Sectional and Ricci Curvature for Three-Dimensional Lie Groups

Formulas for the Riemann and Ricci curvature tensors of an invariant metric on a Lie group are determined.The results are applied to a systematic study of the curvature properties of invariant metrics on three-dimensional Lie groups. In each case the metric is reduced by using the automorphism group of the associated Lie algebra. In particular, the maximum and minimum values of the sectional curvature function are determined.


Introduction
In [1] Milnor investigated curvature properties of invariant metrics on Lie groups particularly in dimension three.A much more recent paper [2] studied the solution of Killing's equations for invariant metrics in dimensions two and three.In certain special cases it was found that some of the metrics corresponded to spaces of constant curvature.In the present paper we reconsider metrics in dimension three and undertake an exhaustive study of sectional and Ricci curvature.Although Milnor used left-invariant metrics we prefer to use right-invariant metrics although of course it is not essential to do so.
In Section 2 we derive formulas in indices for the Riemann and Ricci tensors and scalar curvature of an invariant metric.In Sections 3 and 4 we study in detail curvature properties of invariant metrics in dimensions two and three corresponding to all possible nonabelian Lie algebra.However, in this paper we propose a new technique; that is, we shall reduce the metric using the automorphism group of the algebra.As a result the reduced metric depends only on one or two parameters and it makes it possible to obtain very explicit formulas for the Ricci tensor and sectional curvature function.In each case we find the maximum and minimum values of the sectional curvature function.Since the metrics have been reduced in an equivariant manner, the curvature properties that we find are definitive.
As regards notation, in Section 2 we use the summation convention on repeated indices until it becomes impractical to do so.In Section 4 corresponding to three-dimensional Lie algebra given in [3], we give a matrix Lie group  whose Lie algebra gives the Lie algebra in question as well as the set of left and right-invariant vector fields and one-forms.Furthermore, we specify the Lie algebra of derivations by a 3× 3 matrix with Greek entries.The automorphism group, or at least its identity component, can be found by exponentiating the space of derivations.We shall also find it convenient to use an overall scaling of the metric: it is not absolutely necessary to do so and doing so does not change the curvature tensor nor the signs of the Ricci eigenvalues.
In terms of finding sectional curvature we shall compute it first of all on a collection of planes that are transverse to a given axis (generic case) and then on the residual set of planes that contain the given axis (nongeneric case).At the risk of some confusion, we parametrize the set of two-planes in both cases using the coordinates (, ).

Formulas for Curvature and Ricci Tensors
In this section we derive a formula for the Ricci tensor of a left or right-invariant metric   on a Lie group .It is convenient to keep the metric visible since we cannot assume that we are working in an orthonormal frame.Thus, if   is right-invariant and {  } is a basis for the right-invariant vector 2 Journal of Mathematics fields,   is a constant matrix.The compatibility of   with its Levi-Civita connection Γ   gives whereas the fact that the connection is torsion-free gives From ∇     − ∇     = [  ,   ], we have Γ     − Γ     =      .Now cycle the indices in (1) and use (2) to obtain Notice that in (3) the first two terms are symmetric in  and  and that the third term is skew-symmetric in  and .Now we find that the curvature is given by Notice that if we write (4) invariantly we get the usual formula for curvature: that is, where in the last term we have the Lie algebra bracket.After a very long calculation (4) gives for the Ricci tensor We note that   is symmetric as it must be for a metric connection and also that the fourth term is the Killing form up to a factor.Moreover, the third term vanishes for unimodular Lie algebra and the third and fourth terms vanish for nilpotent Lie algebra.
If we assume that we are in an orthonormal frame so that   =   , where   denotes the Kronecker delta, then we find after very long calculations that the curvature and Ricci tensors and scalar curvature are given by Left-invariant vector fields:   ,     .
The only derivations are inner and the automorphism group coincides with the adjoint representation.As such and including an overall scaling the matrix of the metric may be reduced to the identity.The metric is of constant Gaussian curvature −1.
The space of derivations is given by [ ] and the group of automorphisms by  = [ det()   0  ], where  ∈ (2, R) and  ∈ R 2 is a column vector.
Consider a symmetric matrix  given by  = [      ].The action of the automorphism group on  is given by Since the metric is assumed to be Riemannian,  > 0. As such we may define  fl −(1/) which has the effect of removing  in .Now we continue to normalize  assuming that  = 0 and  = 0.As such  may be reduced to [  0 0  ], where  is the 2 × 2 identity matrix.We find it convenient to replace  by 2 below so that the metric can be reduced to 4.1.1.Sectional Curvature.We shall now investigate the sectional curvature in detail: we wish to find the maximum and minimum values.In the group matrix  above, group coordinates are denoted by (, , ).Now, at the risk of confusion, we are going to use (, ) in a different sense: that is, as coordinates on the space of two-planes in the tangent space at the identity.Note that at the identity the invariant vector fields reduce simply to {  ,   ,   }, which play the role simply of "unit vectors" pointing along the three coordinate axes.As such almost every two-plane has a basis of the form  =   +   ,  =   +   for some ,  ∈ R. Our task now is to compute (, ) = (, (, ))/((, )(, ) − (, ) 2 ).Using the definitions we find that The partial derivatives are given by Clearly (, ) = (0, 0) is the unique critical point.The Hessian of  at (0, 0) is 64 4 > 0 and   (0, 0) = 8 2 > 0 so the origin yields a local and in fact global minimum of (0, 0) = −3/2.The only planes that have been omitted in the (, )parametrization are those planes that contain the -axis.As such, a basis may be chosen of the form { =   +  ,  =   }, where  and  are not both zero and we find that We use  rather  to distinguish the nongeneric from the generic sectional curvature.We shall adhere to the same notation in the remaining cases below.Hence, the sectional curvature varies between /2 and −3/2 inclusive.
The space of derivations is given by [

