}\right)=\frac{1}{2}\left\{ \gamma_{\mu},\gamma_{\nu}\right\} $$

Where I is the identity matrix. This is all pretty standard (Note that I'm NOT using tetrad/verbein frames, thus these are generalized gamma matrices).

For the Minkowski metric $R^{3,1}$ (denoted $\eta_{\mu\nu}$) the commutator of the gamma matrices

$$\sigma_{\mu\nu}=\frac{1}{2}\left(\gamma_{\mu}\gamma_{\nu}-\gamma_{\nu}\gamma_{\mu}\right)=\frac{1}{2}\left[\gamma_{\mu},\gamma_{\nu}\right]$$

turn out to be the infinitesimal generators of the Lorentz transformations (ie elements of the group algebra). In this case this also turns out to be the Lie group $O(3,1)$ associated with the Minkowski space

Given the bilinear form associated with the Minkowski metric, the appropriate group follows directly from the theory ....The appropriate group is O(3,1), in this context called the Lorentz group. Minkowski space-Wikipedia

These can in turn be exponentiated to form the elements of the Lorentz Group (if I'm remembering correctly?):

$$T_{\nu}=exp(\theta^{\mu}\sigma_{\mu\nu})$$

So for Minkowski space we have recovered the group properties of the manifold from examining the properties of the metric tensor (or rather it's decomposition).

What about an $SU(2)$ (represented as the manifold of a 3-sphere) or a metric which is part of a more general Lie group? Does anyone know if the metric tensor still carries these properties? I'd been hoping to utilize this to solve some ever so fun problems I'd been working on.

Ulitimately I am trying to apply this to a specific case of compact Lie group products, namely $SU(2)xU(1)$. According to s. harp's comment below this is sufficient to insure a bi-invariant metric (though I'm honestly not sure). In an attempt to address comments below I'm going to try and illustrate what I've been doing:

I'm just going to focus on $SU(2)$ for the moment. From the Peter-Weyl Theorem, We know that:

$$L^{2}(S^{3})=L^{2}(SU(2))$$

Which allows us to represent objects on the three sphere in terms of irreducible representations of $SU(2)$ in dimension=3 (this is just the generalization of the Fourier type Series expansion to compact groups).

Thus our metric may be represented by tensor harmonics on the three sphere:

$$g_{\mu\nu}=\sum_{k,l,m}^{\infty}A^{klm}Y_{\mu\nu}^{klm}$$

This makes it appear as though any metric that is homeomorphic to (a continuous deformation of ) $S^3$ is represented by a sum of weighted irreducible representations of $SU(2)$. A situation like this is discussed here 4th page 3rd paragraph

Thus I would still expect the group structure of $SU(2)$ to play a central role in properties of the metric tensor (such as the gamma matrix decomposition mentioned above). All I can figure is that for a general metric homeomorphic to $S^3$ each gamma matrix would actually be a sum of different irrreps corresponding to it's series expansion.

Maybe I've over bogged myself in this?

NOTE: I'm just becoming more deeply acquainted with group theory (long overdue). I get that the gamma matrices form a Clifford algebra but I really haven't gotten that far on how that relates to any pertinent Lie groups.

Any answer or direction towards a book that covers this subject would be greatly appreciated.

$\endgroup$
  • $\begingroup$ What are these $\gamma$ matrices that act on the Lie algebra of your (arbitrary) Lie group? Or are you talking explicitly about the Lorentz group here? $\endgroup$ – s.harp Nov 14 '18 at 10:00
  • $\begingroup$ I don't know enough about the topic to give an answer but I'll note that since you can put whatever metric you want on a Lie group, a metric will not give you any useful relations, really. Bi-invariant metrics however do carry useful information. For example the Riemannian exponential map and the Lie exponential map coincide. Bi-invariant metrics are also the ones that are induced via translations from an inner product on the Lie algebra. If your Lie group is compact, then the so-called Killing-Cartan form is nondegenerate, and plays the role of such an inner product. $\endgroup$ – Bence Racskó Nov 14 '18 at 10:33
  • $\begingroup$ @s.harp The generalized gamma matrices are any matrices satisfying the anticommutation relation (equation 1). $\endgroup$ – R. Rankin Nov 14 '18 at 11:36
  • $\begingroup$ @Uldreth Surely I can't put ANY metric on a Lie group? Since Lie groups are by definition smooth manifolds) Consider a space that is the unit three-sphere (which is the manifold of $SU(2)$), the choice of metric is severely constrained to just the choice of coordinate system. thanks for the input I'll see if my metrics have to be Bi-invariant $\endgroup$ – R. Rankin Nov 14 '18 at 11:49
  • 1
    $\begingroup$ There is canonical Killing-Cartan metric on a group manifold: $ds^2=\mathop{\mathrm{tr}}(dG d(G^{-1}))$, which appears in such things as sigma-models, WZW model, string (and branes) theory on group manifolds. $\endgroup$ – A.V.S. Nov 15 '18 at 8:28
2
$\begingroup$

In general there is no reason why a metric $g$ on a Lie group $G$ to manifest any of the group properties. For example, you might know from your study of differential geometry that every smooth manifold $M$ can be endowed with a Riemannian metric $g$ (essentially just by pulling back the metric on $\mathbb R^n$ via charts and gluing them together). There is no reason whatsoever that this generic construction to obey any properties of group multiplication if you choose $M=G$.

