most recent 30 from http://mathoverflow.net 2013-05-19T03:15:28Z http://mathoverflow.net/feeds/user/7466 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/31437/the-lie-algebra-of-the-orthogonal-group-of-an-arbitrary-space-time-metric Matt 2010-07-11T17:57:52Z 2010-07-13T01:03:17Z

<p>Let X ad Y be two vectors in R⁴, and define the inner product of X and Y as:</p> <p>(X*Y) = g_ikX_iY_k (summation convention for repeated indicies)</p> <p>Then we consider the 4x4 matrix g whose components are g_ik. I am of course interested in the case that g is NOT positive definite, because this is the situation when g represents the gravitational field in general relativity.</p> <p>Let A be a 4x4 matrix which satisfies (X*Y) = (AX*AY), then I say that A is an element O(g), the orthogonal group determined by g.</p> <p>I am interested in finding any sort of formula which relates the lie algebra of O(g) to the metric g.</p> <p>In a previous question, it was suggested that I diagonalize the matrix g using the theorem on diagonalizing positive definite matrices. This method works nicely and gives a simple solution for the lie algebra in terms of the transformation matrix which diagonalizes g, but only when g is positive definite. </p> <p>Can I still diagonalize my non positive definite g by finding the roots of the characteristic polynomial? I believe I must first somehow restrict the set of vectors I allow the inner product to work on, to avoid the case (X*Y) = 0. Nevertheless, for arguments sake let's assume that I can diagonalize g.</p> <p>Let B be the transformation matrix, then I assume that I can write:</p> <p>g = B^-1&eta;B, where &eta; is the identity matrix of signature (1,3), i.e. the metric of flat space time.</p> <p>We can characterize the elements of O(g) by realizing that our inner product can be written as:</p> <p>(X*Y) = X^TgY</p> <p>It's clear that if A is an element of O(g), then</p> <p>(AX*AY) = X^TA^TgAY = X^TgY</p> <p>Hence, A^TgA = g</p> <p>This formula can be written as gA^-1g = A^T, since g = g^-1 explicitly.</p> <p>Now applying the transformation matrix B:</p> <p>gA^-1g = B^-1&eta;BA^-1B^-1&eta;B = A^T, which I rearrange as:</p> <p>&eta;BA^-1B^-1&eta; = BA^TB^-1</p> <p>If by some chance (BA^-1B^-1)^-1 = (BA^TB^-1)^T, then I can immediately conclude that A is in O(g) so long as BA^-1B^-1 is in O(1,3) (the group which preserves the metric &eta;). From this step is it quite straightforward to compute the lie algebra, by taking advantage of the formula Be^XB^-1 = e^{BXB^-1}.</p> <p>I am concerned about several steps of this procedure:</p> <p>1) Is it legal to diagonalize g? I believe I need exclude any combination of vectors X, Y for which (X*Y) = 0. Since the squared norm is (X*X), then this amounts to disregarding vectors which lie along geodesic paths. Then I can deal separately for the case (X*X) > 0 and (X*X) &lt; 0.</p> <p>2) How am I supposed to deal with the condition that (BA^-1B^-1)^-1 = (BA^TB^-1)^T? This condition seems quite restrictive.</p>

http://mathoverflow.net/questions/31464/can-i-derive-the-boltzmann-distribution-by-an-invariance-argument/31492#31492 Matt 2010-07-12T01:51:56Z 2010-07-12T06:47:59Z

<p>I'd like to chime in here, as someone with a physics background.</p> <p>I absolutely love the derivation given by Landau in volume 5 on statistical physics, chapter 1. The basic idea is that since the log of the probability distribution function (i.e. the entropy) is an additive constant of the motion, it can be expressed as a linear combination of the 7 fundamental additive constants of the motion, namely the three components of momentum, the three components of angular momentum, and the energy. But since the momentum/angular momentum components can be reduced to zero with an appropriate frame of reference, the log of the distribution function depends only on some multiple of the energy, which turns out to be 1/T. We obtain the partition function naturally by normalizing the probability distribution.</p> <p>I think this answers your question 1 from a physics point of view.</p> <p>EDIT:</p> <p>in view of the comments below, I should point out the the probability distribution I am referring to gives the probability of finding a system of N particles which obey the laws of classical mechanics in the state for which the n^th particle is at position r_n and moving with a velocity v_n</p>

http://mathoverflow.net/questions/31198/curvature-of-a-lie-group Matt 2010-07-09T14:56:20Z 2010-07-11T20:27:15Z

<p>Since a lie group is a manifold with the structure of a continuous group, then each point of the manifold has some scalar curvature R. Is there a nice formula which relates the lie algebra of the group to the scalar curvature at a point of the manifold? </p>

http://mathoverflow.net/questions/31430/the-role-of-the-mean-value-theorem-mvt-in-first-year-calculus/31444#31444 Matt 2010-07-11T18:18:59Z 2010-07-11T18:18:59Z

