Research Article | Open Access

Gerard Thompson, Giriraj Bhattarai, "Sectional and Ricci Curvature for Three-Dimensional Lie Groups", *Journal of Mathematics*, vol. 2016, Article ID 3681529, 10 pages, 2016. https://doi.org/10.1155/2016/3681529

# Sectional and Ricci Curvature for Three-Dimensional Lie Groups

**Academic Editor:**K. F. C. Yiu

#### Abstract

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.

#### 1. 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 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 .

#### 2. 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 fields, is a constant matrix. The compatibility of with its Levi-Civita connection giveswhereas the fact that the connection is torsion-free givesFrom , we have . Now cycle the indices in (1) and use (2) to obtainNotice 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 byNotice 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 tensorWe 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

#### 3. Two-Dimensional Lie Group

##### 3.1. Dimension 2 Nonabelian

Nonzero brackets:

Left-invariant vector fields: . Left-invariant one-forms: . Right-invariant vector fields: . Right-invariant one-forms:

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.

#### 4. Three-Dimensional Lie Groups

##### 4.1. Algebra 3.1 (Heisenberg: Nilpotent)

Nonzero brackets: .

Left-invariant vector fields: . Left-invariant one-forms: , , . Right-invariant vector fields: . Right-invariant one-forms:

The space of derivations is given by and the group of automorphisms by , where and 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, . As such we may define which has the effect of removing in . Now we continue to normalize assuming that and . As such may be reduced to , where is the identity matrix. We find it convenient to replace by 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 . Our task now is to compute . Using the definitions we find thatSo

The partial derivatives are given byClearly is the unique critical point. The Hessian of at is and so the origin yields a local and in fact global minimum of .

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 thatWe 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 and inclusive.

###### 4.1.2. Ricci Curvature

The Ricci tensor as a type tensor is

##### 4.2. Algebra 3.2

Nonzero brackets: , .

Left-invariant vector fields: . Left-invariant one-forms: . Right-invariant vector fields: . Right-invariant one-forms:

The space of derivations is given by

###### 4.2.1. Sectional Curvature

The metric can be reduced to , where . 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 .

Proceeding as we did for Algebra 3.1, almost all tangent planes at the origin have a basis of the form . As suchHence,The partial derivatives are given by

Multiplying (18) by and adding times (19) gives . Now so either or . If (18) gives so we have . If then (19) gives , which is impossible since . Hence, the origin is the only critical point. The Hessian at the origin turns out to be so that the origin gives a saddle point.

This time the exceptional values of the sectional curvature are given byTo find the critical points we have to solve

The critical points are given by for which assumes the values . Hence, the maximum and minimum values are, respectively, and . Note that . Hence, if the values of the sectional curvature will always be negative.

###### 4.2.2. Ricci Curvature

The Ricci tensor as type is . The eigenvalues are . Two eigenvalues are unconditionally negative and the third is negative, zero, or positive according as is less than, equal to, or greater than .

##### 4.3. Algebra 3.3 , 3.4 , 3.5

Nonzero brackets: , .

Left-invariant vector fields: . Left-invariant one-forms: . Right-invariant vector fields: . Right-invariant one-forms:

If the space of derivations is given by The case is very special. It is the three-dimensional case of the algebra discussed in [1] which has the property that, for all , for some . It may be shown that in dimension this algebra is unique up to isomorphism and has nonzero brackets for . Milnor shows that for such an algebra* any* left-invariant metric has constant curvature 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 being in fact the Lie algebra of the affine group of . 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 . 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.

The remainder of this subsection is devoted to the case for which the space of derivations is given by The metric can be reduced to , where and again we have used scaling to simplify the metric.

###### 4.3.1. Sectional Curvature

The generic sectional curvature function is given by

To find the critical points we have to solve

Multiplying (24) by and subtracting times (25) give

Using (26) in (24) gives . However, since then and hence the origin is the only critical point. The Hessian at the origin turns out to be so that if the origin gives a saddle point and if a local extremum. We find also that so that this local extremum is actually a local maximum given by .

In the exceptional case we find that

The partial derivatives are given by

Notice that if and then so we exclude this case in the remainder of the discussion. Note also that the denominator since . Critical points are given by the pair of lines , except that the origin must be excluded, or else the plane on which the sectional curvature is being evaluated degenerates to a line. Furthermore, if the discriminant whereas if then : in fact since we are excluding the case and .

Using the* first* derivative test it follows thatgive local extrema, a local minimum for , and local maximum for . After much algebra, this minimum and maximum turn out to be given simply by . In the case where , the function has a minimum of for and maximum of for . Recall also that or else the algebra does not belong to the given family.

Let us compare the two possible maxima in the case, where . Subtracting gives

Considering the numerator above clearly ; however, since when . Hence, when the maximum actually is .

We can summarize the maxima and minima for algebras 3.3–3.5 as follows:(i) and consequently and sectional curvature is identically .(ii), : maximum , minimum .(iii), : minimum , maximum = .