Sectional Curvature.
The metric can be reduced to where  > 0. Indeed the metric can first of all be diagonalized using the parameters , ,  in the space of derivations introduced above.After diagonalization we have only one parameter left at our disposal but we choose also to use an overall scaling by a nonzero factor  of the metric so as to make it as simple as possible.It is not essential to do so but it does help to simplify some of the formulas.The Riemann and Ricci curvature tensors are unaffected by such a scaling but the scalar, sectional curvatures, and Ricci eigenvalues are multiplied by 1/.
Proceeding as we did for Algebra 3.1, almost all tangent planes at the origin have a basis of the form { =   +   ,  =   +   }.As such The partial derivatives are given by Multiplying (18) by  and adding  times (19) gives 8 2 (2 − ) = 0. Now  > 0 so either  = 0 or  = 2.If  = 0 (18) gives 2( 2 + 1) so we have  =  = 0.If  = 2 then (19) gives  +  2 + 2 2 = 0, which is impossible since  > 0. Hence, the origin is the only critical point.The Hessian at the origin turns out to be −4 so that the origin gives a saddle point.
Two eigenvalues are unconditionally negative and the third is negative, zero, or positive according as  is less than, equal to, or greater than √ 5 − 1.
If  = 1 the space of derivations is given by [ ] .The case  = 1 is very special.It is the three-dimensional case of the algebra g discussed in [1] which has the property that, for all ,  ∈ g, [, ] =  +  for some ,  ∈ R. It may be shown that in dimension  this algebra is unique up to isomorphism and has nonzero brackets [  ,   ] =   for 1 ≤  ≤  − 1. Milnor shows that for such an algebra any leftinvariant metric has constant curvature −1 and in fact that these algebras are the only ones to have this property [1].
From our point of view we can explain the significance of this metric as follows.The space of derivations is given by the algebra of  ×  matrices whose bottom row is zero but is otherwise arbitrary, which is of dimension ( − 1) being in fact the Lie algebra of the affine group of R −1 .Of all algebra of dimension  it has the largest dimensional automorphism group apart from the one-dimensional extension of it obtained by adding multiples of the identity and the abelian algebra R  .Having such a large automorphism group, it is possible to reduce the matrix of a left or right-invariant metric to the identity matrix.Thus, there is essentially only one invariant metric allowing for symmetry.
Let us compare the two possible maxima in the case, where −1 ≤  < 0. Subtracting gives Considering the numerator above clearly ( − 1) ≤ 0; We can summarize the maxima and minima for algebras 3.3-3.5 as follows: (i)  = 1 and consequently  = 0 and sectional curvature is identically −1.
We have to treat  = −1 as a special case for which we obtain a negative or zero eigenvalue according as  is nonzero or zero and we shall have to assume also that  2 < 1.
Assuming then that  ̸ = −1, in order to have a nonnegative eigenvalue we need or In the first case we find that (iv) if  = −1 the third eigenvalue is negative unless  = 0 in which case the second and third eigenvalues are zero.
The scalar curvature is −(3(1 −  2 )( + 1) 2 +(−1) 2 )/2(1 −  2 ) < 0. In the special case  = −1, which is the Lie algebra of (1, 1), the group of motions of the pseudo-Euclidean plane, the eigenvalues simplify to −2/(1− 2 ), ±2/(1− 2 ) so we have either one positive and two negatives or one negative and two zeros if  = 0. ] and again we have used scaling to simplify the metric.For the Lie algebra itself, it may be assumed without loss of generality that  ≥ 0 and for the metric that 0 <  ≤ 1 but not both of these conditions may hold simultaneously so we shall assume that the latter holds.
= −1 and so if  2 satisfies the first equality then necessarily  2 < 1.Notice also that in order to have  2 ≥ 0 we must have that 0 <  < 1.