1
$\begingroup$

I'm pretty new to the mathematicians notation in this area so apologies in advance. Suppose we have some manifold M equipped with a left invariant metric $g(u,v)$ and a full set of $n$ left-invariant vector fields $E_{i}$ which span the space of the manifold. Furthermore, let us take $M=G$, where $G$ is a compact Lie group. We know that the $E_{i}$ form the Lie algebra of G:

$$[E_{i},E_{j}]=\sum_{k=1}^{n}C_{ij}^{k}E_{k}$$ Where the brackets denote the commutator.

Let us perform a diffeomorphism on $M$. From a geometric view I picture this as a distortion of the manifold. Take $S^{1}$ for instance (as a trivial example) we might dilate the unit circle and otherwise warp it such that it is some non-symmetric shape (but of course it's still topologically $S^{1}$).

How does this affect the Lie algebra above? My first thought is to take the diffeomorphism as a map $\phi:M\rightarrow N$ where $N$ is our new manifold. Then supposing we can represent the map as a linear transformation ( I don't know if that is possible in general?) we obtain:

$$[\phi(x)E_{i},\phi(x)E_{j}]=\sum_{k=1}^{n}C_{ij}^{k}\phi(x)E_{k}$$

However that doesn't seem right as we still have a Lie algebra and in general I wouldn't expect the transformed $E_{i}$ to still be left invariant. Therefore, I'm guessing that the structure constants would have to transform in some way?

$\endgroup$
5
$\begingroup$

So there's a couple points of confusion here - for one thing, you start out by letting $M$ be an arbitrary manifold, then you equip $M$ with a "left-invariant" metric, and only then do you come out and say "suppose further $M$ is a compact Lie group". This is... backwards. You can't talk about left-invariant things when there's no additional structure on the manifold to preserve, like a Lie group structure. You have to have the structure there first before you can start talking about things that preserve it. In general you seem to be a bit confused by all the different levels of structure that exist, as you also mention things like arbitrary diffeomorphisms that don't preserve the Lie structure, and then ask how they affect it. Also, linear maps require a vector space structure on their domain and codomains. If your Lie groups are vector spaces, you can go ahead and make them linear. Otherwise, the question doesn't even make any sense.

Let's be a little clearer about our terms. We have $M$ a real compact Lie group, $N$ a smooth manifold. If we have a diffeomorphism $\phi: M \to N$, then we can certainly impose a Lie structure on $N$ which makes $\phi$ into a Lie group isomorphism automatically. So we define a group operation on $N$ by letting $y_1 *_N y_2 = \phi^{-1}(y_1) *_M \phi^{-1}(y_2)$, where $*_M$ is the group operation in $M$, and $y_1, y_2 \in N$. Since $\phi$ is a diffeomorphism, the inverse is defined and the multiplication and inversion operations are smooth. Now that $N$ is a Lie group itself, we can start talking about Lie-theoretic things between $M$ and $N$. See, before we had equipped $N$ with that group structure, there was no Lie algebra associated to $N$ to speak of. The Lie algebra of $N$ is the set of left-invariant vector fields on $N$, and again without a group operation there, there are no left-invariant vector fields to speak of.

Now you ask: how does the diffeomorphism (i.e. Lie group isomorphism) affect the Lie algebras? Well, it gives you an isomorphism of Lie algebras. How does it affect the structure constants? That depends on what basis you pick for your Lie algebra on $N$. We started out with a full set of left-invariant vector fields on $M$, but we did not do the same thing for $N$. We can push forward our chosen vector fields on $M$ to make a full set of left-invariant vector fields for $N$; under this choice the structure constants for $N$ are the same as those for $M$. What if we had picked a different set of vector fields for $N$? Well then that basically boils down to change-of-basis formulae. You take the Lie bracket as a rank-(1,2) tensor, expressed via the structure constants w.r.t. the basis pushed forward from the one on $M$, and then apply your change-of-basis to express the tensor with respect to whatever your new basis is.

A more interesting thing to do is to look at two Lie groups $G_1, G_2$, with $\psi:G_1 \to G_2$ a Lie group homomorphism. When you ask how does this affect the Lie algebras, well the answer is that it induces a Lie algebra homomorphism, no more, no less. If you equip each group with a full set of left-invariant vector fields, then you can compute exactly what this homomorphism does with respect to each basis. Since it is a homomorphism, you can be guaranteed that $[\psi(E_i),\psi(E_j)] = \psi([E_i,E_j])$ for all $i,j$, but what those values are depends on $\psi$. The question is basically too general and vague to have a meaningful answer.

$\endgroup$
9
  • $\begingroup$ Ok So I should've transposed my second and third sentence (since they don't commute!!! lol!) $\endgroup$ – R. Rankin Nov 24 '18 at 6:37
  • $\begingroup$ To be less vague: I'm looking at spacetimes with particular Lie group topologies. Beginning with spaces of constant curvature and (usually) bi-invariant metrics, I'm then looking at how the Group (or it's algebraic) properties manifest once the shape has been distorted "away" from having Left-Right invariant vector fields. I was hoping to use a diffeomorphism for this purpose (as it preserves topology). Anyway, Like I said in sentence 1 I'm new to the mathematician's terminology $\endgroup$ – R. Rankin Nov 24 '18 at 6:58
  • $\begingroup$ "spacetimes with particular Lie group topologies" - ok, where is the group structure coming from? As I understand it, spacetime is a particular Lorentzian 4-manifold, but it seems odd to be putting a Lie group structure on it. I don't know any physics though, so there's not much I can say. $\endgroup$ – silvascientist Nov 24 '18 at 7:01
  • $\begingroup$ Ok, for example a closed universe might have the shape $SU(2)XS^1$ which is very different from a flat universe $R^4$ $\endgroup$ – R. Rankin Nov 24 '18 at 7:03
  • $\begingroup$ Ok, so topologically it has that shape but where is the group structure coming in to play? $\endgroup$ – silvascientist Nov 24 '18 at 7:04

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.