Riemannian metrics – left, right and bi-invariant

Posted on by arkajad

The discussion in this post applies to Riemannian metrics on Lie groups in general, but we will concentrate
on just one case in hand: SL(2,R). Let G be a Lie group. Vectors tangent to paths in G, at identity e\in G form the Lie algebra of the group. Usually it is denoted by \mathrm{Lie}(G). That is a real linear space, endowed with the commutator. For matrix groups the Lie algebra is a space of matrices (for the group SL(2,R) its Lie algebra sl(2,R) consists of matrices of zero trace), and the commutator is realized as the commutator of matrices: [A,B]=AB-BA.

Suppose we have scalar product (\xi,\eta)_e defined on the Lie algebra. For instance, if \xi,\eta are matrices, we may try to define (\xi,\eta)_e=\mathrm{Tr}(\xi\eta). That is a possible natural definition, the scalar product so defined is automatically symmetric. But it is not always nondegenerate, so one needs to be careful. When we have scalar product at identity, we can use group multiplication to propagate it to the whole group space by left translations:

(1)   \begin{equation*}(\xi,\eta)_g=(g^{-1}\xi,g^{-1}\eta)_e,\end{equation*}

if \xi,\eta are tangents to paths at g.

Notice that we are multiplying tangent vectors by group elements. For matrix groups that is easy, we simply multiply matrices. For more general groups is is understood that if \xi is tangent to the path \gamma(t), then g\xi is tangent to g\gamma(t).

The scalar product defined everywhere by Eq. (1) has automatically the property of being left-invariant:

(2)   \begin{equation*}(g\xi,g\eta)=(\xi,\eta).\end{equation*}

The above equation needs explanation, its meaning follows from the context. On the left hand side we have scalar product. So, it is probably calculated at a certain point. We need to give this point some name. The symbol g is used in the formula itself for the shift, so let us choose h. So, we specify the left hand side to mean

    \[(g\xi,g\eta)_h.\]

Now, if g\xi is tangent at h, then \xi itself must be tangent at g^{-1}h. So, the complete formula should be:

(3)   \begin{equation*}(g\xi,g\eta)_h=(\xi,\eta)_{g^{-1}h}.\end{equation*}

That is supposed to hold for any g,h in the group. How do we prove it? We use the definition (1). The left hand side becomes

(4)   \begin{equation*}(g\xi,g\eta)_h=(h^{-1}(g\xi),h^{-1}(g\eta))_e.\end{equation*}

The right hand side becomes

(5)   \begin{equation*}(\xi,\eta)_{g^{-1}h}=((g^{-1}h)^{-1}\xi,(g^{-1}h)^{-1}\eta)_e.\end{equation*}

The left and right hand sides equal because of the associativity of multiplication: (g^{-1}h)^{-1}\xi=h^{-1}(g\xi).

In a similar way we can propagate the scalar product from the Lie algebra to the whole group by right shifts. The so obtained scalar product (aka Riemannian metric) is then right-invariant. But these two Riemannian metrics in general will be different. Let us check under which conditions they would coincide? In order to coincide we would have to have:

    \[(g^{-1}\xi,g^{-1}\eta)_e=(\xi g^{-1},\eta g^{-1})_e,\]

for all \xi,\eta tangent at g. Denoting \xi'=g^{-1}\xi,\eta'=g^{-1}\eta, we would have to have

    \[(\xi',\eta')_e=(g\xi'g^{-1},g\eta'g^{-1})_e\]

for all \xi',\eta' in the Lie algebra. In other words: the scalar product at the identity would have to be Ad-invariant. We recall that “Ad” denotes the adjoint representation of the group on its Lie algebra defined as

    \[ Ad_g:\xi\mapsto g\xi g^{-1}.\]

The metric we have defined using trace

    \[(\xi,\eta)_e=\frac{1}{2}\mathrm{Tr}(\xi\eta)\]

has this property because of the general property of the trace \mathrm{Tr}(AB)=\mathrm{Tr}(BA). But we could easily start with a different scalar product. Each scalar product give rise to different geodesics, different curvature. The group endowed with bi-invariant metric is, so to say, maximally “round”.
When metric is left (or right) invariant, that means that the group universe looks the same way from every point – it is “homogeneous”. But when it is also bi-invariant, that means that at every point it looks maximally the same in all directions – that is it has its natural “isotropy” property.