###### 4.3.2. Ricci Curvature

Ricci as a type tensor which depends on and is given by

The Ricci eigenvalues are , . The first eigenvalue is always negative since . The second and third eigenvalues are real since . Moreover, taking the plus sign the second eigenvalue is clearly positive except in the very special case , .

As regards the third eigenvalue it is negative, zero, or positive according as is negative, zero, or positive.

We have to treat 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 .

Assuming then that , in order to have a nonnegative eigenvalue we need or In the first case we find that

Now which is always nonnegative and positive if which would give contrary to hypothesis.

In the second case we find that or

Now with equality iff and so if satisfies the first equality then necessarily . Notice also that in order to have we must have that .

Similarly, in order to have a nonpositive third eigenvalue or However, since it follows that Hence, if the first inequality is identically satisfied. To summarize,(i)the third eigenvalue is positive if and ,(ii)the third eigenvalue is negative if and ,(iii)the third eigenvalue is zero if and and (iv)if the third eigenvalue is negative unless in which case the second and third eigenvalues are zero.

The scalar curvature is . In the special case , which is the Lie algebra of , the group of motions of the pseudo-Euclidean plane, the eigenvalues simplify to so we have either one positive and two negatives or one negative and two zeros if .

##### 4.4. Algebra 3.6 , 3.7

Nonzero brackets: , .

The space of derivations is given by and the metric can be reduced to and again we have used scaling to simplify the metric. For the Lie algebra itself, it may be assumed without loss of generality that and for the metric that but not both of these conditions may hold simultaneously so we shall assume that the latter holds.

###### 4.4.1. Sectional Curvature

The generic sectional curvature function is given byWe shall assume to begin with that . The critical points of come from solvingNow times (44) − times (45) gives . We shall discuss and later so that . Substituting in (45) gives so that either or . The first of these with gives . The second gives which would give which is impossible. Thus, . Substituting in the first gives . The only possibility is . The Hessian at the origin is and so the origin gives a saddle point.

In the exceptional case we find that

Since we are excluding and we deduce that the critical points come from solving . Note that so that there always two lines of critical points. Using the* first* derivative test it follows that gives a local, hence global and minimum, and gives a local, hence global and maximum. After much algebra, this maximum and minimum turn out to be given simply by .

Finally, if and the space is of constant curvature . If but is not necessarily , the maximum and minimum values just found remain valid: that is, and .

###### 4.4.2. Ricci Curvature

Ricci as a type tensor is and the eigenvalues are , . The scalar curvature is which is negative. Two of the eigenvalues are unconditionally negative but the third eigenvalue can be positive, zero, or negative.

If the third eigenvalue is zero then . However, if we take the minus sign then , so we must take the plus sign and, in so doing, we obtain precisely the condition to have a zero eigenvalue. Indeed rearranging the quadratic condition gives that the third eigenvalue provided that and provided that .

##### 4.5. Algebra 3.8

Nonzero brackets: , , .

We note that is isomorphic to and we rewrite the algebra as , , . We take the Lorentz inner product on as so that the group acts on its Lie algebra as automorphisms. A symmetric matrix can be diagonalized; that is, there exists such that is .

###### 4.5.1. Sectional Curvature

The generic sectional curvature function is given by

Clearly gives a critical point and critical value of . In order to have a second solution for which we require that , : however, these conditions are incompatible.

If then the conditions to obtain a critical point become , and similarly if and a fortiori if .

In the nongeneric case

The critical points are of the form , where and where giving, respectively, critical* values* of and .

Suppose that . Then in the nongeneric case . If we find that the numerators of and both contain a factor of , which is not zero and so we do not get any more critical points.

To summarize, the maximum and minimum sectional curvatures are to be found among , , and .

###### 4.5.2. Ricci Curvature

In this case Ricci as a type tensor is and the scalar curvature is . It is impossible for Ricci to be zero or negative definite: , , are mutually incompatible and hence the metric can never be of constant curvature. Furthermore, if one eigenvalue is zero, two of them are zero and the third is negative.

##### 4.6. Algebra 3.9

Nonzero brackets: , , .

We can diagonalize the metric using an orthogonal transformation so that its matrix is . If the space is of constant curvature .

###### 4.6.1. Sectional Curvature

We proceed as before and find the sectional curvature of a tangent 2-plane that is transverse to the -axis for which a basis consists of . Thus,

The partial derivatives are given by

In order to have a solution other than we need

, hence , .

Then must be distinct. Note that the inequalities are invariant under interchange of and . Hence, it is sufficient to check the cases: , , . Thus, the first inequality gives whereas the second . If the second inequality gives which contradicts . Finally if the first inequality gives which contradicts . Thus, is the only critical point and .

We can find all sectional curvatures by permuting the roles of , , and . The maximum and minimum values will be found among

###### 4.6.2. Ricci Curvature

Ricci is given by . Three eigenvalues can be positive but only if they are all equal and ; otherwise, we can have two zero and one positive or one positive and two negative eigenvalues, respectively. The metric is of constant curvature only in the case .

The scalar curvature is given by