<p>I believe students must be guided by intuition in a first year calculus course. </p> <p>Therefore, it is better to use geometry to explain derivatives in terms of tangent lines and geometric limits of pushing points together. A student cannot appreciate the need for rigor if their intuition is not first sufficiently developed.</p> <p>Since the MVT is mostly useful for rigorous computations, notably for the proof of the fundamental theorem of calculus, I agree that it is out of place in an ideal introduction to calculus.</p> <p>My main objection is that it is a long side track to prove, and it seems pointless to a student who doesn't appreciate rigor, which is likely to turn them off to mathematics. </p> <p>Furthermore, The fundamental theorem can be argued for intuitively by discussing the geometric ideas behind integration. When the student is able to see the flaws in the intuitive argument, they will also be able to appreciate the MVT.</p>

http://mathoverflow.net/questions/31311/the-orthogonal-group-of-a-riemannian-metric Matt 2010-07-10T17:41:43Z 2010-07-10T23:43:05Z

<p>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) = g_ikXⁱY^k; using the summation convention for repeated indicies. </p> <p>Let A be a 4 x 4 matrix which satisfies: (X*Y)=(AX*AY). </p> <p>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?</p> <p>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)</p> <p>Is there a nice formula that determines the matrix lie algebra of this group?</p> <p>EDIT: </p> <p>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.</p>

http://mathoverflow.net/questions/31437/the-lie-algebra-of-the-orthogonal-group-of-an-arbitrary-space-time-metric/31640#31640 Matt 2010-07-14T16:11:45Z 2010-07-14T16:11:45Z

I really like this solution, thanks!

http://mathoverflow.net/questions/31437/the-lie-algebra-of-the-orthogonal-group-of-an-arbitrary-space-time-metric Matt 2010-07-14T16:06:25Z 2010-07-14T16:06:25Z

@Deane, I am confused by your most recent comment. 1) Are you implying that if g is a non degenerate matrix, that it must be the matrix corresponding to the metric of flat space time? 2) g is just a matrix, and the entries of g are functions g(i,k) from R4 -&gt; R. I don't think of g as a function on R4, its just a matrix that can be used to represent the inner product as defined in the OP. Has your issue been cleared up by david speyers excellent answer given below? If not, can you please elaborate your objection because I am interested in the problem you see here.

http://mathoverflow.net/questions/31437/the-lie-algebra-of-the-orthogonal-group-of-an-arbitrary-space-time-metric Matt 2010-07-14T15:55:40Z 2010-07-14T15:55:40Z

@Michael, the entries of the vectors X are real numbers, and the entries of A should be functions of 4 real parameters. This can be seen by analogy with rotations in three space, for which the lie algebra corresponds to the three principle axis, and the three real parameters which elements of O(3) depend on correspond to rotations about the principle axis.

http://mathoverflow.net/questions/31437/the-lie-algebra-of-the-orthogonal-group-of-an-arbitrary-space-time-metric Matt 2010-07-13T02:06:56Z 2010-07-13T02:06:56Z

@michael, yes for sure since that is the case in general relativity.

http://mathoverflow.net/questions/31464/can-i-derive-the-boltzmann-distribution-by-an-invariance-argument/31492#31492 Matt 2010-07-12T05:51:55Z 2010-07-12T05:51:55Z

If you would forgive me for protesting your comment Qiaochu, then I would say that I would not know how to state a &quot;physical principle&quot; to show &quot;that the Boltzmann distribution only depends on the relative energies of the states&quot; without a reference to the energy of particles. The information theoretic approach is useful for quantum mechanics, but,in my opinion, if we want a clear picture in our head of why the Boltzmann distribution is related to the relative energy of states, we must resort to an analogy with the classical mechanics of systems of extremely large numbers of particles.

http://mathoverflow.net/questions/31437/the-lie-algebra-of-the-orthogonal-group-of-an-arbitrary-space-time-metric/31450#31450 Matt 2010-07-11T20:14:46Z 2010-07-11T20:14:46Z

I am a little concerned about transforming g to the identity matrix with signature (3,1). How can we be sure that the eigenvalue's of g are all 1 or -1?

http://mathoverflow.net/questions/31437/the-lie-algebra-of-the-orthogonal-group-of-an-arbitrary-space-time-metric Matt 2010-07-11T19:28:16Z 2010-07-11T19:28:16Z

I don't believe the invere of the transformation matrix P is it's own transpose in general, because this would only be true for positive def matrices, unless I am mistaken.

http://mathoverflow.net/questions/31437/the-lie-algebra-of-the-orthogonal-group-of-an-arbitrary-space-time-metric Matt 2010-07-11T19:03:11Z 2010-07-11T19:03:11Z

I am well aware of the lie algebra of the lorentz group, naturally. Is this the reference you are referring me to? I believe my group O(g) is quite distinct from O(3,1), since O(3,1) is a special case of O(g) when g is the flat space time metric. For example, the spin group and lorentz group have different lie algebras, and both are derived by substiting a specific g into the formulas in my post.

http://mathoverflow.net/questions/31311/the-orthogonal-group-of-a-riemannian-metric/31316#31316 Matt 2010-07-11T04:39:20Z 2010-07-11T04:39:20Z

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)

http://mathoverflow.net/questions/31311/the-orthogonal-group-of-a-riemannian-metric/31316#31316 Matt 2010-07-11T04:33:56Z 2010-07-11T04:33:56Z

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.

http://mathoverflow.net/questions/31311/the-orthogonal-group-of-a-riemannian-metric Matt 2010-07-10T23:20:36Z 2010-07-10T23:20:36Z

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.

http://mathoverflow.net/questions/31311/the-orthogonal-group-of-a-riemannian-metric/31316#31316 Matt 2010-07-10T23:12:51Z 2010-07-10T23:12:51Z

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.