Riemannian metric on SL(2,R)- explicit formula

Posted on by arkajad

Riemannian metric is usually expressed through its metric tensor. For instance in Conformally Euclidean geometry of the upper half-plane we were discussing the SL(2,R) invariant Riemannian metric on the upper half-plane and came out with the formula:

(1)   \begin{equation*}g=\frac{1}{y^2}\begin{bmatrix}1&0\\0&1\end{bmatrix}.\end{equation*}

Riemannian metric in n dimensions is given by n\times n symmetric matrix g_{ij}=g_{ji},\,(i,j=1,...,n). The components of this matrix are, in general, functions of coordinates of the points of the space under consideration. Knowing the metric, we can calculate scalar product of any two tangent vectors at the same point. Knowing the scalar product of any two vectors we can calculate the metric tensor. It goes as follows. Suppose we have coordinates x^1,...,x^n. Sometimes the coordinates have their names, like \theta,r,u, but then it is often convenient to number them, for instance x^1=\theta,x^2=r,x^3=u, and to refer to them through their indices. When we have coordinates, we have vectors tangent to the coordinate lines. They are usually denoted as \frac{\partial}{\partial_i} or even simply \partial_i. This is a standard notation used in differential geometry texts. So, for instance, \partial_1 is vector tangent to the coordinate line \theta, that is the line when \theta is varying while r,u are kept constant. More precisely: \partial_1 is a vector field, since we can draw the coordinate of \theta through every point. Therefore when I write \partial_i I mean the vector field or I mean one particular vector at one particular point that should be evident from the context. When it needs to be specified, we can write (\partial_i)_p – which means that we are specifying particular point p.

If we have scalar product (\xi,\eta)_p defined at every point p of our space, then the metric tensor at a point p is given by the matrix

(2)   \begin{equation*}g_{ij}(p)= (\partial_i,\partial_j)_p.\end{equation*}

Vectors \partial_i tangent to the coordinate lines form a basis in the tangent space at every point. When we say that a given \xi has components \xi^i, that means that \xi=\xi^{i}\partial_i – where we use Einstein convention implying summation over the dummy index. Thus we may write:

(3)   \begin{equation*}(\xi,\eta)=(\xi^{i}\partial_i,\eta^{j}\partial_j)=\xi^{i}\eta^{j}(\partial_{i},\partial_j)=g_{ij}\xi^{i}\eta^j=\xi^Tg\eta.\end{equation*}

We will now calculate explicitly the coordinate expression for the metric on SL(2,R) described in the last post Riemannian metric on SL(2,R).
The coordinates x^1=\theta,x^2=r,x^3=u on the group manifold were introduced in Parametrization of SL(2,R) through

(4)   \begin{equation*} A(\theta,r,u)=\begin{bmatrix} r \cos (\theta )+\frac{u \sin (\theta )}{r} & \frac{\cos (\theta ) u}{r}-r \sin (\theta ) \\ \frac{\sin (\theta )}{r} & \frac{\cos (\theta )}{r}\end{bmatrix}. \end{equation*}

Let us now find the vector fields tangent to the coordinate lines:

(5)   \begin{equation*}\partial_1=\frac{\partial A(\theta,r,u)}{\partial \theta}= \begin{bmatrix} \frac{u \cos (\theta )}{r}-r \sin (\theta ) & -r \cos (\theta )-\frac{u \sin (\theta )}{r} \\ \frac{\cos (\theta )}{r} & -\frac{\sin (\theta )}{r} \end{bmatrix}, \end{equation*}

(6)   \begin{equation*}\partial_2=\frac{\partial A(\theta,r,u)}{\partial r}=\begin{bmatrix} \cos (\theta )-\frac{u \sin (\theta )}{r^2} & -\frac{u \cos (\theta )}{r^2}-\sin (\theta ) \\ -\frac{\sin (\theta )}{r^2} & -\frac{\cos (\theta )}{r^2} \end{bmatrix},\end{equation*}

(7)   \begin{equation*}\partial_3=\frac{\partial A(\theta,r,u)}{\partial u}=\begin{bmatrix} \frac{\sin (\theta )}{r} & \frac{\cos (\theta )}{r} \\ 0 & 0 \end{bmatrix}.\end{equation*}

These are tangent vectors at A. According to the prescription described in Riemannian metric on SL(2,R) we need to shift them to the identity, that is we need to calculate A^{-1}\partial_i, and then take trace of their products:

(8)   \begin{equation*}g_{ij}=\frac{1}{2}\mathrm{Tr}(A^{-1}\partial_iA^{-1}\partial_j).\end{equation*}

I did these calculations using computer. Here are the results:

(9)   \begin{equation*}A^{-1}=\begin{bmatrix} \frac{\cos (\theta )}{r} & r \sin (\theta )-\frac{u \cos (\theta )}{r} \\ -\frac{\sin (\theta )}{r} & r \cos (\theta )+\frac{u \sin (\theta )}{r}, \end{bmatrix}\end{equation*}