However, if the Lie group is nice enough (e.g. compact) you can have metrics, which are for example left- (or even bi-) invariant. For example, what you can do is the following. Assume that you choose your favourite scalar product $\left< \cdot, \cdot \right>$ on $\mathbb R^n$. Then since $T_e G \cong \mathbb R^n$, we have a scalar product on the tangent space at the identity and then define for all elements $x \in G$

$$\forall v,w \in T_x G \quad (v,w)_x := \left<DL_x^{-1} v,DL_x^{-1} w\right> $$

where $L_x$ is multiplication on the left by $x \in G$ and $DL_x$ is the differential of $L_x$. So what you do is pull any vector $v \in T_x G$ by the left multiplication back to $T_e G$, compute the scalar product there. Note that this scalar product will generally not be bi-invariant. If $G$ is compact, then you can "average" this scalar product with the (left invariant) Haar measure to get a bi-invariant scalar product. In fact, bi-invariant scalar products do speak volumes about the group structure of $G$.

$\endgroup$
  • $\begingroup$ Thank you, I think this is what I'm looking for as I'm only dealing with spacetimes with compac topologies. $\endgroup$ – R. Rankin Nov 15 '18 at 1:59
  • $\begingroup$ However, note that a generic metric even on compact Lie groups will have no connection to the group structure whatsoever. $\endgroup$ – Gonenc Mogol Nov 15 '18 at 2:15
  • 1
    $\begingroup$ "However, if the Lie group is nice enough (e.g. compact) you can have metrics, which are for example left- (or even bi-) invariant." - Every Lie group admits left-invariant metrics. $\endgroup$ – s.harp Nov 15 '18 at 9:40
  • $\begingroup$ @s.harp which is also clear from the proof that i gave "nice enough" was referring to bi-invariant metrics. I might rephrase it tho if it is ambiguous. $\endgroup$ – Gonenc Mogol Nov 15 '18 at 14:41
  • $\begingroup$ @GonencMogol Suppose I had a metric space of constant curvature on G, could I then write the riemann tensor in terms of the Lie algebra corresponding to G? $\endgroup$ – R. Rankin Nov 23 '18 at 6:31
1
$\begingroup$

I think I understand what you are asking. Assume you have a Lie group $G$ of dimension $n$ with a left, or right, or bi invariant metric $g_0$ on $G$. This means that either $(L_q)^*g_0 = g_0$ or $(R_q)^*g_0 = g_0$ or both for all $q \in G$.

  1. First, you take a (possibly small) open subset $U$ of $G$.
  2. Then, you take a diffeomorphism $\phi : U \to W \subset \mathbb{R}^n$ that maps the open domain $U$ of $G$ to an open subset $W$ of $ \mathbb{R}^n$ with coordinates $x = (x^1, ..., x^n)$.
  3. And, finally, you send the metric $g_0$ via $\phi$ from the Lie group $G$ to the open set $W$ in $\mathbb{R}^n$. More precisely, you define the (pull back) metric tensor $g(x) = \big(\phi^{-1}\big)_x^*g_0(\phi^{-1}(x))$ on the open set $W$ in $\mathbb{R}^n$ with coordinates $x = (x^1, ..., x^n)$.

Basically, $\big( U, \phi\big)$ is a chart of $G$ and $g(x)$ is the coordinate representation of $g_0$ in that chart.

Now, your question seems to be about the converse. Imagine you give me just a metric tensor $g(x) = \big(\, g_{ij}(x)\,\big)_{i,j = 1}^{n}$ defined on some open domain $W$ of $\mathbb{R}^n$ and you ask me: 'Can you figure out whether the metric tensor $g(x)$ is the coordinate representation of a left, or right or bi-invariant metric on some Lie group, i.e. did the metric tensor $g(x)$ came from a Lie group after the execution of procedure 1-2-3 from above?'

To answer this question, what I would do is to try and find the Lie algebra of Killing vector fields of $g(x)$, i.e. I will look for all vector fields $X(x)$ for which the Lie derivative of $g(x)$ is $L_Xg = 0$. Equivalently, if in coordinates $x = (x^2,...,x^n)$ on $\mathbb{R}^n$, the vector fields $X(x)$ are written as $$X(x) = X^{j}(x^1,...,x^n)\, \frac{\partial}{\partial x^j}\,$$ then the equation $L_Xg = 0$ can be written as the system of equations $$\nabla_{\frac{\partial}{\partial x_k}}\,X_j(x) \, + \,\nabla_{\frac{\partial}{\partial x_j}}\,X_k(x) = 0 $$ So I would try to form the set of all such Killing vector fields $$\mathcal{K} = \big\{\, X(x) \, : \, L_Xg = 0 \, \big\}$$ This set $\mathcal{K}$ is in fact a finite dimensional Lie algebra of vector fields that generate the isometries of the metric tensor $g(x)$. If $\dim{\mathcal{K}} < n$ then there is no chance that $g(x)$ comes from a metric $g_0$ on a Lie group $G$. However, if $\dim{\mathcal{K}} = n$ then it does. Indeed, according to Cartan-Lie theorem (maybe also called Lie's third theorem), every finite dimensional Lie algebra is the Lie algebra of a simply connected finite dimensional Lie group. The appropriate combination of phase flows of a basis of Killing vector fields from $\mathcal{K}$ will generate the diffeomorphism $\phi$ discussed in point 2 above.

Finally, if I am not wrong, I believe that if $\dim{\mathcal{K}} > n$ it's probably because the metric tensor $g(x)$ comes from the metric of a homogeneous space. For example, the two-sphere is a homogeneous space for $SU(2)$, i.e. $SU(2)$ acts nicely on the two sphere and the two-sphere is diffeomorphic to the quotient of $SU(2)$ by a copy of U(1) embedded in $SU(2)$. For example, recall the Hopf fibration map. That is exactly this quotient map from $SU(2)$ to the two-sphere.

$\endgroup$
  • $\begingroup$ Great Answer! and you're right, Killing vectors seem to be the way to go. $\endgroup$ – R. Rankin Dec 22 '18 at 13:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.