So this is a bit of a follow-up to my recent question. I don't mean to inundate the feed with my quandaries, but as I move through the theory I keep hitting stumbling blocks (which y'all so kindly help me through).

As done previously, the group action defined by $h(g)p(z)=p(g^{-1}z)$ where $g\in\rm{SL}(3,\Bbb C)$, $p$ is from the vector space of polynomials of degree $\le2$ in three variables, and $z\in \Bbb C^3$.

I'm now introduced to a new function, $dh(X)=\left.\frac{d}{dt}h(e^{XT}]\right|_{t=0}$, where $X$ resides in the lie algebra of $\rm SL(3,\Bbb C)$ [ie $\mathfrak{sl}(3,\Bbb C)$]. The goal is: show that this is a lie algebra homomorphism $\mathfrak{sl}(3,\Bbb C)\to \mathfrak{gl}(6,\Bbb C)$.

Our basis in our space of polynomials is the standard one. Namely, the degree 2 terms in their various permutations.

I've been looking into the complexification of $\mathfrak{su}(3,\Bbb C)$ as a way of making sense of the polynomial when acted on by h, but I can't seem to get a handle on how to understand the derivative. I have a hunch this probably isn't even remotely the right course of action.

Any and all help is much appreciated. Feel free to assume I know the bare minimum.

Edit: While the induced homomorphism approach is awesome, a direct proof would help me get a better feel for how the matrix exponential is affecting the polynomial. Also, I'm too "machinery illiterate" to understand some of the more general formalisms at this point.

  • $\begingroup$ What is your definition of $e^X$? Is it just a power series? $\endgroup$ – Jason DeVito Apr 26 '12 at 3:24
  • $\begingroup$ @JasonDeVito indeed it is... $\endgroup$ – AsinglePANCAKE Apr 26 '12 at 3:24
  • $\begingroup$ Great. What about derivatives? Do you know the chain rule for maps between manifolds? $\endgroup$ – Jason DeVito Apr 26 '12 at 3:26
  • $\begingroup$ @JasonDeVito Here's where I'm failing...I've no idea how to compute that derivative... $\endgroup$ – AsinglePANCAKE Apr 26 '12 at 3:28
  • 1
    $\begingroup$ In whatever notes/book you're studying, how is the derivative defined? Does it mean "look at $h(e^{tX})$ as a matrix with $t$s in it and take the usual derivative of that."? $\endgroup$ – Jason DeVito Apr 26 '12 at 3:33

$h: SL(3,\mathbb C) \to GL(6,\mathbb C)$ is a homomorphism of Lie groups. Then your map $dh$ is the induced homomorphism of Lie algebras $\mathfrak{sl}(3,\mathbb C) \to \mathfrak{gl}(6,\mathbb C)$. So this is really just the general (and very important/fundamental fact) that the derivative at the identity of a Lie group homomorphism is a Lie algebra homomorphism. Proving this in generality would be a good exercise.

EDIT: Here's a brief outline (you can refer to the book I mentioned in the comments for more details if you need them though if you're familiar with some of these concepts they aren't that hard to prove). Let $\phi : G \to H$ be a Lie group homomorphism. Let $X_e, Y_e \in T_e G$ and denote by $X,Y$ the corresponding left invariant vector fields on $G$. Then one shows that the left-invariant vector field on $H$ that is equal to $\phi_*\vert_e X_e$ at the identity is $\phi$ related to $X$, i.e. for all $g \in G$ we have $\phi_*\vert_g X_g = {L_{\phi(g)}}_* \phi_*\vert_e X_e$. So call this vector field $\phi_* X$ (this is only a slight abuse of notation). Now by properties of Lie brackets, $[\phi_* X, \phi_* Y] = \phi_* [X,Y]$ so that $\phi_*\vert_e$ is a Lie algebra homomorphism.

  • $\begingroup$ So--lamentably--I'm not "allowed" to know that yet. I agree the general proof would be a good idea, but here I could use a great deal of help proving that it's a homomorphism directly... $\endgroup$ – AsinglePANCAKE Apr 26 '12 at 0:25
  • $\begingroup$ Could you perhaps direct me to a place that might offer this more general proof? $\endgroup$ – AsinglePANCAKE Apr 26 '12 at 18:09
  • $\begingroup$ John Lee's book introduction to smooth manifolds goes through this in chapter 4. It requires a non-trivial amount of smooth manifold theory (vector fields, pushforwards, etc.) $\endgroup$ – Eric O. Korman Apr 26 '12 at 18:32

I learned something about this from Lawrence Dresner, formerly of the Magnetics Division at Y-12 in Oak Ridge. He taught a class on how to use Lie Algebras to solve nonlinear ordinary and partial differential equations back in 1992, and I have been pursuing the subject ever since. For more general information you may wish to consult Thm. 4.4.1 of Sophus Lie's Differential Invariant Paper of 1884. This is applied math so it may lack the rigor you need, but is offered in the hopes that seeing an actual application may provide clarity.

Your function dh(X) is what Sophus Lie referred to as an infinitesimal transformation. By finding these functions for an infinite continuous group (Lie Group) and using the method of characteristics (and recalling that constants of integration are stabilizers for differential equations just as constants are polynomial stabilizers) you can find the stabilizers for the Lie Group invariant to the DEQ. These stabilizers (a.k.a. invariants) form an embedded subspace in the DEQ's field or manifold. Invariant solutions to the DEQ (devoid of integration constants) lie at singularities or saddle points and along separatrices in the direction field of the DEQ, and these may be found algebraically using the Lie Algebra between the stabilizers at these points. This works both on linear and nonlinear DEQ's because Lie Groups preserve the structure of the smooth manifold.

Here's an example. y"-(y'/x)-y^2=0 Lie Group Family: X=tx Y=(t^B)y, and for this family of groups the unitary transformation is t=1 (instead of t=0 for the exponential group you mention). Also, you can easily see that Y'=dY/dX=(t^(B-1))y' and Y"=(t^(B-2))y". Using the new function you cite, dh(X)=x, dh(Y)=By, dh(Y')=(B-1)y' and dh(Y")=(B-2)y". Applying this Lie Group Family to our DEQ, B=-2 for invariance (simple substitution gives you this). Using Euler's finite difference algorithm, dt = dx/x = dy/-2y = dy'/-3y' = dy"/-4y", and so on. dx/x = dy/-2y integrates to S1=(x^2)y where S1 is a constant of integration (a stabilizer for a DEQ). dx/x = dy'/-3y' gives S2=(x^3)y' and in similar fashion S3=(x^4)y". Because these stabilizers form an embedded subspace in the DEQ's field (or manifold for nonhomogeneous DEQ's) the DEQ may be rewritten in terms of the group stabilizers as follows: x^4(y"-(y'/x)-y^2)=(x^4)y"-(x^3)y'-((x^2)y)^2=S3-S2-S1^2=0. Differentiating the stabilizers w.r.t.x and setting the results equal to zero picks out the places in the direction field of the transformed DEQ where the slope is both zero and infinity: singularities and saddle points. dS1/dx=(x^-1)(2S1+S2)=0 yields S2=-2S1. dS2/dx=0 yields S3=-3S2=6S1 (by substitution). This simple Lie Algebra is part of the kernel of the map and may be used to find special solutions of the DEQ. S3-S2-S1^2=6S1-(-2S1)-S1^2=8S1-S1^2=0 Discarding the trivial solution and substituting for S1 we have the special solution y=8(x^-2), which is easily checked. For more information on this subject you may consult the works of Dresner or, more recently, Olver.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.