(10)   \begin{equation*}A^{-1}\partial_1=\begin{bmatrix}0&-1\\1&0\end{bmatrix},\end{equation*}

(11)   \begin{equation*}A^{-1}\partial_2=\begin{bmatrix} \frac{\cos (2 \theta )}{r} & -\frac{\sin (2 \theta )}{r} \\ -\frac{\sin (2 \theta )}{r} & -\frac{\cos (2 \theta )}{r}\end{bmatrix},\end{equation*}

(12)   \begin{equation*}A^{-1}\partial_3=\begin{bmatrix} \frac{\cos (\theta ) \sin (\theta )}{r^2} & \frac{\cos ^2(\theta )}{r^2} \\ -\frac{\sin ^2(\theta )}{r^2} & -\frac{\cos (\theta ) \sin (\theta )}{r^2} \end{bmatrix},\end{equation*}

(13)   \begin{equation*}g=\begin{bmatrix} -1 & 0 & \frac{1}{2 r^2} \\ 0 & \frac{1}{r^2} & 0 \\ \frac{1}{2 r^2} & 0 & 0\end{bmatrix}\end{equation*}

The metric is not diagonal. That means the coordinate lines are not all perpendicular to each other. Moreover, the third line is “light-like”. We have (\partial_3,\partial_3)=0. In Minkowski space-time it would mean that it is a trajectory of an object moving with the speed of light. At the group identity we have \theta=u=0, r=1. At this point we have

(14)   \begin{equation*}(\partial_3)_e= \begin{bmatrix}0&1\\0&0\end{bmatrix}.\end{equation*}

In SL(2,R) generators and vector fields on the half-plane we have denoted this generator as Y_-.

Once we have the metric, we can now calculate its geodesics and curvature. So, we have a plan for the following notes.

Riemannian metric on SL(2,R)

Posted on by arkajad

Every Lie group is like a Universe. Now it is time for us to play with the cosmology of SL(2,R). In Real magic – space-time in Lie algebra we have already started this game – we have defined a natural “metric” on the Lie algebra sl(2,R). We have defined scalar product of any two vectors tangent to the group at the group identity. And we have seen that this scalar product is similar to that of Minkowski space-time, except that with only two space (and one time) dimensions.

Riemannian metric, on the other hand, is defined when we have scalar product of tangent vectors at every point. In recent posts we were playing with Riemannian metric on the upper half-plane, which is a homogeneous space for the group, of only two dimensions – the cross section of the torus. Now we want to define Riemannian metric on the whole torus – an interesting extension.

We already have the metric at one point – at the group identity. Can we extend this definition to the whole torus, and do it in a natural way?

Of course we can. Because we are on the group. Here it is how it is being done. Suppose we have two vectors tangent at some group point g. I am using the letter g now, but g here is another notation for a real matrix A of determinant one, an element of SL(2,R). That is we have two paths \gamma_1(t),\gamma_2(t) with \gamma_1(0)=\gamma_2(0)=g. Our two tangent vectors, say \xi_1,\xi_2, are vectors tangent to \gamma_1(t) and \gamma_2(t) at g, they are represented by matrices

(1)   \begin{equation*}\xi_i=\frac{d\gamma_i(t)}{dt}|_{t=0}, \quad (i=1,2).\end{equation*}

The matrices \xi_1,\xi_2 are not in the Lie algebra. For instance, in our case, they will not be (in general) of trace zero. That is because at t=0 the two paths are not at the identity. But we can shift them to the identity. The paths g^{-1}\gamma_i(t) at t=0 are at the identity. Thus even if \xi_i are not in the Lie algebra, g^{-1}\xi_i are. We can use this fact and define the scalar product at g as follows:

(2)   \begin{equation*}(\xi_1,\xi_2)_g=(g^{-1}\xi_1,g^{-1}\xi_2)_e.\end{equation*}

I am using the symbol e rather than the unit matrix I to denote the group identity here, because the construction above is quite general, is being used for any Lie group, not just for SL(2,R).

Thus once we have scalar product defined in the Lie algebra, we have it defined everywhere.

Of course one can ask: why not use another definition, with right shifts? What is wrong with

(3)   \begin{equation*}(\xi_1,\xi_2)_g=(\xi_1g^{-1},\xi_2g^{-1})_e.\end{equation*}

Nothing is wrong. The scalar product at the identity that we have defined in Real magic – space-time in Lie algebra has the property of invariance expressed there in Eq. (11). It is a homework to show that using this property we can prove that the two definitions above lead to the same Riemannian metric!