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Let the inner product of the vectors X and Y on a given four dimensional manifold (EDIT: make this R4) be defined as (X*Y) = gikXiYk; using the summation convention for repeated indicies.

Let A be a 4 x 4 matrix which satisfies: (X*Y)=(AX*AY).

Then the set of all A is a matrix lie group. My question is this, what properties characterize the matrices A which preserve this inner product, and furthermore, what properties characterize the lie algebra of this group?

Is there a nice formula that gives the parametrized components of the orthogonal matrices A, analogous to the case of a euclidean metric? (i.e. the rotation matrix)

Is there a nice formula that determines the matrix lie algebra of this group?

EDIT:

As stated in my comment below, what I really want is an expression for the matrix components of the lie algebra as functions of the components of the metric tensor.

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    $\begingroup$ It seems to me that you are fixing a point on the manifold and looking at the metric on tangent space at that point. If that is right then this question is not really about manifolds and Riemannian-geometry but about classical matrix groups. Maybe you can be more precise. $\endgroup$
    – Michael Bächtold
    Commented Jul 10, 2010 at 18:12
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    $\begingroup$ Maybe to add to my previous comment: at the tangent space of any fixed point, a Riemannian metric (positive definite signature) always looks like the standard euclidean metric , i.e. you can find a basis in which $g_{ik}=\delta_{ik}$. $\endgroup$
    – Michael Bächtold
    Commented Jul 10, 2010 at 18:24
  • $\begingroup$ Let me attempt to rephrase the question as follows: given two vectors X and Y in R4, define the inner product as (X*Y) = g(ik)X(i)Y(k) as before. Now find the group of matrices acting on R4 which preserve this inner product, and the corresponding lie algebra. This way of presenting the question avoids all references to manifolds. $\endgroup$
    – Matt
    Commented Jul 10, 2010 at 23:20
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    $\begingroup$ Judging from your various comments, you do not assume that the product is positive definite? (Looks like you just want a scalar product that is a non-degenerate quadratic form.) Then it is a theorem that $g$ can be diagonalized. So $g = B^T\delta^{p,q}B$ where $B$ is some invertible linear transformation of $\mathbb{R}^4$ to itself, and $\delta^{p,q}$ is the diagonal matrix of signature p,q. Then any 'rotation' $A$ can be written as $B^{-1}Y$, where $Y\in O(p,q)$. $B$ can be constructed using some sort of a Gram-Schmidt process. $\endgroup$
    – Willie Wong
    Commented Jul 10, 2010 at 23:30
  • $\begingroup$ Ack! Typo! Meant to type $B^{-1}Y B$ instead of $B^{-1}Y$. It is of course a conjugation that gives the right answer. The corresponding Lie algebra components can be obtained by conjugating the standard presentation of the $o(p,q)$ Lie algebra by the matrix $B$ also. $\endgroup$
    – Willie Wong
    Commented Jul 11, 2010 at 0:52

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  1. A matrix $A$ preserves this inner product if $g_{ij} \: A_k^i \: A_l^j \; = \;g_{kl}$ (similar to the $AA^T=I$ definition of orthogonal matrices). One interpretation of this equation is that the columns of $A$ must be orthonormal with respect to the given inner product. Similarly, the rows of $A$ must be orthonormal with respect to the dual inner product $g^{ij}$.

  2. The defining equation for matrices $B$ of the Lie algebra is $g_{ik} \: B_j^k \;+\; g_{kj}\:B^k_i \;=\; 0$.

  3. Assuming the inner product is positive definite, the Lie group and Lie algebra in question are in fact isomorphic to the standard orthogonal group and Lie algebra of antisymmetric matrices, respectively.

Is this what you're looking for? I'm not entirely sure what some parts of your question are referring to. (For example, what is "a formula that gives the parameterized components...of a euclidean metric"?)

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  • $\begingroup$ I knew about this formula, but it's not what I am looking for. What I really want is an expression for the matrix components of the lie algebra as functions of the components of the metric tensor. I worked with this formula a lot to find the lie algebra of the Lorentz group and spin group, but I always wanted to generalize it for an arbitrary metric. I expect there may not be such a formula, or else if there is one it only exists for special types of metrics with certain symmetrical features. As for the final question in parenthesis, I simply was referring to the rotation matrix. $\endgroup$
    – Matt
    Commented Jul 10, 2010 at 23:12
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    $\begingroup$ Could you describe what you want in terms of the standard inner product (when the metric $g$ is the identity matrix)? It seems to me that Jim's answer is exactly what you're asking for. What else could there be? $\endgroup$
    – Deane Yang
    Commented Jul 11, 2010 at 0:46
  • $\begingroup$ Perhaps he's looking for the conjugacy that gives a group isomorphism between his isometry group and $O_n$. There are many of these, and computing one amounts to computing an orthonormal basis in the $g_{ij}$-metric. $\endgroup$
    – Ryan Budney
    Commented Jul 11, 2010 at 4:08
  • $\begingroup$ In response to deane, The rotation matrices in R3 have a lie algebra which is well known, the components are all 1, 0, or -1. You can use the same method jim suggested to compute the lie algebra's of all sorts of specific metrics, for example the lorentz and spin groups. I am interested in the most general case of an arbitrary g(i,k). I actually really like Willie's suggestion in the comment to the OP, about diagonalizing the metric and then computing the lie algebra's in terms of the transformation matrix and the standard Lorentz group. $\endgroup$
    – Matt
    Commented Jul 11, 2010 at 4:33
  • $\begingroup$ in case my previous comment was not clear, in the case of a standard metric, what I would be looking for is the lie algebra of O(3) $\endgroup$
    – Matt
    Commented Jul 11, 2010 at 4:39

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